# Introduction and first examples
## What is a partial differential equation?

Let $d\in \mathbb {N}$ be a natural number, and let $B\subseteq \mathbb {R} ^{d}$ be an arbitrary set. A partial differential equation on $B$ looks like this:
    $\forall (x_{1},\ldots ,x_{d})\in B:h(x_{1},\ldots ,x_{d},u(x_{1},\ldots ,x_{d}),\overbrace {\partial _{x_{1}}u(x_{1},\ldots ,x_{d}),\ldots ,\partial _{x_{d}}u(x_{1},\ldots ,x_{d}),\partial _{x_{1}}^{2}u(x_{1},\ldots ,x_{d}),\ldots } ^{{\text{arbitrary and arbitrarily finitely many partial derivatives, }}n{\text{ inputs of }}h{\text{ in total}}})=0$
$h$ is an arbitrary function here, specific to the partial differential equation, which goes from $\mathbb {R} ^{n}$ to $\mathbb {R}$, where $n\in \mathbb {N}$ is a natural number. And a solution to this partial differential equation on $B$ is a function $u:B\to \mathbb {R}$ satisfying the above logical statement. The solutions of some partial differential equations describe processes in nature; this is one reason why they are so important.

## Multiindices

In the whole theory of partial differential equations, multiindices are extremely important. Only with their help we are able to write down certain formulas a lot briefer.
Definitions 1.1:
A $d$-dimensional multiindex is a vector $\alpha \in \mathbb {N} _{0}^{d}$, where $\mathbb {N} _{0}$ are the natural numbers and zero.
If $\alpha =(\alpha _{1},\ldots ,\alpha _{d})$ is a multiindex, then its absolute value $|\alpha |$ is defined by
    $|\alpha |:=\sum _{k=1}^{d}\alpha _{k}$
If $\alpha$ is a $d$-dimensional multiindex, $B\subseteq \mathbb {R} ^{d}$ is an arbitrary set and $u:B\to \mathbb {R}$ is sufficiently often differentiable, we define $\partial _{\alpha }u$, the $\alpha$-th derivative of $u$, as follows:
    $\partial _{\alpha }u:=\partial _{x_{1}}^{\alpha _{1}}\cdots \partial _{x_{d}}^{\alpha _{d}}u$

## Types of partial differential equations

We classify partial differential equations into several types, because for partial differential equations of one type we will need different solution techniques as for differential equations of other types. We classify them into linear and nonlinear equations, and into equations of different orders.
Definitions 1.2:
A linear partial differential equation is an equation of the form
    $\forall x\in B:\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }(x)\partial _{\alpha }u(x)=f(x)$
, where only finitely many of the $a_{\alpha }$s are not the constant zero function. A solution takes the form of a function $u:B\to \mathbb {R}$. We have $B\subseteq \mathbb {R} ^{d}$ for an arbitrary $d\in \mathbb {N}$, $f:B\to \mathbb {R}$ is an arbitrary function and the sum in the formula is taken over all possible $d$-dimensional multiindices. If $f=0$ the equation is called homogenous.
A partial differential equation is called nonlinear iff it is not a linear partial differential equation.
Definition 1.3:
Let $n\in \mathbb {N}$. We say that a partial differential equation has $n$-th order iff $n$ is the smallest number such that it is of the form
    $\forall (x_{1},\ldots ,x_{d})\in B\subseteq \mathbb {R} ^{d}:h(x_{1},\ldots ,x_{d},u(x_{1},\ldots ,x_{d}),\overbrace {\partial _{x_{1}}u(x_{1},\ldots ,x_{d}),\ldots ,\partial _{x_{d}}u(x_{1},\ldots ,x_{d}),\partial _{x_{1}}^{2}u(x_{1},\ldots ,x_{d}),\ldots } ^{{\text{partial derivatives at most up to order }}n})=0$

## First example of a partial differential equation

Now we are very curious what practical examples of partial differential equations look like after all.
Theorem and definition 1.4:
If $g:\mathbb {R} \to \mathbb {R}$ is a differentiable function and $c\in \mathbb {R}$, then the function
    $u:\mathbb {R} ^{2}\to \mathbb {R} ,u(t,x):=g(x+ct)$
solves the one-dimensional homogenous transport equation
    $\forall (t,x)\in \mathbb {R} ^{2}:\partial _{t}u(t,x)-c\partial _{x}u(t,x)=0$
Proof: Exercise 2.
We therefore see that the one-dimensional transport equation has many different solutions; one for each continuously differentiable function in existence. However, if we require the solution to have a specific initial state, the solution becomes unique.
Theorem and definition 1.5:
If $g:\mathbb {R} \to \mathbb {R}$ is a differentiable function and $c\in \mathbb {R}$, then the function
    $u:\mathbb {R} ^{2}\to \mathbb {R} ,u(t,x):=g(x+ct)$
is the unique solution to the initial value problem for the one-dimensional homogenous transport equation
    ${\begin{cases}\forall (t,x)\in \mathbb {R} ^{2}:&\partial _{t}u(t,x)-c\partial _{x}u(t,x)=0\\\forall x\in \mathbb {R} :&u(0,x)=g(x)\end{cases}}$
Proof:
Surely $\forall x\in \mathbb {R} :u(0,x)=g(x+c\cdot 0)=g(x)$. Further, theorem 1.4 shows that also:
    $\forall (t,x)\in \mathbb {R} ^{2}:\partial _{t}u(t,x)-c\partial _{x}u(t,x)=0$
Now suppose we have an arbitrary other solution to the initial value problem. Let's name it $v$. Then for all $(t,x)\in \mathbb {R} ^{2}$, the function
    $\mu _{(t,x)}(\xi ):=v(t-\xi ,x+c\xi )$
is constant:
    ${\frac {d}{d\xi }}v(t-\xi ,x+c\xi )={\begin{pmatrix}\partial _{t}v(t-\xi ,x+c\xi )&\partial _{x}v(t-\xi ,x+c\xi )\end{pmatrix}}{\begin{pmatrix}-1\\c\end{pmatrix}}=-\partial _{t}v(t-\xi ,x+c\xi )+c\partial _{x}v(t-\xi ,x+c\xi )=0$
Therefore, in particular
    $\forall (t,x)\in \mathbb {R} ^{2}:\mu _{(t,x)}(0)=\mu _{(t,x)}(t)$
, which means, inserting the definition of $\mu _{(t,x)}$, that
    $\forall (t,x)\in \mathbb {R} ^{2}:v(t,x)=v(0,x+ct){\overset {\text{initial condition}}{=}}g(x+ct)$
, which shows that $u=v$. Since $v$ was an arbitrary solution, this shows uniqueness.$\Box$
In the next chapter, we will consider the non-homogenous arbitrary-dimensional transport equation.

## Exercises

* Have a look at the definition of an ordinary differential equation (see for example the Wikipedia page on that) and show that every ordinary differential equation is a partial differential equation.
* Prove Theorem 1.4 using direct calculation.
* What is the order of the transport equation?
* Find a function $u:\mathbb {R} ^{2}\to \mathbb {R}$ such that $\partial _{t}u-2\partial _{x}u=0$ and $\forall x\in \mathbb {R} :u(0,x)=x^{3}$.

## Sources

* Martin Brokate (2011/2012), Partielle Differentialgleichungen, Vorlesungsskript (PDF) (in German) {{citation}}: Check date values in: |year= (help)
* Daniel Matthes (2013/2014), Partial Differential Equations, lecture notes {{citation}}: Check date values in: |year= (help)

# The transport equation

In the first chapter, we had already seen the one-dimensional transport equation. In this chapter we will see that we can quite easily generalise the solution method and the uniqueness proof we used there to multiple dimensions. Let $d\in \mathbb {N}$. The inhomogenous $d$-dimensional transport equation looks like this:
    $\forall (t,x)\in \mathbb {R} \times \mathbb {R} ^{d}:\partial _{t}u(t,x)-\mathbf {v} \cdot \nabla _{x}u(t,x)=f(t,x)$
, where $f:\mathbb {R} \times \mathbb {R} ^{d}\to \mathbb {R}$ is a function and $\mathbf {v} \in \mathbb {R} ^{d}$ is a vector.

## Solution

The following definition will become a useful shorthand notation in many occasions. Since we can use it right from the beginning of this chapter, we start with it.
Definition 2.1:
Let $f:\mathbb {R} ^{d}\to \mathbb {R}$ be a function and $n\in \mathbb {N}$. We say that $f$ is $n$ times continuously differentiable iff all the partial derivatives
    $\partial _{\alpha }f,\alpha \in \mathbb {N} _{0}^{d}{\text{ and }}|\alpha |\leq n$
exist and are continuous. We write $f\in {\mathcal {C}}^{n}(\mathbb {R} ^{d})$.
Before we prove a solution formula for the transport equation, we need a theorem from analysis which will play a crucial role in the proof of the solution formula.
Theorem 2.2: (Leibniz' integral rule)
Let $O\subseteq \mathbb {R}$ be open and $B\subseteq \mathbb {R} ^{d}$, where $d\in \mathbb {N}$ is arbitrary, and let $f\in {\mathcal {C}}^{1}(O\times B)$. If the conditions
* for all $x\in O$, $\int _{B}|f(x,y)|dy<\infty$
* for all $x\in O$ and $y\in B$, ${\frac {d}{dx}}f(x,y)$ exists
* there is a function $g:B\to \mathbb {R}$ such that
       $\forall (x,y)\in O\times B:|\partial _{x}f(x,y)|\leq |g(y)|{\text{ and }}\int _{B}|g(y)|dy<\infty$
hold, then
    ${\frac {d}{dx}}\int _{B}f(x,y)dy=\int _{B}{\frac {d}{dx}}f(x,y)$
We will omit the proof.
Theorem 2.3: If $f\in {\mathcal {C}}^{1}(\mathbb {R} \times \mathbb {R} ^{d})$, $g\in {\mathcal {C}}^{1}(\mathbb {R} ^{d})$ and $\mathbf {v} \in \mathbb {R} ^{d}$, then the function
    $u:\mathbb {R} \times \mathbb {R} ^{d}\to \mathbb {R} ,u(t,x):=g(x+\mathbf {v} t)+\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds$
solves the inhomogenous $d$-dimensional transport equation
    $\forall (t,x)\in \mathbb {R} \times \mathbb {R} ^{d}:\partial _{t}u(t,x)-\mathbf {v} \cdot \nabla _{x}u(t,x)=f(t,x)$
Note that, as in chapter 1, that there are many solutions, one for each continuously differentiable $g$ in existence.
Proof:
1.
We show that $u$ is sufficiently often differentiable. From the chain rule follows that $g(x+\mathbf {v} t)$ is continuously differentiable in all the directions $t,x_{1},\ldots ,x_{d}$. The existence of
    $\partial _{x_{n}}\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds,n\in \{1,\ldots ,d\}$
follows from the Leibniz integral rule (see exercise 1). The expression
    $\partial _{t}\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds$
we will later in this proof show to be equal to
    $f(t,x)+\mathbf {v} \cdot \nabla _{x}\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds$,
which exists because
    $\nabla _{x}\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds$
just consists of the derivatives
    $\partial _{x_{n}}\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds,n\in \{1,\ldots ,d\}$
2.
We show that
    $\forall (t,x)\in \mathbb {R} \times \mathbb {R} ^{d}:\partial _{t}u(t,x)-\mathbf {v} \cdot \nabla _{x}u(t,x)=f(t,x)$
in three substeps.
2.1
We show that
    $\partial _{t}g(x+\mathbf {v} t)-\mathbf {v} \cdot \nabla _{x}g(x+\mathbf {v} t)=0~~~~~(*)$
This is left to the reader as an exercise in the application of the multi-dimensional chain rule (see exercise 2).
2.2
We show that
    $\partial _{t}\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds-\mathbf {v} \cdot \nabla _{x}\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds=f(t,x)~~~~~(**)$
We choose
    $F(t,x):=\int _{0}^{t}f(s,x-\mathbf {v} s)ds$
so that we have
    $F(t,x+\mathbf {v} t)=\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds$
By the multi-dimensional chain rule, we obtain
    ${\begin{aligned}{\frac {d}{dt}}F(t,x+\mathbf {v} t)&={\begin{pmatrix}\partial _{t}F(t,x+\mathbf {v} t)&\partial _{x_{1}}F(t,x+\mathbf {v} t)&\cdots &\partial _{x_{d}}F(t,x+\mathbf {v} t)\end{pmatrix}}{\begin{pmatrix}1\\\mathbf {v} \end{pmatrix}}\\&=\partial _{t}F(t,x+\mathbf {v} t)+\mathbf {v} \cdot \nabla _{x}F(t,x+\mathbf {v} t)\end{aligned}}$
But on the one hand, we have by the fundamental theorem of calculus, that $\partial _{t}F(t,x)=f(t,x-\mathbf {v} t)$ and therefore
    $\partial _{t}F(t,x+\mathbf {v} t)=f(t,x)$
and on the other hand
    $\partial _{x_{n}}F(t,x+\mathbf {v} t)=\partial _{x_{n}}\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds$
, seeing that the differential quotient of the definition of $\partial _{x_{n}}$ is equal for both sides. And since on the third hand
    ${\frac {d}{dt}}F(t,x+\mathbf {v} t)=\partial _{t}\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds$
, the second part of the second part of the proof is finished.
2.3
We add $(*)$ and $(**)$ together, use the linearity of derivatives and see that the equation is satisfied. $\Box$

## Initial value problem

Theorem and definition 2.4: If $f\in {\mathcal {C}}^{1}(\mathbb {R} \times \mathbb {R} ^{d})$ and $g\in {\mathcal {C}}^{1}(\mathbb {R} ^{d})$, then the function
    $u:\mathbb {R} \times \mathbb {R} ^{d}\to \mathbb {R} ,u(t,x):=g(x+\mathbf {v} t)+\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds$
is the unique solution of the initial value problem of the transport equation
    ${\begin{cases}\forall (t,x)\in \mathbb {R} \times \mathbb {R} ^{d}:&\partial _{t}u(t,x)-\mathbf {v} \cdot \nabla _{x}u(t,x)=f(t,x)\\\forall x\in \mathbb {R} ^{d}:&u(0,x)=g(x)\end{cases}}$
Proof:
Quite easily, $u(0,x)=g(x+\mathbf {v} \cdot 0)+\int _{0}^{0}f(s,x+\mathbf {v} (t-s))ds=g(x)$. Therefore, and due to theorem 2.3, $u$ is a solution to the initial value problem of the transport equation. So we proceed to show uniqueness.
Assume that $v$ is an arbitrary other solution. We show that $v=u$, thereby excluding the possibility of a different solution.
We define $w:=u-v$. Then
    ${\begin{array}{llll}\forall (t,x)\in \mathbb {R} \times \mathbb {R} ^{d}:&\partial _{t}w(t,x)-\mathbf {v} \cdot \nabla _{x}w(t,x)&=(\partial _{t}u(t,x)-\mathbf {v} \cdot \nabla _{x}u(t,x))-(\partial _{t}v(t,x)-\mathbf {v} \cdot \nabla _{x}v(t,x))&\\&&=f(t,x)-f(t,x)=0&~~~~~(*)\\\forall x\in \mathbb {R} ^{d}:&w(0,x)=u(0,x)-v(0,x)&=g(x)-g(x)=0&~~~~~(**)\end{array}}$
Analogous to the proof of uniqueness of solutions for the one-dimensional homogenous initial value problem of the transport equation in the first chapter, we define for arbitrary $(t,x)\in \mathbb {R} \times \mathbb {R} ^{d}$,
    $\mu _{(t,x)}(\xi ):=w(t-\xi ,x+\mathbf {v} \xi )$
Using the multi-dimensional chain rule, we calculate $\mu _{(t,x)}'(\xi )$:
    ${\begin{aligned}\mu _{(t,x)}'(\xi )&:={\frac {d}{d\xi }}w(t-\xi ,x+\mathbf {v} \xi )&{\text{ by defs. of the }}'{\text{ symbol and }}\mu \\&={\begin{pmatrix}\partial _{t}w(t-\xi ,x+\mathbf {v} \xi )&\partial _{x_{1}}w(t-\xi ,x+\mathbf {v} \xi )&\cdots &\partial _{x_{d}}w(t-\xi ,x+\mathbf {v} \xi )\end{pmatrix}}{\begin{pmatrix}-1\\\mathbf {v} \end{pmatrix}}&{\text{chain rule}}\\&=-\partial _{t}w(t-\xi ,x+\mathbf {v} \xi )+\mathbf {v} \cdot \nabla _{x}w(t-\xi ,x+\mathbf {v} \xi )&\\&=0&(*)\end{aligned}}$
Therefore, for all $(t,x)\in \mathbb {R} \times \mathbb {R} ^{d}$ $\mu _{(t,x)}(\xi )$ is constant, and thus
    $\forall (t,x)\in \mathbb {R} \times \mathbb {R} ^{d}:w(t,x)=\mu _{(t,x)}(0)=\mu _{(t,x)}(t)=w(0,x+\mathbf {v} t){\overset {(**)}{=}}0$
, which shows that $w=u-v=0$ and thus $u=v$.$\Box$

## Exercises

* Let $f\in {\mathcal {C}}^{1}(\mathbb {R} \times \mathbb {R} ^{d})$ and $\mathbf {v} \in \mathbb {R} ^{d}$. Using Leibniz' integral rule, show that for all $n\in \{1,\ldots ,d\}$ the derivative$\partial _{x_{n}}\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds$is equal to$\int _{0}^{t}\partial _{x_{n}}f(s,x+\mathbf {v} (t-s))ds$and therefore exists.
* Let $g\in {\mathcal {C}}^{1}(\mathbb {R} ^{d})$ and $\mathbf {v} \in \mathbb {R} ^{d}$. Calculate $\partial _{t}g(x+\mathbf {v} t)$.
* Find the unique solution to the initial value problem
    ${\begin{cases}\forall (t,x)\in \mathbb {R} \times \mathbb {R} ^{3}:&\partial _{t}u(t,x)-{\begin{pmatrix}2\\3\\4\end{pmatrix}}\cdot \nabla _{x}u(t,x)=t^{5}+x_{1}^{6}+x_{2}^{7}+x_{3}^{8}\\\forall x\in \mathbb {R} ^{3}:&u(0,x)=x_{1}^{9}+x_{2}^{10}+x_{3}^{11}\end{cases}}$.

## Sources

* Martin Brokate (2011/2012), Partielle Differentialgleichungen, Vorlesungsskript (PDF) (in German) {{citation}}: Check date values in: |year= (help)

# Test functions
## Motivation

Before we dive deeply into the chapter, let's first motivate the notion of a test function. Let's consider two functions which are piecewise constant on the intervals $[0,1),[1,2),[2,3),[3,4),[4,5)$ and zero elsewhere; like, for example, these two:
Let's call the left function $f_{1}$, and the right function $f_{2}$.
Of course we can easily see that the two functions are different; they differ on the interval $[4,5)$; however, let's pretend that we are blind and our only way of finding out something about either function is evaluating the integrals
    $\int _{\mathbb {R} }\varphi (x)f_{1}(x)dx$ and $\int _{\mathbb {R} }\varphi (x)f_{2}(x)dx$
for functions $\varphi$ in a given set of functions ${\mathcal {X}}$.
We proceed with choosing ${\mathcal {X}}$ sufficiently clever such that five evaluations of both integrals suffice to show that $f_{1}\neq f_{2}$. To do so, we first introduce the characteristic function. Let $A\subseteq \mathbb {R}$ be any set. The characteristic function of $A$ is defined as
    $\chi _{A}(x):={\begin{cases}1&x\in A\\0&x\notin A\end{cases}}$
With this definition, we choose the set of functions ${\mathcal {X}}$ as
    ${\mathcal {X}}:=\{\chi _{[0,1)},\chi _{[1,2)},\chi _{[2,3)},\chi _{[3,4)},\chi _{[4,5)}\}$
It is easy to see (see exercise 1), that for $n\in \{1,2,3,4,5\}$, the expression
    $\int _{\mathbb {R} }\chi _{[n-1,n)}(x)f_{1}(x)dx$
equals the value of $f_{1}$ on the interval $[n-1,n)$, and the same is true for $f_{2}$. But as both functions are uniquely determined by their values on the intervals $[n-1,n),n\in \{1,2,3,4,5\}$ (since they are zero everywhere else), we can implement the following equality test:
    $f_{1}=f_{2}\Leftrightarrow \forall \varphi \in {\mathcal {X}}:\int _{\mathbb {R} }\varphi (x)f_{1}(x)dx=\int _{\mathbb {R} }\varphi (x)f_{2}(x)dx$
This obviously needs five evaluations of each integral, as $\#{\mathcal {X}}=5$.
Since we used the functions in ${\mathcal {X}}$ to test $f_{1}$ and $f_{2}$, we call them test functions. What we ask ourselves now is if this notion generalises from functions like $f_{1}$ and $f_{2}$, which are piecewise constant on certain intervals and zero everywhere else, to continuous functions. The following chapter shows that this is true.

## Bump functions

In order to write down the definition of a bump function more shortly, we need the following two definitions:
Definition 3.1:
Let $B\subseteq \mathbb {R} ^{d}$, and let $f:B\to \mathbb {R}$. We say that $f$ is smooth if all the partial derivatives
    $\partial _{\alpha }f,\alpha \in \mathbb {N} _{0}^{d}$
exist in all points of $B$ and are continuous. We write $f\in {\mathcal {C}}^{\infty }(B)$.
Definition 3.2:
Let $f:\mathbb {R} ^{d}\to \mathbb {R}$. We define the support of $f$, ${\text{supp }}f$, as follows:
    ${\text{supp }}f:={\overline {\{x\in \mathbb {R} ^{d}|f(x)\neq 0\}}}$
Now we are ready to define a bump function in a brief way:
Definition 3.3:
$\varphi :\mathbb {R} ^{d}\to \mathbb {R}$ is called a bump function iff $\varphi \in {\mathcal {C}}^{\infty }(\mathbb {R} ^{d})$ and ${\text{supp }}\varphi$ is compact. The set of all bump functions is denoted by ${\mathcal {D}}(O)$.
These two properties make the function really look like a bump, as the following example shows:
Example 3.4: The standard mollifier $\eta$, given by
    $\eta :\mathbb {R} ^{d}\to \mathbb {R} ,\eta (x)={\frac {1}{c}}{\begin{cases}e^{-{\frac {1}{1-\|x\|^{2}}}}&{\text{ if }}\|x\|_{2}<1\\0&{\text{ if }}\|x\|_{2}\geq 1\end{cases}}$
, where $c:=\int _{B_{1}(0)}e^{-{\frac {1}{1-\|x\|^{2}}}}dx$, is a bump function (see exercise 2).

## Schwartz functions

As for the bump functions, in order to write down the definition of Schwartz functions shortly, we first need two helpful definitions.
Definition 3.5:
Let $X$ be an arbitrary set, and let $f:X\to \mathbb {R}$ be a function. Then we define the supremum norm of $f$ as follows:
    $\|f\|_{\infty }:=\sup \limits _{x\in X}|f(x)|$
Definition 3.6:
For a vector $x=(x_{1},\ldots ,x_{d})\in \mathbb {R} ^{d}$ and a $d$-dimensional multiindex $\alpha \in \mathbb {N} _{0}^{d}$ we define $x^{\alpha }$, $x$ to the power of $\alpha$, as follows:
    $x^{\alpha }:=x_{1}^{\alpha _{1}}\cdots x_{d}^{\alpha _{d}}$
Now we are ready to define a Schwartz function.
Definition 3.7:
We call $\phi :\mathbb {R} ^{d}\to \mathbb {R}$ a Schwartz function iff the following two conditions are satisfied:
* $\phi \in {\mathcal {C}}^{\infty }(\mathbb {R} ^{d})$
* $\forall \alpha ,\beta \in \mathbb {N} _{0}^{d}:\|x^{\alpha }\partial _{\beta }\phi \|_{\infty }<\infty$
By $x^{\alpha }\partial _{\beta }\phi$ we mean the function $x\mapsto x^{\alpha }\partial _{\beta }\phi (x)$.
Example 3.8: The function
    $f:\mathbb {R} ^{2}\to \mathbb {R} ,f(x,y)=e^{-x^{2}-y^{2}}$
is a Schwartz function.
Theorem 3.9:
Every bump function is also a Schwartz function.
This means for example that the standard mollifier is a Schwartz function.
Proof:
Let $\varphi$ be a bump function. Then, by definition of a bump function, $\varphi \in {\mathcal {C}}^{\infty }(\mathbb {R} ^{d})$. By the definition of bump functions, we choose $R>0$ such that
    ${\text{supp }}\varphi \subseteq {\overline {B_{R}(0)}}$
, as in $\mathbb {R} ^{d}$, a set is compact iff it is closed & bounded. Further, for $\alpha ,\beta \in \mathbb {N} _{0}^{d}$ arbitrary,
    ${\begin{aligned}\|x^{\alpha }\partial _{\beta }\varphi (x)\|_{\infty }&:=\sup _{x\in \mathbb {R} ^{d}}|x^{\alpha }\partial _{\beta }\varphi (x)|&\\&=\sup _{x\in {\overline {B_{R}(0)}}}|x^{\alpha }\partial _{\beta }\varphi (x)|&{\text{supp }}\varphi \subseteq {\overline {B_{R}(0)}}\\&=\sup _{x\in {\overline {B_{R}(0)}}}\left(|x^{\alpha }||\partial _{\beta }\varphi (x)|\right)&{\text{rules for absolute value}}\\&\leq \sup _{x\in {\overline {B_{R}(0)}}}\left(R^{|\alpha |}|\partial _{\beta }\varphi (x)|\right)&\forall i\in \{1,\ldots ,d\},(x_{1},\ldots ,x_{d})\in {\overline {B_{R}(0)}}:|x_{i}|\leq R\\&<\infty &{\text{Extreme value theorem}}\end{aligned}}$
$\Box$

## Convergence of bump and Schwartz functions

Now we define what convergence of a sequence of bump (Schwartz) functions to a bump (Schwartz) function means.
Definition 3.10:
A sequence of bump functions $(\varphi _{i})_{i\in \mathbb {N} }$ is said to converge to another bump function $\varphi$ iff the following two conditions are satisfied:
* There is a compact set $K\subset \Omega$ such that $\forall i\in \mathbb {N} :{\text{supp }}\varphi _{i}\subseteq K$
* $\forall \alpha \in \mathbb {N} _{0}^{d}:\lim _{i\rightarrow \infty }\|\partial _{\alpha }\varphi _{i}-\partial _{\alpha }\varphi \|_{\infty }=0$
Definition 3.11:
We say that the sequence of Schwartz functions $(\phi _{i})_{i\in \mathbb {N} }$ converges to $\phi$ iff the following condition is satisfied:
    $\forall \alpha ,\beta \in \mathbb {N} _{0}^{d}:\|x^{\alpha }\partial _{\beta }\phi _{i}-x^{\alpha }\partial _{\beta }\phi \|_{\infty }\to 0,i\to \infty$
Theorem 3.12:
Let $(\varphi _{i})_{i\in \mathbb {N} }$ be an arbitrary sequence of bump functions. If $\varphi _{i}\to \varphi$ with respect to the notion of convergence for bump functions, then also $\varphi _{i}\to \varphi$ with respect to the notion of convergence for Schwartz functions.
Proof:
Let $O\subseteq \mathbb {R} ^{d}$ be open, and let $(\varphi _{l})_{l\in \mathbb {N} }$ be a sequence in ${\mathcal {D}}(O)$ such that $\varphi _{l}\to \varphi \in {\mathcal {D}}(O)$ with respect to the notion of convergence of ${\mathcal {D}}(O)$. Let thus $K\subset \mathbb {R} ^{d}$ be the compact set in which all the ${\text{supp }}\varphi _{l}$ are contained. From this also follows that ${\text{supp }}\varphi \subseteq K$, since otherwise $\|\varphi _{l}-\varphi \|_{\infty }\geq |c|$, where $c\in \mathbb {R}$ is any nonzero value $\varphi$ takes outside $K$; this would contradict $\varphi _{l}\to \varphi$ with respect to our notion of convergence.
In $\mathbb {R} ^{d}$, ‘compact’ is equivalent to ‘bounded and closed’. Therefore, $K\subset B_{R}(0)$ for an $R>0$. Therefore, we have for all multiindices $\alpha ,\beta \in \mathbb {N} _{0}^{d}$:
    ${\begin{aligned}\|x^{\alpha }\partial _{\beta }\varphi _{l}-x^{\alpha }\partial _{\beta }\varphi \|_{\infty }&=\sup _{x\in \mathbb {R} ^{d}}\left|x^{\alpha }\partial _{\beta }\varphi _{l}(x)-x^{\alpha }\partial _{\beta }\varphi (x)\right|&{\text{ definition of the supremum norm}}\\&=\sup _{x\in B_{R}(0)}\left|x^{\alpha }\partial _{\beta }\varphi _{l}(x)-x^{\alpha }\partial _{\beta }\varphi (x)\right|&{\text{ as }}{\text{supp }}\varphi _{l},{\text{supp }}\varphi \subseteq K\subset B_{R}(0)\\&\leq R^{|\alpha |}\sup _{x\in B_{R}(0)}\left|\partial _{\beta }\varphi _{l}(x)-\partial _{\beta }\varphi (x)\right|&\forall i\in \{1,\ldots ,d\},(x_{1},\ldots ,x_{d})\in {\overline {B_{R}(0)}}:|x_{i}|\leq R\\&=R^{|\alpha |}\sup _{x\in \mathbb {R} ^{d}}\left|\partial _{\beta }\varphi _{l}(x)-\partial _{\beta }\varphi (x)\right|&{\text{ as }}{\text{supp }}\varphi _{l},{\text{supp }}\varphi \subseteq K\subset B_{R}(0)\\&=R^{|\alpha |}\left\|\partial _{\beta }\varphi _{l}(x)-\partial _{\beta }\varphi (x)\right\|_{\infty }&{\text{ definition of the supremum norm}}\\&\to 0,l\to \infty &{\text{ since }}\varphi _{l}\to \varphi {\text{ in }}{\mathcal {D}}(O)\end{aligned}}$
Therefore the sequence converges with respect to the notion of convergence for Schwartz functions.$\Box$

## The ‘testing’ property of test functions

In this section, we want to show that we can test equality of continuous functions $f,g$ by evaluating the integrals
    $\int _{\mathbb {R} ^{d}}f(x)\varphi (x)dx$ and $\int _{\mathbb {R} ^{d}}g(x)\varphi (x)dx$
for all $\varphi \in {\mathcal {D}}(O)$ (thus, evaluating the integrals for all $\varphi \in {\mathcal {S}}(\mathbb {R} ^{d})$ will also suffice as ${\mathcal {D}}(O)\subset {\mathcal {S}}(\mathbb {R} ^{d})$ due to theorem 3.9).
But before we are able to show that, we need a modified mollifier, where the modification is dependent of a parameter, and two lemmas about that modified mollifier.
Definition 3.13:
For $R\in \mathbb {R} _{>0}$, we define
    $\eta _{R}:\mathbb {R} ^{d}\to \mathbb {R} ,\eta _{R}(x)=\eta \left({\frac {x}{R}}\right){\big /}R^{d}$.
Lemma 3.14:
Let $R\in \mathbb {R} _{>0}$. Then
    ${\text{supp }}\eta _{R}={\overline {B_{R}(0)}}$.
Proof:
From the definition of $\eta$ follows
    ${\text{supp }}\eta ={\overline {B_{1}(0)}}$.
Further, for $R\in \mathbb {R} _{>0}$
    ${\begin{aligned}{\frac {x}{R}}\in {\overline {B_{1}(0)}}&\Leftrightarrow \left\|{\frac {x}{R}}\right\|\leq 1\\&\Leftrightarrow \|x\|\leq R\\&\Leftrightarrow x\in {\overline {B_{R}(0)}}\end{aligned}}$
Therefore, and since
    $x\in {\text{supp }}\eta _{R}\Leftrightarrow {\frac {x}{R}}\in {\text{supp }}\eta$
, we have:
    $x\in {\text{supp }}\eta _{R}\Leftrightarrow x\in {\overline {B_{R}(0)}}$$\Box$
In order to prove the next lemma, we need the following theorem from integration theory:
Theorem 3.15: (Multi-dimensional integration by substitution)
If $O,U\subseteq \mathbb {R} ^{d}$ are open, and $\psi :U\to O$ is a diffeomorphism, then
    $\int _{O}f(x)dx=\int _{U}f(\psi (x))|\det J_{\psi }(x)|dx$
We will omit the proof, as understanding it is not very important for understanding this wikibook.
Lemma 3.16:
Let $R\in \mathbb {R} _{>0}$. Then
    $\int _{\mathbb {R} ^{d}}\eta _{R}(x)dx=1$.
Proof:
    ${\begin{aligned}\int _{\mathbb {R} ^{d}}\eta _{R}(x)dx&=\int _{\mathbb {R} ^{d}}\eta \left({\frac {x}{R}}\right){\big /}R^{d}dx&{\text{Def. of }}\eta _{R}\\&=\int _{\mathbb {R} ^{d}}\eta (x)dx&{\text{integration by substitution using }}x\mapsto Rx\\&=\int _{B_{1}(0)}\eta (x)dx&{\text{Def. of }}\eta \\&={\frac {\int _{B_{1}(0)}e^{-{\frac {1}{1-\|x\|}}}dx}{\int _{B_{1}(0)}e^{-{\frac {1}{1-\|x\|}}}dx}}&{\text{Def. of }}\eta \\&=1\end{aligned}}$$\Box$
Now we are ready to prove the ‘testing’ property of test functions:
Theorem 3.17:
Let $f,g:\mathbb {R} ^{d}\to \mathbb {R}$ be continuous. If
    $\forall \varphi \in {\mathcal {D}}(O):\int _{\mathbb {R} ^{d}}\varphi (x)f(x)dx=\int _{\mathbb {R} ^{d}}\varphi (x)g(x)dx$,
then $f=g$.
Proof:
Let $x\in \mathbb {R} ^{d}$ be arbitrary, and let $\epsilon \in \mathbb {R} _{>0}$. Since $f$ is continuous, there exists a $\delta \in \mathbb {R} _{>0}$ such that
    $\forall y\in {\overline {B_{\delta }(x)}}:|f(x)-f(y)|<\epsilon$
Then we have
    ${\begin{aligned}\left|f(x)-\int _{\mathbb {R} ^{d}}f(y)\eta _{\delta }(x-y)dy\right|&=\left|\int _{\mathbb {R} ^{d}}(f(x)-f(y))\eta _{\delta }(x-y)dy\right|&{\text{lemma 3.16}}\\&\leq \int _{\mathbb {R} ^{d}}|f(x)-f(y)|\eta _{\delta }(x-y)dy&{\text{triangle ineq. for the }}\int {\text{ and }}\eta _{\delta }\geq 0\\&=\int _{\overline {B_{\delta }(0)}}|f(x)-f(y)|\eta _{\delta }(x-y)dy&{\text{lemma 3.14}}\\&\leq \int _{\overline {B_{\delta }(0)}}\epsilon \eta _{\delta }(x-y)dy&{\text{monotony of the }}\int \\&\leq \epsilon &{\text{lemma 3.16 and }}\eta _{\delta }\geq 0\end{aligned}}$
Therefore, $\int _{\mathbb {R} ^{d}}f(y)\eta _{\delta }(x-y)dy\to f(x),\delta \to 0$. An analogous reasoning also shows that $\int _{\mathbb {R} ^{d}}g(y)\eta _{\delta }(x-y)dy\to g(x),\delta \to 0$. But due to the assumption, we have
    $\forall \delta \in \mathbb {R} _{>0}:\int _{\mathbb {R} ^{d}}g(y)\eta _{\delta }(x-y)dy=\int _{\mathbb {R} ^{d}}f(y)\eta _{\delta }(x-y)dy$
As limits in the reals are unique, it follows that $f(x)=g(x)$, and since $x\in \mathbb {R} ^{d}$ was arbitrary, we obtain $f=g$.$\Box$
Remark 3.18: Let $f,g:\mathbb {R} ^{d}\to \mathbb {R}$ be continuous. If
    $\forall \varphi \in {\mathcal {S}}(\mathbb {R} ^{d}):\int _{\mathbb {R} ^{d}}\varphi (x)f(x)dx=\int _{\mathbb {R} ^{d}}\varphi (x)g(x)dx$,
then $f=g$.
Proof:
This follows from all bump functions being Schwartz functions, which is why the requirements for theorem 3.17 are met.$\Box$

