proof of Riesz representation theorem for separable Hilbert spaces

Let $\{{𝐞}_0,{𝐞}_1,{𝐞}_2,\mathrm{\dots }\}$ be an orthonormal basis for the Hilbert space $\mathscr{H}$. Define

$$c_i=f({𝐞}_i)\mathit{\hspace{1em}\hspace{1em}}\text{ and }\mathit{\hspace{1em}\hspace{1em}}v=\sum _{k=0}^n{\overline{c}}_i{𝐞}_i.$$

The linear map (http://planetmath.org/ContinuousLinearMapping) $f$ is continuous if and only if it is bounded, i.e. there exists a constant $C$ such that $|f(v)|\le C\parallel v\parallel$. Then

$$f(v)=\sum _{k=0}^n{\overline{c}}_{k}f({𝐞}_k)=\sum _{k=0}^n{|c_k|}^2\le C\sqrt{\sum _{k=0}^n{|c_k|}^2}.$$

Simplifying, ${\sum }_{k=0}^n{|c_k|}^2\le C^2$. Hence ${\sum }_{k=0}^{\mathrm{\infty }}{c}_k{𝐞}_k$ converges to an element $u$ in $H$.

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

$$f(v)=⟨u,v⟩\text{ if }v\text{ is a finite superposition of basis vectors.}$$

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

$$f(v)=⟨u,v⟩\text{ for all }v\in \mathscr{H}$$

It is easy to see that $u$ is unique. Suppose there existed two vectors $u_1$ and $u_2$ such that $f(v)=⟨u_1,v⟩=⟨u_2,v⟩$. Then $⟨u_1-u_2,v⟩=0$ for all vectors $v\in \mathscr{H}$. But then, $⟨u_1-u_2,u_1-u_2⟩=0$ which is only possible if $u_1-u_2=0$, i.e. if $u_1=u_2$.