# proof of Riesz representation theorem

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

Suppose now $f\neq 0$, i.e. $Kerf\neq\mathcal{H}$.

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

 $\mathcal{H}=Kerf\oplus(Kerf)^{\perp}$

and as $Kerf\neq\mathcal{H}$ we can find $z\in(Kerf)^{\perp}$ such that $\|z\|=1$.

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

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

and since $z\in(Kerf)^{\perp}$

 $\displaystyle\quad\quad\quad\quad\quad\quad 0$ $\displaystyle=$ $\displaystyle\langle f(x)z-f(z)x,z\rangle$ $\displaystyle=$ $\displaystyle f(x)\langle z,z\rangle-f(z)\langle x,z\rangle$ $\displaystyle=$ $\displaystyle f(x)\|z\|^{2}-\langle x,\overline{f(z)}z\rangle$ $\displaystyle=$ $\displaystyle f(x)-\langle x,\overline{f(z)}z\rangle$

which implies

 $f(x)=\langle x,\overline{f(z)}z\rangle\;.$

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

Uniqueness - Suppose there were $u_{1},u_{2}\in\mathcal{H}$ such that for every $x\in\mathcal{H}$

 $f(x)=\langle x,u_{1}\rangle=\langle x,u_{2}\rangle.$

Then $\langle x,u_{1}-u_{2}\rangle=0$ for every $x\in\mathcal{H}$. Taking $x=u_{1}-u_{2}$ we obtain $\|u_{1}-u_{2}\|^{2}=0$, which implies $u_{1}=u_{2}$. $\square$

Title proof of Riesz representation theorem ProofOfRieszRepresentationTheorem 2013-03-22 17:32:37 2013-03-22 17:32:37 asteroid (17536) asteroid (17536) 5 asteroid (17536) Proof msc 46C99