## Exercises

* Let $b\in \mathbb {R}$ and $f:\mathbb {R} \to \mathbb {R}$ be constant on the interval $[b-1,b)$. Show that
    $\forall y\in [b-1,b):\int _{\mathbb {R} }\chi _{[b-1,b)}(x)f(x)dx=f(y)$
* Prove that the standard mollifier as defined in example 3.4 is a bump function by proceeding as follows:
** Prove that the functionis contained in ${\mathcal {C}}^{\infty }(\mathbb {R} )$.
    $x\mapsto {\begin{cases}e^{-{\frac {1}{x}}}&x>0\\0&x\leq 0\end{cases}}$
** Prove that the functionis contained in ${\mathcal {C}}^{\infty }(\mathbb {R} ^{d})$.
    $x\mapsto 1-\|x\|$
** Conclude that $\eta \in {\mathcal {C}}^{\infty }(\mathbb {R} ^{d})$.
** Prove that ${\text{supp }}\eta$ is compact by calculating ${\text{supp }}\eta$ explicitly.
* Let $O\subseteq \mathbb {R} ^{d}$ be open, let $\varphi \in {\mathcal {D}}(O)$ and let $\phi \in {\mathcal {S}}(\mathbb {R} ^{d})$. Prove that if $\alpha ,\beta \in \mathbb {N} _{0}^{d}$, then $\partial _{\alpha }\varphi \in {\mathcal {D}}(O)$ and $x^{\alpha }\partial _{\beta }\phi \in {\mathcal {S}}(\mathbb {R} ^{d})$.
* Let $O\subseteq \mathbb {R} ^{d}$ be open, let $\varphi _{1},\ldots ,\varphi _{n}\in {\mathcal {D}}(O)$ be bump functions and let $c_{1},\ldots ,c_{n}\in \mathbb {R}$. Prove that $\sum _{j=1}^{n}c_{j}\varphi _{j}\in {\mathcal {D}}(O)$.
* Let $\phi _{1},\ldots ,\phi _{n}$ be Schwartz functions functions and let $c_{1},\ldots ,c_{n}\in \mathbb {R}$. Prove that $\sum _{j=1}^{n}c_{j}\phi _{j}$ is a Schwartz function.
* Let $\alpha \in \mathbb {N} _{0}^{d}$, let $p(x):=\sum _{\varsigma \leq \alpha }c_{\varsigma }x^{\varsigma }$ be a polynomial, and let $\phi _{l}\to \phi$ in the sense of Schwartz functions. Prove that $p\phi _{l}\to p\phi$ in the sense of Schwartz functions.

# Distributions
## Distributions and tempered distributions

Definition 4.1:
Let $O\subseteq \mathbb {R} ^{d}$ be open, and let ${\mathcal {T}}:{\mathcal {D}}(O)\to \mathbb {R}$ be a function. We call ${\mathcal {T}}$ a distribution iff
* ${\mathcal {T}}$ is linear ($\forall \varphi ,\vartheta \in {\mathcal {D}}(O),b,c\in \mathbb {R} :{\mathcal {T}}(b\varphi +c\vartheta )=b{\mathcal {T}}(\varphi )+c{\mathcal {T}}(\vartheta )$)
* ${\mathcal {T}}$ is sequentially continuous (if $\varphi _{l}\to \varphi$ in the notion of convergence of bump functions, then ${\mathcal {T}}(\varphi _{l})\to {\mathcal {T}}(\varphi )$ in the reals)
The set of all distributions for ${\mathcal {D}}(O)$ we denote by ${\mathcal {D}}(O)^{*}$
Definition 4.2:
Let ${\mathcal {T}}:{\mathcal {S}}(\mathbb {R} ^{d})\to \mathbb {R}$ be a function. We call ${\mathcal {T}}$ a tempered distribution iff
* ${\mathcal {T}}$ is linear ($\forall \varphi ,\vartheta \in {\mathcal {S}}(\mathbb {R} ^{d}),b,c\in \mathbb {R} :{\mathcal {T}}(b\varphi +c\vartheta )=b{\mathcal {T}}(\varphi )+c{\mathcal {T}}(\vartheta )$)
* ${\mathcal {T}}$ is sequentially continuous (if $\varphi _{l}\to \varphi$ in the notion of convergence of Schwartz functions, then ${\mathcal {T}}(\varphi _{l})\to {\mathcal {T}}(\varphi )$ in the reals)
The set of all tempered distributions we denote by ${\mathcal {S}}(\mathbb {R} ^{d})$.
Theorem 4.3:
Let ${\mathcal {T}}$ be a tempered distribution. Then the restriction of ${\mathcal {T}}$ to bump functions is a distribution.
Proof:
Let ${\mathcal {T}}$ be a tempered distribution, and let $O\subseteq \mathbb {R} ^{d}$ be open.
1.
We show that ${\mathcal {T}}(\varphi )$ has a well-defined value for $\varphi \in {\mathcal {D}}(O)$.
Due to theorem 3.9, every bump function is a Schwartz function, which is why the expression
    ${\mathcal {T}}(\varphi )$
makes sense for every $\varphi \in {\mathcal {D}}(O)$.
2.
We show that the restriction is linear.
Let $a,b\in \mathbb {R}$ and $\varphi ,\vartheta \in {\mathcal {D}}(O)$. Since due to theorem 3.9 $\varphi$ and $\vartheta$ are Schwartz functions as well, we have
    $\forall a,b\in \mathbb {R} ,\varphi ,\vartheta \in {\mathcal {D}}(O):{\mathcal {T}}(a\varphi +b\vartheta )=a{\mathcal {T}}(\varphi )+b{\mathcal {T}}(\vartheta )$
due to the linearity of ${\mathcal {T}}$ for all Schwartz functions. Thus ${\mathcal {T}}$ is also linear for bump functions.
3.
We show that the restriction of ${\mathcal {T}}$ to ${\mathcal {D}}(O)$ is sequentially continuous. Let $\varphi _{l}\to \varphi$ in the notion of convergence of bump functions. Due to theorem 3.11, $\varphi _{l}\to \varphi$ in the notion of convergence of Schwartz functions. Since ${\mathcal {T}}$ as a tempered distribution is sequentially continuous, ${\mathcal {T}}(\varphi _{l})\to {\mathcal {T}}(\varphi )$.$\Box$

## The convolution

Definition 4.4:
Let $f,g:\mathbb {R} ^{d}\to \mathbb {R}$. The integral
    $f*g:\mathbb {R} ^{d}\to \mathbb {R} ,(f*g)(y):=\int _{\mathbb {R} ^{d}}f(x)g(y-x)dx$
is called convolution of $f$ and $g$ and denoted by $f*g$ if it exists.
The convolution of two functions may not always exist, but there are sufficient conditions for it to exist:
Theorem 4.5:
Let $p,q\in [1,\infty ]$ such that ${\frac {1}{p}}+{\frac {1}{q}}=1$ and let $f\in L^{p}(\mathbb {R} ^{d})$ and $g\in L^{q}(\mathbb {R} ^{d})$. Then for all $y\in O$, the integral
    $\int _{\mathbb {R} ^{d}}f(x)g(y-x)dx$
has a well-defined real value.
Proof:
Due to Hölder's inequality,
    $\int _{\mathbb {R} ^{d}}|f(x)g(y-x)|dx\leq \left(\int _{\mathbb {R} ^{d}}|f(x)|^{p}dx\right)^{1/p}\left(\int _{\mathbb {R} ^{d}}|g(y-x)|^{q}dx\right)^{1/q}<\infty$.$\Box$
We shall now prove that the convolution is commutative, i. e. $f*g=g*f$.
Theorem 4.6:
Let $p,q\in [1,\infty ]$ such that ${\frac {1}{p}}+{\frac {1}{q}}=1$ (where ${\frac {1}{\infty }}=0$) and let $f\in L^{p}(\mathbb {R} ^{d})$ and $g\in L^{q}(\mathbb {R} ^{d})$. Then for all $y\in \mathbb {R} ^{d}$:
    $\forall y\in \mathbb {R} ^{d}:(f*g)(y)=(g*f)(y)$
Proof:
We apply multi-dimensional integration by substitution using the diffeomorphism $x\mapsto y-x$ to obtain
    $(f*g)(y)=\int _{\mathbb {R} ^{d}}f(x)g(y-x)dx=\int _{\mathbb {R} ^{d}}f(y-x)g(x)dx=(g*f)(y)$.$\Box$
Lemma 4.7:
Let $O\subseteq \mathbb {R} ^{d}$ be open and let $f\in L^{1}(\mathbb {R} ^{d})$. Then $f*\eta _{\delta }\in {\mathcal {C}}^{\infty }(\mathbb {R} ^{d})$.
Proof:
Let $\alpha \in \mathbb {N} _{0}^{d}$ be arbitrary. Then, since for all $y\in \mathbb {R} ^{d}$
    $\int _{\mathbb {R} ^{d}}|f(x)\partial _{\alpha }\eta _{\delta }(y-x)|dx\leq \|\partial _{\alpha }\eta _{\delta }\|_{\infty }\int _{\mathbb {R} ^{d}}|f(x)|dx$
and further
    $|f(x)\partial _{\alpha }\eta _{\delta }(y-x)|\leq |f(x)|$,
Leibniz' integral rule (theorem 2.2) is applicable, and by repeated application of Leibniz' integral rule we obtain
    $\partial _{\alpha }f*\eta _{\delta }=f*\partial _{\alpha }\eta _{\delta }$.$\Box$

## Regular distributions

In this section, we shortly study a class of distributions which we call regular distributions. In particular, we will see that for certain kinds of functions there exist corresponding distributions.
Definition 4.8:
Let $O\subseteq \mathbb {R} ^{d}$ be an open set and let ${\mathcal {T}}\in {\mathcal {D}}(O)^{*}$. If for all $\varphi \in {\mathcal {D}}(O)$ ${\mathcal {T}}(\varphi )$ can be written as
    ${\mathcal {T}}(\varphi )=\int _{O}f(x)\varphi (x)dx$
for a function $f:O\to \mathbb {R}$ which is independent of $\varphi$, then we call ${\mathcal {T}}$ a regular distribution.
Definition 4.9:
Let ${\mathcal {T}}\in {\mathcal {S}}(\mathbb {R} ^{d})^{*}$. If for all $\phi \in {\mathcal {S}}(\mathbb {R} ^{d})$ ${\mathcal {T}}(\phi )$ can be written as
    ${\mathcal {T}}(\phi )=\int _{\mathbb {R} ^{d}}f(x)\phi (x)dx$
for a function $f:\mathbb {R} ^{d}\to \mathbb {R}$ which is independent of $\phi$, then we call ${\mathcal {T}}$ a regular tempered distribution.
Two questions related to this definition could be asked: Given a function $f:\mathbb {R} ^{d}\to \mathbb {R}$, is ${\mathcal {T}}_{f}:{\mathcal {D}}(O)\to \mathbb {R}$ for $O\subseteq \mathbb {R} ^{d}$ open given by
    ${\mathcal {T}}_{f}(\varphi ):=\int _{O}f(x)\varphi (x)dx$
well-defined and a distribution? Or is ${\mathcal {T}}_{f}:{\mathcal {S}}(\mathbb {R} ^{d})\to \mathbb {R}$ given by
    ${\mathcal {T}}_{f}(\phi ):=\int _{\mathbb {R} ^{d}}f(x)\phi (x)dx$
well-defined and a tempered distribution? In general, the answer to these two questions is no, but both questions can be answered with yes if the respective function $f$ has the respectively right properties, as the following two theorems show. But before we state the first theorem, we have to define what local integrability means, because in the case of bump functions, local integrability will be exactly the property which $f$ needs in order to define a corresponding regular distribution:
Definition 4.10:
Let $O\subseteq \mathbb {R} ^{d}$ be open, $f:O\to \mathbb {R}$ be a function. We say that $f$ is locally integrable iff for all compact subsets $K$ of $O$
    $-\infty <\int _{K}f(x)dx<\infty$
We write $f\in L_{\text{loc}}^{1}(O)$.
Now we are ready to give some sufficient conditions on $f$ to define a corresponding regular distribution or regular tempered distribution by the way of
    ${\mathcal {T}}_{f}:{\mathcal {D}}(O)\to \mathbb {R} ,{\mathcal {T}}_{f}(\varphi ):=\int _{O}f(x)\varphi (x)dx$
or
    ${\mathcal {T}}_{f}:{\mathcal {S}}(\mathbb {R} ^{d})\to \mathbb {R} ,{\mathcal {T}}_{f}(\phi ):=\int _{\mathbb {R} ^{d}}f(x)\phi (x)dx$:
Theorem 4.11:
Let $O\subseteq \mathbb {R} ^{d}$ be open, and let $f:O\to \mathbb {R}$ be a function. Then
    ${\mathcal {T}}_{f}:{\mathcal {D}}(O)\to \mathbb {R} ,{\mathcal {T}}_{f}(\varphi ):=\int _{O}f(x)\varphi (x)dx$
is a regular distribution iff $f\in L_{\text{loc}}^{1}(O)$.
Proof:
1.
We show that if $f\in L_{\text{loc}}^{1}(O)$, then ${\mathcal {T}}_{f}:{\mathcal {D}}(O)\to \mathbb {R}$ is a distribution.
Well-definedness follows from the triangle inequality of the integral and the monotony of the integral:
    ${\begin{aligned}\left|\int _{U}\varphi (x)f(x)dx\right|\leq \int _{U}|\varphi (x)f(x)|dx=\int _{{\text{supp }}\varphi }|\varphi (x)f(x)|dx\\\leq \int _{{\text{supp }}\varphi }\|\varphi \|_{\infty }|f(x)|dx=\|\varphi \|_{\infty }\int _{{\text{supp }}\varphi }|f(x)|dx<\infty \end{aligned}}$
In order to have an absolute value strictly less than infinity, the first integral must have a well-defined value in the first place. Therefore, ${\mathcal {T}}_{f}$ really maps to $\mathbb {R}$ and well-definedness is proven.
Continuity follows similarly due to
    $|T_{f}\varphi _{l}-T_{f}\varphi |=\left|\int _{K}(\varphi _{l}-\varphi )(x)f(x)dx\right|\leq \|\varphi _{l}-\varphi \|_{\infty }\underbrace {\int _{K}|f(x)|dx} _{{\text{independent of }}l}\to 0,l\to \infty$
, where $K$ is the compact set in which all the supports of $\varphi _{l},l\in \mathbb {N}$ and $\varphi$ are contained (remember: The existence of a compact set such that all the supports of $\varphi _{l},l\in \mathbb {N}$ are contained in it is a part of the definition of convergence in ${\mathcal {D}}(O)$, see the last chapter. As in the proof of theorem 3.11, we also conclude that the support of $\varphi$ is also contained in $K$).
Linearity follows due to the linearity of the integral.
2.
We show that ${\mathcal {T}}_{f}$ is a distribution, then $f\in L_{\text{loc}}^{1}(O)$ (in fact, we even show that if ${\mathcal {T}}_{f}(\varphi )$ has a well-defined real value for every $\varphi \in {\mathcal {D}}(O)$, then $f\in L_{\text{loc}}^{1}(O)$. Therefore, by part 1 of this proof, which showed that if $f\in L_{\text{loc}}^{1}(O)$ it follows that ${\mathcal {T}}_{f}$ is a distribution in ${\mathcal {D}}^{*}(O)$, we have that if ${\mathcal {T}}_{f}(\varphi )$ is a well-defined real number for every $\varphi \in {\mathcal {D}}(O)$, ${\mathcal {T}}_{f}$ is a distribution in ${\mathcal {D}}(O)$.
Let $K\subset U$ be an arbitrary compact set. We define
    $\mu :K\to \mathbb {R} ,\mu (\xi ):=\inf _{x\in \mathbb {R} ^{d}\setminus O}\|\xi -x\|$
$\mu$ is continuous, even Lipschitz continuous with Lipschitz constant $1$: Let $\xi ,\iota \in \mathbb {R} ^{d}$. Due to the triangle inequality, both
    $\forall (x,y)\in \mathbb {R} ^{2}:\|\xi -x\|\leq \|\xi -\iota \|+\|\iota -y\|+\|y-x\|~~~~~(*)$
and
    $\forall (x,y)\in \mathbb {R} ^{2}:\|\iota -y\|\leq \|\iota -\xi \|+\|\xi -x\|+\|x-y\|~~~~~(**)$
, which can be seen by applying the triangle inequality twice.
We choose sequences $(x_{l})_{l\in \mathbb {N} }$ and $(y_{m})_{m\in \mathbb {N} }$ in $\mathbb {R} ^{d}\setminus O$ such that $\lim _{l\to \infty }\|\xi -x_{l}\|=\mu (\xi )$ and $\lim _{m\to \infty }\|\iota -y_{m}\|=\mu (\iota )$ and consider two cases. First, we consider what happens if $\mu (\xi )\geq \mu (\iota )$. Then we have
    ${\begin{aligned}|\mu (\xi )-\mu (\iota )|&=\mu (\xi )-\mu (\iota )&\\&=\inf _{x\in \mathbb {R} ^{d}\setminus O}\|\xi -x\|-\inf _{y\in \mathbb {R} ^{d}\setminus O}\|\iota -y\|&\\&=\inf _{x\in \mathbb {R} ^{d}\setminus O}\|\xi -x\|-\lim _{m\to \infty }\|\iota -y_{m}\|&\\&=\lim _{m\to \infty }\inf _{x\in \mathbb {R} ^{d}\setminus O}\left(\|\xi -x\|-\|\iota -y_{m}\|\right)&\\&\leq \lim _{m\to \infty }\inf _{x\in \mathbb {R} ^{d}\setminus O}\left(\|\xi -\iota \|+\|x-y_{m}\|\right)&(*){\text{ with }}y=y_{m}\\&=\|\xi -\iota \|&\end{aligned}}$.
Second, we consider what happens if $\mu (\xi )\leq \mu (\iota )$:
    ${\begin{aligned}|\mu (\xi )-\mu (\iota )|&=\mu (\iota )-\mu (\xi )&\\&=\inf _{y\in \mathbb {R} ^{d}\setminus O}\|\iota -y\|-\inf _{x\in \mathbb {R} ^{d}\setminus O}\|\xi -x\|&\\&=\inf _{y\in \mathbb {R} ^{d}\setminus O}\|\iota -y\|-\lim _{l\to \infty }\|\xi -x_{l}\|&\\&=\lim _{l\to \infty }\inf _{y\in \mathbb {R} ^{d}\setminus O}\left(\|\iota -y\|-\|\xi -x_{l}\|\right)&\\&\leq \lim _{l\to \infty }\inf _{y\in \mathbb {R} ^{d}\setminus O}\left(\|\xi -\iota \|+\|y-x_{l}\|\right)&(**){\text{ with }}x=x_{l}\\&=\|\xi -\iota \|&\end{aligned}}$
Since always either $\mu (\xi )\geq \mu (\iota )$ or $\mu (\xi )\leq \mu (\iota )$, we have proven Lipschitz continuity and thus continuity. By the extreme value theorem, $\mu$ therefore has a minimum $\kappa \in \mathbb {R} ^{d}$. Since $\mu (\kappa )=0$ would mean that $\|\xi -x_{l}\|\to 0,l\to \infty$ for a sequence $(x_{l})_{l\in \mathbb {N} }$ in $\mathbb {R} ^{d}\setminus O$ which is a contradiction as $\mathbb {R} ^{d}\setminus O$ is closed and $\kappa \in K\subset O$, we have $\mu (\kappa )>0$.
Hence, if we define $\delta :=\mu (\kappa )$, then $\delta >0$. Further, the function
    $\vartheta :\mathbb {R} ^{d}\to \mathbb {R} ,\vartheta (x):=(\chi _{K+B_{\delta /4}(0)}*\eta _{\delta /4})(x)=\int _{\mathbb {R} ^{d}}\eta _{\delta /4}(y)\chi _{K+B_{\delta /4}(0)}(x-y)dy=\int _{B_{\delta /4}(0)}\eta _{\delta /4}(y)\chi _{K+B_{\delta /4}(0)}(x-y)dy$
has support contained in $O$, is equal to $1$ within $K$ and further is contained in ${\mathcal {C}}^{\infty }(\mathbb {R} ^{d})$ due to lemma 4.7. Hence, it is also contained in ${\mathcal {D}}(O)$. Since therefore, by the monotonicity of the integral
    $\int _{K}|f(x)|dx=\int _{O}|f(x)|\chi _{K}(x)dx\leq \int _{\mathbb {R} ^{d}}|f(x)|\vartheta (x)dx$
, $f$ is indeed locally integrable.$\Box$
Theorem 4.12:
Let $f\in L^{2}(\mathbb {R} ^{d})$, i. e.
    $\int _{\mathbb {R} ^{d}}|f(x)|^{2}dx<\infty$
Then
    ${\mathcal {T}}_{f}:{\mathcal {S}}(\mathbb {R} ^{d})\to \mathbb {R} ,{\mathcal {T}}_{f}(\phi ):=\int _{\mathbb {R} ^{d}}f(x)\phi (x)dx$
is a regular tempered distribution.
Proof:
From Hölder's inequality we obtain
    $\int _{\mathbb {R} ^{d}}|\phi (x)||f(x)|dx\leq \|\phi \|_{L^{2}}\|f\|_{L^{2}}<\infty$.
Hence, ${\mathcal {T}}_{f}$ is well-defined.
Due to the triangle inequality for integrals and Hölder's inequality, we have
    $|T_{f}(\phi _{l})-T_{f}(\phi )|\leq \int _{\mathbb {R} ^{d}}|(\phi _{l}-\phi )(x)||f(x)|dx\leq \|\phi _{l}-\phi \|_{L^{2}}\|f\|_{L^{2}}$
Furthermore
    ${\begin{aligned}\|\phi _{l}-\phi \|_{L^{2}}^{2}&\leq \|\phi _{l}-\phi \|_{\infty }\int _{\mathbb {R} ^{d}}|(\phi _{l}-\phi )(x)|dx\\&=\|\phi _{l}-\phi \|_{\infty }\int _{\mathbb {R} ^{d}}\prod _{j=1}^{d}(1+x_{j}^{2})|(\phi _{l}-\phi )(x)|{\frac {1}{\prod _{j=1}^{d}(1+x_{j}^{2})}}dx\\&\leq \|\phi _{l}-\phi \|_{\infty }\left\|\prod _{j=1}^{d}(1+x_{j}^{2})(\phi _{l}-\phi )\right\|_{\infty }\underbrace {\int _{\mathbb {R} ^{d}}{\frac {1}{\prod _{j=1}^{d}(1+x_{j}^{2})}}dx} _{=\pi ^{d}}\end{aligned}}$.
If $\phi _{l}\to \phi$ in the notion of convergence of the Schwartz function space, then this expression goes to zero. Therefore, continuity is verified.
Linearity follows from the linearity of the integral.$\Box$

## Equicontinuity

We now introduce the concept of equicontinuity.
Definition 4.13:
Let $M$ be a metric space equipped with a metric which we shall denote by $d$ here, let $X\subseteq M$ be a set in $M$, and let ${\mathcal {Q}}$ be a set of continuous functions mapping from $X$ to the real numbers $\mathbb {R}$. We call this set ${\mathcal {Q}}$ equicontinuous if and only if
    $\forall x\in X:\exists \delta \in \mathbb {R} _{>0}:\forall y\in X:d(x,y)<\delta \Rightarrow \forall f\in {\mathcal {Q}}:|f(x)-f(y)|<\epsilon$.
So equicontinuity is in fact defined for sets of continuous functions mapping from $X$ (a set in a metric space) to the real numbers $\mathbb {R}$.
Theorem 4.14:
Let $M$ be a metric space equipped with a metric which we shall denote by $d$, let $Q\subseteq M$ be a sequentially compact set in $M$, and let ${\mathcal {Q}}$ be an equicontinuous set of continuous functions from $Q$ to the real numbers $\mathbb {R}$. Then follows: If $(f_{l})_{l\in \mathbb {N} }$ is a sequence in ${\mathcal {Q}}$ such that $f_{l}(x)$ has a limit for each $x\in Q$, then for the function $f(x):=\lim _{l\to \infty }f_{l}(x)$, which maps from $Q$ to $\mathbb {R}$, it follows $f_{l}\to f$ uniformly.
Proof:
In order to prove uniform convergence, by definition we must prove that for all $\epsilon >0$, there exists an $N\in \mathbb {N}$ such that for all $l\geq N:\forall x\in Q:|f_{l}(x)-f(x)|<\epsilon$.
So let's assume the contrary, which equals by negating the logical statement
    $\exists \epsilon >0:\forall N\in \mathbb {N} :\exists l\geq N:\exists x\in Q:|f_{l}(x)-f(x)|\geq \epsilon$.
We choose a sequence $(x_{m})_{m\in \mathbb {N} }$ in $Q$. We take $x_{1}$ in $Q$ such that $|f_{l_{1}}(x_{1})-f(x_{1})|\geq \epsilon$ for an arbitrarily chosen $l_{1}\in \mathbb {N}$ and if we have already chosen $x_{k}$ and $l_{k}$ for all $k\in \{1,\ldots ,m\}$, we choose $x_{m+1}$ such that $|f_{l_{m+1}}(x_{m+1})-f(x_{m+1})|\geq \epsilon$, where $l_{m+1}$ is greater than $l_{m}$.
As $Q$ is sequentially compact, there is a convergent subsequence $(x_{m_{j}})_{j\in \mathbb {N} }$ of $(x_{m})_{m\in \mathbb {N} }$. Let us call the limit of that subsequence sequence $x$.
As ${\mathcal {Q}}$ is equicontinuous, we can choose $\delta \in \mathbb {R} _{>0}$ such that
    $\|x-y\|<\delta \Rightarrow \forall f\in {\mathcal {Q}}:|f(x)-f(y)|<{\frac {\epsilon }{4}}$.
Further, since $x_{m_{j}}\to x$ (if $j\to \infty$ of course), we may choose $J\in \mathbb {N}$ such that
    $\forall j\geq J:\|x_{m_{j}}-x\|<\delta$.
But then follows for $j\geq J$ and the reverse triangle inequality:
    $|f_{l_{m_{j}}}(x)-f(x)|\geq \left||f_{l_{m_{j}}}(x)-f(x_{m_{j}})|-|f(x_{m_{j}})-f(x)|\right|$
Since we had $|f(x_{m_{j}})-f(x)|<{\frac {\epsilon }{4}}$, the reverse triangle inequality and the definition of t
    $|f_{l_{m_{j}}}(x)-f(x_{m_{j}})|\geq \left||f_{l_{m_{j}}}(x_{m_{j}})-f(x_{m_{j}})|-|f_{l_{m_{j}}}(x)-f_{l_{m_{j}}}(x_{m_{j}})|\right|\geq \epsilon -{\frac {\epsilon }{4}}$
, we obtain:
    ${\begin{aligned}|f_{l_{m_{j}}}(x)-f(x)|&\geq \left||f_{l_{m_{j}}}(x)-f(x_{m_{j}})|-|f(x_{m_{j}})-f(x)|\right|\\&=|f_{l_{m_{j}}}(x)-f(x_{m_{j}})|-|f(x_{m_{j}})-f(x)|\\&\geq \epsilon -{\frac {\epsilon }{4}}-{\frac {\epsilon }{4}}\\&\geq {\frac {\epsilon }{2}}\end{aligned}}$
Thus we have a contradiction to $f_{l}(x)\to f(x)$.$\Box$
Theorem 4.15:
Let ${\mathcal {Q}}$ be a set of differentiable functions, mapping from the convex set $X\subseteq \mathbb {R} ^{d}$ to $\mathbb {R}$. If we have, that there exists a constant $b\in \mathbb {R} _{>0}$ such that for all functions in ${\mathcal {Q}}$, $\forall x\in X:\|\nabla f(x)\|\leq b$ (the $\nabla f$ exists for each function in ${\mathcal {Q}}$ because all functions there were required to be differentiable), then ${\mathcal {Q}}$ is equicontinuous.
Proof: We have to prove equicontinuity, so we have to prove
    $\forall x\in X:\exists \delta \in \mathbb {R} _{>0}:\forall y\in X:\|x-y\|<\delta \Rightarrow \forall f\in {\mathcal {Q}}:|f(x)-f(y)|<\epsilon$.
Let $x\in X$ be arbitrary.
We choose $\delta :={\frac {\epsilon }{b}}$.
Let $y\in X$ such that $\|x-y\|<\delta$, and let $f\in {\mathcal {Q}}$ be arbitrary. By the mean-value theorem in multiple dimensions, we obtain that there exists a $\lambda \in [0,1]$ such that:
    $f(x)-f(y)=\nabla f(\lambda x+(1-\lambda )y)\cdot (x-y)$
The element $\lambda x+(1-\lambda )y$ is inside $X$, because $X$ is convex. From the Cauchy-Schwarz inequality then follows:
    $|f(x)-f(y)|=|\nabla f(\lambda x+(1-\lambda )y)\cdot (x-y)|\leq \|\nabla f(\lambda x+(1-\lambda )y)\|\|x-y\|<b\delta ={\frac {b}{b}}\epsilon =\epsilon$$\Box$

