L-2 Inner Product is Inner Product

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Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $\map {\mathcal L^2} {X, \Sigma, \mu}$ be the Lebesgue $2$-space of $\struct {X, \Sigma, \mu}$.

Let $\map {L^2} {X, \Sigma, \mu}$ be the $L^2$ space of $\struct {X, \Sigma, \mu}$.

Let $\innerprod \cdot \cdot$ be the $L^2$ inner product.


Then $\innerprod \cdot \cdot$ is an inner product on $\map {L^2} {X, \Sigma, \mu}$.


Proof

Proof of Symmetry

Let $E, F \in \map {L^2} {X, \Sigma, \mu}$.

Let $E = \eqclass f \sim$ and $F = \eqclass g \sim$.

Then:

\(\ds \innerprod E F\) \(=\) \(\ds \int \paren {f \cdot g} \rd \mu\)
\(\ds \) \(=\) \(\ds \int \paren {g \cdot f} \rd \mu\) Real Multiplication is Commutative
\(\ds \) \(=\) \(\ds \innerprod F E\)

$\Box$

Proof of Linearity in First Argument

Let $E, F, G \in \map {L^2} {X, \Sigma, \mu}$ and $\alpha \in \R$.

Let $E = \eqclass f \sim$, $F = \eqclass g \sim$ and $G = \eqclass h \sim$.

Then:

\(\ds \innerprod {E + \alpha F} G\) \(=\) \(\ds \innerprod {\eqclass f \sim + \alpha \eqclass g \sim} {\eqclass h \sim}\)
\(\ds \) \(=\) \(\ds \innerprod {\eqclass f \sim + \eqclass {\alpha g} \sim} {\eqclass h \sim}\) Definition of Pointwise Scalar Multiplication on Space of Real-Valued Measurable Functions Identified by A.E. Equality
\(\ds \) \(=\) \(\ds \innerprod {\eqclass {f + \alpha g} \sim} {\eqclass h \sim}\) Definition of Pointwise Addition on Space of Real-Valued Measurable Functions Identified by A.E. Equality
\(\ds \) \(=\) \(\ds \int \paren {f + \alpha g} \cdot h \rd \mu\)
\(\ds \) \(=\) \(\ds \int \paren {f \cdot h} \rd \mu + \int \paren {\alpha g} \cdot h \rd \mu\) Integral of Integrable Function is Additive
\(\ds \) \(=\) \(\ds \int \paren {f \cdot h} \rd \mu + \alpha \int \paren {g \cdot h} \rd \mu\) Integral of Integrable Function is Homogeneous
\(\ds \) \(=\) \(\ds \innerprod E G + \alpha \innerprod F G\)

$\Box$

Proof of Non-Negative Definiteness and Positivity

Let $E \in \map {L^2} {X, \Sigma, \mu}$.

Let $E = \eqclass f \sim$.

Then, we have:

$\ds \innerprod E E = \int \size f^2 \rd \mu$

By the definition of the $\mu$-integral of a measurable function, we then have $\innerprod E E \ge 0$.

We have $\innerprod E E = 0$ if and only if:

$\ds \int \size f^2 \rd \mu = 0$

From Measurable Function Zero A.E. iff Absolute Value has Zero Integral, we then have:

$\size f^2 = 0$ $\mu$-almost everywhere.

From Pointwise Exponentiation preserves A.E. Equality, we have:

$\size f = 0$ $\mu$-almost everywhere.

So:

$f = 0$ $\mu$-almost everywhere.

So:

$\eqclass f \sim = \eqclass 0 \sim = \mathbf 0_{\map {L^2} {X, \Sigma, \mu} }$

so:

$E = \mathbf 0_{\map {L^2} {X, \Sigma, \mu} }$

as required.

$\blacksquare$

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