Existence and Uniqueness of Distributional Primitive

== Theorem ==
Let $T \in \map {\DD'} \R$ be a distribution.
Then there exists a distribution $S \in \map {\DD'} \R$ such that in the distributional sense:
:$S' = T$
Furthermore, $S$ is unique up to an arbitrary constant.
== Proof ==
Let $\mathbf 0 : \R \to 0$ be the zero mapping.
Let $\phi_0 \in \map \DD \R \setminus \set {\mathbf 0}$ be a test function.
Let $\psi \in \map \DD \R$ be a test function.
Let $\phi : \R \to \R$ be a real function such that:
:$\ds \phi := \psi - \frac {\int_{-\infty}^\infty \map \psi x \rd x} {\int_{-\infty}^\infty \map {\phi_0} x \rd x} \phi_0$
=== Existence of Unique Primitive of $\phi$ ===
Since $\phi_0, \psi \in \map \DD \R$, by the closure of the test function vector space $\phi \in \map \DD \R$ too.
Moreover:
Thus:
:$\ds \phi \in Y := \set {\phi \in \map \DD \R : \int_{-\infty}^\infty \map \phi x \rd x = 0}$
By Characterization of Derivative of Test Function, there exists a unique $\Phi \in \map \DD \R$ such that $\Phi' = \phi$.
Let $T \in \map {\DD'} \R$ be a distribution.
Let $S: \map \DD \R \to \C$ be a mapping such that $\map S \psi = - \map T \Phi$.
=== Linearity of $S$ ===
Let $\psi_1, \psi_2 \in \map \DD \R$.
Let $\Phi_1, \Phi_2 \in \map \DD \R$ be such that:
:$\ds \Phi_1' = \phi_1 := \psi_1 - \frac {\int_{-\infty}^\infty \map {\psi_1} x \rd x} {\int_{-\infty}^\infty \map {\phi_0} x \rd x} \phi_0$
:$\ds \Phi_2' = \phi_2 := \psi_2 - \frac {\int_{-\infty}^\infty \map {\psi_2} x \rd x} {\int_{-\infty}^\infty \map {\phi_0} x \rd x} \phi_0$
Then:
:$\ds \paren {\Phi_1 + \Phi_2}' = \psi_1 + \psi_2 - \frac {\int_{-\infty}^\infty \paren {\map {\psi_1} x + \map {\psi_2} x} \rd x} {\int_{-\infty}^\infty \map {\phi_0} x \rd x} \phi_0$
Thus:
Similarly:
=== Continuity of $S$ ===
Let $\sequence {\psi_n}_{n \mathop \in \N}$ be a sequence in $\map \DD \R$ such that $\psi_n \stackrel {\DD} {\longrightarrow} \mathbf 0$.
By definition of a convergent sequence in the test function space, there exists $a \in \R_{> 0}$ such that:
:$\forall n \in \N : \forall x \notin \closedint {-a} a : \map {\psi_n} x = 0$
and $\sequence {\psi_n}_{n \mathop \in \N}$ converges uniformly to $\mathbf 0$.
Hence:
Let:
:$\ds \phi_n := \psi_n - \frac {\int_{-\infty}^\infty \map {\psi_n} x \rd x} {\int_{-\infty}^\infty \map {\phi_0} x \rd x} \phi_0$
$\phi_n$ is composed of $\psi_n$ and $\phi_0$, which have compact supports.
Then $\phi_n$ also has a compact support.
Thus:
:$\forall n \in \N : \exists b \in \R_{> 0} : \forall x \notin \closedint {-b} b : \map {\phi_n} x = 0$
For $k \in \N_{> 0}$ we have that:
:$\ds \phi_n^{\paren k} = \psi_n^{\paren k} - \frac {\int_{-\infty}^\infty \map {\psi_n} x \rd x} {\int_{-\infty}^\infty \map {\phi_0} x \rd x} \phi_0^{\paren k}$
Since $\sequence {\psi_n}_{n \mathop \in \N}$ converges in the test function space, it holds that, for all $k \in \N_{>0}$:
:$\sequence {\psi_n^{\paren k}}_{n \mathop \in \N}$ converges uniformly to $\mathbf 0$.
Also, for all $k \in \N_{> 0}$:
:$\phi_0^{\paren k}$ is fixed.
Then:
Thus, for all $k \in \N_{> 0}$ the sequence $\sequence {\phi_n^{\paren k}}_{n \mathop \in \N}$ converges uniformly to the same mapping as $\sequence {\psi_n^{\paren k}}_{n \mathop \in \N}$, that is, $\mathbf 0$.
Hence, $\phi_n$ converges in the test function space to $\mathbf 0$:
:$\phi_n \stackrel {\DD} {\longrightarrow} \mathbf 0$
For every $n \in \N$ let $\Phi_n$ be the unique element in $\map \DD \R$ such that $\Phi'_n = \phi_n$.
From Conditions for Preservation of Covergence in Test Function Space under Differentiation we conclude that:
:$\Phi_n \stackrel {\DD} {\longrightarrow} \mathbf 0$
Consequently:
Thus, $S \in \map {\DD'} \R$.
=== $S' = T$ ===
Let $\Phi \in \map \DD \R$.
Then:
Thus:
Hence, $S' = T$.
=== $S$ is unique up to a constant ===
Suppose $S, R \in \map {\DD'} \R$ are distributions such that for some test function $\rho \in \map \DD \R$ we have:
:$\map {S'} \rho = \map T \rho$
:$\map {R'} \rho = \map T \rho$
Then:
We have that:
:Distributional Derivative on Distributions is Linear Operator
:Distribution Space with Pointwise Addition and Pointwise Scalar Multiplication forms Vector Space
By Vanishing Distributional Derivative of Distribution implies Distribution is Constant, $S - R$ must be a constant distribution $F_c$ with $c \in \R$.
== Sources ==
* : Chapter $\S 6.2$: Derivatives in the distributional sense
Category:Distributional Derivatives