## The generalised product rule

Definition 4.16:
If $\alpha =(\alpha _{1},\ldots ,\alpha _{d}),\beta =(\beta _{1},\ldots ,\beta _{d})\in \mathbb {N} _{0}^{d}$ are two $d$-dimensional multiindices, we define the binomial coefficient of $\alpha$ over $\beta$ as
    ${\binom {\alpha }{\beta }}:={\binom {\alpha _{1}}{\beta _{1}}}{\binom {\alpha _{2}}{\beta _{2}}}\cdots {\binom {\alpha _{d}}{\beta _{d}}}$.
We also define less or equal relation on the set of multi-indices.
Definition 4.17:
Let $\alpha =(\alpha _{1},\ldots ,\alpha _{d}),\beta =(\beta _{1},\ldots ,\beta _{d})\in \mathbb {N} _{0}^{d}$ be two $d$-dimensional multiindices. We define $\beta$ to be less or equal than $\alpha$ if and only if
    $\beta \leq \alpha :\Leftrightarrow \forall n\in \{1,\ldots ,d\}:\beta _{n}\leq \alpha _{n}$.
For $d\geq 2$, there are vectors $\alpha ,\beta \in \mathbb {N} _{0}^{d}$ such that neither $\alpha \leq \beta$ nor $\beta \leq \alpha$. For $d=2$, the following two vectors are examples for this:
    $\alpha =(1,0),\beta =(0,1)$
This example can be generalised to higher dimensions (see exercise 6).
With these multiindex definitions, we are able to write down a more general version of the product rule. But in order to prove it, we need another lemma.
Lemma 4.18:
If $n\in \{1,\ldots ,d\}$ and $e_{n}:=(0,\ldots ,0,1,0,\ldots ,0)$, where the $1$ is at the $n$-th place, we have
    ${\binom {\alpha -e_{n}}{\beta -e_{n}}}+{\binom {\alpha -e_{n}}{\beta }}={\binom {\alpha }{\beta }}$
for arbitrary multiindices $\alpha ,\beta \in \mathbb {N} _{0}^{d}$.
Proof:
For the ordinary binomial coefficients for natural numbers, we had the formula
    ${\binom {n-1}{k-1}}+{\binom {n-1}{k}}={\binom {n}{k}}$.
Therefore,
    ${\begin{aligned}{\binom {\alpha -e_{n}}{\beta -e_{n}}}+{\binom {\alpha -e_{n}}{\beta }}&={\binom {\alpha _{1}}{\beta _{1}}}\cdots {\binom {\alpha _{n}-1}{\beta _{n}-1}}\cdots {\binom {\alpha _{d}}{\beta _{d}}}+{\binom {\alpha _{1}}{\beta _{1}}}\cdots {\binom {\alpha _{n}-1}{\beta _{n}}}\cdots {\binom {\alpha _{d}}{\beta _{d}}}\\&={\binom {\alpha _{1}}{\beta _{1}}}\cdots \left({\binom {\alpha _{n}-1}{\beta _{n}-1}}+{\binom {\alpha _{n}-1}{\beta _{n}}}\right)\cdots {\binom {\alpha _{d}}{\beta _{d}}}\\&={\binom {\alpha _{1}}{\beta _{1}}}\cdots {\binom {\alpha _{n}}{\beta _{n}}}\cdots {\binom {\alpha _{d}}{\beta _{d}}}={\binom {\alpha }{\beta }}\end{aligned}}$$\Box$
This is the general product rule:
Theorem 4.19:
Let $f\in {\mathcal {C}}^{n}(\mathbb {R} ^{d})$ and let $|\alpha |\leq n$. Then
    $\partial _{\alpha }(fg)=\sum _{\beta \leq \alpha }{\binom {\alpha }{\beta }}\partial _{\beta }f\partial _{\alpha -\beta }g$
Proof:
We prove the claim by induction over $|\alpha |$.
1.
We start with the induction base $|\alpha |=0$. Then the formula just reads
    $f(x)g(x)=f(x)g(x)$
, and this is true. Therefore, we have completed the induction base.
2.
Next, we do the induction step. Let's assume the claim is true for all $\alpha \in \mathbb {N} _{0}^{d}$ such that $|\alpha |=n$. Let now $\alpha \in \mathbb {N} _{0}^{d}$ such that $|\alpha |=n+1$. Let's choose $k\in \{1,\ldots ,d\}$ such that $\alpha _{k}>0$ (we may do this because $|\alpha |=k+1>0$). We define again $e_{k}=(0,\ldots ,0,1,0,\ldots ,0)$, where the $1$ is at the $k$-th place. Due to Schwarz' theorem and the ordinary product rule, we have
    $\partial _{\alpha }fg=\partial _{\alpha -e_{k}}\left(\partial _{x_{k}}fg\right)=\partial _{\alpha -e_{k}}\left(\partial _{x_{k}}fg+f\partial _{x_{k}}g\right)$.
By linearity of derivatives and induction hypothesis, we have
    ${\begin{aligned}\partial _{\alpha -e_{k}}\left(\partial _{x_{k}}fg+f\partial _{x_{k}}g\right)&=\partial _{\alpha -e_{k}}\left(\partial _{x_{k}}fg\right)+\partial _{\alpha -e_{k}}\left(f\partial _{x_{k}}g\right)\\&=\sum _{\varsigma \leq \alpha -e_{k}}{\binom {\alpha -e_{k}}{\varsigma }}\partial _{\varsigma }\partial _{x_{k}}f\partial _{\alpha -e_{k}-\varsigma }g+\sum _{\varsigma \leq \alpha -e_{k}}{\binom {\alpha -e_{k}}{\varsigma }}\partial _{\varsigma }f\partial _{\alpha -e_{k}-\varsigma }\partial _{x_{k}}g\end{aligned}}$.
Since
    $\partial _{\alpha -e_{k}-\varsigma }=\partial _{\alpha -(\varsigma +e_{k})}$
and
    $\{\varsigma \in \mathbb {N} _{0}^{d}|0\leq \varsigma \leq \alpha -e_{k}\}=\{\varsigma -e_{k}\in \mathbb {N} _{0}^{d}|e_{k}\leq \varsigma \leq \alpha \}$,
we are allowed to shift indices in the first of the two above sums, and furthermore we have by definition
    $\partial _{\varsigma }\partial _{x_{k}}=\partial _{\varsigma +e_{k}}$.
With this, we obtain
    $\sum _{\varsigma \leq \alpha -e_{k}}{\binom {\alpha -e_{k}}{\varsigma }}\partial _{\varsigma }\partial _{x_{k}}f\partial _{\alpha -e_{k}-\varsigma }g+\sum _{\varsigma \leq \alpha -e_{k}}{\binom {\alpha -e_{k}}{\varsigma }}\partial _{\varsigma }f\partial _{\alpha -e_{k}-\varsigma }\partial _{x_{k}}g=\sum _{e_{k}\leq \varsigma \leq \alpha }{\binom {\alpha -e_{k}}{\varsigma -e_{k}}}\partial _{\varsigma }f\cdot \partial _{\alpha -\varsigma }g+\sum _{\varsigma \leq \alpha -e_{k}}{\binom {\alpha -e_{k}}{\varsigma }}\partial _{\varsigma }f\partial _{\alpha -\varsigma }g$
Due to lemma 4.18,
    ${\binom {\alpha -e_{k}}{\beta -e_{i}}}+{\binom {\alpha -e_{k}}{\beta }}={\binom {\alpha }{\beta }}$.
Further, we have
    ${\binom {\alpha -e_{i}}{0}}={\binom {\alpha }{0}}=1$ where $0=(0,\ldots ,0)$ in $\mathbb {N} _{0}^{d}$,
and
    ${\binom {\alpha -e_{k}}{\alpha -e_{k}}}={\binom {\alpha }{\alpha }}=1$
(these two rules may be checked from the definition of ${\binom {\alpha }{\beta }}$). It follows
    ${\begin{aligned}\partial _{\alpha }(fg)&=\sum _{e_{k}\leq \varsigma \leq \alpha }{\binom {\alpha -e_{k}}{\varsigma -e_{k}}}\partial _{\varsigma }f\cdot \partial _{\alpha -\varsigma }g+\sum _{\varsigma \leq \alpha -e_{k}}{\binom {\alpha -e_{k}}{\varsigma }}\partial _{\varsigma }f\partial _{\alpha -\varsigma }g\\&={\binom {\alpha -e_{k}}{0}}f\partial _{\alpha }g+\sum _{e_{k}\leq \varsigma \leq \alpha -e_{k}}\left[{\binom {\alpha -e_{k}}{\varsigma -e_{k}}}+{\binom {\alpha -e_{k}}{\varsigma }}\right]\partial _{\varsigma }f\partial _{\alpha -\varsigma }g+{\binom {\alpha -e_{k}}{\alpha -e_{k}}}f\partial _{\alpha }g\\&=\sum _{\varsigma \leq \alpha }{\binom {\alpha }{\varsigma }}\partial _{\varsigma }f\partial _{\alpha -\varsigma }\end{aligned}}$.$\Box$

## Operations on Distributions

For $\varphi ,\vartheta \in {\mathcal {D}}(\mathbb {R} ^{d})$ there are operations such as the differentiation of $\varphi$, the convolution of $\varphi$ and $\vartheta$ and the multiplication of $\varphi$ and $\vartheta$. In the following section, we want to define these three operations (differentiation, convolution with $\vartheta$ and multiplication with $\vartheta$) for a distribution ${\mathcal {T}}$ instead of $\varphi$.
Lemma 4.20:
Let $O,U\subseteq \mathbb {R} ^{d}$ be open sets and let $L:{\mathcal {D}}(O)\to L_{\text{loc}}^{1}(U)$ be a linear function. If there is a linear and sequentially continuous (in the sense of definition 4.1) function ${\mathcal {L}}:{\mathcal {D}}(U)\to {\mathcal {D}}(O)$ such that
    $\forall \varphi \in {\mathcal {D}}(O),\vartheta \in {\mathcal {D}}(U):\int _{O}\varphi (x){\mathcal {L}}(\vartheta )(x)dx=\int _{U}L(\varphi )(x)\vartheta (x)dx$
, then for every distribution ${\mathcal {T}}\in {\mathcal {D}}(O)^{*}$, the function $\varphi \mapsto {\mathcal {T}}({\mathcal {L}}(\varphi ))$ is a distribution. Therefore, the function
    $\Lambda :{\mathcal {D}}(O)^{*}\to {\mathcal {D}}(U)^{*},\Lambda ({\mathcal {T}}):={\mathcal {T}}\circ {\mathcal {L}}$
really maps to ${\mathcal {D}}(U)^{*}$. This function has the property
    $\forall \varphi \in {\mathcal {D}}(O):\Lambda ({\mathcal {T}}_{\varphi })={\mathcal {T}}_{L(\varphi )}$
Proof:
We have to prove two claims: First, that the function $\varphi \mapsto {\mathcal {T}}({\mathcal {L}}(\varphi ))$ is a distribution, and second that $\Lambda$ as defined above has the property
    $\forall \varphi \in {\mathcal {D}}(O):\Lambda ({\mathcal {T}}_{\varphi })={\mathcal {T}}_{L(\varphi )}$
1.
We show that the function $\varphi \mapsto {\mathcal {T}}({\mathcal {L}}(\varphi ))$ is a distribution.
${\mathcal {T}}({\mathcal {L}}(\varphi ))$ has a well-defined value in $\mathbb {R}$ as ${\mathcal {L}}$ maps to ${\mathcal {D}}(O)$, which is exactly the preimage of ${\mathcal {T}}$. The function $\varphi \mapsto {\mathcal {T}}({\mathcal {L}}(\varphi ))$ is continuous since it is the composition of two continuous functions, and it is linear for the same reason (see exercise 2).
2.
We show that $\Lambda$ has the property
    $\forall \varphi \in {\mathcal {D}}(O):\Lambda ({\mathcal {T}}_{\varphi })={\mathcal {T}}_{L(\varphi )}$
For every $\vartheta \in {\mathcal {D}}(U)$, we have
    $\Lambda ({\mathcal {T}}_{\varphi })(\vartheta ):=({\mathcal {T}}_{\varphi }\circ {\mathcal {L}})(\vartheta ):=\int _{O}\varphi (x){\mathcal {L}}(\vartheta )(x)dx{\overset {\text{by assumption}}{=}}\int _{U}L(\varphi )(x)\vartheta (x)dx=:{\mathcal {T}}_{L(\varphi )}(\vartheta )$
Since equality of two functions is equivalent to equality of these two functions evaluated at every point, this shows the desired property.$\Box$
We also have a similar lemma for Schwartz distributions:
Lemma 4.21:
Let $L:{\mathcal {S}}(\mathbb {R} ^{d})\to L_{\text{loc}}^{1}(\mathbb {R} ^{d})$ be a linear function. If there is a linear and sequentially continuous (in the sense of definition 4.2) function ${\mathcal {L}}:{\mathcal {S}}(\mathbb {R} ^{d})\to {\mathcal {S}}(\mathbb {R} ^{d})$ such that
    $\forall \phi ,\theta \in {\mathcal {S}}(\mathbb {R} ^{d}):\int _{\mathbb {R} ^{d}}\phi (x){\mathcal {L}}(\theta )(x)dx=\int _{\mathbb {R} ^{d}}L(\phi )(x)\theta (x)dx$
, then for every distribution ${\mathcal {T}}\in S(\mathbb {R} ^{d})^{*}$, the function $\phi \mapsto {\mathcal {T}}({\mathcal {L}}(\phi ))$ is a distribution. Therefore, we may define a function
    $\Lambda :{\mathcal {S}}(\mathbb {R} ^{d})^{*}\to {\mathcal {S}}(\mathbb {R} ^{d})^{*},\Lambda ({\mathcal {T}}):={\mathcal {T}}\circ {\mathcal {L}}$
This function has the property
    $\forall \phi \in {\mathcal {S}}(\mathbb {R} ^{d}):\Lambda ({\mathcal {T}}_{\phi })={\mathcal {T}}_{L(\phi )}$
The proof is exactly word-for-word the same as the one for lemma 4.20.
Noting that multiplication, differentiation and convolution are linear, we will define these operations for distributions by taking $L$ in the two above lemmas as the respective of these three operations.
Theorem and definitions 4.22:
Let $f\in {\mathcal {C}}^{\infty }(\mathbb {R} ^{d})$, and let $O\subseteq \mathbb {R} ^{d}$ be open. Then for all $\varphi \in {\mathcal {D}}(O)$, the pointwise product $f\varphi$ is contained in ${\mathcal {D}}(O)$, and if further $f$ and all of it's derivatives are bounded by polynomials, then for all $\phi \in {\mathcal {S}}(\mathbb {R} ^{d})$ the pointwise product $f\phi$ is contained in ${\mathcal {S}}(\mathbb {R} ^{d})$. Also, if $\varphi _{l}\to \varphi$ in the sense of bump functions, then $f\varphi _{l}\to f\varphi$ in the sense of bump functions, and if $f$ and all of it's derivatives are bounded by polynomials, then $\phi _{l}\to \phi$ in the sense of Schwartz functions implies $f\phi _{l}\to f\phi$ in the sense of Schwartz functions. Further:
* Let ${\mathcal {T}}:{\mathcal {D}}(O)\to \mathbb {R}$ be a distribution. If we definethen the expression on the right hand side is well-defined and for all $\vartheta \in {\mathcal {D}}(O)$ we haveand $f{\mathcal {T}}$ is a distribution.
    $f{\mathcal {T}}:{\mathcal {D}}(O)\to \mathbb {R} ,f{\mathcal {T}}(\varphi ):={\mathcal {T}}(f\varphi )$,
    $f{\mathcal {T}}_{\vartheta }={\mathcal {T}}_{f\vartheta }$,
* Assume that $f$ and all of it's derivatives are bounded by polynomials. Let ${\mathcal {T}}:{\mathcal {S}}(\mathbb {R} ^{d})\to \mathbb {R}$ be a tempered distribution. If we definethen the expression on the right hand side is well-defined and for all $\theta \in {\mathcal {S}}(\mathbb {R} ^{d})$ we haveand $f{\mathcal {T}}$ is a tempered distribution.
    $f{\mathcal {T}}:{\mathcal {S}}(\mathbb {R} ^{d})\to \mathbb {R} ,f{\mathcal {T}}(\phi ):={\mathcal {T}}(f\phi )$,
    $f{\mathcal {T}}_{\theta }={\mathcal {T}}_{f\theta }$,
Proof:
The product of two ${\mathcal {C}}^{\infty }$ functions is again ${\mathcal {C}}^{\infty }$, and further, if $\varphi (x)=0$, then also $(f\varphi )(x)=f(x)\varphi (x)=0$. Hence, $f\varphi \in {\mathcal {D}}(O)$.
Also, if $\varphi _{l}\to \varphi$ in the sense of bump functions, then, if $K\subset \mathbb {R} ^{d}$ is a compact set such that ${\text{supp }}\varphi _{n}\subseteq K$ for all $n\in \mathbb {N}$,
    ${\begin{aligned}\|\partial _{\alpha }(f(\varphi _{l}-\varphi ))\|_{\infty }&=\left\|\sum _{\varsigma \leq \alpha }{\binom {\alpha }{\varsigma }}\partial _{\varsigma }f\partial _{\alpha -\varsigma }(\varphi _{l}-\varphi )\right\|_{\infty }\\&\leq \sum _{\varsigma \leq \alpha }\|\partial _{\varsigma }f\partial _{\alpha -\varsigma }(\varphi _{l}-\varphi )\|_{\infty }\\&\leq \sum _{\varsigma \leq \alpha }\max _{x\in K}|\partial _{\varsigma }f|\|\partial _{\alpha -\varsigma }(\varphi _{l}-\varphi )\|_{\infty }\to 0,l\to \infty \end{aligned}}$.
Hence, $f\varphi _{l}\to f\varphi$ in the sense of bump functions.
Further, also $f\phi \in {\mathcal {C}}^{\infty }(\mathbb {R} ^{d})$. Let $\alpha ,\beta \in \mathbb {N} _{0}^{d}$ be arbitrary. Then
    $\partial _{\beta }f\phi =\sum _{\varsigma \leq \beta }{\binom {\beta }{\varsigma }}\partial _{\varsigma }f\partial _{\beta -\varsigma }\phi$.
Since all the derivatives of $f$ are bounded by polynomials, by the definition of that we obtain
    $\forall x\in \mathbb {R} ^{d}:|\partial _{\varsigma }f(x)|\leq |p_{\varsigma }(x)|$
, where $p_{\varsigma },\varsigma \in \mathbb {N} _{0}^{d}$ are polynomials. Hence,
    $\|x^{\alpha }\partial _{\beta }f\phi \|_{\infty }\leq \sum _{\varsigma \leq \beta }\|x^{\alpha }p_{\varsigma }\partial _{\beta -\varsigma }\phi \|_{\infty }<\infty$.
Similarly, if $\phi _{l}\to \phi$ in the sense of Schwartz functions, then by exercise 3.6
    $\|x^{\alpha }\partial _{\beta }f(\phi -\phi _{l})\|_{\infty }\leq \sum _{\varsigma \leq \beta }\|x^{\alpha }p_{\varsigma }\partial _{\beta -\varsigma }(\phi -\phi _{l})\|_{\infty }\to 0,l\to \infty$
and hence $f\phi _{l}\to f\phi$ in the sense of Schwartz functions.
If we define $L(\varphi ):={\mathcal {L}}(\varphi ):=f\varphi$, from lemmas 4.20 and 4.21 follow the other claims.$\Box$
Theorem and definitions 4.23:
Let $O\subseteq \mathbb {R} ^{d}$ be open. We define
    $L:{\mathcal {S}}(\mathbb {R} ^{d})\to {\mathcal {C}}^{\infty }(\mathbb {R} ^{d}),L(\phi ):=\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }\phi$
, where $a_{\alpha }\in {\mathcal {C}}^{\infty }(\mathbb {R} ^{d})$ such that only finitely many of the $a_{\alpha }$ are different from the zero function (such a function is also called a linear partial differential operator), and further we define
    ${\mathcal {L}}:{\mathcal {S}}(\mathbb {R} ^{d})\to {\mathcal {C}}^{\infty }(\mathbb {R} ^{d}),{\mathcal {L}}(\phi ):=\sum _{|\alpha |\leq k}(-1)^{|\alpha |}\partial _{\alpha }(a_{\alpha }\phi )$.
* Let ${\mathcal {T}}:{\mathcal {D}}(O)\to \mathbb {R}$ be a distribution. If we definethen the expression on the right hand side is well-defined, for all $\vartheta \in {\mathcal {D}}(O)$ we haveand $\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }{\mathcal {T}}$ is a distribution.
    $\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }{\mathcal {T}}:{\mathcal {D}}(O)\to \mathbb {R} ,\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }{\mathcal {T}}(\varphi ):={\mathcal {T}}({\mathcal {L}}(\varphi ))$,
    $\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }{\mathcal {T}}_{\vartheta }={\mathcal {T}}_{L(\vartheta )}$,
* Assume that all $a_{\alpha }$s and all their derivatives are bounded by polynomials. Let ${\mathcal {T}}:{\mathcal {S}}(\mathbb {R} ^{d})\to \mathbb {R}$ be a tempered distribution. If we definethen the expression on the right hand side is well-defined, for all $\vartheta \in {\mathcal {D}}(O)$ we haveand $\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }{\mathcal {T}}$ is a tempered distribution.
    $\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }{\mathcal {T}}:{\mathcal {D}}(O)\to \mathbb {R} ,\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }{\mathcal {T}}(\varphi ):={\mathcal {T}}({\mathcal {L}}(\varphi ))$,
    $\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }{\mathcal {T}}_{\vartheta }={\mathcal {T}}_{L(\vartheta )}$,
Proof:
We want to apply lemmas 4.20 and 4.21. Hence, we prove that the requirements of these lemmas are met.
Since the derivatives of bump functions are again bump functions, the derivatives of Schwartz functions are again Schwartz functions (see exercise 3.3 for both), and because of theorem 4.22, we have that $L$ and ${\mathcal {L}}$ map ${\mathcal {D}}(O)$ to ${\mathcal {D}}(O)$, and if further all $a_{\alpha }$ and all their derivatives are bounded by polynomials, then $L$ and ${\mathcal {L}}$ map ${\mathcal {S}}(\mathbb {R} ^{d})$ to ${\mathcal {S}}(\mathbb {R} ^{d})$.
The sequential continuity of ${\mathcal {L}}$ follows from theorem 4.22.
Further, for all $\phi ,\theta \in {\mathcal {S}}(\mathbb {R} ^{d})$,
    $\int _{\mathbb {R} ^{d}}\phi (x){\mathcal {L}}(\theta )(x)dx=\sum _{\alpha \in \mathbb {N} _{0}^{d}}(-1)^{|\alpha |}\int _{\mathbb {R} ^{d}}\phi (x)\partial _{\alpha }(a_{\alpha }\theta )(x)dx$.
Further, if we single out an $\alpha \in \mathbb {N} _{0}^{d}$, by Fubini's theorem and integration by parts we obtain
    ${\begin{aligned}\int _{\mathbb {R} ^{d}}\phi (x)\partial _{\alpha }(a_{\alpha }\theta )(x)dx&=\int _{\mathbb {R} ^{d-1}}\int _{\mathbb {R} }\phi (x)\partial _{\alpha }(a_{\alpha }\theta )(x)dx_{1}d(x_{2},\ldots ,x_{d})\\&=\int _{\mathbb {R} ^{d-1}}\int _{\mathbb {R} }\phi (x)\partial _{\alpha }(a_{\alpha }\theta )(x)dx_{1}d(x_{2},\ldots ,x_{d})\\&=\int _{\mathbb {R} ^{d-1}}(-1)^{\alpha _{1}}\int _{\mathbb {R} }\partial _{(\alpha _{1},0,\ldots ,0)}\phi (x)\partial _{\alpha -(\alpha _{1},0,\ldots ,0)}(a_{\alpha }\theta )(x)dx_{1}d(x_{2},\ldots ,x_{d})\\&=\cdots =(-1)^{|\alpha |}\int _{\mathbb {R} ^{d}}\partial _{\alpha }\phi (x)a_{\alpha }(x)\theta (x)dx\end{aligned}}$.
Hence,
    $\int _{\mathbb {R} ^{d}}\phi (x){\mathcal {L}}(\theta )(x)dx=\int _{\mathbb {R} ^{d}}L(\phi )(x)\theta (x)dx$
and the lemmas are applicable.$\Box$
Definition 4.24:
Let ${\mathcal {T}}\in {\mathcal {D}}(\mathbb {R} ^{d})^{*}$ and let $\varphi \in {\mathcal {D}}(\mathbb {R} ^{d})$. Then we define the function
    ${\mathcal {T}}*\varphi (x):={\mathcal {T}}(\varphi (x-\cdot ))$.
This function is called the convolution of ${\mathcal {T}}$ and $\varphi$.
Theorem 4.25:
Let ${\mathcal {T}}\in {\mathcal {D}}(\mathbb {R} ^{d})^{*}$ and let $\varphi \in {\mathcal {D}}(\mathbb {R} ^{d})$. Then
* ${\mathcal {T}}*\varphi$ is continuous,
* $\forall \alpha \in \mathbb {N} _{0}^{d}:\partial _{\alpha }({\mathcal {T}}*\varphi )={\mathcal {T}}*(\partial _{\alpha }\varphi )$ and
* ${\mathcal {T}}*\varphi \in {\mathcal {C}}^{\infty }(\mathbb {R} ^{d})$.
Proof:
1.
Let $x\in \mathbb {R} ^{d}$ be arbitrary, and let $(x_{l})_{l\in \mathbb {N} }$ be a sequence converging to $x$ and let $N\in \mathbb {N}$ such that $\forall n\geq N:\|x_{n}-x\|\leq 1$. Then
    $K:={\overline {\bigcup _{n\geq N}{\text{supp }}\varphi (x_{n}-\cdot )\cup \bigcup _{n<N}{\text{supp }}\varphi (x_{n}-\cdot )}}$
is compact. Hence, if $\beta \in \mathbb {N} _{0}^{d}$ is arbitrary, then $\partial _{\beta }\varphi (x_{l}-\cdot )|_{K}\to \partial _{\beta }\varphi (x-\cdot )|_{K}$ uniformly. But outside $K$, $\partial _{\beta }\varphi (x_{l}-\cdot )-\partial _{\beta }\varphi (x-\cdot )=0$. Hence, $\partial _{\beta }\varphi (x_{l}-\cdot )\to \partial _{\beta }\varphi (x-\cdot )$ uniformly. Further, for all $n\in \mathbb {N}$ ${\text{supp }}\varphi (x_{n}-\cdot )\subseteq K$. Hence, $\varphi (x_{l}-\cdot )\to \varphi ,l\to \infty$ in the sense of bump functions. Thus, by continuity of ${\mathcal {T}}$,
    $({\mathcal {T}}*\varphi )(x_{l})={\mathcal {T}}(\varphi (x_{l}-\cdot ))\to {\mathcal {T}}(\varphi (x-\cdot ))=({\mathcal {T}}*\varphi )(x),l\to \infty$.
2.
We proceed by induction on $|\alpha |$.
The induction base $|\alpha |=0$ is obvious, since $\partial _{(0,\ldots ,0)}f=f$ for all functions $f:\mathbb {R} ^{d}\to \mathbb {R}$ by definition.
Let the statement be true for all $\alpha \in \mathbb {N} _{0}^{d}$ such that $|\alpha |=n$. Let $\beta \in \mathbb {N} _{0}^{d}$ such that $|\beta |=n+1$. We choose $k\in \{1,\ldots ,d\}$ such that $\beta _{k}>0$ (this is possible since otherwise $\beta =\mathbf {0}$). Further, we define
    $e_{k}:=(0,\ldots ,0,\overbrace {1} ^{k{\text{th place}}},0,\ldots ,0)$.
Then $|\beta -e_{k}|=n$, and hence $\partial _{\beta -e_{k}}({\mathcal {T}}*\varphi )={\mathcal {T}}*(\partial _{\beta -e_{k}}\varphi )$.
Furthermore, for all $\vartheta \in {\mathcal {D}}(\mathbb {R} ^{d})$,
    $\lim _{\lambda \to 0}{\frac {{\mathcal {T}}*\vartheta (x+\lambda e_{k})-{\mathcal {T}}*\vartheta (x)}{\lambda }}=\lim _{\lambda \to 0}{\mathcal {T}}\left({\frac {\vartheta (x+\lambda e_{k}-\cdot )-\vartheta (x-\cdot )}{\lambda }}\right)$.
But due to Schwarz' theorem, ${\frac {\vartheta (x+\lambda e_{k}-\cdot )-\vartheta (x-\cdot )}{\lambda }}\to \partial _{x_{k}}\vartheta ,\lambda \to 0$ in the sense of bump functions, and thus
    $\lim _{\lambda \to 0}{\mathcal {T}}\left({\frac {\vartheta (x+\lambda e_{k}-\cdot )-\vartheta (x-\cdot )}{\lambda }}\right)={\mathcal {T}}(\vartheta (x-\cdot ))$.
Hence, $\partial _{\beta }({\mathcal {T}}*\varphi )=\partial _{e_{k}}{\mathcal {T}}*(\partial _{\beta -e_{k}}\varphi )={\mathcal {T}}*(\partial _{\beta }\varphi )$, since $\partial _{\beta -e_{k}}\varphi$ is a bump function (see exercise 3.3).
3.
This follows from 1. and 2., since $\partial _{\beta }\varphi$ is a bump function for all $\beta \in \mathbb {N} _{0}^{d}$ (see exercise 3.3).$\Box$

## Exercises

* Let ${\mathcal {T}}_{1},\ldots ,{\mathcal {T}}_{n}$ be (tempered) distributions and let $c_{1},\ldots ,c_{n}\in \mathbb {R}$. Prove that also $\sum _{j=1}^{n}c_{j}{\mathcal {T}}_{j}$ is a (tempered) distribution.
* Let $f:\mathbb {R} ^{d}\to \mathbb {R}$ be essentially bounded. Prove that ${\mathcal {T}}_{f}$ is a tempered distribution.
* Prove that if ${\mathcal {Q}}$ is a set of differentiable functions which go from $[0,1]^{d}$ to $\mathbb {R}$, such that there exists a $c\in \mathbb {R} _{>0}$ such that for all $g\in {\mathcal {Q}}$ it holds $\forall x\in \mathbb {R} ^{d}:\|\nabla g(x)\|<c$, and if $(f_{l})_{l\in \mathbb {N} }$ is a sequence in ${\mathcal {Q}}$ for which the pointwise limit $\lim _{l\to \infty }f_{l}(x)$ exists for all $x\in \mathbb {R} ^{d}$, then $f_{l}$ converges to a function uniformly on $[0,1]^{d}$ (hint: $[0,1]^{d}$ is sequentially compact; this follows from the Bolzano–Weierstrass theorem).
* Let $f:\mathbb {R} ^{d}\to \mathbb {R}$ such that ${\mathcal {T}}_{f}$ is a distribution. Prove that for all $\varphi \in {\mathcal {D}}(O)$ ${\mathcal {T}}_{f}*\varphi =f*\varphi$.
* Prove that for $x\in \mathbb {R} ^{d}$ the function $\delta _{x}:{\mathcal {S}}(\mathbb {R} ^{d})\to \mathbb {R} ,\delta (\phi ):=\phi (x)$ is a tempered distribution (this function is called the Dirac delta distribution after Paul Dirac).
* For each $d\in \mathbb {N}$, find $\alpha _{d},\beta _{d}\in \mathbb {N} _{0}^{d}$ such that neither $\alpha _{d}\leq \beta _{d}$ nor $\beta _{d}\leq \alpha _{d}$.

## Sources

* Rudin, Walter (1991). Functional Analysis (2nd ed.). McGraw-Hill. ISBN 9780070542365.
* Daniel Matthes (2013/2014), Partial Differential Equations, lecture notes {{citation}}: Check date values in: |year= (help)
* Hasse Carlsson (2011), Lecture notes on Distributions (PDF)
* Ivan F. Wilde, DISTRIBUTION THEORY (GENERALIZED FUNCTIONS) NOTES (PDF)

# Fundamental solutions, Green's functions and Green's kernels

In the last two chapters, we have studied test function spaces and distributions. In this chapter we will demonstrate a method to obtain solutions to linear partial differential equations which uses test function spaces and distributions.

