L-2 Inner Product is Inner Product

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

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

Then:

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:

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 $\innerprod E E = 0$ :


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

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.