PlanetMath: proof of Riesz representation theorem for separable Hilbert spaces

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proof of Riesz representation theorem for separable Hilbert spaces

(Proof)

Let $\lbrace {\bf e}_0, {\bf e}_1, {\bf e}_2, \ldots \rbrace$ be an orthonormal basis for the Hilbert space $\mathcal{H}$ . Define

$\displaystyle c_i = f({\bf e}_i)$ and $\displaystyle \qquad v = \sum_{k=0}^n {\bar c}_i {\bf e}_i.$

The linear map

is continuous if and only if it is bounded, i.e. there exists a constant

such that $\vert f(v)\vert \le C \Vert v\Vert$ . Then

$\displaystyle f(v) = \sum_{k=0}^n {\bar c}_k f({\bf e}_k) = \sum_{k=0}^n \vert c_k\vert^2 \le C \sqrt{\sum_{k=0}^n \vert c_k\vert^2}.$

Simplifying, $\sum_{k=0}^n \vert c_k\vert^2 \le C^2$ . Hence $\sum_{k=0}^\infty c_k {\bf e}_k$ converges to an element

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

$\displaystyle f(v) = \langle u, v \rangle$ if

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,

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

It is easy to see that is unique. Suppose there existed two vectors and 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 , i.e. if .