## Distributional and fundamental solutions

In the last chapter, we had defined multiplication of a distribution with a smooth function and derivatives of distributions. Therefore, for a distribution ${\mathcal {T}}$, we are able to calculate such expressions as
    $a\cdot \partial _{\alpha }{\mathcal {T}}$
for a smooth function $a:\mathbb {R} ^{d}\to \mathbb {R}$ and a $d$-dimensional multiindex $\alpha \in \mathbb {N} _{0}^{d}$. We therefore observe that in a linear partial differential equation of the form
    $\forall x\in O:\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }(x)\partial _{\alpha }u(x)=f(x)$
we could insert any distribution ${\mathcal {T}}$ instead of $u$ in the left hand side. However, equality would not hold in this case, because on the right hand side we have a function, but the left hand side would give us a distribution (as finite sums of distributions are distributions again due to exercise 4.1; remember that only finitely many $a_{\alpha }$ are allowed to be nonzero, see definition 1.2). If we however replace the right hand side by ${\mathcal {T}}_{f}$ (the regular distribution corresponding to $f$), then there might be distributions ${\mathcal {T}}$ which satisfy the equation. In this case, we speak of a distributional solution. Let's summarise this definition in a box.
Definition 5.1:
Let $O\subseteq \mathbb {R} ^{d}$ be open, let
    $\forall x\in O:\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }(x)\partial _{\alpha }u(x)=f(x)$
be a linear partial differential equation, and let ${\mathcal {T}}\in {\mathcal {D}}(O)^{*}$. ${\mathcal {T}}$ is called a distributional solution to the above linear partial differential equation if and only if
    $\forall \varphi \in {\mathcal {D}}(O):\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }{\mathcal {T}}(\varphi )={\mathcal {T}}_{f}(\varphi )$.
Definition 5.2:
Let $O\subseteq \mathbb {R} ^{d}$ be open and let
    $\forall x\in O:\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }(x)\partial _{\alpha }u(x)=f(x)$
be a linear partial differential equation. If $F:O\to {\mathcal {D}}(O)^{*}$ has the two properties
* $\forall \varphi \in {\mathcal {D}}(O):x\mapsto F(x)(\varphi )$ is continuous and
* $\forall x\in O:\forall \varphi \in {\mathcal {D}}(O):\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }F(x)(\varphi )=\delta _{x}(\varphi )$,
we call $F$ a fundamental solution for that partial differential equation.
For the definition of $\delta _{x}$ see exercise 4.5.
Lemma 5.3:
Let $O\subseteq \mathbb {R} ^{d}$ be open and let $\{{\mathcal {T}}_{x}|x\in S\}\subseteq {\mathcal {D}}(O)^{*}$ be a set of distributions, where $S\subseteq \mathbb {R} ^{d}$. Let's further assume that for all $\varphi \in {\mathcal {D}}(O)$, the function $S\to \mathbb {R} ,x\mapsto {\mathcal {T}}_{x}(\varphi )$ is continuous and bounded, and let $f\in L^{1}(S)$ be compactly supported. Then
    ${\mathcal {T}}(\varphi ):=\int _{S}f(x){\mathcal {T}}_{x}(\varphi )dx$
is a distribution.
Proof:
Let $C\subset \mathbb {R} ^{d}$ be the support of $f$. For $\varphi \in {\mathcal {D}}(O)$, let us denote the supremum norm of the function $C\to \mathbb {R} ,x\mapsto {\mathcal {T}}_{x}(\varphi )$ by
    $\|{\mathcal {T}}_{\cdot }(\varphi )\|_{\infty }$.
For $\|f\|_{L_{1}}=0$ or $\|{\mathcal {T}}_{\cdot }(\varphi )\|_{\infty }=0$, ${\mathcal {T}}$ is identically zero and hence a distribution. Hence, we only need to treat the case where both $\|f\|_{L_{1}}\neq 0$ and $\|{\mathcal {T}}_{\cdot }(\varphi )\|_{\infty }\neq 0$.
For each $n\in \mathbb {N}$, ${\overline {B_{n}(0)}}$ is a compact set since it is bounded and closed. Therefore, we may cover ${\overline {B_{n}(0)}}\cap S$ by finitely many pairwise disjoint sets $Q_{n,1},\ldots ,Q_{n,k_{n}}$ with diameter at most $1/n$ (for convenience, we choose these sets to be subsets of ${\overline {B_{n}(0)}}\cap S$). Furthermore, we choose $x_{n,1}\in Q_{n,1},\ldots ,x_{n,k_{n}}\in Q_{n,k_{n}}$.
For each $n\in \mathbb {N}$, we define
    ${\mathcal {T}}_{n}(\varphi ):=\sum _{j=1}^{k_{n}}\int _{Q_{n,j}}f(x){\mathcal {T}}_{x_{n,j}}(\varphi )dx$
, which is a finite linear combination of distributions and therefore a distribution (see exercise 4.1).
Let now $\vartheta \in {\mathcal {D}}(O)$ and $\epsilon >0$ be arbitrary. We choose $N_{1}\in \mathbb {N}$ such that for all $n\geq N_{1}$
    $\forall x\in B_{R_{n}}(0)\cap S:y\in B_{1/n}(x)\Rightarrow |{\mathcal {T}}_{x}(\varphi )-{\mathcal {T}}_{y}(\varphi )|<{\frac {\epsilon }{2\|f\|_{L^{1}}}}$.
This we may do because continuous functions are uniformly continuous on compact sets. Further, we choose $N_{2}\in \mathbb {N}$ such that
    $\int _{S\setminus B_{n}(0)}|f(x)|dx<{\frac {\epsilon }{2\|{\mathcal {T}}_{\cdot }(\varphi )\|_{\infty }}}$.
This we may do due to dominated convergence. Since for $n\geq N:=\max\{N_{1},N_{2}\}$
    $|{\mathcal {T}}_{n}(\varphi )-{\mathcal {T}}(\varphi )|<\sum _{j=1}^{k_{n}}\int _{Q{n,j}}|f(x)||{\mathcal {T}}_{\lambda _{x_{n,j}}}(\varphi )-{\mathcal {T}}_{x}(\varphi )|dx+{\frac {\epsilon \|{\mathcal {T}}_{\cdot }(\varphi )\|_{\infty }}{2\|T_{\cdot }(\varphi )\|_{\infty }}}<\epsilon$,
$\forall \varphi \in {\mathcal {D}}(O):{\mathcal {T}}_{l}(\varphi )\to {\mathcal {T}}(\varphi )$. Thus, the claim follows from theorem AI.33.$\Box$
Theorem 5.4:
Let $O\subseteq \mathbb {R} ^{d}$ be open, let
    $\forall x\in O:\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }(x)\partial _{\alpha }u(x)=f(x)$
be a linear partial differential equation such that $f$ is integrable and has compact support. Let $F$ be a fundamental solution of the PDE. Then
    ${\mathcal {T}}:{\mathcal {D}}(O)\to \mathbb {R} ,{\mathcal {T}}(\varphi ):=\int _{\mathbb {R} ^{d}}f(x)F(x)(\varphi )dx$
is a distribution which is a distributional solution for the partial differential equation.
Proof: Since by the definition of fundamental solutions the function $x\mapsto F(x)(\varphi )$ is continuous for all $\varphi \in {\mathcal {D}}(O)$, lemma 5.3 implies that ${\mathcal {T}}$ is a distribution.
Further, by definitions 4.16,
    ${\begin{aligned}\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }{\mathcal {T}}(\varphi )&={\mathcal {T}}\left(\sum _{\alpha \in \mathbb {N} _{0}^{d}}\partial _{\alpha }(a_{\alpha }\varphi )\right)\\&=\int _{\mathbb {R} ^{d}}f(x)F(x)\left(\sum _{\alpha \in \mathbb {N} _{0}^{d}}\partial _{\alpha }(a_{\alpha }\varphi )\right)dx\\&=\int _{\mathbb {R} ^{d}}f(x)\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }F(x)(\varphi )dx\\&=\int _{\mathbb {R} ^{d}}f(x)\delta _{x}(\varphi )dx\\&=\int _{\mathbb {R} ^{d}}f(x)\varphi (x)dx\\&={\mathcal {T}}_{f}(\varphi )\end{aligned}}$.$\Box$
Lemma 5.5:
Let $\varphi \in {\mathcal {D}}(\mathbb {R} ^{d})$, $f\in {\mathcal {C}}^{\infty }(\mathbb {R} ^{d})$, $\alpha \in \mathbb {N} _{0}^{d}$ and ${\mathcal {T}}\in {\mathcal {D}}(\mathbb {R} ^{d})^{*}$. Then
    $f\partial _{\alpha }({\mathcal {T}}*\varphi )=(f\partial _{\alpha }{\mathcal {T}})*\varphi$.
Proof:
By theorem 4.21 2., for all $x\in \mathbb {R} ^{d}$
    ${\begin{aligned}f\partial _{\alpha }({\mathcal {T}}*\varphi )(x)&=f{\mathcal {T}}*(\partial _{\alpha }\varphi )(x)\\&=f{\mathcal {T}}((\partial _{\alpha }\varphi )(x-\cdot ))\\&=f{\mathcal {T}}\left((-1)^{|\alpha |}\partial _{\alpha }(\varphi (x-\cdot ))\right)\\&=f(\partial _{\alpha }{\mathcal {T}})(\varphi (x-\cdot ))\\&=(\partial _{\alpha }{\mathcal {T}})(f\varphi (x-\cdot ))\\&=(f\partial _{\alpha }{\mathcal {T}})(\varphi (x-\cdot ))=(f\partial _{\alpha }{\mathcal {T}})*\varphi (x)\\\end{aligned}}$.$\Box$
Theorem 5.6:
Let ${\mathcal {T}}$ be a solution of the equation
    $\forall \varphi \in {\mathcal {D}}(\mathbb {R} ^{d}):\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }{\mathcal {T}}(\varphi )=\delta _{0}$,
where only finitely many $a_{\alpha }$ are nonzero, and let $\vartheta \in {\mathcal {D}}(\mathbb {R} ^{d})$. Then $u:={\mathcal {T}}*\vartheta$ solves
    $\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }u=\vartheta$.
Proof:
By lemma 5.5, we have
    ${\begin{aligned}\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }u(x)&=\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }({\mathcal {T}}*\vartheta )(x)\\&=\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }(\partial _{\alpha }{\mathcal {T}})*\vartheta (x)\\&=\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }{\mathcal {T}}(\vartheta (x-\cdot ))\\&=\delta _{0}(\vartheta (x-\cdot ))=\vartheta (x)\end{aligned}}$.$\Box$

## Partitions of unity

In this section you will get to know a very important tool in mathematics, namely partitions of unity. We will use it in this chapter and also later in the book. In order to prove the existence of partitions of unity (we will soon define what this is), we need a few definitions first.
Definitions 5.7:
Let $S\subseteq \mathbb {R} ^{d}$ be a set. We define:
* $\partial S:=\left\{x\in \mathbb {R} {\big |}\forall \epsilon >0:B_{\epsilon }(x)\cap S\neq \emptyset \wedge B_{\epsilon }(x)\cap (\mathbb {R} ^{d}\setminus S)\neq \emptyset \right\}$
* ${\overset {\circ }{S}}:=S\setminus \partial S$
$\partial S$ is called the boundary of $S$ and ${\overset {\circ }{S}}$ is called the interior of $S$. Further, if $x\in \mathbb {R} ^{d}$, we define
    ${\text{dist}}(S,x):=\inf _{y\in S}\|x-y\|$.
We also need definition 3.13 in the proof, which is why we restate it now:
Definition 3.13:
For $R\in \mathbb {R} _{>0}$, we define
    $\eta _{R}:\mathbb {R} ^{d}\to \mathbb {R} ,\eta _{R}(x)=\eta \left({\frac {x}{R}}\right){\big /}R^{d}$.
Theorem and definitions 5.8: Let $O\subseteq \mathbb {R} ^{d}$ be an open set, and let $U_{\upsilon },\upsilon \in \Upsilon$ be open subsets of $\mathbb {R} ^{d}$ such that $\bigcup _{\upsilon \in \Upsilon }U_{\upsilon }=O$ (i. e. the sets $U_{\upsilon },\upsilon \in \Upsilon$ form an open cover of $O$). Then there exists a sequence of functions $(\eta _{l})_{l\in \mathbb {N} }$ in ${\mathcal {D}}(\mathbb {R} ^{d})$ such that the following conditions are satisfied:
* $\forall n\in \mathbb {N} :\forall x\in O:0\leq \eta _{n}(x)\leq 1$
* $\forall n\in \mathbb {N} :\exists \upsilon \in \Upsilon :{\text{supp }}\eta _{n}\subseteq U_{\upsilon }$
* $\forall x\in O:|\{n\in \mathbb {N} |\eta _{n}(x)\neq 0\}|<\infty$
* $\forall x\in O:\sum _{i=0}^{\infty }\eta _{i}(x)=1$
The sequence $(\eta _{l})_{l\in \mathbb {N} }$ is called a partition of unity for $O$ with respect to $U_{\upsilon },\upsilon \in \Upsilon$.
Proof: We will prove this by explicitly constructing such a sequence of functions.
1. First, we construct a sequence of open balls $(B_{l})_{l\in \mathbb {N} }$ with the properties
* $\forall n\in \mathbb {N} :\exists \upsilon \in \Upsilon :{\overline {B_{n}}}\subseteq U_{\upsilon }$
* $\forall x\in O:|\{n\in \mathbb {N} |x\in {\overline {B_{n}}}\}|<\infty$
* $\bigcup _{j\in \mathbb {N} }B_{j}=O$.
In order to do this, we first start with the definition of a sequence compact sets; for each $n\in \mathbb {N}$, we define
    $K_{n}:=\left\{x\in O{\big |}{\text{dist}}(\partial O,x)\geq {\frac {1}{n}},\|x\|\leq n\right\}$.
This sequence has the properties
* $\bigcup _{j\in \mathbb {N} }K_{j}=O$
* $\forall n\in \mathbb {N} :K_{n}\subset {\overset {\circ }{K_{n+1}}}$.
We now construct $(B_{l})_{l\in \mathbb {N} }$ such that
* $K_{1}\subset \bigcup _{1\leq j\leq k_{1}}B_{j}\subseteq {\overset {\circ }{K_{2}}}$ and
* $\forall n\in \mathbb {N} :K_{n+1}\setminus {\overset {\circ }{K_{n}}}\subset \bigcup _{k_{n}<j\leq k_{n+1}}B_{j}\subseteq {\overset {\circ }{K_{n+2}}}\setminus K_{n-1}$
for some $k_{1},k_{2},\ldots \in \mathbb {N}$. We do this in the following way: To meet the first condition, we first cover $K_{1}$ with balls by choosing for every $x\in K_{1}$ a ball $B_{x}$ such that $B_{x}\subseteq U_{\upsilon }\cap {\overset {\circ }{K_{2}}}$ for an $\upsilon \in \Upsilon$. Since these balls cover $K_{1}$, and $K_{1}$ is compact, we may choose a finite subcover $B_{1},\ldots B_{k_{1}}$.
To meet the second condition, we proceed analogously, noting that for all $n\in \mathbb {N} _{\geq 2}$ $K_{n+1}\setminus {\overset {\circ }{K_{n}}}$ is compact and ${\overset {\circ }{K_{n+2}}}\setminus K_{n-1}$ is open.
This sequence of open balls has the properties which we wished for.
2. We choose the respective functions. Since each $B_{n}$, $n\in \mathbb {N}$ is an open ball, it has the form
    $B_{n}=B_{R_{n}}(x_{n})$
where $R_{n}\in \mathbb {R}$ and $x_{n}\in \mathbb {R} ^{d}$.
It is easy to prove that the function defined by
    ${\tilde {\eta }}_{n}(x):=\eta _{R_{n}}(x-x_{n})$
satisfies ${\tilde {\eta }}_{n}(x)=0$ if and only if $x\in B_{n}$. Hence, also ${\text{supp }}{\tilde {\eta }}_{n}={\overline {B_{n}}}$. We define
    $\eta (x):=\sum _{j=1}^{\infty }{\tilde {\eta }}_{j}(x)$
and, for each $n\in \mathbb {N}$,
    $\eta _{n}:={\frac {{\tilde {\eta }}_{n}}{\eta }}$.
Then, since $\eta$ is never zero, the sequence $(\eta _{l})_{l\in \mathbb {N} }$ is a sequence of ${\mathcal {D}}(\mathbb {R} ^{d})$ functions and further, it has the properties 1. - 4., as can be easily checked.$\Box$

## Green's functions and Green's kernels

Definition 5.9:
Let
    $\forall x\in O:\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }(x)\partial _{\alpha }u(x)=f(x)$
be a linear partial differential equation. A function $G:\mathbb {R} ^{d}\times \mathbb {R} ^{d}\to \mathbb {R}$ such that for all $x\in \mathbb {R} ^{d}$ ${\mathcal {T}}_{G(\cdot ,x)}$ is well-defined and
    $F(x):={\mathcal {T}}_{G(\cdot ,x)}$
is a fundamental solution of that partial differential equation is called a Green's function of that partial differential equation.
Definition 5.10:
Let
    $\forall x\in O:\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }(x)\partial _{\alpha }u(x)=f(x)$
be a linear partial differential equation. A function $K:\mathbb {R} ^{d}\to \mathbb {R}$ such that the function
    $G(y,x):=K(y-x)$
is a Greens function for that partial differential equation is called a Green's kernel of that partial differential equation.
Theorem 5.11:
Let
    $\forall x\in O:\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }(x)\partial _{\alpha }u(x)=f(x)$
be a linear partial differential equation (in the following, we will sometimes abbreviate PDE for partial differential equation) such that $f\in {\mathcal {C}}(\mathbb {R} ^{d})$, and let $K$ be a Green's kernel for that PDE. If
    $u:=f*K$
exists and $\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }u$ exists and is continuous, then $u$ solves the partial differential equation.
Proof:
We choose $(\eta _{l})_{l\in \mathbb {N} }$ to be a partition of unity of $O$, where the open cover of $O$ shall consist only of the set $O$. Then by definition of partitions of unity
    $f=\sum _{j\in \mathbb {N} }\eta _{j}f$.
For each $n\in \mathbb {N}$, we define
    $f_{n}:=\eta _{n}f$
and
    $u_{n}:=f_{n}*K$.
By Fubini's theorem, for all $\varphi \in {\mathcal {D}}(\mathbb {R} ^{d})$ and $n\in \mathbb {N}$
    ${\begin{aligned}\int _{\mathbb {R} ^{d}}T_{K(\cdot -y)}(\varphi )f_{n}(y)dy&=\int _{\mathbb {R} ^{d}}\int _{\mathbb {R} ^{d}}K(x-y)\varphi (x)dxf_{n}(y)dy\\&=\int _{\mathbb {R} ^{d}}\int _{\mathbb {R} ^{d}}f_{n}(y)K(x-y)\varphi (x)dydx\\&=\int _{\mathbb {R} ^{d}}(f_{n}*K)(x)\varphi (x)dx\\&={\mathcal {T}}_{u_{n}}(\varphi )\end{aligned}}$.
Hence, ${\mathcal {T}}_{u_{n}}$ as given in theorem 4.11 is a well-defined distribution.
Theorem 5.4 implies that ${\mathcal {T}}_{u_{n}}$ is a distributional solution to the PDE
    $\forall x\in O:\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }(x)\partial _{\alpha }u_{n}(x)=f_{n}(x)$.
Thus, for all $\varphi \in {\mathcal {D}}(\mathbb {R} ^{d})$ we have, using theorem 4.19,
    ${\begin{aligned}\int _{\mathbb {R} ^{d}}\left(\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }u_{n}\right)(x)\varphi (x)dx&={\mathcal {T}}_{\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }u_{n}}(\varphi )\\&=\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }{\mathcal {T}}_{u_{n}}(\varphi )\\&=T_{f_{n}}(\varphi )=\int _{\mathbb {R} ^{d}}f_{n}(x)\varphi (x)dx\end{aligned}}$.
Since $\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }u_{n}$ and $f_{n}$ are both continuous, they must be equal due to theorem 3.17. Summing both sides of the equation over $n$ yields the theorem.$\Box$
Theorem 5.12:
Let $K\in L_{\text{loc}}^{1}$ and let $O\subseteq \mathbb {R} ^{d}$ be open. Then for all $\varphi \in {\mathcal {D}}(O)$, the function $x\mapsto {\mathcal {T}}_{K(\cdot -x)}(\varphi )$ is continuous.
Proof:
If $x_{l}\to x,l\to \infty$, then
    ${\begin{aligned}{\mathcal {T}}_{K(\cdot -x_{l})}(\varphi )-{\mathcal {T}}_{K(\cdot -x)}(\varphi )&=\int _{\mathbb {R} ^{d}}K(y-x_{l})\varphi (y)dy-\int _{\mathbb {R} ^{d}}K(y-x)\phi (y)dy\\&=\int _{\mathbb {R} ^{d}}K(y)(\varphi (y+x_{l})-\varphi (y+x))dy\\&\leq \max _{y\in \mathbb {R} ^{d}}|\varphi (y+x_{l})-\varphi (y+x)|\underbrace {\int _{{\text{supp }}\varphi +B_{1}(x)}K(y)dy} _{\text{constant}}\end{aligned}}$
for sufficiently large $l$, where the maximum in the last expression converges to $0$ as $l\to \infty$, since the support of $\varphi$ is compact and therefore $\varphi$ is uniformly continuous by the Heine–Cantor theorem.$\Box$
The last theorem shows that if we have found a locally integrable function $K$ such that
    $\forall x\in \mathbb {R} ^{d}:\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }\partial _{\alpha }{\mathcal {T}}_{K(\cdot -x)}=\delta _{x}$,
we have found a Green's kernel $K$ for the respective PDEs. We will rely on this theorem in our procedure to get solutions to the heat equation and Poisson's equation.

## Sources

* Hasse Carlsson (2011), Lecture notes on Distributions (PDF)
* Daniel Matthes (2013/2014), Partial Differential Equations, lecture notes {{citation}}: Check date values in: |year= (help)

# The heat equation

This chapter is about the heat equation, which looks like this:
    $\forall (t,x)\in \mathbb {R} \times \mathbb {R} ^{d}:\partial _{t}u(t,x)-\Delta _{x}u(t,x)=f(t,x)$
for some $f:\mathbb {R} \times \mathbb {R} ^{d}\to \mathbb {R}$. Using distribution theory, we will prove an explicit solution formula (if $f$ is often enough differentiable), and we even prove a solution formula for the initial value problem.

## Green's kernel and solution

Lemma 6.1:
    $\int _{\mathbb {R} }e^{-x^{2}}dx={\sqrt {\pi }}$
Proof:
    ${\begin{aligned}\left(\int _{\mathbb {R} }e^{-x^{2}}\right)^{2}&=\left(\int _{\mathbb {R} }e^{-x^{2}}\right)\cdot \left(\int _{\mathbb {R} }e^{-y^{2}}\right)&\\&=\int _{\mathbb {R} }\int _{\mathbb {R} }e^{-(x^{2}+y^{2})}dxdy&\\&=\int _{\mathbb {R} ^{2}}e^{-\|(x,y)\|^{2}}d(x,y)&{\text{Fubini}}\\&=\int _{0}^{\infty }\int _{0}^{2\pi }re^{-r^{2}}d\varphi dr&{\text{ integration by substitution using spherical coordinates}}\\&=2\pi \int _{0}^{\infty }re^{-r^{2}}dr&\\&=2\pi \int _{0}^{\infty }{\frac {1}{2{\sqrt {r}}}}{\sqrt {r}}e^{-r}dr&{\text{integration by substitution using }}r\mapsto {\sqrt {r}}\\&=\pi &\end{aligned}}$
Taking the square root on both sides finishes the proof.$\Box$
Lemma 6.2:
    $\int _{\mathbb {R} ^{d}}e^{-\|x\|^{2}/2}dx={\sqrt {2\pi }}^{d}$
Proof:
    ${\begin{aligned}\int _{\mathbb {R} ^{d}}e^{-{\frac {\|x\|^{2}}{2}}}dx&=\overbrace {\int _{-\infty }^{\infty }\cdots \int _{-\infty }^{\infty }} ^{d{\text{ times}}}e^{-{\frac {x_{1}^{2}+\cdots +x_{d}^{2}}{2}}}dx_{1}\cdots dx_{d}&{\text{Fubini's theorem}}\\&=\int _{-\infty }^{\infty }e^{-{\frac {x_{d}^{2}}{2}}}\cdots \int _{-\infty }^{\infty }e^{-{\frac {x_{1}^{2}}{2}}}dx_{1}\,\cdots dx_{d}&{\text{pulling the constants out of the integrals}}\end{aligned}}$
By lemma 6.1,
    $\int _{\mathbb {R} }e^{-x^{2}}dx={\sqrt {\pi }}$.
If we apply to this integration by substitution (theorem 5.5) with the diffeomorphism $x\mapsto {\frac {x}{\sqrt {2}}}$, we obtain
    ${\sqrt {\pi }}=\int _{\mathbb {R} }{\frac {1}{\sqrt {2}}}e^{-{\frac {x^{2}}{2}}}dx$
and multiplying with ${\sqrt {2}}$
    ${\sqrt {2\pi }}=\int _{\mathbb {R} }e^{-{\frac {x^{2}}{2}}}dx$
Therefore, calculating the innermost integrals first and then pulling out the resulting constants,
    $\overbrace {\int _{-\infty }^{\infty }e^{-{\frac {x_{d}^{2}}{2}}}\cdots \int _{-\infty }^{\infty }e^{-{\frac {x_{1}^{2}}{2}}}} ^{d{\text{ times}}}dx_{1}\cdots dx_{d}={\sqrt {2\pi }}^{d}$$\Box$
Theorem 6.3:
The function
    $E:\mathbb {R} \times \mathbb {R} ^{d}\to \mathbb {R} ,E(t,x)={\begin{cases}{\sqrt {4\pi t}}^{-d}e^{-{\frac {\|x\|^{2}}{4t}}}&t>0\\0&t\leq 0\end{cases}}$
is a Green's kernel for the heat equation.
Proof:
1.
We show that $E$ is locally integrable.
Let $K\subset \mathbb {R} \times \mathbb {R} ^{d}$ a compact set, and let $T>0$ such that $K\subset (-T,T)\times \mathbb {R} ^{d}$. We first show that the integral
    $\int _{(-T,T)\times \mathbb {R} ^{d}}E(s,y)d(s,y)$
exists:
    ${\begin{aligned}\int _{(-T,T)\times \mathbb {R} ^{d}}E(s,y)d(s,y)&=\int _{(0,T)\times \mathbb {R} ^{d}}E(s,y)d(s,y)&\forall s\leq 0:E(s,y)=0\\&=\int _{0}^{T}\int _{\mathbb {R} ^{d}}{\frac {1}{{\sqrt {4\pi s}}^{d}}}e^{-{\frac {\|y\|^{2}}{4s}}}dyds&{\text{Fubini's theorem}}\end{aligned}}$
By transformation of variables in the inner integral using the diffeomorphism $y\mapsto {\sqrt {2s}}y$, and lemma 6.2, we obtain:
    $=\int _{0}^{T}\int _{\mathbb {R} ^{d}}{\frac {{\sqrt {2s}}^{d}}{{\sqrt {4\pi s}}^{d}}}e^{-{\frac {\|y\|^{2}}{2}}}dyds=\int _{0}^{T}1ds=T$
Therefore the integral
    $\int _{(-T,T)\times \mathbb {R} ^{d}}E(s,y)d(s,y)$
exists. But since
    $\forall (s,y)\in \mathbb {R} \times \mathbb {R} ^{d}:|\chi _{K}(s,y)E(s,y)|\leq |E(s,y)|$
, where $\chi _{K}$ is the characteristic function of $K$, the integral
    $\int _{(-T,T)\times \mathbb {R} ^{d}}\chi _{K}(s,y)E(s,y)d(s,y)=\int _{K}E(s,y)d(s,y)$
exists. Since $K$ was an arbitrary compact set, we thus have local integrability.
2.
We calculate $\partial _{t}E$ and $\Delta _{x}E$ (see exercise 1).
    $\partial _{t}E(t,x)=\left({\frac {\|x\|^{2}}{4t^{2}}}-{\frac {d}{4t}}\right)E(t,x)$
    $\Delta _{x}E(t,x)=\left({\frac {\|x\|^{2}}{4t^{2}}}-{\frac {d}{4t}}\right)E(t,x)$
3.
We show that
    $\forall \varphi \in {\mathcal {D}}(\mathbb {R} \times \mathbb {R} ^{d}),(t,x)\in \mathbb {R} \times \mathbb {R} ^{d}:(\partial _{t}-\Delta _{x})T_{E(\cdot -(t,x))}(\varphi )=\delta _{(t,x)}(\varphi )$
Let $\varphi \in {\mathcal {D}}(\mathbb {R} \times \mathbb {R} ^{d}),(t,x)\in \mathbb {R} \times \mathbb {R} ^{d}$ be arbitrary.
In this last step of the proof, we will only manipulate the term $(\partial _{t}-\Delta _{x})T_{E(\cdot -(t,x))}(\varphi )$.
    ${\begin{aligned}(\partial _{t}-\Delta _{x})T_{E(\cdot -(t,x))}(\varphi )&=T_{E(\cdot -(t,x))}((-\partial _{t}-\Delta _{x})\varphi )&{\text{by definition of distribution derivation}}\\&=\int _{\mathbb {R} \times \mathbb {R} ^{d}}(-\partial _{t}-\Delta _{x})\varphi (s,y)E(s-t,y-x)d(s,y)&\\&=\int _{(t,\infty )\times \mathbb {R} ^{d}}(-\partial _{t}-\Delta _{x})\varphi (s,y)E(s-t,y-x)d(s,y)&\forall t\leq 0:E(t,x)=0\\\end{aligned}}$
If we choose $R>0$ and $T>0$ such that
    ${\text{supp }}\varphi \subset (-\infty ,t+T)\times B_{R}(x)$
, we have even
    $(\partial _{t}-\Delta _{x})T_{E(\cdot -(t,x))}(\varphi )=\int _{(t,t+T)\times B_{R}(x)}(-\partial _{t}-\Delta _{x})\varphi (s,y)E(s-t,y-x)d(s,y)$
Using the dominated convergence theorem (theorem 5.1), we can rewrite the term again:
    ${\begin{aligned}(\partial _{t}-\Delta _{x})T_{E(\cdot -(t,x))}(\varphi )&=\int _{(t,t+T)\times B_{R}(x)}(-\partial _{t}-\Delta _{x})\varphi (s,y)E(s-t,y-x)d(s,y)\\&=\lim _{\epsilon \downarrow 0}\int _{(t,t+T)\times B_{R}(x)}(-\partial _{t}-\Delta _{x})\varphi (s,y)E(s-t,y-x)(1-\chi _{[t,t+\epsilon ]}(s))d(s,y)\\&=\lim _{\epsilon \downarrow 0}\int _{(t+\epsilon ,t+T)\times B_{R}(x)}(-\partial _{t}-\Delta _{x})\varphi (s,y)E(s-t,y-x)d(s,y)\end{aligned}}$
, where $\chi _{[t,t+\epsilon ]}$ is the characteristic function of $[t,t+\epsilon ]$.
We split the limit term in half to manipulate each summand separately:
    ${\begin{aligned}\int _{(t+\epsilon ,t+T)\times B_{R}(x)}(-\partial _{t}-\Delta _{x})\varphi (s,y)E(s-t,y-x)d(s,y)\\=-\int _{(t+\epsilon ,t+T)\times B_{R}(x)}\Delta _{x}\varphi (s,y)E(s-t,y-x)d(s,y)\\-\int _{(t+\epsilon ,t+T)\times B_{R}(x)}\partial _{t}\varphi (s,y)E(s-t,y-x)d(s,y)\\\end{aligned}}$
The last integrals are taken over $(t+\epsilon ,t+T)\times B_{R}(x)$ for $\epsilon >0$. In this area and its boundary, $E(s-t,y-x)$ is differentiable. Therefore, we are allowed to integrate by parts.
${\begin{aligned}\int _{(t+\epsilon ,t+T)\times B_{R}(x)}\Delta _{x}\varphi (s,y)E(s-t,y-x)d(s,y)&=\int _{t+\epsilon }^{t+T}\int _{B_{R}(x)}\Delta _{x}\varphi (s,y)E(s-t,y-x)dyds&{\text{Fubini}}\\=\int _{t+\epsilon }^{t+T}\int _{\partial B_{R}(x)}E(s,y)n(y)\cdot \underbrace {\nabla _{x}\varphi (s,y)} _{=0}dyds&-\int _{t+\epsilon }^{t+T}\int _{B_{R}(x)}\nabla _{x}\varphi (s,y)\cdot \nabla _{x}E(s-t,y-x)dyds&{\text{integration by parts in }}y\\=\int _{t+\epsilon }^{t+T}\int _{B_{R}(x)}\varphi (s,y)\Delta _{x}E(s-t,y-x)dyds&-\int _{t+\epsilon }^{t+T}\int _{\partial B_{R}(x)}\underbrace {\varphi (s,y)} _{=0}n(y)\cdot \nabla _{x}E(s-t,y-x)dyds&{\text{integration by parts in }}y\end{aligned}}$
In the last two manipulations, we used integration by parts where $\varphi$ and $f$ exchanged the role of the function in theorem 5.4, and $\nabla _{x}f$ and $\nabla _{x}\varphi$ exchanged the role of the vector field. In the latter manipulation, we did not apply theorem 5.4 directly, but instead with subtracted boundary term on both sides.
Let's also integrate the other integral by parts.
    ${\begin{aligned}\int _{(t+\epsilon ,t+T)\times B_{R}(x)}\partial _{t}\varphi (s,y)E(s-t,y-x)d(s,y)&=\int _{B_{R}(x)}\int _{t+\epsilon }^{t+T}\partial _{t}\varphi (s,y)E(s-t,y-x)dsdy&{\text{Fubini}}\\=\int _{B_{R}(x)}\underbrace {\varphi (s,y)E(s-t,y-x){\big |}_{s=t+\epsilon }^{s=t+T}} _{=-\varphi (t+\epsilon ,y)E(\epsilon ,y-x)}dy&-\int _{B_{R}(x)}\int _{t+\epsilon }^{t+T}\varphi (s,y)\partial _{t}E(s-t,y-x)dsdy&{\text{integration by parts in }}s\end{aligned}}$
Now we add the two terms back together and see that
    ${\begin{aligned}(\partial _{t}-\Delta _{x})T_{E(\cdot -(t,x))}(\varphi )&=\lim _{\epsilon \downarrow 0}-\int _{B_{R}(x)}-\varphi (t+\epsilon ,y)E(\epsilon ,y-x)dy\\+\int _{B_{R}(x)}\int _{t+\epsilon }^{t+T}\varphi (s,y)\partial _{t}E(s-t,y-x)dsdy&-\int _{t+\epsilon }^{t+T}\int _{B_{R}(x)}\varphi (s,y)\Delta _{x}E(s-t,y-x)dyds\end{aligned}}$
The derivative calculations from above show that $\partial _{t}E=\Delta _{x}E$, which is why the last two integrals cancel and therefore
    $(\partial _{t}-\Delta _{x})T_{E(\cdot -(t,x))}(\varphi )=\lim _{\epsilon \downarrow 0}\int _{B_{R}(x)}\varphi (t+\epsilon ,y)E(\epsilon ,y-x)dy$
Using that ${\text{supp }}\varphi (t+\epsilon ,\cdot )\subset B_{R}(x)$ and with multi-dimensional integration by substitution with the diffeomorphism $y\mapsto x+{\sqrt {2\epsilon }}y$ we obtain:
    $\int _{B_{R}(x)}\varphi (t+\epsilon ,y)E(\epsilon ,y-x)dy=\int _{\mathbb {R} ^{d}}\varphi (t+\epsilon ,y)E(\epsilon ,y-x)dy$ $=\int _{\mathbb {R} ^{d}}\varphi (t+\epsilon ,y){\frac {1}{{\sqrt {4\pi \epsilon }}^{d}}}e^{-{\frac {\|y-x\|^{2}}{4\epsilon }}}dy$ $=\int _{\mathbb {R} ^{d}}\varphi (t+\epsilon ,x+{\sqrt {2\epsilon }}y){\frac {{\sqrt {2\epsilon }}^{d}}{{\sqrt {4\pi \epsilon }}^{d}}}e^{-{\frac {\|y\|^{2}}{2}}}dy={\frac {1}{{\sqrt {2\pi }}^{d}}}\int _{\mathbb {R} ^{d}}\varphi (t+\epsilon ,x+{\sqrt {2\epsilon }}y)e^{-{\frac {\|y\|^{2}}{2}}}dy$
Since $\varphi$ is continuous (even smooth), we have
    $\forall x\in \mathbb {R} ^{d}:\lim _{\epsilon \to 0}\varphi (t+\epsilon ,x+{\sqrt {2\epsilon }}y)=\varphi (t,x)$
Therefore
    ${\begin{aligned}(\partial _{t}-\Delta _{x})T_{E(\cdot -(t,x))}(\varphi )&=\lim _{\epsilon \downarrow 0}{\frac {1}{{\sqrt {2\pi }}^{d}}}\int _{\mathbb {R} ^{d}}\varphi (t+\epsilon ,x+{\sqrt {2\epsilon }}y)e^{-{\frac {\|y\|^{2}}{2}}}dy&\\&={\frac {1}{{\sqrt {2\pi }}^{d}}}\int _{\mathbb {R} ^{d}}\varphi (t,x)e^{-{\frac {\|y\|^{2}}{2}}}dy&{\text{dominated convergence}}\\&=\varphi (t,x)&{\text{lemma 6.2}}\\&=\delta _{(t,x)}(\varphi )&\end{aligned}}$$\Box$
Theorem 6.4: If $f:\mathbb {R} \times \mathbb {R} ^{d}$ is bounded, once continuously differentiable in the $t$-variable and twice continuously differentiable in the $x$-variable, then
    $u(t,x):=(E*f)(t,x)$
solves the heat equation
    $\forall (t,x)\in \mathbb {R} \times \mathbb {R} ^{d}:\partial _{t}u(t,x)-\Delta _{x}u(t,x)=f(t,x)$
Proof:
1.
We show that $(E*f)(t,x)$ is sufficiently often differentiable such that the equations are satisfied.
2.
We invoke theorem 5.?, which states exactly that a convolution with a Green's kernel is a solution, provided that the convolution is sufficiently often differentiable (which we showed in part 1 of the proof).$\Box$

