$L^2$-spaces are Hilbert spacesL2SpacesAreHilbertSpaces2007-09-14 21:11:432009-01-28 21:08:36TheoremHilbert spacelinear space of square integrable functionssequilinearityHilbert spacesLp spaceBanach spacessequilinearitylinearity of the Lebesgue integralconjugate symmetry$L^p$-norm% this is the default PlanetMath preamble. as your knowledge
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Let $(X, \mathfrak{B}, \mu)$ be a measure space. Let $L^2(X)$ denote the \PMlinkname{$L^2$-space}{LpSpace} associated with this measure space, i.e. $L^2(X)$ consists of measurable functions $f:X \longrightarrow \mathbb{C}$ such that
\begin{displaymath}
\|f\|_2 := \left (\int_X |f|^2 d\mu \right)^{\frac{1}{2}} < \infty
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identified up to equivalence almost everywhere.
It is known that all \PMlinkname{$L^p$-spaces}{LpSpace}, with $1\leq p \leq \infty$, are Banach spaces with respect to the \PMlinkname{$L^p$-norm}{LpSpace} $\;\|\cdot\|_p$. For $L^2(X)$ we can say \PMlinkescapetext{even} more:
{\bf Theorem -} $L^2(X)$ is an Hilbert Space with respect to the inner product $\langle \cdot, \cdot \rangle$ defined by
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\langle f, g \rangle = \int_X f\overline{g} \;d\mu
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\emph{Proof:}
\emph{Sequilinearity} follows from the \PMlinkname{linearity of the Lebesgue integral}{PropertiesOfTheLebesgueIntegralOfLebesgueIntegrableFunctions} (that is, the inner product defined above is linear in the second argument and conjugate linear in the first one). The conjugate symmetry is evident.
Positive definiteness holds by construction: If $\int_X |f|^2 d\mu = 0$, then $|f|^2$ (and therefore $f$) is zero almost everywhere, thus the equivalence class of $f$ is the equivalence class of the zero function (which is the additive neutral element of the space).
Completeness is proved for the general case of $L^p$-spaces in \PMlinkname{this article}{ProofThatLpSpacesAreComplete}.$\square$
\subsubsection{Remarks}
\begin{itemize}
\item The spaces $\mathbb{C}^n$ or $\mathbb{R}^n$ with the usual inner product are particular examples of $L^2(X)$, choosing $X = \{1, \dots, n\}$ with the counting measure.
\item Choosing appropriate spaces $X$ it can be shown that all Hilbert spaces are isometrically isomorphic to a $L^2$-space.
\end{itemize}