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 $S$: $d(x,S)=\text{inf}\{d(x,s)\text{ such that }s\in S\}$. Now, $r>0$ because $S$ is closed. Next, we consider $b\in S$ such that

This vector $b$ exists: as then

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

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

Now, define

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

Trivially,

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

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

$$\parallel s-x_{\alpha }\parallel =\parallel s-\frac{x-b}{\parallel x-b\parallel }\parallel =\frac{1}{\parallel x-b\parallel }\cdot \parallel \parallel x-b\parallel \cdot s+b-x\parallel \ge \frac{r}{\parallel x-b\parallel }$$

because

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

But

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

QED.