# proof of Riesz’ Lemma

Let’s consider $x\in E-S$ and let $r=d(x,S)$. Recall that $r$ is the distance between $x$ and $S$: $d(x,S)=\textrm{inf}\{d(x,s)\textrm{ such that }s\in S\}$. Now, $r>0$ because $S$ is closed. Next, we consider $b\in S$ such that

 $\|x-b\|<\frac{r}{\alpha}$

This vector $b$ exists: as $0<\alpha<1$ then

 $\frac{r}{\alpha}>r$

But then the definition of infimum implies there is $b\in S$ such that

 $\|x-b\|<\frac{r}{\alpha}$

Now, define

 $x_{\alpha}=\frac{x-b}{\|x-b\|}$

Trivially,

 $\|x_{\alpha}\|=1$

Notice that $x_{\alpha}\in E-S$, because if $x_{\alpha}\in S$ then $x-b\in S$, and so $(x-b)+b=x\in S$, an absurd. Plus, for every $s\in S$ we have

 $\|s-x_{\alpha}\|=\|s-\frac{x-b}{\|x-b\|}\|=\frac{1}{\|x-b\|}\cdot\|\|x-b\|% \cdot s+b-x\|\geq\frac{r}{\|x-b\|}$

because

 $\|x-b\|\cdot s+b\in S$

But

 $\frac{r}{\|x-b\|}>\frac{\alpha}{r}\cdot r=\alpha$

QED.

Title proof of Riesz’ Lemma ProofOfRieszLemma 2013-03-22 14:56:14 2013-03-22 14:56:14 gumau (3545) gumau (3545) 6 gumau (3545) Proof msc 54E35 msc 15A03