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Homeproof of Riesz representation theorem for separable Hilbert spaces

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# 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 $f$ 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 $u_{1}=u_{2}$.

## Mathematics Subject Classification

46C99*no label found*

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## Comments

## generalisation

I wanted to add a proof for the general case, but you might like to change your entry to have only one proof for the theorem.

## Re: generalisation

I suggest you post your proof as an addition to the entry --- the proof I give is specific to Hilbert spaces insofar as it makes use of the inner product in an essential way. I think it would be good to have both proofs here, your proof for the general case and my proof using a method adapted to this special case.