proof of Riesz representation theorem

Existence - If $f=0$ we can just take $u=0$ and thereby have $f(x)=0=⟨x,0⟩$ for all $x\in \mathscr{H}$.

Suppose now $f\ne 0$, i.e. $Kerf\ne \mathscr{H}$.

Recall that, since $f$ is continuous (http://planetmath.org/ContinuousMap), $Kerf$ is a closed subspace of $\mathscr{H}$ (continuity of $f$ implies that $f^{-1}(0)$ is closed in $\mathscr{H}$). It then follows from the orthogonal decomposition theorem that

$$\mathscr{H}=Kerf\oplus {(Kerf)}^{⟂}$$

and as $Kerf\ne \mathscr{H}$ we can find $z\in {(Kerf)}^{⟂}$ such that $\parallel z\parallel =1$.

It follows easily from the linearity of $f$ that for every $x\in \mathscr{H}$ we have

$$f(x)z-f(z)x\in Kerf$$

and since $z\in {(Kerf)}^{⟂}$

$\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}0$$=$$⟨f(x)z-f(z)x,z⟩$

$=$$f(x)⟨z,z⟩-f(z)⟨x,z⟩$

$=$$f(x){\parallel z\parallel }^2-⟨x,\overline{f(z)}z⟩$

$=$$f(x)-⟨x,\overline{f(z)}z⟩$

which implies

$$f(x)=⟨x,\overline{f(z)}z⟩.$$

The theorem then follows by taking $u=\overline{f(z)}z$.

Uniqueness - Suppose there were $u_1,u_2\in \mathscr{H}$ such that for every $x\in \mathscr{H}$

$$f(x)=⟨x,u_1⟩=⟨x,u_2⟩.$$

Then $⟨x,u_1-u_2⟩=0$ for every $x\in \mathscr{H}$. Taking $x=u_1-u_2$ we obtain ${\parallel u_1-u_2\parallel }^2=0$, which implies $u_1=u_2$. $\mathrm{\square }$