## Initial Value Problem

Definition 6.5: Let $\mu :\mathbb {R} \times \mathbb {R} ^{d}\to \mathbb {R}$ and $f:\mathbb {R} ^{d}\to \mathbb {R}$ be two functions. The spatial convolution of $\mu$ and $f$ is given by:
    $(\mu *_{x}f)(t,x):=\int _{\mathbb {R} ^{d}}\mu (t,y-x)f(y)dy$
Theorem and definition 6.6: Let $f:[0,\infty )\times \mathbb {R} ^{d}\to \mathbb {R}$ be bounded, once continuously differentiable in the $t$-variable and twice continuously differentiable in the $x$-variable, and let $E:\mathbb {R} ^{d}\to \mathbb {R}$ be continuous and bounded. If we define
    ${\tilde {f}}:\mathbb {R} \times \mathbb {R} ^{d}\to \mathbb {R} ,{\tilde {f}}(t,x)={\begin{cases}f(t,x)&t\geq 0\\0&t<0\end{cases}}$
, then the function
    $u:[0,\infty )\times \mathbb {R} ^{d}\to \mathbb {R} ,u(t,x)={\begin{cases}(E*_{x}g)(t,x)+({\tilde {f}}*E)(t,x)&t>0\\g(x)&t=0\end{cases}}$
is a continuous solution of the initial value problem for the heat equation, that is
    ${\begin{cases}\forall (t,x)\in (0,\infty )\times \mathbb {R} ^{d}:&\partial _{t}u(t,x)-\Delta _{x}u(t,x)=f(t,x)\\\forall x\in \mathbb {R} ^{d}:&u(0,x)=g(x)\end{cases}}$
Note that if we do not require the solution to be continuous, we may just take any solution and just set it to $g$ at $t=0$.
Proof:
1.
We show
    $\forall (t,x)\in (0,\infty )\times \mathbb {R} ^{d}:\partial _{t}u(t,x)-\Delta _{x}u(t,x)=f(t,x)~~~~~(*)$
From theorem 7.4, we already know that ${\tilde {f}}*E$ solves
    $\forall (t,x)\in (0,\infty )\times \mathbb {R} ^{d}:\partial _{t}({\tilde {f}}*E)(t,x)-\Delta _{x}({\tilde {f}}*E)(t,x)={\tilde {f}}(t,x){\overset {t>0}{=}}f(t,x)$
Therefore, we have for $\forall (t,x)\in (0,\infty )\times \mathbb {R} ^{d}$,
    ${\begin{aligned}\partial _{t}u(t,x)-\Delta _{x}u(t,x)=&\partial _{t}(E*_{x}g)(t,x)+\partial _{t}({\tilde {f}}*E)(t,x)\\&-\Delta _{x}(E*_{x}g)(t,x)-\Delta _{x}({\tilde {f}}*E)(t,x)\\=&f(t,x)+\partial _{t}(E*_{x}g)(t,x)-\Delta _{x}(E*_{x}g)(t,x)\end{aligned}}$
which is why $(*)$ would follow if
    $\forall (t,x)\in (0,\infty )\times \mathbb {R} ^{d}:\partial _{t}(E*_{x}g)(t,x)-\Delta _{x}(E*_{x}g)(t,x)=0$
This we shall now check.
By definition of the spatial convolution, we have
    $\partial _{t}(E*_{x}g)(t,x)=\partial _{t}\int _{\mathbb {R} ^{d}}E(t,x-y)g(y)dy$
and
    $\Delta _{x}(E*_{x}g)(t,x)=\Delta _{x}\int _{\mathbb {R} ^{d}}E(t,x-y)g(y)dy$
By applying Leibniz' integral rule (see exercise 2) we find that
    ${\begin{aligned}\partial _{t}(E*_{x}g)(t,x)-\Delta _{x}(E*_{x}g)(t,x)&=\partial _{t}\int _{\mathbb {R} ^{d}}E(t,x-y)g(y)dy-\Delta _{x}\int _{\mathbb {R} ^{d}}E(t,x-y)g(y)dy&\\&=\int _{\mathbb {R} ^{d}}\partial _{t}E(t,x-y)g(y)dy-\int _{\mathbb {R} ^{d}}\Delta _{x}E(t,x-y)g(y)dy&{\text{ Leibniz' integral rule}}\\&=\int _{\mathbb {R} ^{d}}\left(\partial _{t}E(t,x-y)-\Delta _{x}E(t,x-y)\right)g(y)dy&{\text{ linearity of the integral}}\\&=0&{\text{ exercise 1}}\end{aligned}}$
for all $(t,x)\in (0,\infty )\times \mathbb {R} ^{d}$.
2.
We show that $u$ is continuous.
It is clear that $u$ is continuous on $(0,\infty )\times \mathbb {R} ^{d}$, since all the first-order partial derivatives exist and are continuous (see exercise 2). It remains to be shown that $u$ is continuous on $\{0\}\times \mathbb {R} ^{d}$.
To do so, we first note that for all $(t,x)\in (0,\infty )\times \mathbb {R} ^{d}$
    ${\begin{aligned}\int _{\mathbb {R} ^{d}}E(t,x-y)dy&=\int _{\mathbb {R} ^{d}}E(t,y)dy&{\text{ integration by substitution using }}y\mapsto x-y\\&=\int _{\mathbb {R} ^{d}}{\sqrt {4\pi t}}^{-d}e^{\frac {\|y\|^{2}}{4t}}dy&\\&=\int _{\mathbb {R} ^{d}}{\sqrt {2\pi }}^{-d}e^{\frac {\|y\|^{2}}{2}}dy&{\text{ integration by substitution using }}y\mapsto {\sqrt {2t}}y\\&=1&{\text{ lemma 6.2}}\end{aligned}}$
Furthermore, due to the continuity of $g$, we may choose for arbitrary $\epsilon >0$ and any $x\in \mathbb {R} ^{d}$ a $\delta >0$ such that
    $\forall y\in B_{\delta }(x):|g(y)-g(x)|<\epsilon$.
From these last two observations, we may conclude:
    ${\begin{aligned}|g(x)-(E*_{x}g)(t,x)|&=\left|1\cdot g(x)-\int _{\mathbb {R} ^{d}}E(t,x-y)g(x)dy\right|&\\&=\left|\int _{\mathbb {R} ^{d}}E(t,x-y)g(x)dy-\int _{\mathbb {R} ^{d}}E(t,x-y)g(x)dy\right|\\&=\left|\int _{B_{\delta }(x)}E(t,x-y)(g(y)-g(x))dy+\int _{\mathbb {R} ^{d}\setminus B_{\delta }(x)}E(t,x-y)(g(y)-g(x))dy\right|&\\&\leq \left|\int _{B_{\delta }(x)}E(t,x-y)(g(y)-g(x))dy\right|+\left|\int _{\mathbb {R} ^{d}\setminus B_{\delta }(x)}E(t,x-y)(g(y)-g(x))dy\right|&{\text{triangle ineq. in }}\mathbb {R} \\&\leq \int _{B_{\delta }(x)}|E(t,x-y)|\underbrace {|g(y)-g(x)|} _{<\epsilon }dy+\int _{\mathbb {R} ^{d}\setminus B_{\delta }(x)}|E(t,x-y)(g(y)-g(x))|dy&{\text{ triangle ineq. for }}\int \\&<\int _{\mathbb {R} ^{d}}|E(t,x-y)|\epsilon dy+\int _{\mathbb {R} ^{d}\setminus B_{\delta }(x)}|E(t,x-y)|\underbrace {(|g(y)|+|g(x)|)} _{\leq 2\|g\|_{\infty }}dy&{\text{ monotony of the }}\int \\&=\epsilon +2\|g\|_{\infty }\left|\int _{\mathbb {R} ^{d}\setminus B_{\delta }(x)}E(t,x-y)dy\right|\end{aligned}}$
But due to integration by substitution using the diffeomorphism $x\mapsto {\sqrt {2t}}x$, we obtain
    $\int _{\mathbb {R} ^{d}\setminus B_{\delta }(x)}E(t,x-y)dy=\int _{\mathbb {R} ^{d}\setminus B_{\delta }(0)}E(t,x)dy=\int _{\mathbb {R} ^{d}\setminus B_{\frac {\delta }{\sqrt {2t}}}(0)}{\frac {1}{{\sqrt {2\pi }}^{d}}}e^{-{\frac {\|x\|^{2}}{2}}}dy\to 0,t\to 0$
which is why
    $\lim _{t\to 0}|g(x)-(E*_{x}g)(t,x)|<\epsilon$
Since $\epsilon >0$ was arbitrary, continuity is proven.$\Box$

## Sources
# Poisson's equation

This chapter deals with Poisson's equation
    $\forall x\in \mathbb {R} ^{d}:-\Delta u(x)=f(x)$
Provided that $f\in {\mathcal {C}}^{2}(\mathbb {R} ^{d})$, we will through distribution theory prove a solution formula, and for domains with boundaries satisfying a certain property we will even show a solution formula for the boundary value problem. We will also study solutions of the homogenous Poisson's equation
    $\forall x\in \mathbb {R} ^{d}:-\Delta u(x)=0$
The solutions to the homogenous Poisson's equation are called harmonic functions.

## Important theorems from multi-dimensional integration

In section 2, we had seen Leibniz' integral rule, and in section 4, Fubini's theorem. In this section, we repeat the other theorems from multi-dimensional integration which we need in order to carry on with applying the theory of distributions to partial differential equations. Proofs will not be given, since understanding the proofs of these theorems is not very important for the understanding of this wikibook. The only exception will be theorem 6.3, which follows from theorem 6.2. The proof of this theorem is an exercise.
Theorem 6.2: (Divergence theorem)
Let $K\subset \mathbb {R} ^{d}$ a compact set with smooth boundary. If $\mathbf {V} :K\to \mathbb {R} ^{d}$ is a vector field, then
    $\int _{K}\nabla \cdot \mathbf {V} (x)dx=\int _{\partial K}\nu (x)\cdot \mathbf {V} (x)dx$
, where $\nu :\partial K\to \mathbb {R} ^{d}$ is the outward normal vector.
Theorem 6.3: (Multi-dimensional integration by parts)
Let $K\subset \mathbb {R} ^{d}$ a compact set with smooth boundary. If $f:K\to \mathbb {R}$ is a function and $\mathbf {W} :K\to \mathbb {R} ^{d}$ is a vector field, then
    $\int _{K}f(x)\nabla \cdot \mathbf {W} (x)dx=\int _{\partial K}f(x)\nu (x)\cdot \mathbf {W} (x)dx-\int _{K}\mathbf {W} (x)\cdot \nabla f(x)dx$
, where $\nu :\partial K\to \mathbb {R} ^{d}$ is the outward normal vector.
Proof: See exercise 1.

## The volume and surface area of d-dimensional spheres

Definition 6.5:
The Gamma function $\Gamma :\mathbb {R} _{>0}\to \mathbb {R}$ is defined by
    $\Gamma (x):=\int _{0}^{\infty }s^{x-1}e^{-s}ds$
The Gamma function satisfies the following equation:
Theorem 6.6:
    $\forall x\in \mathbb {R} _{>0}:\Gamma (x+1)=x\Gamma (x)$
Proof:
    $\Gamma (x+1)=\int _{0}^{\infty }s^{x}e^{-s}ds{\overset {\text{integration by parts}}{=}}\underbrace {-s^{x}e^{-s}{\big |}_{s=0}^{s=\infty }} _{=0}-\int _{0}^{\infty }-xs^{x-1}e^{-s}ds=x\Gamma (x)$
$\Box$
If the Gamma function is shifted by 1, it is an interpolation of the factorial (see exercise 2):
As you can see, in the above plot the Gamma function also has values on negative numbers. This is because what is plotted above is some sort of a natural continuation of the Gamma function which one can construct using complex analysis.
Definition and theorem 6.7:
The $d$-dimensional spherical coordinates, given by $\Psi :(0,\infty )\times (0,2\pi )\times (-\pi /2,\pi /2)^{d-2}\to \mathbb {R} ^{d}\setminus \{(x_{1},\ldots ,x_{d})\in \mathbb {R} ^{d}:x_{1}\geq 0\wedge x_{2}=0\}$
    $\Psi (r,\Phi ,\Theta _{1},\ldots ,\Theta _{d-2})={\begin{pmatrix}r\cos(\Phi )\cos(\Theta _{1})\cdots \cos(\Theta _{d-2})\\r\sin(\Phi )\cos(\Theta _{1})\cdots \cos(\Theta _{d-2})\\r\sin(\Theta _{1})\cos(\Theta _{2})\cdots \cos(\Theta _{d-2})\\\vdots \\r\sin(\Theta _{d-3})\cos(\Theta _{d-2})\\r\sin(\Theta _{d-2})\\\end{pmatrix}}$
are a diffeomorphism. The determinant of the Jacobian matrix of $\Psi$, $\det J_{\Psi }$, is given by
    $\det J_{\Psi }(r,\Phi ,\Theta _{1},\ldots ,\Theta _{d-2})=r^{d-1}\cos(\Theta _{1})\cos(\Theta _{2})^{2}\cdots \cos(\Theta _{d-2})^{d-2}$
Proof:
Theorem 6.8:
The volume of the $d$-dimensional ball with radius $R\in \mathbb {R} _{>0}$, $B_{R}(0)$ is given by
    $V_{d}(R):={\frac {\pi ^{d/2}}{\Gamma (d/2+1)}}R^{d}$
Proof:
Theorem 6.9:
The area of the surface of the $d$-dimensional ball with radius $R\in \mathbb {R} _{>0}$ (i. e. the area of $\partial B_{R}(0)$) is given by
    $A_{d}(R):={\frac {d\pi ^{d/2}}{\Gamma (d/2+1)}}R^{d-1}$
The surface area and the volume of the $d$-dimensional ball with radius $R\in \mathbb {R} _{>0}$ are related to each other "in a differential way" (see exercise 3).
Proof:
    ${\begin{aligned}A_{d}(R)&:={\frac {d\pi ^{d/2}}{\Gamma (d/2+1)}}R^{d-1}&\\&={\frac {d}{R}}V_{d}(R)&\\&={\frac {d}{R}}\int _{B_{R}(0)}1dx&{\text{ theorem 6.8}}\\&=\int _{B_{R}(0)}{\frac {1}{R}}\nabla \cdot xdx&\\&=\int _{\partial B_{R}(0)}{\frac {1}{R}}{\frac {x}{R}}\cdot xdx&{\text{ divergence theorem}}\\&=\int _{\partial B_{R}(0)}1dx&\end{aligned}}$$\Box$

## Green's kernel

We recall a fact from integration theory:
Lemma 6.11: $f$ is integrable $\Leftrightarrow$ $|f|$ is integrable.
We omit the proof.
Theorem 6.12:
The function $P:\mathbb {R} ^{d}\to \mathbb {R}$, given by
    $P(x):={\begin{cases}-{\frac {1}{2}}|x|&d=1\\-{\frac {1}{2\pi }}\ln(\|x\|)&d=2\\{\frac {1}{(d-2)A_{d}(1)\|x\|^{d-2}}}&d\geq 3\end{cases}}$
is a Green's kernel for Poisson's equation.
We only prove the theorem for $d\geq 2$. For $d=1$ see exercise 4.
Proof:
1.
We show that $P$ is locally integrable. Let $K\subseteq \mathbb {R} ^{d}$ be compact. We have to show that
    $\int _{K}P(x)dx$
is a real number, which by lemma 6.11 is equivalent to
    $\int _{K}|P(x)|dx$
is a real number. As compact in $\mathbb {R} ^{d}$ is equivalent to bounded and closed, we may choose an $R>0$ such that $K\subset B_{R}(0)$. Without loss of generality we choose $R>1$, since if it turns out that the chosen $R$ is $\leq 1$, any $R>1$ will do as well. Then we have
    $\int _{K}|P(x)|dx\leq \int _{B_{R}(0)}|P(x)|dx$
For $d=2$, ${\begin{aligned}\int _{B_{R}(0)}|P(x)|dx&=\int _{B_{1}(0)}-{\frac {1}{2\pi }}\ln(\|x\|)dx+\int _{B_{R}(0)\setminus B_{1}(0)}{\frac {1}{2\pi }}\ln(\|x\|)dx&\\&=\int _{B_{1}(0)}-{\frac {1}{2\pi }}\ln(\|x\|)dx+\int _{B_{R}(0)}{\frac {1}{2\pi }}\ln(\|x\|)dx-\int _{B_{1}(0)}{\frac {1}{2\pi }}\ln(\|x\|)dx&\\&=\int _{0}^{1}-{\frac {r}{\pi }}\ln(r)dr+\int _{0}^{R}{\frac {r}{2\pi }}\ln(r)dr&{\text{int. by subst. using spherical coords.}}\\&={\frac {1}{4\pi }}+{\frac {R^{2}}{4\pi }}\left(\ln(R)-{\frac {1}{2}}R\right)<\infty \end{aligned}}$
For $d\geq 3$,
    ${\begin{aligned}\int _{B_{R}(0)}|P(x)|dx&=\int _{B_{R}(0)}{\frac {1}{(d-2)A_{d}(1)\|x\|^{d-2}}}\\&=\int _{0}^{R}\int _{0}^{2\pi }\underbrace {\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cdots \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}} _{d-2{\text{ times}}}\overbrace {|r^{d-1}\cos(\Theta _{1})\cdots \cos(\Theta _{d-2})^{d-2}|} ^{\leq r^{d-1}}{\frac {1}{(d-2)A_{d}(1)r^{d-2}}}d\Theta _{1}\cdots d\Theta _{d-2}d\Phi dr\\&\leq \int _{0}^{R}2{\frac {r^{d-1}}{(d-2)A_{d}(1)r^{d-2}}}\pi ^{d-1}dr\\&={\frac {\pi ^{d-1}}{(d-2)A_{d}(1)}}R^{2}\end{aligned}}$
, where we applied integration by substitution using spherical coordinates from the first to the second line.
2.
We calculate some derivatives of $P$ (see exercise 5):
For $d=2$, we have
    $\forall x\in \mathbb {R} ^{d}\setminus \{0\}:\nabla P(x)=-{\frac {x}{2\pi \|x\|^{2}}}$
For $d\geq 3$, we have
    $\forall x\in \mathbb {R} ^{d}\setminus \{0\}:\nabla P(x)={\frac {x}{A_{d}(1)\|x\|^{d}}}$
For all $d\geq 2$, we have $\forall x\in \mathbb {R} ^{d}\setminus \{0\}:\Delta P(x)=0$
3.
We show that
    $\forall x\in \mathbb {R} ^{d}:-\Delta {\mathcal {T}}_{P(\cdot -x)}=\delta _{x}$
Let $x\in \mathbb {R} ^{d}$ and $\varphi \in {\mathcal {D}}(\mathbb {R} ^{d})$ be arbitrary. In this last step of the proof, we will only manipulate the term $-\Delta {\mathcal {T}}_{P(\cdot -x)}(\varphi )$. Since $\varphi \in {\mathcal {D}}(\mathbb {R} ^{d})$, $\varphi$ has compact support. Let's define
    $K:={\text{supp }}\varphi \cup {\overline {B_{1}(x)}}$
Since the support of
    ${\begin{aligned}-\Delta {\mathcal {T}}_{P(\cdot -x)}(\varphi )&={\mathcal {T}}_{P(\cdot -x)}(-\Delta \varphi )&\\&=\int _{\mathbb {R} ^{d}}P(y-x)(-\Delta \varphi (y))dy&\\&=\int _{K}P(y-x)(-\Delta \varphi (y))dy&\\&=\lim _{\epsilon \downarrow 0}\int _{K}P(y-x)(-\Delta \varphi (y))\chi _{K\setminus B_{\epsilon }(x)}(y)dy&{\text{dominated convergence}}\\&=\lim _{\epsilon \downarrow 0}\int _{K\setminus B_{\epsilon }(x)}P(y-x)(-\Delta \varphi (y))dy&\\\end{aligned}}$
, where $\chi _{K\setminus B_{\epsilon }(0)}$ is the characteristic function of $K\setminus B_{\epsilon }(0)$.
The last integral is taken over $K\setminus B_{\epsilon }(x)$ (which is bounded and as the intersection of the closed sets $K$ and $\mathbb {R} ^{d}\setminus B_{\epsilon }(x)$ closed and thus compact as well). In this area, due to the above second part of this proof, $P(y-x)$ is continuously differentiable. Therefore, we are allowed to integrate by parts. Thus, noting that ${\frac {x-y}{\|y-x\|}}$ is the outward normal vector in $y\in \partial B_{\epsilon }(x)$ of $K\setminus B_{\epsilon }(x)$, we obtain
    $\int _{K\setminus B_{\epsilon }(x)}P(y-x)\overbrace {(-\Delta \varphi (y))} {=-\nabla \cdot \nabla \varphi (y)}dy=\int _{\partial B_{\epsilon }(x)}P(y-x)\nabla \varphi (y)\cdot {\frac {x-y}{\|y-x\|}}dy-\int _{\mathbb {R} ^{d}\setminus B_{\epsilon }(x)}\nabla \varphi (y)\cdot \nabla P(y-x)dy$
Let's furthermore choose $w(x)={\tilde {G}}(x-\xi )\nabla \varphi (x)$. Then
    $\nabla \cdot w(x)=\Delta \varphi (x){\tilde {G}}(x-\xi )+\langle \nabla {\tilde {G}}(x-\xi ),\nabla \varphi (x)\rangle$.
From Gauß' theorem, we obtain
    $\int _{\mathbb {R} ^{d}\setminus B_{R}(\xi )}\Delta \varphi (x){\tilde {G}}(x-\xi )+\langle \nabla {\tilde {G}}(x-\xi ),\nabla \varphi (x)\rangle dx=-\int _{\partial B_{R}(\xi )}\langle {\tilde {G}}(x-\xi )\nabla \varphi (x),{\frac {x-\xi }{\|x-\xi \|}}\rangle dx$
, where the minus in the right hand side occurs because we need the inward normal vector. From this follows immediately that
    $\int _{\mathbb {R} ^{d}\setminus B_{R}(\xi )}-\Delta \varphi (x){\tilde {G}}(x-\xi )=\underbrace {\int _{\partial B_{R}(\xi )}\langle {\tilde {G}}(x-\xi )\nabla \varphi (x),{\frac {x-\xi }{\|x-\xi \|}}\rangle dx} _{:=J_{1}(R)}-\underbrace {\int _{\mathbb {R} ^{d}\setminus B_{R}(\xi )}\langle \nabla {\tilde {G}}(x-\xi ),\nabla \varphi (x)\rangle dx} _{:=J_{2}(R)}$
We can now calculate the following, using the Cauchy-Schwartz inequality:
    $|J_{1}(R)|\leq \int _{\partial B_{R}(\xi )}\|{\tilde {G}}(x-\xi )\nabla \varphi (x)\|\overbrace {\|{\frac {x-\xi }{\|x-\xi \|}}\|} ^{=1}dx$
    $={\begin{cases}\displaystyle \int _{\partial B_{R}(\xi )}-{\frac {1}{2\pi }}\ln |x-\xi |\|\nabla \varphi (x)\|dx=\int _{\partial B_{1}(\xi )}-R{\frac {1}{2\pi }}\ln \|R(x-\xi )\|\|\nabla \varphi (Rx)\|dx&d=2\\\displaystyle \int _{\partial B_{R}(\xi )}{\frac {1}{(d-2)c}}{\frac {1}{|x-\xi |^{d-2}}}\|\nabla \varphi (x)\|dx=\int _{\partial B_{1}(\xi )}R^{d-1}{\frac {1}{(d-2)c}}{\frac {1}{|R(x-\xi )|^{d-2}}}&d\geq 3\end{cases}}$
    $\leq {\begin{cases}\displaystyle \max \limits _{x\in \mathbb {R} ^{d}}\|\nabla \varphi (Rx)\|\int _{\partial B_{1}(\xi )}-R{\frac {1}{2\pi }}\ln R^{2}dx=-\max \limits _{x\in \mathbb {R} ^{d}}\|\nabla \varphi (Rx)\|{\frac {c}{2\pi }}R\ln R^{2}\to 0,R\to 0&d=2\\\displaystyle \max \limits _{x\in \mathbb {R} ^{d}}\|\nabla \varphi (Rx)\|\int _{\partial B_{1}(\xi )}{\frac {1}{(d-2)c}}Rdx=\max \limits _{x\in \mathbb {R} ^{d}}\|\nabla \varphi (Rx)\|{\frac {R}{d-2}}\to 0,R\to 0&d\geq 3\end{cases}}$
Now we define $v(x)=\varphi (x)\nabla {\tilde {G}}(x-\xi )$, which gives:
    $\nabla \cdot v(x)=\varphi (x)\underbrace {\Delta {\tilde {G}}(x-\xi )} _{=0,x\neq \xi }+\langle \nabla \varphi (x),\nabla {\tilde {G}}(x-\xi )\rangle$
Applying Gauß' theorem on $v$ gives us therefore
    $J_{2}(R)=\int _{\partial B_{R}(\xi )}\varphi (x)\langle \nabla {\tilde {G}}(x-\xi ),{\frac {x-\xi }{\|x-\xi \|}}\rangle dx$
    $=\int _{\partial B_{R}(\xi )}\varphi (x)\langle -{\frac {x-\xi }{c\|x-\xi \|^{d}}},{\frac {x-\xi }{\|x-\xi \|}}\rangle dx=-{\frac {1}{c}}\int _{\partial B_{R}(\xi )}{\frac {1}{R^{d-1}}}\varphi (x)dx$
, noting that $d=2\Rightarrow c=2\pi$.
We furthermore note that
    $\varphi (\xi )={\frac {1}{c}}\int _{\partial B_{1}(\xi )}\varphi (\xi )dx={\frac {1}{c}}\int _{\partial B_{R}(\xi )}{\frac {1}{R^{d-1}}}\varphi (\xi )dx$
Therefore, we have
    $\lim _{R\to 0}|-J_{2}(R)-\varphi (\xi )|\leq {\frac {1}{c}}\lim _{R\to 0}\int _{\partial B_{R}(\xi )}{\frac {1}{R^{d-1}}}|\varphi (\xi )-\varphi (x)|dx\leq \lim _{R\to 0}{\frac {1}{c}}\max _{x\in B_{R}(\xi )}|\varphi (x)-\varphi (\xi )|\int _{\partial B_{1}(\xi )}1dx$
    $=\lim _{R\to 0}\max _{x\in B_{R}(\xi )}|\varphi (x)-\varphi (\xi )|=0$
due to the continuity of $\varphi$.
Thus we can conclude that
    $\forall \Omega {\text{ domain of }}\mathbb {R} ^{d}:\forall \varphi \in {\mathcal {D}}(\Omega ):-\Delta T_{{\tilde {G}}(\cdot -\xi )}(\varphi )=\lim _{R\to 0}J_{0}(R)=\lim _{R\to 0}J_{1}(R)-J_{2}(R)=0+\varphi (\xi )=\delta _{\xi }(\varphi )$.
Therefore, ${\tilde {G}}$ is a Green's kernel for the Poisson's equation for $d\geq 2$.
QED.

## Integration over spheres

Theorem 6.12:
Let $f:\mathbb {R} ^{d}\to \mathbb {R}$ be a function. $\int _{\partial B_{R}(0)}f(x)dx=R^{d-1}\int _{0}^{2\pi }\underbrace {\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cdots \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}} _{d-2{\text{ times}}}f\left(\Psi (r,\Phi ,\Theta _{1},\ldots ,\Theta _{d-2})\right)\cos(\Theta _{1})\cos(\Theta _{2})^{2}\cdots \cos(\Theta _{d-2})^{d-2}d\Theta _{1}d\Theta _{2}\cdots d\Theta _{d-2}d\Phi$
Proof: We choose as an orientation the border orientation of the sphere. We know that for $\partial B_{r}(0)$, an outward normal vector field is given by $\nu (x)={\frac {x}{r}}$. As a parametrisation of $B_{r}(0)$, we only choose the identity function, obtaining that the basis for the tangent space there is the standard basis, which in turn means that the volume form of $B_{r}(0)$ is
    $\omega _{B_{r}(0)}(x)=e_{1}^{*}\wedge \cdots \wedge e_{d}^{*}$
Now, we use the normal vector field to obtain the volume form of $\partial B_{r}(0)$:
    $\omega _{\partial B_{r}(0)}(x)(v_{1},\ldots ,v_{d-1})=\omega _{B_{r}(0)}(x)(\nu (x),v_{1},\ldots ,v_{d-1})$
We insert the formula for $\omega _{B_{r}(0)}(x)$ and then use Laplace's determinant formula:
    $=e_{1}^{*}\wedge \cdots \wedge e_{d}^{*}(\nu (x),v_{1},\ldots ,v_{d-1})={\frac {1}{r}}\sum _{i=1}^{d}(-1)^{i+1}x_{i}e_{1}^{*}\cdots \wedge e_{i-1}^{*}\wedge e_{i+1}^{*}\wedge \cdots \wedge e_{d}^{*}(v_{1},\ldots ,v_{d-1})$
As a parametrisation of $\partial B_{r}(x)$ we choose spherical coordinates with constant radius $r$.
We calculate the Jacobian matrix for the spherical coordinates:
    $J=\left({\begin{smallmatrix}\cos(\varphi )\cos(\vartheta _{1})\cdots \cos(\vartheta _{d-2})&r-\sin(\varphi )\cos(\vartheta _{1})\cdots \cos(\vartheta _{d-2})&-r\cos(\varphi )\sin(\vartheta _{1})\cdots \cos(\vartheta _{d-2})&\cdots &\cdots &-r\cos(\varphi )\cos(\vartheta _{1})\cdots \sin(\vartheta _{d-2})\\\sin(\varphi )\cos(\vartheta _{1})\cdots \cos(\vartheta _{d-2})&r\cos(\varphi )\cos(\vartheta _{1})\cdots \cos(\vartheta _{d-2})&-r\sin(\varphi )\sin(\vartheta _{1})\cdots \cos(\vartheta _{d-2})&\cdots &\cdots &-r\sin(\varphi )\cos(\vartheta _{1})\cdots \sin(\vartheta _{d-2})\\\vdots &0&\ddots &\ddots &\ddots &\vdots \\\vdots &\vdots &\ddots &\ddots &\ddots &\\\sin(\vartheta _{d-3})\cos(\vartheta _{d-2})&0&\cdots &0&r\cos(\vartheta _{d-3})\cos(\vartheta _{d-2})&r\sin(\vartheta _{d-3})\cos(\vartheta _{d-2})\\\sin(\vartheta _{d-2})&0&\cdots &\cdots &0&r\cos(\vartheta _{d-2})\end{smallmatrix}}\right)$
We observe that in the first column, we have only the spherical coordinates divided by $r$. If we fix $r$, the first column disappears. Let's call the resulting matrix $J'$ and our parametrisation, namely spherical coordinates with constant $r$, $\Psi$. Then we have:
    $\Psi ^{*}\omega _{\partial B_{r}(0)}(x)(v_{1},\ldots ,v_{d-1})=\omega _{\partial B_{r}(0)}(\Psi (x))(J'v_{1},\ldots ,J'v_{d-1})$
    $={\frac {1}{r}}\sum _{i=1}^{d}(-1)^{i+1}\Psi (x)_{i}e_{1}^{*}\cdots \wedge e_{i-1}^{*}\wedge e_{i+1}^{*}\wedge \cdots \wedge e_{d}^{*}(J'v_{1},\ldots ,J'v_{d-1})$
    $={\frac {1}{r}}\sum _{i=1}^{d}(-1)^{i+1}\Psi (x)_{i}\det(e_{j}^{*}(J'v_{k}))_{j\neq i}=\det J\cdot \det(v_{1},\ldots ,v_{d-1})$
Recalling that
    $\det J=r^{d-1}\cos(\phi _{1})^{n-2}\cos(\phi _{2})^{d-3}\cdots \cos(\phi _{d-2})$
, the claim follows using the definition of the surface integral.
Theorem 6.13:
Let $f:\mathbb {R} ^{d}\to \mathbb {R}$ be a function. Then
    $\int _{\mathbb {R} ^{d}}f(x)dx=\int _{0}^{\infty }r^{d-1}\int _{\partial B_{1}(0)}f(rx)dxdr$
Proof:
We have $r\Psi (1,\Phi ,\Theta _{1},\ldots ,\Theta _{d-2})=\Psi (r,\Phi ,\Theta _{1},\ldots ,\Theta _{d-2})$, where $\Psi$ are the spherical coordinates. Therefore, by integration by substitution, Fubini's theorem and the above formula for integration over the unit sphere,
    ${\begin{aligned}\int _{\mathbb {R} ^{d}}f(x)dx&=\int _{(0,\infty )\times (0,2\pi )\times (-\pi /2,\pi /2)^{d-2}}f(\Psi (x))|\det J_{\Psi }(x)|dx\\&=\int _{0}^{\infty }\int _{0}^{2\pi }\underbrace {\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cdots \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}} _{d-2{\text{ times}}}f(\Psi (r,\Phi ,\Theta _{1},\ldots ,\Theta _{d-2}))r^{d-1}\cos(\Theta _{1})\cdots \cos(\Theta _{d-2})^{d-2}d\Theta _{1}\cdots d\Theta _{d-2}d\Phi dr\\&=\int _{0}^{\infty }r^{d-1}\int _{0}^{2\pi }\underbrace {\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cdots \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}} _{d-2{\text{ times}}}f(r\Psi (1,\Phi ,\Theta _{1},\ldots ,\Theta _{d-2}))\cos(\Theta _{1})\cdots \cos(\Theta _{d-2})^{d-2}d\Theta _{1}\cdots d\Theta _{d-2}d\Phi dr\\&=\int _{0}^{\infty }r^{d-1}\int _{\partial B_{1}(0)}f(rx)dr\end{aligned}}$$\Box$

