ProofOfRieszRepresentationTheoremForSeparableHilbertSpaces 2004-09-03 02:58:14 2007-10-30 02:44:29 Proof Riesz representation theorem 0 Hilbert Riesz representation % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Let $\lbrace {\bf e}_0, {\bf e}_1, {\bf e}_2, \ldots \rbrace$ 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 \PMlinkname{linear map}{ContinuousLinearMapping} $f$ is continuous if and only if it is bounded, i.e. there exists a constant $C$ such that $|f(v)| \le C \|v\|$. Then $$f(v) = \sum_{k=0}^n {\bar c}_k f({\bf e}_k) = \sum_{k=0}^n |c_k|^2 \le C \sqrt{\sum_{k=0}^n |c_k|^2}.$$ Simplifying, $\sum_{k=0}^n |c_k|^2 \le C^2$. 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 $u_1 = u_2$.