# Measure Theory/L2 Inner Product Space
## Inner Products

We have defined, in Lesson 0, the inner product for $L^{2}(E)$. However, when calling something by the name "inner product" we mean to communicate a few properties. We should ensure that our inner product,
    $\langle f,g\rangle =\int _{E}fg$
deserves the name.
Besides justifying the name, out of a sense of principle, these properties will also be useful common manipulations in the proofs of the theorems that we care about -- the most important, for this section, being completeness.
Definition: inner product
Let V be any vector space over ${\mathbb {R}}$, and let $\langle \cdot ,\cdot \rangle :V\to {\mathbb {R}}$ be any binary function. In all of the definitions below, let ${\vec {v}},{\vec {w}},{\vec {x}}\in V$ be arbitrary vectors and $a\in {\mathbb {R}}$ an arbitrary real number.
We define $\langle \cdot ,\cdot \rangle$ to be symmetric if
    $\langle {\vec {v}},{\vec {w}}\rangle =\langle {\vec {w}},{\vec {v}}\rangle$
We define it to be linear in the first coordinate (or just linear for short) if
    $\langle a{\vec {v}}+{\vec {w}},{\vec {x}}\rangle =a\langle {\vec {v}},{\vec {x}}\rangle +\langle {\vec {w}},{\vec {x}}\rangle$
We define it to be positive definite if
    $\langle {\vec {v}},{\vec {v}}\rangle =\|{\vec {v}}\|\geq 0$ with equality holding if and only if ${\vec {v}}={\vec {0}}$.
We define $\langle \cdot ,\cdot \rangle$ to be an inner product on V if it is symmetric, linear, and positive definite.
Recall that you have already decided what ${\vec {0}}$ is when the vector space in question is $L^{2}(E)$, in the previous lesson, Lesson 1.
* Exercise 1. L$^2$ Inner Product Is an Inner Product
* Show that the $L^{2}(E)$ inner product is symmetric and linear. (This should be trivial.)
* Also show that for every ${\vec {v}}\in V$, we have $\|{\vec {v}}\|_{2}\geq 0$.
* Also show that $\|{\vec {0}}\|_{2}=0$.
* Now consider proving that if $\|{\vec {v}}\|=0$ then ${\vec {v}}={\vec {0}}$. First show that, in fact, there is a function $f\in {\mathcal {L}}^{2}(E)$ for which $\int _{E}f^{2}=0$ even though $f\neq 0$.
* Next recall that, technically, $L^{2}(E)$ is a set of equivalence classes. Recalling also a result from a previous section, prove that if $\|{\vec {v}}\|_{2}=0$ then as an equality of equivalence classes, it follows that ${\vec {v}}={\vec {0}}$.
* Conclude that the $L^{2}(E)$ inner product, is an inner product on $L^{2}(E)$.

## Normed Space

Definition: norm
Let V be a vector space and $\|\cdot \|:V\to {\mathbb {R}}$ a function. Let ${\vec {v}}\in V$ be an arbitrary vector and $a\in {\mathbb {R}}$ an arbitrary real number, throughout this definition. We say that $\|\cdot \|$ is positive definite if
    $\|{\vec {v}}\|\geq 0$ with equality holding if and only if ${\vec {v}}={\vec {0}}$
We call it homogeneous if
    $\|a{\vec {v}}\|=|a|\|{\vec {v}}\|$
We call it subadditive if
    $\|{\vec {v}}+{\vec {w}}\|\leq \|{\vec {v}}\|+\|{\vec {w}}\|$
(We call the inequality itself, $\|{\vec {v}}+{\vec {w}}\|\leq \|{\vec {v}}\|+\|{\vec {w}}\|$, the triangle inequality.)
We call any function a norm if it is positive definite, homogeneous, and subadditive. We call V a normed vector space if it has a norm.
* Exercise 2. L$^2$ Is a Normed Metric Space
* Show that every inner product space is a normed space, and infer that $L^{2}(E)$ is a normed space with norm $\|{\vec {v}}\|_{2}=\langle {\vec {v}},{\vec {v}}\rangle$.

## Metric Space

Definition: metric space
Let X be a set and $d:X\times X\to {\mathbb {R}}$. Let $v,w,x\in X$. We say that d is positive definite if
    $d(v,w)\geq 0$ with equality if and only if $v=w$
We say that d is symmetric if
    $d(v,w)=d(w,v)$
We call the inequality
    $d(v,w)\leq d(v,x)+d(x,w)$
the triangle inequality. (If we wish, we may distinguish between the norm and distance "triangle inequality".)
The function d is called a distance metric if it is positive definite, symmetric, and satisfies the triangle inequality. The space X is called a metric space if it has a distance metric.
* Exercise 3. L$^2$ Is a Metric Space
* Prove that every normed space is a metric space, and infer that $L^{2}(E)$ is a metric space with metric $d({\vec {v}},{\vec {w}})=\|{\vec {v}}-{\vec {w}}\|_{2}$.