## Harmonic functions

Definition 6.14: Let $O\subseteq \mathbb {R} ^{d}$ be open and let $u:O\to \mathbb {R}$ be a function. If $u\in {\mathcal {C}}^{2}(O)$ and
    $\forall x\in O:\Delta u(x)=0$
$u$ is called a harmonic function.
Theorem 6.15:
Let $O\subseteq \mathbb {R} ^{d}$ be open and let $u\in {\mathcal {C}}^{2}(O)$. The following conditions are equivalent:
* $u$ is harmonic
* $\forall x\in O:\forall R{\text{ such that }}{\overline {B_{R}(0)}}\subset O:u(x)={\frac {1}{A_{d}(R)}}\int _{\partial B_{R}(x)}u(y)dy$
* $\forall x\in O:\forall R{\text{ such that }}{\overline {B_{R}(0)}}\subset O:u(x)={\frac {1}{V_{d}(R)}}\int _{B_{R}(x)}u(y)dy$
Proof: Let's define the following function:
    $\phi (r)={\frac {1}{r^{d-1}}}\int _{\partial B_{r}(x)}u(y)dy$
From first coordinate transformation with the diffeomorphism $y\mapsto x+y$ and then applying our formula for integration on the unit sphere twice, we obtain:
    $\phi (r)={\frac {1}{r^{d-1}}}\int _{\partial B_{r}(0)}u(y+x)dy=\int _{\partial B_{1}(0)}u(x+ry)dy$
From first differentiation under the integral sign and then Gauss' theorem, we know that
    $\phi '(r)=\int _{\partial B_{1}(0)}\langle \nabla u(x+ry),y\rangle dy=\int _{B_{1}(0)}\Delta u(x+ry)dy=0$
Case 1: If $u$ is harmonic, then we have
    $\int _{B_{1}(0)}\Delta u(x+ry)dy=0$
, which is why $\phi$ is constant. Now we can use the dominated convergence theorem for the following calculation:
    $\lim _{r\to 0}\phi (r)=\int _{\partial B_{1}(0)}\lim _{r\to 0}u(x+ry)dy=c(1)u(x)$
Therefore $\phi (r)=c(1)u(x)$ for all $r$.
With the relationship
    $r^{d-1}c(1)=c(r)$
, which is true because of our formula for $c(x),x\in \mathbb {R} _{>0}$, we obtain that
    $u(x)={\frac {\phi (r)}{c(1)}}={\frac {1}{c(1)}}{\frac {1}{r^{d-1}}}\int _{\partial B_{r}(x)}u(y)dy={\frac {1}{c(r)}}\int _{\partial B_{r}(x)}u(y)dy$
, which proves the first formula.
Furthermore, we can prove the second formula by first transformation of variables, then integrating by onion skins, then using the first formula of this theorem and then integration by onion skins again:
    $\int _{B_{r}(x)}u(y)dy=\int _{B_{r}(0)}u(y+x)dy=\int _{0}^{r}s^{d-1}\int _{\partial B_{1}(0)}u(y+sx)dxds=\int _{0}^{r}s^{d-1}u(x)\int _{\partial B_{1}(0)}1dxds=u(x)d(r)$
This shows that if $u$ is harmonic, then the two formulas for calculating $u$, hold.
Case 2: Suppose that $u$ is not harmonic. Then there exists an $x\in \Omega$ such that $-\Delta u(x)\neq 0$. Without loss of generality, we assume that $-\Delta u(x)>0$; the proof for $-\Delta u(x)<0$ will be completely analogous exept that the direction of the inequalities will interchange. Then, since as above, due to the dominated convergence theorem, we have
    $\lim _{r\to 0}\phi '(r)=\int _{B_{1}(0)}\lim _{r\to 0}\Delta u(x+ry)dy>0$
Since $\phi '$ is continuous (by the dominated convergence theorem), this is why $\phi$ grows at $0$, which is a contradiction to the first formula.
The contradiction to the second formula can be obtained by observing that $\phi '$ is continuous and therefore there exists a $\sigma \in \mathbb {R} _{>0}$
    $\forall r\in [0,\sigma ):\phi '(r)>0$
This means that since
    $\lim _{r\to 0}\phi (r)=\int _{\partial B_{1}(0)}\lim _{r\to 0}u(x+ry)dy=c(1)u(x)$
and therefore
    $\phi (0)=c(1)u(x)$
, that
    $\forall r\in (0,\sigma ):\phi (r)>c(1)u(x)$
and therefore, by the same calculation as above,
    $\int _{B_{r}(x)}u(y)dy=\int _{B_{r}(0)}u(y+x)dy=\int _{0}^{r}s^{d-1}\int _{\partial B_{1}(0)}u(y+sx)dxds>\int _{0}^{r}s^{d-1}u(x)\int _{\partial B_{1}(0)}1dxds=u(x)d(r)$
This shows (by proof with contradiction) that if one of the two formulas hold, then $u\in C^{2}(\Omega )$ is harmonic.
Definition 6.16:
A domain is an open and connected subset of $\mathbb {R} ^{d}$.
For the proof of the next theorem, we need two theorems from other subjects, the first from integration theory and the second from topology.
Theorem 6.17:
Let $B\subseteq \mathbb {R} ^{d}$ and let $f:B\to \mathbb {R}$ be a function. If
    $\int _{B}|f(x)|dx=0$
then $f(x)=0$ for almost every $x\in B$.
Theorem 6.18:
In a connected topological space, the only simultaneously open and closed sets are the whole space and the empty set.
We will omit the proofs.
Theorem 6.19:
Let $\Omega \subseteq \mathbb {R} ^{d}$ be a domain and let $u:\Omega \to \mathbb {R}$ be harmonic. If there exists an $x\in \Omega$ such that
    $u(x)=\sup _{y\in \Omega }u(y)$
, then $u$ is constant.
Proof:
We choose
    $B:=\left\{x\in \Omega :u(x)=\sup _{y\in \Omega }u(y)\right\}$
Since $\Omega$ is open by assumption and $B\subseteq \Omega$, for every $x\in B$ exists an $R\in \mathbb {R} _{>0}$ such that
    ${\overline {B_{R}(x)}}\subseteq \Omega$
By theorem 6.15, we obtain in this case:
    $\sup _{y\in \Omega }u(y)=u(x)={\frac {1}{V_{d}(R)}}\int _{B_{R}(x)}u(z)dz$
Further,
    $\sup _{y\in \Omega }u(y)=\sup _{y\in \Omega }u(y){\frac {V_{d}(R)}{V_{d}(R)}}={\frac {1}{V_{d}(R)}}\int _{B_{r}(x)}\sup _{y\in \Omega }u(y)dz$
, which is why
    ${\begin{aligned}{\frac {1}{V_{d}(R)}}\int _{B_{R}(x)}u(z)dz=\int _{B_{R}(x)}\sup _{y\in \Omega }u(y)dz\\\Leftrightarrow {\frac {1}{V_{d}(R)}}\int _{B_{R}(x)}(u(z)-\sup _{y\in \Omega }u(y))dz=0\end{aligned}}$
Since
    $\forall z\in \Omega :\sup _{y\in \Omega }u(y)\geq u(z)$
, we have even
    $0={\frac {1}{V_{d}(R)}}\int _{B_{R}(x)}(u(z)-\sup _{y\in \Omega }u(y))dz=-{\frac {1}{V_{d}(R)}}\int _{B_{R}(x)}|u(z)-\sup _{y\in \Omega }u(y)|dz$
By theorem 6.17 we conclude that
    $u(z)=\sup _{y\in \Omega }u(y)$
almost everywhere in $B_{R}(x)$, and since
    $z\mapsto u(z)-\sup _{y\in \Omega }u(y)$
is continuous, even
    $u(z)=\sup _{y\in \Omega }u(y)$
really everywhere in $B_{R}(x)$ (see exercise 6). Therefore $B_{R}(0)\subseteq B$, and since $x\in B$ was arbitrary, $B$ is open.
Also,
    $B=u^{-1}\left(\left\{\sup _{y\in \Omega }u(y)\right\}\right)$
and $u$ is continuous. Thus, as a one-point set is closed, lemma 3.13 says $B$ is closed in $\Omega$. Thus $B$ is simultaneously open and closed. By theorem 6.18, we obtain that either $B=\emptyset$ or $B=\Omega$. And since by assumtion $B$ is not empty, we have $B=\Omega$.$\Box$
Theorem 6.18:
Let $\Omega \subseteq \mathbb {R} ^{d}$ be a domain and let $u:\Omega \to \mathbb {R}$ be harmonic. If there exists an $x\in \Omega$ such that
    $u(x)=\inf _{y\in \Omega }u(y)$
, then $u$ is constant.
Proof: See exercise 7.
Corollary 6.20:
Let $\Omega \subseteq \mathbb {R} ^{d}$ be a bounded domain and let $u:{\overline {\Omega }}\to \mathbb {R}$ be harmonic on $\Omega$ and continuous on ${\overline {\Omega }}$. Then
    $\forall x\in {\overline {\Omega }}:\inf _{y\in \partial \Omega }u(y)\leq u(x)\leq \sup _{y\in \partial \Omega }u(y)$
Proof:
Theorem 6.20:
Let $O\subseteq \mathbb {R} ^{d}$ be open and $u:O\to \mathbb {R}$ be a harmonic function, let $x_{0}\in O$ and let $R>0$ such that ${\overline {B_{R}(x_{0})}}\subset O$. Then
    $\forall x\in B_{R}(x_{0}):u(x)={\frac {R^{2}+\|x-x_{0}\|^{2}}{RA_{d}(1)}}\int _{\partial B_{R}(x_{0})}{\frac {u(y)}{\|x-y\|^{d}}}dy$
Proof:
What we will do next is showing that every harmonic function $u\in {\mathcal {C}}^{2}(O)$ is in fact automatically contained in ${\mathcal {C}}^{\infty }(O)$.
Theorem 6.25: Let $O\subseteq \mathbb {R} ^{d}$ be open, and let $u:O\to \mathbb {R}$ be harmonic. Then $u\in {\mathcal {C}}^{\infty }(O)$. Furthermore, for all $n\in \mathbb {N}$, there is a constant $C_{d,n}$ depending only on the dimension $d$ and $n$ such that for all $x_{0}\in O$ and $R>0$ such that $B_{R}(x_{0})\subseteq O$
    $\forall \alpha \in \mathbb {N} ^{d}{\text{ with }}|\alpha |=n:\partial _{\alpha }u(x)\leq {\frac {C_{d,n}}{R^{d+n}}}\int _{B_{R}(x_{0})}|u(y)|dy$
Proof:
Definition 6.26:
Let $(u_{l})_{l\in \mathbb {N} }$ be a sequence of harmonic functions, and let $u:O\to \mathbb {R}$ be a function. $(u_{l})_{l\in \mathbb {N} }$ converges locally uniformly to $u$ iff
Theorem 6.27:
Let $O\subseteq \mathbb {R} ^{d}$ be open and let $u_{l}:O\to \mathbb {R} ,l\in \mathbb {N}$ be harmonic functions such that the sequence $(u_{l})_{l\in \mathbb {N} }$ converges locally uniformly to a function $u:O\to \mathbb {R}$. Then also $u$ is harmonic.
Proof:
Definition 6.28:
Theorem 6.29: (Arzelà-Ascoli) Let $F$ be a set of continuous functions, which are defined on a compact set $K$. Then the following two statements are equivalent:
* ${\overline {F}}$ (the closure of $F$) is compact
* $F$ is bounded and equicontinuous
Proof:
Definition 6.30:
Theorem 6.31:
Let $(u_{l})_{l\in \mathbb {N} }$ be a locally uniformly bounded sequence of harmonic functions. Then it has a locally uniformly convergent subsequence.
Proof:

## Boundary value problem

The dirichlet problem for the Poisson equation is to find a solution for
    ${\begin{cases}-\Delta u(x)=f(x)&x\in \Omega \\u(x)=g(x)&x\in \partial \Omega \end{cases}}$

### Uniqueness of solutions

If $\Omega$ is bounded, then we can know that if the problem
    ${\begin{cases}-\Delta u(x)=f(x)&x\in \Omega \\u(x)=g(x)&x\in \partial \Omega \end{cases}}$
has a solution $u_{1}$, then this solution is unique on $\Omega$.
Proof: Let $u_{2}$ be another solution. If we define $u=u_{1}-u_{2}$, then $u$ obviously solves the problem
    ${\begin{cases}-\Delta u(x)=0&,x\in \Omega \\u(x)=0&x\in \partial \Omega \end{cases}}$
, since $-\Delta (u_{1}(x)-u_{2}(x))=-\Delta u_{1}(x)-(-\Delta u_{2}(x))=f(x)-f(x)=0$ for $x\in \Omega$ and $u_{1}(x)-u_{2}(x)=g(x)-g(x)=0$ for $x\in \partial \Omega$.
Due to the above corollary from the minimum and maximum principle, we obtain that $u$ is constantly zero not only on the boundary, but on the whole domain $\Omega$. Therefore $u_{1}(x)-u_{2}(x)=0\Leftrightarrow u_{1}(x)=u_{2}(x)$ on $\Omega$. This is what we wanted to prove.

### Green's functions of the first kind

Let $\Omega \subseteq \mathbb {R} ^{d}$ be a domain. Let ${\tilde {G}}$ be the Green's kernel of Poisson's equation, which we have calculated above, i.e.
    ${\tilde {G}}(x):={\begin{cases}-{\frac {1}{2}}|x|&d=1\\-{\frac {1}{2\pi }}\ln \|x\|&d=2\\{\frac {1}{(d-2)c}}{\frac {1}{\|x\|^{d-2}}}&d\geq 3\end{cases}}$
, where $c:=\int _{\partial B_{1}(0)}1dz$ denotes the surface area of $B_{1}(0)\subset \mathbb {R} ^{d}$.
Suppose there is a function $h:\Omega \times \Omega \to \mathbb {R}$ which satisfies
    ${\begin{cases}-\Delta h(x,\xi )=0&x\in \Omega \\h(x,\xi )={\tilde {G}}(x-\xi )&x\in \partial \Omega \end{cases}}$
Then the Green's function of the first kind for $-\Delta$ for $\Omega$ is defined as follows:
    ${\tilde {G}}_{\Omega }(x,\xi ):={\tilde {G}}(x-\xi )-h(x,\xi )$
${\tilde {G}}(x-\xi )-h(x,\xi )$ is automatically a Green's function for $-\Delta$. This is verified exactly the same way as veryfying that ${\tilde {G}}$ is a Green's kernel. The only additional thing we need to know is that $h$ does not play any role in the limit processes because it is bounded.
A property of this function is that it satisfies
    ${\begin{cases}-\Delta {\tilde {G}}_{\Omega }(x,\xi )=0&x\in \Omega \setminus \{\xi \}\\{\tilde {G}}_{\Omega }(x,\xi )=0&x\in \partial \Omega \end{cases}}$
The second of these equations is clear from the definition, and the first follows recalling that we calculated above (where we calculated the Green's kernel), that $\Delta {\tilde {G}}(x)=0$ for $x\neq 0$.

### Representation formula

Let $\Omega \subseteq \mathbb {R} ^{d}$ be a domain, and let $u\in C^{2}(\Omega )$ be a solution to the Dirichlet problem
    ${\begin{cases}-\Delta u(x)=f(x)&x\in \Omega \\u(x)=g(x)&x\in \partial \Omega \end{cases}}$
. Then the following representation formula for $u$ holds:
    $u(\xi )=\int _{\Omega }-\Delta u(y){\tilde {G}}_{\Omega }(y,\xi )dy-\int _{\partial \Omega }u(y)\nu (y)\nabla _{y}{\tilde {G}}_{\Omega }(y,\xi )dy$
, where ${\tilde {G}}_{\Omega }$ is a Green's function of the first kind for $\Omega$.
Proof: Let's define
    $J(\epsilon ):=\int _{\Omega \setminus B_{\epsilon }(\xi )}-\Delta u(y){\tilde {G}}_{\Omega }(y,\xi )dy$
. By the theorem of dominated convergence, we have that
    $\lim _{\epsilon \to 0}J(\epsilon )=\int _{\Omega }-\Delta u(y){\tilde {G}}_{\Omega }(y,\xi )dy$
Using multi-dimensional integration by parts, it can be obtained that:
    $J(\epsilon )=-\int _{\partial \Omega }\underbrace {{\tilde {G}}_{\Omega }(y,\xi )} _{=0}\langle \nabla u(y),\nu (y)\rangle dy+\int _{\partial B_{\epsilon }(\xi )}{\tilde {G}}_{\Omega }(y,\xi )\langle \nabla u(y),{\frac {y-\xi }{\|y-\xi \|}}\rangle dy+\int _{\Omega \setminus B_{\epsilon }(\xi )}\langle \nabla u(y),\nabla _{x}{\tilde {G}}_{\Omega }(y,\xi )\rangle dy$
    $=\underbrace {\int _{\partial B_{\epsilon }(\xi )}{\tilde {G}}_{\Omega }(y,\xi )\langle \nabla u(y),{\frac {y-\xi }{\|y-\xi \|}}\rangle dy} _{:=J_{1}(\epsilon )}-\int _{\Omega \setminus B_{\epsilon }(\xi )}\Delta {\tilde {G}}_{\Omega }(y,\xi )u(y)dy$
    $-\underbrace {\int _{\partial B_{\epsilon }(\xi )}u(y)\langle \nabla {\tilde {G}}_{\Omega }(y,\xi ),{\frac {y-\xi }{\|y-\xi \|}}\rangle dy} _{:=J_{2}(\epsilon )}-\int _{\partial \Omega }u(y)\langle \nabla {\tilde {G}}_{\Omega }(y,\xi ),\nu (y)\rangle dy$
When we proved the formula for the Green's kernel of Poisson's equation, we had already shown that
    $\lim _{\epsilon \to 0}-J_{2}(\epsilon )=u(\xi )$ and
    $\lim _{\epsilon \to 0}J_{1}(\epsilon )=0$
The only additional thing which is needed to verify this is that $h\in C^{\infty }(\Omega )$, which is why it stays bounded, while ${\tilde {G}}$ goes to infinity as $\epsilon \to 0$, which is why $h$ doesn't play a role in the limit process.
This proves the formula.

### Harmonic functions on the ball: A special case of the Dirichlet problem
#### Green's function of the first kind for the ball

Let's choose
    $h(x,\xi )={\tilde {G}}\left({\frac {\|\xi \|}{r}}\left(x-{\frac {r^{2}}{\|\xi \|^{2}}}\xi \right)\right)$
Then
    ${\tilde {G}}_{B_{r}(x_{0})}(x,\xi ):={\tilde {G}}(x-\xi )-h(x-x_{0},\xi -x_{0})$
is a Green's function of the first kind for $B_{r}(x_{0})$.
Proof: Since $\xi -x_{0}\in B_{r}(0)\Rightarrow {\frac {r^{2}}{\|\xi -x_{0}\|^{2}}}(\xi -x_{0})\notin B_{r}(0)$ and therefore
    $\forall x,\xi \in B_{r}(0):-\Delta _{x}h(x-x_{0},\xi -x_{0})=0$
Furthermore, we obtain:
    $\int _{B_{r}(x_{0})}-\Delta \varphi (x){\tilde {G}}_{\Omega }(x,\xi )dx=\int _{B_{r}(x_{0})}-\Delta \varphi (x){\tilde {G}}(x-\xi )dx+\int _{B_{r}(x_{0})}\varphi (x)-\Delta h(x,\xi )dx=\varphi (\xi )+0$
, which is why ${\tilde {G}}_{\Omega }(x,\xi )$ is a Green's function.
The property for the boundary comes from the following calculation:
    $\forall x\in \partial B_{r}(0):\|x-\xi \|^{2}=\langle x-\xi ,x-\xi \rangle =r^{2}+\|\xi \|^{2}-2\langle x,\xi \rangle ={\frac {\|\xi \|^{2}}{r^{2}}}(\langle x-{\frac {r^{2}}{\|\xi \|^{2}}}\xi ,x-{\frac {r^{2}}{\|\xi \|^{2}}}\xi \rangle )={\frac {\|\xi \|^{2}}{r^{2}}}\|x-{\frac {r^{2}}{\|\xi \|^{2}}}\xi \|^{2}$
, which is why $x\in \partial B_{r}(0)\Rightarrow h(x,\xi )={\tilde {G}}(x,\xi )$, since ${\tilde {G}}$ is radially symmetric.

#### Solution formula

Let's consider the following problem:
    ${\begin{cases}-\Delta u(x)=0&x\in B_{r}(0)\\u(x)=\varphi (x)&x\in \partial B_{r}(0)\end{cases}}$
Here $\varphi$ shall be continuous on $\partial B_{r}(0)$. Then the following holds: The unique solution $u\in C({\overline {B_{r}(0)}})\cap C^{2}(B_{r}(0))$ for this problem is given by:
    $u(\xi )={\begin{cases}\int _{\partial B_{r}(0)}\langle -\nu (y),\nabla _{y}{\tilde {G}}_{B_{r}(0)}(y,\xi )\rangle \varphi (y)dy&\xi \in B_{r}(0)\\\varphi (\xi )&\xi \in \partial B_{r}(0)\end{cases}}$
Proof: Uniqueness we have already proven; we have shown that for all Dirichlet problems for $-\Delta$ on bounded domains (and the unit ball is of course bounded), the solutions are unique.
Therefore, it only remains to show that the above function is a solution to the problem. To do so, we note first that
    $-\Delta \int _{\partial B_{r}(0)}\langle -\nu (y),{\tilde {\nabla }}_{y}G_{B_{r}(0)}(y,\xi )\rangle \varphi (y)dy=-\Delta \int _{\partial B_{r}(0)}\langle -\nu (y),\nabla _{y}({\tilde {G}}(y-\xi )-h(y,\xi ))\rangle \varphi (y)dy$
Let $0<s<r$ be arbitrary. Since ${\tilde {G}}_{B_{r}(0)}$ is continuous in $B_{s}(0)$, we have that on $B_{s}(0)$ it is bounded. Therefore, by the fundamental estimate, we know that the integral is bounded, since the sphere, the set over which is integrated, is a bounded set, and therefore the whole integral must be always below a certain constant. But this means, that we are allowed to differentiate under the integral sign on $B_{s}(0)$, and since $r>s>0$ was arbitrary, we can directly conclude that on $B_{r}(0)$,
    $-\Delta u(\xi )=\int _{\partial B_{r}(0)}\overbrace {-\Delta (\langle -\nu (y),{\tilde {\nabla }}_{y}{\tilde {G}}(x-\xi )-h(x,\xi )\rangle \varphi (y))} ^{=0}dy=0$
Furthermore, we have to show that $\forall x\in \partial B_{r}(0):\lim _{y\to x}u(y)=\varphi (x)$, i. e. that $u$ is continuous on the boundary.
To do this, we notice first that
    $\int _{\partial B_{r}(0)}\langle -\nu (y),\nabla _{y}{\tilde {G}}_{B_{r}(0)}(y,\xi )\rangle dy=1$
This follows due to the fact that if $u\equiv 1$, then $u$ solves the problem
    ${\begin{cases}-\Delta u(x)=0&x\in B_{r}(0)\\u(x)=1&x\in \partial B_{r}(0)\end{cases}}$
and the application of the representation formula.
Furthermore, if $\|x-x^{*}\|<{\frac {1}{2}}\delta$ and $\|y-x^{*}\|\geq \delta$, we have due to the second triangle inequality:
    $\|x-y\|\geq |\|y-x^{*}\|-\|x^{*}-x\||\geq {\frac {1}{2}}\delta$
In addition, another application of the second triangle inequality gives:
    $(r^{2}-\|x\|^{2})=(r+\|x\|)(r-\|x\|)=(r+\|x\|)(\|x^{*}\|-\|x\|)\leq 2r\|x^{*}-x\|$
Let then $\epsilon >0$ be arbitrary, and let $x^{*}\in \partial B_{r}(0)$. Then, due to the continuity of $\varphi$, we are allowed to choose $\delta >0$ such that
    $\|x-x^{*}\|<\delta \Rightarrow |\varphi (x)-\varphi (x^{*})|<{\frac {\epsilon }{2}}$.
In the end, with the help of all the previous estimations we have made, we may unleash the last chain of inequalities which shows that the representation formula is true:
    $|u(x)-u(x^{*})|=|u(x)-1\cdot \varphi (x^{*})|=\left|\int _{\partial B_{r}(0)}\langle -\nu (y),\nabla _{y}{\tilde {G}}_{B_{r}(0)}(y,x)\rangle (\varphi (x)-\varphi (x^{*}))dy\right|$
    $\leq {\frac {\epsilon }{2}}\int _{\partial B_{r}(0)\cap B_{\delta }(x^{*})}|\langle -\nu (y),\nabla _{y}{\tilde {G}}_{B_{r}(0)}(y,x)\rangle |dy+2\|\varphi \|_{\infty }\int _{\partial B_{r}(0)\setminus B_{\delta }(x^{*})}|\langle -\nu (y),\nabla _{y}{\tilde {G}}_{B_{r}(0)}(y,x)\rangle |dy$
    $\leq {\frac {\epsilon }{2}}+2\|\varphi \|_{\infty }\int _{\partial B_{r}(0)\setminus B_{\delta }(x^{*})}{\frac {r^{2}-\|x\|^{2}}{rc(1)\left({\frac {\delta }{2}}\right)^{d}}}dy\leq {\frac {\epsilon }{2}}+2\|\varphi \|_{\infty }r^{d-2}{\frac {r^{2}-\|x\|^{2}}{\left({\frac {\delta }{2}}\right)^{d}}}$
Since $x\to x^{*}$ implies $r^{2}-\|x\|^{2}\to 0$, we might choose $x$ close enough to $x^{*}$ such that
    $2\|\varphi \|_{\infty }r^{d-2}{\frac {r^{2}-\|x\|^{2}}{\left({\frac {\delta }{2}}\right)^{d}}}<{\frac {\epsilon }{2}}$. Since $\epsilon >0$ was arbitrary, this finishes the proof.

### Barriers

Let $\Omega \subset \mathbb {R} ^{d}$ be a domain. A function $b:\mathbb {R} ^{d}\to \mathbb {R}$ is called a barrier with respect to $y\in \partial \Omega$ if and only if the following properties are satisfied:
* $b$ is continuous
* $b$ is superharmonic on $\Omega$
* $b(y)=0$
* $\forall x\in \mathbb {R} ^{d}\setminus \Omega :b(x)>0$

### Exterior sphere condition

Let $\Omega \subseteq \mathbb {R} ^{d}$ be a domain. We say that it satisfies the exterior sphere condition, if and only if for all $x\in \partial \Omega$ there is a ball $B_{r}(z)\subseteq \mathbb {R} ^{d}\setminus \Omega$ such that $x\in \partial B_{r}(z)$ for some $z\in \mathbb {R} ^{d}\setminus \Omega$ and $r\in \mathbb {R} _{\geq 0}$.

### Subharmonic and superharmonic functions

Let $\Omega \subseteq \mathbb {R} ^{d}$ be a domain and $v\in C(\Omega )$.
We call $v$ subharmonic if and only if:
    $v(x)\leq {\frac {1}{d(r)}}\int _{B_{r}(x)}v(y)dy$
We call $v$ superharmonic if and only if:
    $v(x)\geq {\frac {1}{d(r)}}\int _{B_{r}(x)}v(y)dy$
From this definition we can see that a function is harmonic if and only if it is subharmonic and superharmonic.

### Minimum principle for superharmonic functions

A superharmonic function $u$ on $\Omega$ attains it's minimum on $\Omega$'s border $\partial \Omega$.
Proof: Almost the same as the proof of the minimum and maximum principle for harmonic functions. As an exercise, you might try to prove this minimum principle yourself.

### Harmonic lowering

Let $u\in {\mathcal {S}}_{\varphi }(\Omega )$, and let $B_{r}(x_{0})\subset \Omega$. If we define
    ${\tilde {u}}(x)={\begin{cases}u(x)&x\notin B_{r}(x_{0})\\\int _{\partial B_{r}(0)}\langle -\nu (y),\nabla _{y}{\tilde {G}}_{B_{r}(0)}(y,x)\rangle \varphi (y)dy&x\in B_{r}(x_{0})\end{cases}}$
, then ${\tilde {u}}\in {\mathcal {S}}_{\varphi }(\Omega )$.
Proof: For this proof, the very important thing to notice is that the formula for ${\tilde {u}}$ inside $B_{r}(x_{0})$ is nothing but the solution formula for the Dirichlet problem on the ball. Therefore, we immediately obtain that ${\tilde {u}}$ is superharmonic, and furthermore, the values on $\partial \Omega$ don't change, which is why ${\tilde {u}}\in {\mathcal {S}}_{\varphi }(\Omega )$. This was to show.

### Definition 3.1

Let $\varphi \in C(\partial \Omega )$. Then we define the following set:
    ${\mathcal {S}}_{\varphi }(\Omega ):=\{u\in C({\overline {\Omega }}):u{\text{ superharmonic and }}x\in \partial \Omega \Rightarrow u(x)\geq \varphi (x)\}$

### Lemma 3.2

${\mathcal {S}}_{\varphi }(\Omega )$ is not empty and
    $\forall u\in {\mathcal {S}}_{\varphi }(\Omega ):\forall x\in \Omega :u(x)\geq \min _{y\in \partial \Omega }\varphi (y)$
Proof: The first part follows by choosing the constant function $u(x)=\max _{y\in \partial \Omega }\varphi (y)$, which is harmonic and therefore superharmonic. The second part follows from the minimum principle for superharmonic functions.

### Lemma 3.3

Let $u_{1},u_{2}\in {\mathcal {S}}_{\varphi }(\Omega )$. If we now define $u(x)=\min\{u_{1}(x),u_{2}(x)\}$, then $u\in {\mathcal {S}}_{\varphi }(\Omega )$.
Proof: The condition on the border is satisfied, because
    $\forall x\in \partial \Omega :u_{1}(x)\geq \varphi (x)\wedge u_{2}(x)\geq \varphi (x)$
$u$ is superharmonic because, if we (without loss of generality) assume that $u(x)=u_{1}(x)$, then it follows that
    $u(x)=u_{1}(x)\geq {\frac {1}{d(r)}}\int _{B_{r}(x)}u_{1}(y)dy\geq {\frac {1}{d(r)}}\int _{B_{r}(x)}u(y)dy$
, due to the monotony of the integral. This argument is valid for all $x\in \Omega$, and therefore $u$ is superharmonic.

### Lemma 3.4

If $\Omega \subset \mathbb {R} ^{d}$ is bounded and $\varphi \in C(\partial \Omega )$, then the function
    $u(x)=\inf\{v(x)|v\in {\mathcal {S}}_{\varphi }(\Omega )\}$
is harmonic.
Proof:

### Lemma 3.5

If $\Omega$ satisfies the exterior sphere condition, then for all $y\in \partial \Omega$ there is a barrier function.

### Existence theorem of Perron

Let $\Omega \subset \mathbb {R} ^{d}$ be a bounded domain which satisfies the exterior sphere condition. Then the Dirichlet problem for the Poisson equation, which is, writing it again:
    ${\begin{cases}-\Delta u(x)=f(x)&x\in \Omega \\u(x)=g(x)&x\in \partial \Omega \end{cases}}$
has a solution $u\in C^{\infty }(\Omega )\cap C({\overline {\Omega }})$.
Proof:
Let's summarise the results of this section.
Corollary 6.last:
Let $\Omega \subset \mathbb {R} ^{d}$ be a domain satisfying the exterior sphere condition, let $f\in {\mathcal {C}}^{2}(\mathbb {R} ^{d})$, let $g:\partial \Omega \to \mathbb {R}$ be continuous and let $P_{\Omega }$ be a Green's function of the first kind for $\Omega$. Then
    $u(x)=\int _{\Omega }f(y)P_{\Omega }(y,x)dy-\int _{\partial \Omega }g(y)\nu (y)\cdot \nabla _{y}P_{\Omega }(y,x)dy$
is the unique continuous solution to the boundary value problem
    ${\begin{cases}\forall x\in \Omega :&-\Delta u(x)=f(x)\\\forall x\in \partial \Omega :&u(x)=g(x)\end{cases}}$
In the next chapter, we will have a look at the heat equation.

