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), $K e r f$ 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
Canonical name	ProofOfRieszRepresentationTheorem
Date of creation	2013-03-22 17:32:37
Last modified on	2013-03-22 17:32:37
Owner	asteroid (17536)
Last modified by	asteroid (17536)
Numerical id	5
Author	asteroid (17536)
Entry type	Proof
Classification	msc 46C99