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

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

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

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

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