## Exercises

* Prove theorem 6.3 using theorem 6.2 (Hint: Choose $\mathbf {V} (x)=\mathbf {W} (x)f(x)$ in theorem 6.2).
* Prove that $\forall n\in \mathbb {N} :\Gamma (n+1)=n!$, where $n!$ is the factorial of $n$.
* Calculate $V_{d}'(R)$. Have you seen the obtained function before?
* Prove that for $d=1$, the function $P_{d}$ as defined in theorem 6.11 is a Green's kernel for Poisson's equation (hint: use integration by parts twice).
* For all $d\geq 2$ and $x\in \mathbb {R} ^{d}\setminus \{0\}$, calculate $\nabla P_{d}(x)$ and $\Delta P_{d}(x)$.
* Let $O\subseteq \mathbb {R} ^{d}$ be open and $f:O\to \mathbb {R} ^{d}$ be continuous. Prove that $f(x)=0$ almost everywhere in $O$ implies $f(x)=0$ everywhere in $O$.
* Prove theorem 6.20 by modelling your proof on the proof of theorem 6.19.
* For all dimensions $d\geq 2$, give an example for vectors $\alpha ,\beta \in \mathbb {N} _{0}^{d}$ such that neither $\alpha \leq \beta$ nor $\beta \leq \alpha$.

## Sources
# The Fourier transform

In this chapter, we introduce the Fourier transform. The Fourier transform transforms functions into other functions. It can be used to solve certain types of linear differential equations.

## Definition and calculation rules

Definition 8.1:
Let $f\in L^{1}(\mathbb {R} ^{d})$. Then the Fourier transform of $f$ is defined as follows:
    ${\hat {f}}:\mathbb {R} ^{d}\to \mathbb {R} ,{\hat {f}}(y):=\int _{\mathbb {R} ^{d}}f(x)e^{-2\pi ix\cdot y}dx$
We recall that $f$ is integrable $\Leftrightarrow$ $|f|$ is integrable.
Now we're ready to prove the next theorem:
Theorem 8.2: The Fourier transform of an integrable $f$ is well-defined.
Proof: Since $f$ is integrable, lemma 8.2 tells us that $|f|$ is integrable. But
    $\forall x,y\in \mathbb {R} ^{d}:|f(x)e^{-2\pi ix\cdot y}|=|f(x)|\cdot \overbrace {|e^{-2\pi ix\cdot y}|} ^{=1}=|f(x)|$
, and therefore $x\mapsto |f(x)e^{-2\pi ix\cdot y}|$ is integrable. But then, $x\mapsto f(x)e^{-2\pi ix\cdot y}$ is integrable, which is why
    $\int _{\mathbb {R} ^{d}}f(x)e^{-2\pi ix\cdot y}dx={\hat {f}}(y)$
has a unique complex value, by definition of integrability.$\Box$
Theorem 8.3: Let $f\in L^{1}(\mathbb {R} ^{d})$. Then the Fourier transform of $f$, ${\hat {f}}$, is bounded.
Proof:
    ${\begin{aligned}\left|\int _{\mathbb {R} ^{d}}f(x)e^{-2\pi ix\cdot y}dx\right|&\leq \int _{\mathbb {R} ^{d}}\left|f(x)e^{-2\pi ix\cdot y}\right|dx&{\text{triangle ineq. for the }}\int \\&=\int _{\mathbb {R} ^{d}}|f(x)|dx&\left|e^{-2\pi ix\cdot y}\right|=1\\&\in \mathbb {R} &f\in L^{1}(\mathbb {R} ^{d})\end{aligned}}$$\Box$
Once we have calculated the Fourier transform ${\tilde {f}}$ of a function $f$, we can easily find the Fourier transforms of some functions similar to $f$. The following calculation rules show examples how you can do this. But just before we state the calculation rules, we recall a definition from chapter 2, namely the power of a vector to a multiindex, because it is needed in the last calculation rule.
Definition 2.6:
For a vector $x=(x_{1},\ldots ,x_{d})\in \mathbb {R} ^{d}$ and a $d$-dimensional multiindex $\alpha \in \mathbb {N} _{0}^{d}$ we define $x^{\alpha }$, $x$ to the power of $\alpha$, as follows:
    $x^{\alpha }:=x_{1}^{\alpha _{1}}\cdots x_{d}^{\alpha _{d}}$
Now we write down the calculation rules, using the following notation:
Notation 8.4:
We write
    $f(x)\rightarrow g(y)$
to mean the sentence 'the function $y\mapsto g(y)$ is the Fourier transform of the function $x\mapsto f(x)$'.
Theorem 8.5:
Let ${\hat {f}}$ be the Fourier transform of $f\in L^{1}(\mathbb {R} ^{d})$. Then the following calculation rules hold:
* $f(x)e^{-2\pi ih\cdot x}\rightarrow {\hat {f}}(y+h)$ for arbitrary $h\in \mathbb {R} ^{d}$
* $f(x+h)\rightarrow {\hat {f}}(y)e^{2\pi ih\cdot y}$ for arbitrary $h\in \mathbb {R} ^{d}$
* $f(\delta x)\rightarrow \delta ^{-d}{\hat {f}}(\delta ^{-1}y)$ for arbitrary $\delta >0$
If additionally ${\hat {g}}$ is the Fourier transform of $g\in L^{1}(\mathbb {R} ^{d})$, we have
    4. $\int _{\mathbb {R} ^{d}}{\hat {g}}(x)f(x)dx=\int _{\mathbb {R} ^{d}}g(y){\hat {f}}(y)dy$
Proof: To prove the first rule, we only need one of the rules for the exponential function (and the symmetry of the standard dot product):
1.
    $f(x)e^{-2\pi ih\cdot x}\rightarrow \int _{\mathbb {R} ^{d}}f(x)e^{-2\pi ih\cdot x}e^{-2\pi ix\cdot y}dx=\int _{\mathbb {R} ^{d}}f(x)e^{-2\pi ix\cdot (y+h)}dx={\hat {f}}(y+h)$
For the next two rules, we apply the general integration by substitution rule, using the diffeomorphisms $x\mapsto x-h$ and $x\mapsto \delta ^{-1}x$, which are bijections from $\mathbb {R} ^{d}$ to itself.
2.
    $f(x+h)\rightarrow \int _{\mathbb {R} ^{d}}f(x+h)e^{-2\pi ix\cdot y}dx=\int _{\mathbb {R} ^{d}}f(x)e^{-2\pi i(x-h)\cdot y}dx=e^{2\pi ih\cdot y}\int _{\mathbb {R} ^{d}}f(x)e^{-2\pi ix\cdot y}dx={\hat {f}}(y)e^{2\pi ih\cdot y}$
3.
    $f(\delta x)\rightarrow \int _{\mathbb {R} ^{d}}f(\delta x)e^{-2\pi ix\cdot y}dx=\int _{\mathbb {R} ^{d}}\delta ^{-d}f(x)e^{-2\pi i(\delta ^{-1}x)\cdot y}dx=\delta ^{-d}\int _{\mathbb {R} ^{d}}f(x)e^{-2\pi ix\cdot (\delta ^{-1}y)}dx=\delta ^{-d}{\hat {f}}(\delta ^{-1}y)$
4.
    ${\begin{aligned}\int _{\mathbb {R} ^{d}}{\hat {g}}(x)f(x)dx&=\int _{\mathbb {R} ^{d}}\int _{\mathbb {R} ^{d}}g(y)e^{2\pi ix\cdot y}dyf(x)dx&{\text{Def. of the Fourier transform}}\\&=\int _{\mathbb {R} ^{d}}\int _{\mathbb {R} ^{d}}f(x)g(y)e^{2\pi ix\cdot y}dydx&{\text{putting a constant inside the integral}}\\&=\int _{\mathbb {R} ^{d}}\int _{\mathbb {R} ^{d}}f(x)g(y)e^{2\pi ix\cdot y}dxdy&{\text{Fubini}}\\&=\int _{\mathbb {R} ^{d}}g(y)\int _{\mathbb {R} ^{d}}f(x)e^{2\pi ix\cdot y}dxdy&{\text{pulling a constant out of the integral}}\\&=\int _{\mathbb {R} ^{d}}g(y){\hat {f}}(y)dy&{\text{Def. of the Fourier transform}}\\\end{aligned}}$
$\Box$

## The Fourier transform of Schwartz functions

In order to proceed with further rules for the Fourier transform which involve Schwartz functions, we first need some further properties of Schwartz functions.
Theorem 8.6:
Let $\phi :\mathbb {R} ^{d}\to \mathbb {R}$ be a Schwartz function and let $\alpha ,\beta \in \mathbb {N} _{0}^{d}$. Then the function
    $x\mapsto x^{\alpha }\partial _{\beta }\phi (x)$
is a Schwartz function as well.
Proof:
Let $\varrho ,\varsigma \in \mathbb {N} _{0}^{d}$. Due to the general product rule, we have:
    $\partial _{\varsigma }x^{\alpha }\partial _{\beta }\phi (x)=\sum _{\varepsilon \in \mathbb {N} _{0}^{d} \atop \varepsilon \leq \varsigma }{\binom {\varsigma }{\varepsilon }}\partial _{\varepsilon }(x^{\alpha })\partial _{\varsigma -\varepsilon }\partial _{\beta }\phi (x)$
We note that for all $\alpha$ and $\varepsilon$, $\partial _{\varepsilon }(x^{\alpha })$ equals to $x$ to some multiindex power. Since $\phi$ is a Schwartz function, there exist constants $c_{\varepsilon }$ such that:
    $\|x^{\varrho }\partial _{\varepsilon }(x^{\alpha })\partial _{\varsigma -\varepsilon }\partial _{\beta }\phi \|_{\infty }\leq c_{\varepsilon }$
Hence, the triangle inequality for $\|\cdot \|_{\infty }$ implies:
    $\|x^{\varrho }\partial _{\varsigma }x^{\alpha }\partial _{\beta }\phi \|_{\infty }\leq \sum _{\varepsilon \in \mathbb {N} _{0}^{d} \atop \varepsilon \leq \varsigma }{\binom {\varsigma }{\varepsilon }}c_{\varepsilon }$$\Box$
Theorem 8.7:
Every Schwartz function is integrable.
Proof:
We use that if the absolute value of a function is almost everywhere smaller than the value of an integrable function, then the first function is integrable.
Let $\phi$ be a Schwartz function. Then there exist $b,c\in \mathbb {R} _{>0}$ such that for all $x\in \mathbb {R} ^{d}$:
    $|\phi (x)|\leq \min \left\{b\prod _{j=1}^{d}|x_{j}|^{-2},c\right\}$
The latter function is integrable, and integrability of $\phi$ follows.$\Box$
Now we can prove all three of the following rules for the Fourier transform involving Schwartz functions.
Theorem 8.8:
If ${\hat {\phi }}$ is the Fourier transform of a function $\phi$ in the Schwartz space ${\mathcal {S}}(\mathbb {R} ^{d})$, in addition to the rules in theorem 8.4, also the following rules hold:
* $\partial _{\alpha }\phi (x)\rightarrow (2\pi iy)^{\alpha }{\hat {\phi }}(y)$ for arbitrary $\alpha \in \mathbb {N} _{0}^{d}$
* $(-2\pi ix)^{\alpha }\phi (x)\rightarrow \partial _{\alpha }{\hat {\phi }}(y)$ for arbitrary $\alpha \in \mathbb {N} _{0}^{d}$
If additionally ${\hat {\theta }}$ is the Fourier transform of a $\theta \in {\mathcal {S}}(\mathbb {R} ^{d})$, then
    3. ${\widehat {\phi *\theta }}(y)={\hat {\phi }}(y)\cdot {\hat {\theta }}(y)$
Proof:
1.
For the first rule, we use induction over $|\alpha |$.
It is clear that the claim is true for $|\alpha |=0$ (then the rule states that the Fourier transform of $\phi$ is the Fourier transform of $\phi$).
We proceed to the induction step: Let $n\in \mathbb {N} _{0}$, and assume that the claim is true for all $\alpha \in \mathbb {N} _{0}^{d}$ such that $|\alpha |=n$. Let $\beta \in \mathbb {N} _{0}^{d}$ such that $|\beta |=n+1$. We show that the claim is also true for $\beta$.
Remember that we have $\partial _{\beta }\phi :=\partial _{x_{1}}^{\beta _{1}}\cdots \partial _{x_{d}}^{\beta _{d}}\phi$. We choose $k\in \{1,\ldots ,d\}$ such that $\beta _{k}>0$ (this is possible since otherwise $|\beta |=0$), define
    $e_{k}:=(0,\ldots ,0,\overbrace {1} ^{k{\text{th entry}}},0,\ldots ,0)$
    $\alpha :=\beta -e_{k}$
and obtain
    $\partial _{\beta }\phi =\partial _{x_{k}}\partial _{\alpha }\phi$
by Schwarz' theorem, which implies that one may interchange the order of partial derivation arbitrarily.
Let $R\in \mathbb {R} _{>0}$ be an arbitrary positive real number. From Fubini's theorem and integration by parts, we obtain:
    ${\begin{aligned}\int _{[-R,R]^{d}}\partial _{x_{k}}\partial _{\alpha }\phi (x)e^{-2\pi ix\cdot y}dx&=\int _{[-R,R]^{d-1}}\int _{-R}^{R}\partial _{x_{k}}\partial _{\alpha }\phi (x)e^{-2\pi ix\cdot y}dx_{k}d(x_{1},\ldots ,x_{k-1},x_{k+1},\ldots ,x_{d})\\&=\int _{[-R,R]^{d-1}}\left(\left(\partial _{\alpha }\phi (x)e^{-2\pi ix\cdot y}\right){\big |}_{x_{k}=-R}^{x_{k}=R}-\int _{-R}^{R}\partial _{\alpha }\phi (x)(-2\pi iy_{k})e^{-2\pi ix\cdot y}dx_{k}\right)d(x_{1},\ldots ,x_{k-1},x_{k+1},\ldots ,x_{d})\end{aligned}}$
Due to the dominated convergence theorem (with dominating function $x\mapsto \partial _{x_{k}}\partial _{\alpha }\phi (x)$), the integral on the left hand side of this equation converges to
    $\int _{\mathbb {R} ^{d}}\partial _{x_{k}}\partial _{\alpha }\phi (x)e^{-2\pi ix\cdot y}dx$
as $R\to \infty$. Further, since $\phi$ is a Schwartz function, there are $b,c\in \mathbb {R} _{>0}$ such that:
    $|\partial _{\alpha }\phi (x)|<\min \left\{b\prod _{j=1 \atop j\neq k}^{d}|x_{j}|^{-2},c\right\}$
Hence, the function within the large parentheses in the right hand sinde of the last line of the last equation is dominated by the $L^{1}(\mathbb {R} ^{d-1})$ function
    $(x_{1},\ldots ,x_{k-1},x_{k+1},\ldots ,x_{d})\mapsto \min \left\{b\prod _{j=1 \atop j\neq k}^{d}|x_{j}|^{-2},c\right\}+\int _{-\infty }^{\infty }|\partial _{\alpha }\phi (x)(-2\pi iy_{k})|dx_{k}$
and hence, by the dominated convergence theorem, the integral over that function converges, as $R\to \infty$, to:
    ${\begin{aligned}\int _{\mathbb {R} ^{d-1}}\left(\int _{-\infty }^{\infty }\partial _{\alpha }\phi (x)(2\pi iy_{k})e^{-2\pi ix\cdot y}dx_{k}\right)d(x_{1},\ldots ,x_{k-1},x_{k+1},\ldots ,x_{d})&=\int _{\mathbb {R} ^{d}}\partial _{\alpha }\phi (x)(2\pi iy_{k})e^{-2\pi ix\cdot y}dx&{\text{Fubini}}\\&=\int _{\mathbb {R} ^{d}}(2\pi iy^{\beta })\phi (x)e^{-2\pi ix\cdot y}dx&{\text{induction hypothesis}}\end{aligned}}$
From the uniqueness of limits of real sequences we obtain 1.
2.
We use again induction on $|\alpha |$, note that the claim is trivially true for $|\alpha |=0$, assume that the claim is true for all $\alpha \in \mathbb {N} _{0}^{d}$ such that $|\alpha |=n$, choose $\beta \in \mathbb {N} _{0}^{d}$ such that $|\beta |=n+1$ and $k\in \{1,\ldots ,d\}$ such that $\beta _{k}>0$ and define $\alpha :=\beta -e_{k}$.
Theorems 8.6 and 8.7 imply that
* for all $y\in \mathbb {R} ^{d}$, $\int _{\mathbb {R} ^{d}}|\phi (x)e^{-2\pi ix\cdot y}|dx<\infty$ and
* for all $y\in \mathbb {R} ^{d}$, $\int _{\mathbb {R} ^{d}}|\phi (x)\partial _{y_{k}}e^{-2\pi ix\cdot y}|dx<\infty$.
Further,
* $\partial _{y_{k}}(\phi (x)e^{-2\pi ix\cdot y})$ exists for all $(x,y)\in \mathbb {R} ^{d}\times \mathbb {R} ^{d}$.
Hence, Leibniz' integral rule implies:
    ${\begin{aligned}\partial _{\beta }{\hat {\phi }}(y)&=\partial _{x_{k}}\partial _{\alpha }{\hat {\phi }}(y)&\\&=\partial _{y_{k}}\int _{\mathbb {R} ^{d}}(2\pi ix)^{\alpha }\phi (x)e^{-2\pi ix\cdot y}dx&{\text{induction hypothesis}}\\&=\int _{\mathbb {R} ^{d}}\partial _{y_{k}}(2\pi ix)^{\alpha }\phi (x)e^{-2\pi ix\cdot y}dx&\\&=\int _{\mathbb {R} ^{d}}(2\pi ix)^{\beta }\phi (x)e^{-2\pi ix\cdot y}dx\end{aligned}}$
3.
    ${\begin{aligned}{\widehat {\phi *\theta }}(y)&:=\int _{\mathbb {R} ^{d}}(\phi *\theta )(x)e^{-2\pi ix\cdot y}dx&{\text{Def. of Fourier transform}}\\&:=\int _{\mathbb {R} ^{d}}\int _{\mathbb {R} ^{d}}\phi (z)\theta (x-z)dze^{-2\pi ix\cdot y}dx&{\text{Def. of convolution}}\\&=\int _{\mathbb {R} ^{d}}\int _{\mathbb {R} ^{d}}e^{-2\pi ix\cdot y}\phi (z)\theta (x-z)dzdx&{\text{linearity of the integral}}\\&=\int _{\mathbb {R} ^{d}}\int _{\mathbb {R} ^{d}}e^{-2\pi ix\cdot y}\phi (z)\theta (x-z)dxdz&{\text{Fubini}}\\&=\int _{\mathbb {R} ^{d}}\int _{\mathbb {R} ^{d}}e^{-2\pi ix\cdot y}e^{-2\pi iz\cdot y}\phi (z)\theta (x)dxdz&{\text{Integration by substitution using }}x\mapsto x+z{\text{ and }}\forall b,c\in \mathbb {R} :e^{b+c}=e^{b}e^{c}\\&=\int _{\mathbb {R} ^{d}}\overbrace {\int _{\mathbb {R} ^{d}}e^{-2\pi ix\cdot y}\theta (x)dx} ^{={\hat {\theta }}(y)}\phi (z)e^{-2\pi iz\cdot y}dz&{\text{pulling a constant out of the integral}}\\&={\hat {\theta }}(y)\overbrace {\int _{\mathbb {R} ^{d}}\phi (z)e^{-2\pi iz\cdot y}dz} ^{={\hat {\phi }}(y)}&{\text{pulling a constant out of the integral}}\end{aligned}}$$\Box$
Theorem 8.9:
Let $\phi \in {\mathcal {S}}(\mathbb {R} ^{d})$, and let ${\hat {\phi }}$ be the Fourier transform of $\phi$. Then ${\hat {\phi }}\in {\mathcal {S}}(\mathbb {R} ^{d})$.
Proof:
Let $\alpha ,\beta \in \mathbb {N} _{0}^{d}$ be two arbitrary $d$-dimensional multiindices, and let $\phi \in {\mathcal {S}}(\mathbb {R} ^{d})$. By theorem 8.6 $\partial _{\alpha }((-2\pi \mathrm {i} x)^{\beta }\phi )$ is a Schwartz function as well. Theorem 8.8 implies:
    $x^{\alpha }\partial _{\beta }{\hat {\phi }}={\widehat {\partial _{\alpha }((-2\pi \mathrm {i} x)^{\beta }\phi )}}$
By theorem 8.3, ${\widehat {\partial _{\alpha }((-2\pi \mathrm {i} x)^{\beta }\phi )}}$ is bounded. Since $\alpha ,\beta \in \mathbb {N} _{0}^{d}$ were arbitrary, this shows that ${\hat {\phi }}\in {\mathcal {S}}(\mathbb {R} ^{d})$.$\Box$
Definitions 8.10:
We define the Fourier transform on the Schwartz space to be the function
    ${\mathcal {F}}:{\mathcal {S}}(\mathbb {R} ^{d})\to {\mathcal {S}}(\mathbb {R} ^{d}),{\mathcal {F}}(\phi ):={\hat {\phi }}$.
Theorem 8.9 assures that this function really maps to ${\mathcal {S}}(\mathbb {R} ^{d})$. Furthermore, we define the inverse Fourier transform on the Schwartz space to be the function
    ${\mathcal {F}}^{-1}:{\mathcal {S}}(\mathbb {R} ^{d})\to {\mathcal {S}}(\mathbb {R} ^{d}),{\mathcal {F}}(\phi ):=x\mapsto \int _{\mathbb {R} ^{d}}\phi (y)e^{2\pi ix\cdot y}dy$.
This function maps to ${\mathcal {S}}(\mathbb {R} ^{d})$ since ${\mathcal {F}}^{-1}(\phi )(x)={\hat {\phi }}(-x)$.
Both the Fourier transform and the inverse Fourier transform are sequentially continuous:
Theorem 8.11:
Let $\phi \in {\mathcal {S}}(\mathbb {R} ^{d})$ and let $(\phi _{l})_{l\in \mathbb {N} }$ be a sequence of Schwartz functions such that $\phi _{l}\to \phi ,l\to \infty$. Then ${\mathcal {F}}(\phi _{l})\to {\mathcal {F}}(\phi ),l\to \infty$ and ${\mathcal {F}}^{-1}(\phi _{l})\to {\mathcal {F}}^{-1}(\phi ),l\to \infty$, both in the sense of Schwartz function convergence as defined in definition 3.11.
Proof:
1. We prove ${\mathcal {F}}(\phi _{l})\to {\mathcal {F}}(\phi ),l\to \infty$.
Let $\alpha ,\beta \in \mathbb {N} _{0}^{d}$. Due to theorem 8.8 1. and 2. and the linearity of derivatives, integrals and multiplication, we have
    $x^{\alpha }\partial _{\beta }({\mathcal {F}}(\phi _{l})(x)-{\mathcal {F}}(\phi )(x))=\int _{\mathbb {R} ^{d}}\partial _{\alpha }((-2\pi iy)^{\beta }(\phi _{l}(y)-\phi (y)))e^{-2\pi ix\cdot y}dy$.
As in the proof of theorem 8.3, we hence obtain
    $\left|x^{\alpha }\partial _{\beta }({\mathcal {F}}(\phi _{l})(x)-{\mathcal {F}}(\phi )(x))\right|\leq \|\partial _{\alpha }((-2\pi ix)^{\beta }(\phi _{l}-\phi )))\|_{L^{1}}$.
Due to the multi-dimensional product rule,
    $\partial _{\alpha }((-2\pi ix)^{\beta }(\phi _{l}(x)-\phi (x)))=\sum _{\varsigma \in \mathbb {N} _{0}^{d} \atop \varsigma \leq \alpha }{\binom {\varsigma }{\alpha }}\partial _{\varsigma }((-2\pi ix)^{\beta })\partial _{\alpha -\varsigma }(\phi _{l}(x)-\phi (x))$.
Let now $\epsilon >0$ be arbitrary. Since $\phi _{l}\to \phi$ as defined in definition 3.11, for each $n\in \{1,\ldots ,d\}$ we may choose $N_{1}\in \mathbb {N}$ such that
    $\forall k\geq N_{1}:\sum _{\varsigma \in \mathbb {N} _{0}^{d} \atop \varsigma \leq \alpha }{\binom {\varsigma }{\alpha }}\|x_{n}^{2}\partial _{\varsigma }((-2\pi ix)^{\beta })\partial _{\alpha -\varsigma }(\phi _{l}(x)-\phi (x))\|_{\infty }<\epsilon$.
Further, we may choose $N_{2}\in \mathbb {N}$ such that
    $\forall k\geq N_{2}:\sum _{\varsigma \in \mathbb {N} _{0}^{d} \atop \varsigma \leq \alpha }{\binom {\varsigma }{\alpha }}\|\partial _{\varsigma }((-2\pi ix)^{\beta })\partial _{\alpha -\varsigma }(\phi _{l}(x)-\phi (x))\|_{\infty }<\epsilon$.
Hence follows for $k\geq N:=\max\{N_{1},N_{2}\}$:
    ${\begin{aligned}\|\partial _{\alpha }((-2\pi ix)^{\beta }(\phi _{l}-\phi )))\|_{L^{1}}&:=\int _{\mathbb {R} ^{d}}|\partial _{\alpha }((-2\pi ix)^{\beta }(\phi _{l}(x)-\phi (x)))|dx\\&\leq \int _{\mathbb {R} ^{d}}\epsilon \min\{x_{1}^{-2},\ldots ,x_{d}^{-2},1\}dx\\&=\epsilon \int _{\mathbb {R} ^{d}}\min\{x_{1}^{-2},\ldots ,x_{d}^{-2},1\}dx\end{aligned}}$
Since $\epsilon >0$ was arbitrary, we obtain ${\mathcal {F}}(\phi _{l})\to {\mathcal {F}}(\phi ),l\to \infty$.
2. From 1., we deduce ${\mathcal {F}}^{-1}(\phi _{l})\to {\mathcal {F}}^{-1}(\phi ),l\to \infty$.
If $\phi _{l}\to \phi$ in the sense of Schwartz functions, then also $\theta _{l}\to \theta$ in the sense of Schwartz functions, where we define
    $\theta _{l}(x):=\phi _{l}(-x)$ and $\theta (x):=\phi (-x)$.
Therefore, by 1. and integration by substitution using the diffeomorphism $x\mapsto -x$, ${\mathcal {F}}^{-1}(\phi _{l})={\mathcal {F}}(\theta _{l})\to {\mathcal {F}}(\theta )={\mathcal {F}}^{-1}(\phi )$.$\Box$
In the next theorem, we prove that ${\mathcal {F}}^{-1}$ is the inverse function of the Fourier transform. But for the proof of that theorem (which will be a bit long, and hence to read it will be a very good exercise), we need another two lemmas:
Lemma 8.12:
If we define the function
    $G:\mathbb {R} ^{d}\to \mathbb {R} ,G(x):=e^{-\pi \|x\|^{2}}$,
then ${\mathcal {F}}(G)=G$ and ${\mathcal {F}}^{-1}(G)=G$.
Proof:
1. ${\mathcal {F}}^{-1}(G)=G$:
We define
    $\mu :\mathbb {R} ^{d}\to \mathbb {R} ,\mu (\xi ):=e^{\pi \|\xi \|^{2}}{\hat {G}}(\xi )$.
By the product rule, we have for all $n\in \{1,\ldots ,d\}$
    $\partial _{\xi _{n}}\mu (\xi )=2\pi \xi _{n}e^{\pi \|\xi \|^{2}}{\hat {G}}(\xi )+e^{\pi \|\xi \|^{2}}\partial _{\xi _{n}}{\hat {G}}(\xi )$.
Due to 1. of theorem 8.8, we have
    $2\pi \xi _{n}{\hat {G}}(\xi )=-i{\widehat {\partial _{x_{n}}G}}(\xi )=-i\int _{\mathbb {R} ^{d}}(2\pi x_{n})e^{-\pi \|x\|^{2}}e^{2\pi i\xi \cdot x}dx$;
from 2. of theorem 8.8 we further obtain
    $\partial _{\xi _{n}}{\hat {G}}(\xi )=\int _{\mathbb {R} ^{d}}(-2\pi ix_{n})e^{-\pi \|x\|^{2}}e^{2\pi i\xi \cdot x}dx$.
Hence, $\mu$ is constant. Further,
    ${\begin{aligned}\mu (0)&=\int _{\mathbb {R} ^{d}}e^{-\pi \|x\|^{2}}dx&\\&=\int _{\mathbb {R} ^{d}}{\frac {1}{{\sqrt {2\pi }}^{d}}}e^{-\|x\|^{2}/2}dx&{\text{substitution using }}x\mapsto {\frac {1}{{\sqrt {2\pi }}^{d}}}x\\&=1&{\text{lemma 6.2}}\end{aligned}}$.
2. ${\mathcal {F}}^{-1}(G)=G$:
By substitution using the diffeomorphism $x\mapsto -x$,
    $\forall x\in \mathbb {R} ^{d}:{\mathcal {F}}^{-1}(G)(x)={\mathcal {F}}(G)(x)=G(x)$.$\Box$
For the next lemma, we need example 3.4 again, which is why we restate it:
Example 3.4: The standard mollifier $\eta$, given by
    $\eta :\mathbb {R} ^{d}\to \mathbb {R} ,\eta (x)={\frac {1}{c}}{\begin{cases}e^{-{\frac {1}{1-\|x\|^{2}}}}&{\text{ if }}\|x\|_{2}<1\\0&{\text{ if }}\|x\|_{2}\geq 1\end{cases}}$
, where $c:=\int _{B_{1}(0)}e^{-{\frac {1}{1-\|x\|^{2}}}}dx$, is a bump function (see exercise 3.2).
Lemma 8.13:
Let $\phi \in {\mathcal {S}}(\mathbb {R} ^{d})$, and for each $n\in \mathbb {N}$ define $\phi _{n}(x):=\eta (x/n)\phi (x)$. Then $\phi _{n}\to \phi ,n\to \infty$ in the sense of Schwartz functions.
Proof:
Let $\alpha ,\beta \in \mathbb {N} _{0}^{d}$ be arbitrary. Due to the generalised product rule,
    $\partial _{\beta }(\phi _{n}-\phi )=\sum _{\varsigma \in \mathbb {N} _{0}^{d} \atop \varsigma \leq \beta }{\binom {\beta }{\varsigma }}{\frac {1}{n^{|\beta -\varsigma |}}}\partial _{\beta -\varsigma }\eta (x/n)\partial _{\varsigma }\phi (x)-\partial _{\beta }\phi (x)$.
By the triangle inequality, we may hence deduce
    $|x^{\alpha }\partial _{\beta }(\phi _{n}-\phi )|\leq {\frac {1}{n}}\sum _{\varsigma \in \mathbb {N} _{0}^{d} \atop \varsigma <\beta }{\binom {\beta }{\varsigma }}{\frac {1}{n^{|\beta -\varsigma |-1}}}|x^{\alpha }\partial _{\varsigma }\phi (x)||\partial _{\beta -\varsigma }\eta (x/n)|+|x^{\alpha }\partial _{\beta }\phi (x)||1-\eta (x/n)|$.
Since both $\phi$ and $\eta$ are Schwartz functions (see exercise 3.2 and theorem 3.9), for each $\varrho \in \mathbb {N} _{0}^{d}$ we may choose $b_{\varrho },c_{\varrho }\in \mathbb {R}$ such that
    $\|\partial _{\beta -\varrho }\eta \|_{\infty }<b_{\varrho }$ and $\|x^{\alpha }\partial _{\varrho }\phi \|_{\infty }<c_{\varrho }$.
Further, for each $k\in \{1,\ldots ,d\}$, we may choose $c_{k}\in \mathbb {R} ^{d}$ such that
    $\|x_{k}x^{\alpha }\partial _{\beta }\phi \|_{\infty }<c_{k}$.
Let now $\epsilon >0$ be arbitrary. We choose $N_{1}\in \mathbb {N}$ such that for all $n\geq N_{1}$
    ${\frac {1}{n}}\sum _{\varsigma \in \mathbb {N} _{0}^{d} \atop \varsigma <\beta }{\binom {\beta }{\varsigma }}{\frac {1}{n^{|\beta -\varsigma |-1}}}b_{\varsigma }c_{\varsigma }<\epsilon /2$.
Further, we choose $R\in \mathbb {R} _{>0}$ such that
    $\|x\|>R\Rightarrow |x^{\alpha }\partial _{\beta }\phi (x)|<\epsilon /2$.
This is possible since
    $|x^{\alpha }\partial _{\beta }\phi (x)|\leq \min \left\{{\frac {c_{1}}{|x_{1}|}},\ldots ,{\frac {c_{d}}{|x_{d}|}}\right\}$
due to our choice of $c_{1},\ldots ,c_{d}$.
Then we choose $N_{2}\in \mathbb {N}$ such that for all $n\geq N_{2}$
    $\forall x\in B_{R}(0):|1-\phi (x/n)|<1/c_{\beta }$.
Inserting all this in the above equation gives $|x^{\alpha }\partial _{\beta }(\phi _{n}-\phi )|<\epsilon$ for $n\geq N:=\max\{N_{1},N_{2}\}$. Since $\alpha$, $\beta$ and $\epsilon$ were arbitrary, this proves $\phi _{n}\to \phi$ in the sense of Schwartz functions.$\Box$
Theorem 8.14:
Let $\phi \in {\mathcal {S}}(\mathbb {R} ^{d})$. Then ${\mathcal {F}}({\mathcal {F}}^{-1}(\phi ))=\phi$ and ${\mathcal {F}}^{-1}({\mathcal {F}}(\phi ))=\phi$.
Proof:
1. We prove that if $\phi$ is a Schwartz function vanishing at the origin (i. e. $\phi (0)=0$), then ${\mathcal {F}}^{-1}({\mathcal {F}}(\phi ))(0)=0$.
So let $\phi$ be a Schwartz function vanishing at the origin. By the fundamental theorem of calculus, the multi-dimensional chain rule and the linearity of the integral, we have
    $\phi (x)=\phi (x)-\phi (0)=\int _{0}^{1}{\frac {d}{dt}}\phi (tx)dx=\sum _{j=1}^{d}x_{j}\int _{0}^{1}\partial _{x_{j}}\phi (tx)dt$.
Defining $\phi _{n}(x):=\eta (x/n)\phi (x)$,
    $\theta _{j,n}(x):=\eta (x/n)\int _{0}^{1}\partial _{x_{j}}\phi (tx)dt$,
and multiplying both sides of the above equation by $\eta (x/n)$, we obtain
    $\phi _{n}(x)=\sum _{j=1}^{d}x_{j}\theta _{j,n}(x)$.
Since by repeated application of Leibniz' integral rule for all $\alpha \in \mathbb {N} _{0}^{d}$
    $\partial _{\alpha }\theta _{j,n}(x)=\sum _{\varsigma \leq \alpha }{\binom {\alpha }{\varsigma }}{\frac {1}{n^{|\varsigma |}}}\partial _{\varsigma }\eta (x/n)\int _{0}^{1}\partial _{\alpha -\varsigma }\partial _{x_{j}}\phi (tx)dt$,
all the $\theta _{j,n}$ are bump functions (due to theorem 4.15 and exercise 3.?), and hence Schwartz functions (theorem 3.9). Hence, by theorem 8.8 and the linearity of the Fourier transform (which follows from the linearity of the integral),
    ${\mathcal {F}}(\phi _{n})(y)=\sum _{j=1}^{d}{\mathcal {F}}(x_{j}\theta _{j,n})(y)={\frac {1}{-2\pi i}}\sum _{j=1}^{d}\partial _{y_{j}}{\mathcal {F}}(\theta _{j,n})(y)$.
Hence,
    ${\mathcal {F}}^{-1}({\mathcal {F}}(\phi _{n}))(0)=\int _{\mathbb {R} ^{d}}{\mathcal {F}}(\phi _{n})(y)\overbrace {e^{-2\pi i0\cdot y}} ^{=1}dy={\frac {1}{-2\pi i}}\sum _{j=1}^{d}\int _{\mathbb {R} ^{d}}\partial _{y_{j}}{\mathcal {F}}(\theta _{j,n})(y)dy$.
Let $k\in \{1,\ldots ,d\}$. By Fubini's theorem, the fundamental theorem of calculus and since $\theta _{k,n}$ is a bump function, we have
    $\int _{\mathbb {R} ^{d}}\partial _{y_{k}}{\mathcal {F}}(\theta _{k,n})(y)dy=\int _{\mathbb {R} ^{d-1}}\int _{-\infty }^{\infty }\partial _{y_{k}}{\mathcal {F}}(\theta _{k,n})(y)dy_{k}d(y_{1},\ldots ,y_{k-1},y_{k+1},\ldots ,y_{d})=0$.
If we let $n\to \infty$, theorem 8.11 and lemma 8.13 give the claim.
2. We deduce from 1. that if $\phi$ is an arbitrary Schwartz function, then ${\mathcal {F}}^{-1}({\mathcal {F}}(\phi ))(0)=\phi (0)$.
As in lemma 8.12, we define
    $G:\mathbb {R} ^{d}\to \mathbb {R} ,G(x):=e^{-\pi \|x\|^{2}}$.
Let now $\phi$ be any Schwartz function. Then $\phi -\phi (0)G$ is also a Schwartz function (see exercise 3.?). Further, since $G(0)=1$, it vanishes at the origin. Hence, by 1.,
    ${\mathcal {F}}^{-1}({\mathcal {F}}(\phi -\phi (0)G))(0)=0$.
Further, due to lemma 8.12 and the linearity of the Fourier transform,
    $0={\mathcal {F}}^{-1}({\mathcal {F}}(\phi -\phi (0)G))(0)={\mathcal {F}}^{-1}({\mathcal {F}}(\phi ))-\phi (0)G(0)$.
3. We deduce from 2. that if $\phi$ is a Schwartz function and $x\in \mathbb {R} ^{d}$ is arbitrary, then ${\mathcal {F}}^{-1}({\mathcal {F}}(\phi ))(x)=\phi (x)$ (i. e. ${\mathcal {F}}^{-1}({\mathcal {F}}(\phi ))=\phi$.
Let $\phi \in {\mathcal {S}}(\mathbb {R} ^{d})$ and $x\in \mathbb {R} ^{d}$ be arbitrary. Due to the definition of ${\mathcal {F}}^{-1}$,
    ${\mathcal {F}}^{-1}({\mathcal {F}}(\phi ))(x)=\int _{\mathbb {R} ^{d}}{\mathcal {F}}(\phi )(y)e^{2\pi ix\cdot y}dy$.
Further, if we define $\theta (z):=\phi (z+x)$,
    ${\mathcal {F}}(\phi )(y)e^{2\pi ix\cdot y}=\int _{\mathbb {R} ^{d}}\phi (z)e^{-2\pi iz\cdot y}e^{2\pi ix\cdot y}dz=\int _{\mathbb {R} ^{d}}\theta (z)dze^{-2\pi iz\cdot y}={\mathcal {F}}(\theta )(y)$.
Hence, by 2.,
    ${\mathcal {F}}^{-1}({\mathcal {F}}(\phi ))(x)=\int _{\mathbb {R} ^{d}}{\mathcal {F}}(\theta )(y)dy={\mathcal {F}}^{-1}({\mathcal {F}}(\theta ))(0)=\theta (0)=\phi (x)$.
4. We deduce from 3. that for any Schwartz function $\phi$ we have ${\mathcal {F}}({\mathcal {F}}^{-1}(\phi ))=\phi$.
Let $\phi \in {\mathcal {S}}(\mathbb {R} ^{d})$ and $x\in \mathbb {R} ^{d}$ be arbitrary. Then we have
    ${\mathcal {F}}({\mathcal {F}}^{-1}(\phi ))(y)={\mathcal {F}}^{-1}({\mathcal {F}}^{-1}(\phi ))(-y)=\int _{\mathbb {R} ^{d}}{\mathcal {F}}^{-1}(\phi )(x)e^{-2\pi ix\cdot y}dx=\int _{\mathbb {R} ^{d}}{\mathcal {F}}(\phi )(-x)e^{-2\pi ix\cdot y}dx={\mathcal {F}}^{-1}({\mathcal {F}}(\phi ))(y)=\phi (y)$.$\Box$

