# Mathematics Prelims

## July 17, 2008

### Riesz’s Lemma

Filed under: Analysis,Functional Analysis — cjohnson @ 1:35 pm

If $X$ is a normed space (of any dimension), $Z$ is a subspace of $X$ and $Y$ is a closed proper subspace of $Z$, then for every $\theta \in [0, 1]$ there exists a $z \in Z$ such that $\|z\| = 1$ and $\|z - y\| \geq \theta$ for every $y \in Y$.

Proof: Let $v \in Z \setminus Y$ and let $a = \inf_{y \in Y} \| v - y\|$.  As $Y$ is closed and $v \notin Y$, we have $a > 0$.  Now let $\theta \in (0, 1)$ and note that there exists a $y_0 \in Y$ such that $a\leq \| v - y_0 \| \leq \frac{a}{\theta}$ (as $\theta < 1$, we have $\frac{a}{\theta} > a$).  Let

$\displaystyle z = \frac{v - y_0}{\| v - y_0 \|}$

Obviously $z \in Z$ and $\| z \| = 1$.  Let $y$ be any element of $Y$.  We have the following.

$\displaystyle \|z-y\| = \left\| \frac{1}{\|v - y_0\|} (v - y_0) - y\right\|$

$\displaystyle \qquad = \frac{1}{\|v - y_0\|} \| v - y_0 - (\|v - y_0\|) y\|$

$\displaystyle \qquad \geq \frac{a}{\| v - y_0 \|}$

$\displaystyle \qquad \geq \frac{a}{a/\theta}$

$\displaystyle \qquad = \theta$