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 linear map $f$ 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 $u_1 = u_2$ .