$L^2$-spaces are Hilbert spaces

Let $(X,𝔅,\mu )$ be a measure space. Let $L^2(X)$ denote the $L^2$-space (http://planetmath.org/LpSpace) associated with this measure space, i.e. $L^2(X)$ consists of measurable functions $f:X⟶ℂ$ such that

identified up to equivalence almost everywhere.

It is known that all $L^p$-spaces (http://planetmath.org/LpSpace), with $1\le p\le \mathrm{\infty }$, are Banach spaces with respect to the $L^p$-norm (http://planetmath.org/LpSpace) $\parallel \cdot {\parallel }_p$. For $L^2(X)$ we can say more:

Theorem - $L^2(X)$ is an Hilbert Space with respect to the inner product $⟨\cdot ,\cdot ⟩$ defined by

$$⟨f,g⟩={\int }_{X}f\overline{g}𝑑\mu$$

Proof:

Sesquilinearity follows from the linearity of the Lebesgue integral (http://planetmath.org/PropertiesOfTheLebesgueIntegralOfLebesgueIntegrableFunctions) (that is, the inner product defined above is linear in the first argument and conjugate linear in the second one). The conjugate symmetry is evident.

Positive definiteness holds by construction: If ${\int }_X{|f|}^2𝑑\mu =0$, 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 this article (http://planetmath.org/ProofThatLpSpacesAreComplete).$\mathrm{\square }$

0.0.1 Remarks

• The spaces ${ℂ}^n$ or ${ℝ}^n$ with the usual inner product are particular examples of $L^2(X)$, choosing $X=\{1,\mathrm{\dots },n\}$ with the counting measure.

• Choosing appropriate spaces $X$ it can be shown that all Hilbert spaces are isometrically isomorphic to a $L^2$-space.