Inner Product/Examples/Lebesgue 2-Space

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Example of Inner Product

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

Let $\map {L^2} \mu$ be the Lebesgue $2$-space of $\mu$.


Let $\innerprod \cdot \cdot: \map {L^2} \mu \times \map {L^2} \mu \to \C$ be the mapping defined by:

$\ds \innerprod f g = \int f \, \overline g \rd \mu$


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


Proof

First of all, by Hölder's Inequality for Integrals with $p = q = 2$, it follows that:

$\ds \int f \, \overline g \rd \mu$

is defined.


Now checking the axioms for an inner product in turn:


$(1)$ Conjugate Symmetry

\(\ds \innerprod f g\) \(=\) \(\ds \int f \, \overline g \rd \mu\)
\(\ds \) \(=\) \(\ds \int \overline {\overline f \, g} \rd \mu\) Complex Conjugation is Involution
\(\ds \) \(=\) \(\ds \overline {\int g \, \overline f \rd \mu}\)
\(\ds \) \(=\) \(\ds \overline {\innerprod g f}\)

$\Box$


$(2)$ Sesquilinearity

\(\ds \innerprod {\lambda f + g} h\) \(=\) \(\ds \int \paren {\lambda f + g} \overline h \rd \mu\)
\(\ds \) \(=\) \(\ds \int \paren {\lambda f \, \overline h} + \paren {g \, \overline h} \rd \mu\)
\(\ds \) \(=\) \(\ds \int \lambda f \, \overline h \rd \mu + \int g \, \overline h \rd \mu\)
\(\ds \) \(=\) \(\ds \lambda \int f \, \overline h \rd \mu + \int g \, \overline h \rd \mu\)
\(\ds \) \(=\) \(\ds \lambda \innerprod f h + \innerprod g h\)

$\Box$


$(3)$ Non-Negative Definiteness

\(\ds \innerprod f f\) \(=\) \(\ds \int f \, \overline f \rd \mu\)
\(\ds \) \(=\) \(\ds \int \cmod f^2 \rd \mu\)
\(\ds \) \(\in\) \(\ds \R_{\ge 0}\) Integral of Positive Function is Positive

$\Box$


$(4)$ Positivity

Suppose that $\innerprod f f = 0$.

That is:

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

Hence:

$\ds \int \cmod {f - 0}^2 \rd \mu = 0$

which is to say that $f = 0$ in $\map {L^2} \mu$ by definition of Lebesgue space.

$\Box$


Having verified all the axioms, we conclude $\innerprod \cdot \cdot$ is an inner product.

$\blacksquare$


Sources

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