## The Fourier transform of tempered distributions

Definition 8.15:
Let ${\mathcal {T}}$ be a tempered distribution. We define
    ${\mathcal {F}}({\mathcal {T}})(\phi ):={\mathcal {T}}({\mathcal {F}}(\phi ))$.
Theorem 8.16:
${\mathcal {F}}({\mathcal {T}})$ is a tempered distribution.
Proof:
1. Sequential continuity follows from the sequential continuity of ${\mathcal {T}}$ and ${\mathcal {F}}$ (theorem 8.11) and that the composition of two sequentially continuous functions is sequentially continuous again.
2. Linearity follows from the linearity of ${\mathcal {T}}$ and ${\mathcal {F}}$ and that the composition of two linear functions is linear again.$\Box$
Definition 8.17:
Let ${\mathcal {T}}$ be a tempered distribution. We define
    ${\mathcal {F}}^{-1}({\mathcal {T}})(\phi ):={\mathcal {T}}({\mathcal {F}}^{-1}(\phi ))$.

## Sources

* Stein, Elias M.; Shakarchi, Rami (2003). Fourier Analysis: An Introduction. Analysis courses 2000/2001. Princeton University Press. ISBN 000-0-000-00000-0. {{cite book}}: Check |isbn= value: invalid prefix (help)
* Jerry Shurman (2008), Fourier inversion (PDF)

# The Malgrange-Ehrenpreis theorem
## Vandermonde's matrix

Definition 10.1:
Let $n\in \mathbb {N}$ and let $x_{1},\ldots ,x_{n}\in \mathbb {R}$. Then the Vandermonde matrix associated to $x_{1},\ldots ,x_{n}$ is defined to be the matrix
    ${\begin{pmatrix}x_{1}&\cdots &x_{n}\\x_{1}^{2}&\cdots &x_{n}^{2}\\\vdots &\ddots &\vdots \\x_{1}^{n}&\cdots &x_{n}^{n}\end{pmatrix}}$.
For $x_{1},\ldots ,x_{n}$ pairwise different (i. e. $x_{k}\neq x_{m}$ for $k\neq m$) matrix is invertible, as the following theorem proves:
Theorem 10.2:
Let $\mathbf {A}$ be the Vandermonde matrix associated to the pairwise different points $x_{1},\ldots ,x_{n}$. Then the matrix $\mathbf {B}$ whose $k,m$-th entry is given by
    $\mathbf {b} _{k,m}:={\begin{cases}{\frac {\sum _{1\leq l_{1}<\cdots <l_{n-m}\leq n \atop l_{1},\ldots ,l_{n-m}\neq k}(-1)^{m-1}x_{l_{1}}\cdots x_{l_{n-m}}}{x_{k}\prod _{1\leq l\leq n \atop l\neq k}(x_{l}-x_{k})}}&m<n\\{\frac {1}{x_{k}\prod _{1\leq l\leq n \atop l\neq k}(x_{l}-x_{k})}}&m=n\end{cases}}$
is the inverse matrix of $\mathbf {A}$.
Proof:
We prove that $\mathbf {B} \mathbf {A} =\mathbf {I} _{n}$, where $\mathbf {I} _{n}$ is the $n\times n$ identity matrix.
Let $1\leq k,m\leq n$. We first note that, by direct multiplication,
    $x_{m}\prod _{1\leq l\leq n \atop l\neq k}(x_{l}-x_{m})=\sum _{j=1}^{n}x_{m}^{j}{\begin{cases}\sum _{1\leq l_{1}<\cdots <l_{n-j}\leq n \atop l_{1},\ldots ,l_{n-j}\neq k}(-1)^{j-1}x_{l_{1}}\cdots x_{l_{n-j}}&j<n\\1&j=n\end{cases}}$.
Therefore, if $\mathbf {c} _{k,m}$ is the $k,m$-th entry of the matrix $\mathbf {B} \mathbf {A}$, then by the definition of matrix multiplication
    $\mathbf {c} _{k,m}=\sum _{j=1}^{n}{\frac {x_{m}^{j}{\begin{cases}\sum _{1\leq l_{1}<\cdots <l_{n-j}\leq n \atop l_{1},\ldots ,l_{n-j}\neq k}(-1)^{j-1}x_{l_{1}}\cdots x_{l_{n-j}}&j<n\\1&j=n\end{cases}}}{x_{k}\prod _{1\leq l\leq n \atop l\neq k}(x_{l}-x_{k})}}={\frac {x_{m}\prod _{1\leq l\leq n \atop l\neq k}(x_{l}-x_{m})}{x_{k}\prod _{1\leq l\leq n \atop l\neq k}(x_{l}-x_{k})}}={\begin{cases}1&k=m\\0&k\neq m\end{cases}}$.$\Box$

## The Malgrange-Ehrenpreis theorem

Lemma 10.3:
Let $x_{1},\ldots ,x_{n}\in \mathbb {R}$ be pairwise different. The solution to the equation
    ${\begin{pmatrix}x_{1}&\cdots &x_{n}\\x_{1}^{2}&\cdots &x_{n}^{2}\\\vdots &\ddots &\vdots \\x_{1}^{n}&\cdots &x_{n}^{n}\end{pmatrix}}{\begin{pmatrix}y_{1}\\\vdots \\y_{n}\end{pmatrix}}={\begin{pmatrix}0\\\vdots \\0\\1\end{pmatrix}}$
is given by
    $y_{k}={\frac {1}{x_{k}\prod _{1\leq l\leq n \atop l\neq k}(x_{l}-x_{k})}}$, $k\in \{1,\ldots ,n\}$.
Proof:
We multiply both sides of the equation by $\mathbf {B}$ on the left, where $\mathbf {B}$ is as in theorem 10.2, and since $\mathbf {B}$ is the inverse of
    ${\begin{pmatrix}x_{1}&\cdots &x_{n}\\x_{1}^{2}&\cdots &x_{n}^{2}\\\vdots &\ddots &\vdots \\x_{1}^{n}&\cdots &x_{n}^{n}\end{pmatrix}}$,
we end up with the equation
    ${\begin{pmatrix}y_{1}\\\vdots \\y_{n}\end{pmatrix}}=\mathbf {B} {\begin{pmatrix}0\\\vdots \\0\\1\end{pmatrix}}$.
Calculating the last expression directly leads to the desired formula.$\Box$

## Sources
# Sobolev spaces

There are some partial differential equations which have no solution. However, some of them have something like ‘almost a solution’, which we call a weak solution. Among these there are partial differential equations whose weak solutions model processes in nature, just like solutions of partial differential equations which have a solution.
These weak solutions will be elements of the so-called Sobolev spaces. By proving properties which elements of Sobolev spaces in general have, we will thus obtain properties of weak solutions to partial differential equations, which therefore are properties of some processes in nature.
In this chapter we do show some properties of elements of Sobolev spaces. Furthermore, we will show that Sobolev spaces are Banach spaces (this will help us in the next section, where we investigate existence and uniqueness of weak solutions).

## The fundamental lemma of the calculus of variations

But first we shall repeat the definition of the standard mollifier defined in chapter 3.
Example 3.4: The standard mollifier $\eta$, given by
    $\eta :\mathbb {R} ^{d}\to \mathbb {R} ,\eta (x)={\frac {1}{c}}{\begin{cases}e^{-{\frac {1}{1-\|x\|^{2}}}}&{\text{ if }}\|x\|_{2}<1\\0&{\text{ if }}\|x\|_{2}\geq 1\end{cases}}$
, where $c:=\int _{B_{1}(0)}e^{-{\frac {1}{1-\|x\|^{2}}}}dx$, is a bump function (see exercise 3.2).
Definition 3.13:
For $R\in \mathbb {R} _{>0}$, we define
    $\eta _{R}:\mathbb {R} ^{d}\to \mathbb {R} ,\eta _{R}(x)=\eta \left({\frac {x}{R}}\right){\big /}R^{d}$.
Lemma 12.1: (to be replaced by characteristic function version)
Let $g\in L^{p}$ be a simple function, i. e.
    $g=\sum _{j=1}^{n}b_{j}\chi _{I_{j}}$,
where $I_{j}$ are intervals and $\chi$ is the indicator function. If
    $\epsilon <1/2{\text{diam}}(I_{j})$,
then $\|g*\eta _{\epsilon }-g\|_{p}\leq 2\epsilon \max _{k\in \{1,\ldots ,n\}}b_{k}$.
The following lemma, which is important for some theorems about Sobolev spaces, is known as the fundamental lemma of the calculus of variations:
Lemma 12.2:
Let $S\subseteq \mathbb {R} ^{d}$ and let $f,g:S\to \mathbb {R}$ be functions such that $f,g\in L_{\text{loc}}^{1}(S)$ and ${\mathcal {T}}_{f}={\mathcal {T}}_{g}$. Then $f=g$ almost everywhere.
Proof:
We define
    $h:\mathbb {R} ^{d}\to \mathbb {R} ,h(x):={\begin{cases}f(x)-g(x)&x\in S\\0&x\notin S\end{cases}}$

## Weak derivatives

Definition 12.1:
Let $S\subseteq \mathbb {R} ^{d}$ be a set, $p\in [1,\infty ]$ and $f\in L^{p}(S)$. If $\alpha \in \mathbb {N} _{0}^{d}$ is a $d$-dimensional multiindex and $g\in L^{p}(S)$ such that
    $\partial _{\alpha }{\mathcal {T}}_{f}={\mathcal {T}}_{g}$
, we call $g$ an $\alpha$th-weak derivative of $f$.
Remarks 12.2: If $f\in L^{p}(S)$ is a function and $\alpha \in \mathbb {N} _{0}^{d}$ is a $d$-dimensional multiindex, any two $\alpha$th-weak derivatives of $f$ are equal except on a null set. Furthermore, if $\partial _{\alpha }f$ exists, it also is an $\alpha$th-weak derivative of $f$.
Proof:
1. We prove that any two $\alpha$th-weak derivatives are equal except on a nullset.
Let $g,h\in L^{p}(S)$ be two $\alpha$th-weak derivatives of $f$. Then we have
    ${\mathcal {T}}_{g}=\partial _{\alpha }{\mathcal {T}}_{f}={\mathcal {T}}_{h}$
Notation 12.3 If it exists, we denote the $\alpha$th-weak derivative of $f$ by $\partial _{\alpha }f$, which is of course the same symbol as for the ordinary derivative.
Theorem 12.4:
Let $O\subseteq \mathbb {R} ^{d}$ be open, $p\in [1,\infty ]$, $f,g\in L^{p}(O)$ and $\alpha \in \mathbb {N} _{0}^{d}$. Assume that $f,g$ have $\alpha$-weak derivatives, which we - consistent with notation 12.3 - denote by $\partial _{\alpha }f$ and $\partial _{\alpha }g$. Then for all $b,c\in \mathbb {R}$:
    $\partial _{\alpha }(bf+cg)=b\partial _{\alpha }f+c\partial _{\alpha }g$
Proof:

## Definition and first properties of Sobolev spaces

Definition and theorem 12.6:
Let $O\subseteq \mathbb {R} ^{d}$ be open, $p\in [1,\infty ]$, $f,g\in L^{p}(O)$ and $n\in \mathbb {N} _{0}$. The Sobolev space ${\mathcal {W}}^{n,p}(O)$ is defined as follows:
    ${\mathcal {W}}^{n,p}(O):=\{f\in L^{p}(O):\forall \alpha \in \mathbb {N} _{0}^{d}{\text{ such that }}|\alpha |\leq n:\partial _{\alpha }f{\text{ exists}}\}$
A norm on ${\mathcal {W}}^{n,p}(O)$ is defined as follows:
    $\|f\|_{{\mathcal {W}}^{n,p}(O)}:=\sum _{|\alpha |\leq n}\left\|\partial _{\alpha }f\right\|_{L^{p}(O)}$
With respect to this norm, ${\mathcal {W}}^{n,p}(O)$ is a Banach space.
In the above definition, $\partial _{\alpha }f$ denotes the $\alpha$th-weak derivative of $f$.
Proof:
1.
We show that
    $\|f\|_{{\mathcal {W}}^{n,p}(O)}=\sum _{|\alpha |\leq n}\left\|\partial _{\alpha }f\right\|_{L^{p}(O)}$
is a norm.
We have to check the three defining properties for a norm:
* $\|f\|_{{\mathcal {W}}^{n,p}(O)}=0\Leftrightarrow f=0$ (definiteness)
* $\|cf\|_{{\mathcal {W}}^{n,p}(O)}=|c|\|f\|_{{\mathcal {W}}^{n,p}(O)}$ for every $c\in \mathbb {R}$ (absolute homogeneity)
* $\|f+g\|_{{\mathcal {W}}^{n,p}(O)}\leq \|f\|_{{\mathcal {W}}^{n,p}(O)}+\|g\|_{{\mathcal {W}}^{n,p}(O)}$ (triangle inequality)
We start with definiteness: If $f=0$, then $\|f\|_{{\mathcal {W}}^{n,p}(O)}=0$, since all the directional derivatives of the constant zero function are again the zero function. Furthermore, if $\|f\|_{{\mathcal {W}}^{n,p}(O)}=0$, then it follows that $\|f\|_{L^{p}(O)}=0$ implying that $f=0$ as $\|f\|_{L^{p}(O)}$ is a norm.
We proceed to absolute homogeneity. Let $c\in \mathbb {R}$.
    ${\begin{aligned}\|cf\|_{{\mathcal {W}}^{n,p}(O)}&:=\sum _{|\alpha |\leq n}\left\|\partial _{\alpha }cf\right\|_{L^{p}(O)}&\\&=\sum _{|\alpha |\leq n}\left\|c\partial _{\alpha }f\right\|_{L^{p}(O)}&{\text{ theorem 12.4}}\\&=\sum _{|\alpha |\leq n}|c|\left\|\partial _{\alpha }f\right\|_{L^{p}(O)}&{\text{ by absolute homogeneity of }}\|\cdot \|_{L^{p}(O)}\\&=|c|\sum _{|\alpha |\leq n}\left\|\partial _{\alpha }f\right\|_{L^{p}(O)}&\\&=:|c|\|f\|_{{\mathcal {W}}^{n,p}(O)}\end{aligned}}$
And the triangle inequality has to be shown:
    ${\begin{aligned}\|f+g\|_{{\mathcal {W}}^{n,p}(O)}&:=\sum _{|\alpha |\leq n}\left\|\partial _{\alpha }(f+g)\right\|_{L^{p}(O)}&\\&=\sum _{|\alpha |\leq n}\left\|\partial _{\alpha }f+\partial _{\alpha }g\right\|_{L^{p}(O)}&{\text{ theorem 12.4}}\\&\leq \sum _{|\alpha |\leq n}\left(\left\|\partial _{\alpha }f\right\|_{L^{p}(O)}+\left\|\partial _{\alpha }g\right\|_{L^{p}(O)}\right)&{\text{ by triangle inequality of }}\|\cdot \|_{L^{p}(O)}\\&=\|f\|_{{\mathcal {W}}^{n,p}(O)}+\|g\|_{{\mathcal {W}}^{n,p}(O)}\end{aligned}}$
2.
We prove that ${\mathcal {W}}^{n,p}(O)$ is a Banach space.
Let $(f_{l})_{l\in \mathbb {N} }$ be a Cauchy sequence in ${\mathcal {W}}^{n,p}(O)$. Since for all $d$-dimensional multiindices $\alpha \in \mathbb {N} _{0}^{d}$ with $|\alpha |\leq n$ and $m,l\in \mathbb {N}$
    $\|\partial _{\alpha }f_{l}-\partial _{\alpha }f_{m})\|_{L^{p}(O)}=\|\partial _{\alpha }(f_{l}-f_{m})\|_{L^{p}(O)}\leq \sum _{|\alpha |\leq n}\left\|\partial _{\alpha }(f_{l}-f_{m})\right\|_{L^{p}(O)}$
since we only added non-negative terms, we obtain that for all $d$-dimensional multiindices $\alpha \in \mathbb {N} _{0}^{d}$ with $|\alpha |\leq n$, $(\partial _{\alpha }f_{l})_{l\in \mathbb {N} }$ is a Cauchy sequence in $L^{p}(O)$. Since $L^{p}(O)$ is a Banach space, this sequence converges to a limit in $L^{p}(O)$, which we shall denote by $f_{\alpha }$.
We show now that $f:=f_{(0,\ldots ,0)}\in {\mathcal {W}}^{n,p}(O)$ and $f_{l}\to f,l\to \infty$ with respect to the norm $\|\cdot \|_{{\mathcal {W}}^{n,p}(O)}$, thereby showing that ${\mathcal {W}}^{n,p}(O)$ is a Banach space.
To do so, we show that for all $d$-dimensional multiindices $\alpha \in \mathbb {N} _{0}^{d}$ with $|\alpha |\leq n$ the $\alpha$th-weak derivative of $f$ is given by $f_{\alpha }$. Convergence then automatically follows, as
    ${\begin{aligned}f_{l}\to f,l\to \infty &\Leftrightarrow \|f_{l}-f\|_{{\mathcal {W}}^{n,p}(O)}\to 0,l\to \infty &\\&\Leftrightarrow \sum _{|\alpha |\leq n}\left\|\partial _{\alpha }(f_{l}-f)\right\|_{L^{p}(O)}\to 0,l\to \infty &\\&\Leftrightarrow \sum _{|\alpha |\leq n}\left\|\partial _{\alpha }f_{l}-\partial _{\alpha }f\right\|_{L^{p}(O)}\to 0,l\to \infty &{\text{by theorem 12.4}}\\\end{aligned}}$
where in the last line all the summands converge to zero provided that $\partial _{\alpha }f=f_{\alpha }$ for all $d$-dimensional multiindices $\alpha \in \mathbb {N} _{0}^{d}$ with $|\alpha |\leq n$.
Let $\varphi \in {\mathcal {D}}(O)$. Since $\partial _{\alpha }f_{l}\to f_{\alpha }$ and by the second triangle inequality
    $\|\partial _{\alpha }f-f_{\alpha }\|\geq |\|\partial _{\alpha }f\|-\|f_{\alpha }\||$
, the sequence $(\varphi \partial _{\alpha }f_{l})_{l\in \mathbb {N} }$ is, for large enough $l$, dominated by the function $2\|\varphi \|_{\infty }f_{\alpha }$, and the sequence $(\partial _{\alpha }\varphi f_{l})_{l\in \mathbb {N} }$ is dominated by the function $2\|\partial _{\alpha }\varphi \|_{\infty }f$.
Therefore
    ${\begin{aligned}\int _{\mathbb {R} ^{d}}\partial _{\alpha }\varphi (x)f(x)dx=&\lim _{l\to \infty }\int _{\mathbb {R} ^{d}}\partial _{\alpha }\varphi (x)f_{l}(x)dx&{\text{ dominated convergence}}\\&=\lim _{l\to \infty }(-1)^{|\alpha |}\int _{\mathbb {R} ^{d}}\varphi (x)\partial _{\alpha }f_{l}(x)dx&\\&=(-1)^{|\alpha |}\int _{\mathbb {R} ^{d}}\varphi (x)f_{\alpha }(x)dx&{\text{ dominated convergence}}\end{aligned}}$
, which is why $f_{\alpha }$ is the $\alpha$th-weak derivative of $f$ for all $d$-dimensional multiindices $\alpha \in \mathbb {N} _{0}^{d}$ with $|\alpha |\leq n$.$\Box$

## Approximation by smooth functions

We shall now prove that for any $L^{p}$ function, we can find a sequence of bump functions converging to that function in $L^{p}$ norm.
Let $\Omega \subset \mathbb {R} ^{d}$ be a domain, let $r>0$, and $U\subset \Omega$, such that $U+B_{r}(0)\subseteq \Omega$. Let furthermore $u\in {\mathcal {W}}^{m,p}(U)$. Then $\mu _{\epsilon }*f$ is in $C^{\infty }(U)$ for $\epsilon <r$ and $\lim _{\epsilon \to 0}\|\mu _{\epsilon }*f-f\|_{W^{m,p}(U)}=0$.
Proof: The first claim, that $\mu _{\epsilon }*f\in C^{\infty }(U)$, follows from the fact that if we choose
    ${\tilde {f}}(x)={\begin{cases}f(x)&x\in U\\0&x\notin U\end{cases}}$
Then, due to the above section about mollifying $L^{p}$-functions, we know that the first claim is true.
The second claim follows from the following calculation, using the one-dimensional chain rule:
    ${\frac {\partial ^{\alpha }}{\partial x^{\alpha }}}(\mu _{\epsilon }*f)(y)=\int _{\mathbb {R} ^{d}}{\frac {\partial ^{\alpha }}{\partial x^{\alpha }}}\mu _{\epsilon }(y-x)f(x)dx=(-1)^{|\alpha |}\int _{\mathbb {R} ^{d}}{\frac {\partial ^{\alpha }}{\partial y^{\alpha }}}\mu _{\epsilon }(y-x)f(x)dx$
    $=\int _{\mathbb {R} ^{d}}\mu _{\epsilon }(y-x){\frac {\partial ^{\alpha }}{\partial y^{\alpha }}}f(x)dx=(\mu _{\epsilon }*{\frac {\partial ^{\alpha }}{\partial y^{\alpha }}}f)(y)$
Due to the above secion about mollifying $L^{p}$-functions, we immediately know that $\lim _{\epsilon \to 0}\|\mu _{\epsilon }*{\frac {\partial ^{\alpha }}{\partial y^{\alpha }}}f-f\|=0$, and the second statement therefore follows from the definition of the $W^{m,p}(U)$-norm.
Let $\Omega \subseteq \mathbb {R} ^{d}$ be an open set. Then for all functions $v\in W^{m,p}(\Omega )$, there exists a sequence of functions in $C^{\infty }(\Omega )\cap W^{m,p}(\Omega )$ approximating it.
Proof:
Let's choose
    $U_{i}:=\{x\in \Omega :{\text{dist}}(\partial \Omega ,x)>{\frac {1}{i}}\wedge \|x\|<i\}$
and
    $V_{i}={\begin{cases}U_{3}&i=0\\U_{i+3}\setminus {\overline {U_{i+1}}}&i>0\end{cases}}$
One sees that the $V_{i}$ are an open cover of $\Omega$. Therefore, we can choose a sequence of functions $({\tilde {\eta }}_{i})_{i\in \mathbb {N} }$ (partition of the unity) such that
* $\forall i\in \mathbb {N} :\forall x\in \Omega :0\leq {\tilde {\eta }}_{i}(x)\leq 1$
* $\forall x\in \Omega :\exists {\text{ only finitely many }}i\in \mathbb {N} :{\tilde {\eta }}_{i}(x)\neq 0$
* $\forall i\in \mathbb {N} :\exists j\in \mathbb {N} :{\text{supp }}{\tilde {\eta }}_{i}\subseteq V_{j}$
* $\forall x\in \Omega :\sum _{i=0}^{\infty }{\tilde {\eta }}_{i}(x)=1$
By defining $\mathrm {H} _{i}:=\{{\tilde {\eta }}_{j}\in \{{\tilde {\eta }}_{m}\}_{m\in \mathbb {N} }:{\text{supp }}{\tilde {\eta }}_{j}\subseteq V_{i}\}$ and
    $\eta _{i}(x):=\sum _{\eta \in \mathrm {H} _{i}}\eta (x)$, we even obtain the properties
* $\forall i\in \mathbb {N} :\forall x\in \Omega :0\leq \eta _{i}(x)\leq 1$
* $\forall x\in \Omega :\exists {\text{ only finitely many }}i\in \mathbb {N} :\eta _{i}(x)\neq 0$
* $\forall i\in \mathbb {N} :{\text{supp }}\eta _{i}\subseteq V_{i}$
* $\forall x\in \Omega :\sum _{i=0}^{\infty }{\tilde {\eta }}_{i}(x)=1$
where the properties are the same as before except the third property, which changed. Let $|\alpha |=1$, $\varphi$ be a bump function and $(v_{j})_{j\in \mathbb {N} }$ be a sequence which approximates $v$ in the $L^{p}(\Omega )$-norm. The calculation
    $\int _{\Omega }\eta _{i}(x)v_{j}(x){\frac {\partial ^{\alpha }}{\partial x^{\alpha }}}\varphi (x)dx=-\int _{\Omega }\left({\frac {\partial ^{\alpha }}{\partial x^{\alpha }}}\eta _{i}(x)v_{j}(x)+\eta _{i}(x){\frac {\partial ^{\alpha }}{\partial x^{\alpha }}}v_{j}(x)\right)\varphi (x)dx$
reveals that, by taking the limit $j\to \infty$ on both sides, $v\in W^{m,p}(\Omega )$ implies $\eta _{i}v\in W^{m,p}(\Omega )$, since the limit of $\eta _{i}(x){\frac {\partial ^{\alpha }}{\partial x^{\alpha }}}v_{j}(x)$ must be in $L^{p}(\Omega )$ since we may choose a sequence of bump functions $\varphi _{k}$ converging to 1.
Let's choose now
    $W_{i}={\begin{cases}U_{i+4}\setminus {\overline {U_{i}}}&i\geq 1\\U_{4}&i=0\end{cases}}$
We may choose now an arbitrary $\delta >0$ and $\epsilon _{i}$ so small, that
* $\|\eta _{\epsilon _{i}}*(\eta _{i}v)-\eta _{i}v\|_{W^{m,p}(\Omega )}<\delta \cdot 2^{-(j+1)}$
* ${\text{supp }}(\eta _{\epsilon _{i}}*(\eta _{i}v))\subset W_{i}$
Let's now define
    $w(x):=\sum _{i=0}^{\infty }\eta _{\epsilon _{i}}*(\eta _{i}v)(x)$
This function is infinitely often differentiable, since by construction there are only finitely many elements of the sum which do not vanish on each $W_{i}$, and also since the elements of the sum are infinitely differentiable due to the Leibniz rule of differentiation under the integral sign. But we also have:
    $\|w-v\|_{W^{m,p}(\Omega )}=\left\|\sum _{i=0}^{\infty }\eta _{\epsilon _{i}}*(\eta _{i}v)-\sum _{i=0}^{\infty }(\eta _{i}v)\right\|_{W^{m,p}(\Omega )}\leq \sum _{i=0}^{\infty }\|\eta _{\epsilon _{i}}*(\eta _{i}v)-\eta _{i}v\|_{W^{m,p}(\Omega )}<\delta \sum _{i=0}^{\infty }2^{-(j+1)}=\delta$
Since $\delta$ was arbitrary, this finishes the proof.
Let $\Omega$ be a bounded domain, and let $\partial \Omega$ have the property, that for every point $x\in \partial \Omega$, there is a neighbourhood ${\mathcal {U}}_{x}$ such that
    $\Omega \cap {\mathcal {U}}_{x}=\{(x_{1},\ldots ,x_{d})\in \mathbb {R} ^{d}:x_{i}<f(x_{1},\ldots ,x_{i-1},x_{i+1},\ldots ,x_{d-1})\}$
for a continuous function $f$. Then every function in $W^{m,p}(\Omega )$ can be approximated by $C^{\infty }({\overline {\Omega }})$-functions in the $W^{m,p}(\Omega )$-norm.
Proof: