proof of Riesz representation theorem for separable Hilbert spaces

Let $\{{\bf e}_{0},{\bf e}_{1},{\bf e}_{2},\ldots\}$ be an orthonormal basis for the Hilbert space $\mathcal{H}$ . Define

c_{i}=f({\bf e}_{i})\qquad\mbox{ and }\qquad v=\sum_{k=0}^{n}{\bar{c}}_{i}{\bf e% }_{i}.

The linear map (http://planetmath.org/ContinuousLinearMapping) $f$ is continuous if and only if it is bounded, i.e. there exists a constant $C$ such that $|f(v)|\leq C\|v\|$ . Then

f(v)=\sum_{k=0}^{n}{\bar{c}}_{k}f({\bf e}_{k})=\sum_{k=0}^{n}|c_{k}|^{2}\leq C% \sqrt{\sum_{k=0}^{n}|c_{k}|^{2}}.

Simplifying, $\sum_{k=0}^{n}|c_{k}|^{2}\leq C^{2}$ . Hence $\sum_{k=0}^{\infty}c_{k}{\bf e}_{k}$ converges to an element $u$ in $H$ .

For every basis element, $f({\bf e}_{i})=c_{k}=\langle u,{\bf e}_{i}\rangle$ . By linearity, it will also be true that

f(v)=\langle u,v\rangle\mbox{ if $v$ is a finite superposition of basis % vectors.}

Any vector in the Hilbert space can be written as the limit of a sequence of finite superpositions of basis vectors hence, by continuity,

f(v)=\langle u,v\rangle\mbox{ for all }v\in\mathcal{H}

It is easy to see that $u$ is unique. Suppose there existed two vectors $u_{1}$ and $u_{2}$ such that $f(v)=\langle u_{1},v\rangle=\langle u_{2},v\rangle$ . Then $\langle u_{1}-u_{2},v\rangle=0$ for all vectors $v\in\mathcal{H}$ . But then, $\langle u_{1}-u_{2},u_{1}-u_{2}\rangle=0$ which is only possible if $u_{1}-u_{2}=0$ , i.e. if $u_{1}=u_{2}$ .

Title	proof of Riesz representation theorem for separable Hilbert spaces
Canonical name	ProofOfRieszRepresentationTheoremForSeparableHilbertSpaces
Date of creation	2013-03-22 14:34:20
Last modified on	2013-03-22 14:34:20
Owner	asteroid (17536)
Last modified by	asteroid (17536)
Numerical id	6
Author	asteroid (17536)
Entry type	Proof
Classification	msc 46C99