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Here's a link to a proove of Riesz's Lemma. I think I understand the proof, but can't answere my following question:

If I take these Conditions:

$ ( X,||.||) $a normed Vektorspace, $ dim X>1 $

$ \{ 0 \} \neq U \subset X $ closed vector subspace, $ \delta \in (0,1)$.

How can I proove that there exist an $ x \in X , ||x|| =1 $ so that $ inf \{ || x-u || :u \in U \}= 1- \delta$

Any Hints very appreciated :-) !

pastapesto23
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  • Why do you say that your version is stronger? They seem equivalent to me. – Lorenzo Q Nov 03 '18 at 14:15
  • does it? :o maybe I still don't really understand it...I don't see Issues in showing $ inf{||x-u||:u \in U} \geq 1-\delta $ or $inf{||x-u||:u \in U} \geq \delta $ .But I don't know how to reach that equivalency $inf{||x-u||:u \in U} = 1-\delta $ – pastapesto23 Nov 03 '18 at 15:22
  • No, my bad, I did not notice the equality sign :) – Lorenzo Q Nov 03 '18 at 16:12

1 Answers1

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This follows from a simple continuity argument. Let $$S:=\left\{x\in X:\|x\|=1\right\} $$ and $$f(x):=\inf_{u\in U}\|x-u\|\qquad \forall x\in S $$ then $f$ is continuous in $S$. In fact, it is a Lipschitz function with Lipschitz constant $L=1$: \begin{align*}|f(x)-f(y)|&=\left|\inf_{u\in U}\|x-u\|-\inf_{v\in U}\|y-v\|\right|\leq\|x-y\| \end{align*}

Now from the usual version of the Riesz Lemma you obtain $x_0\in S$ such that $$f(x_0)\geq 1-\delta $$ Moreover since $U$ is a subspace we have $U\cap S\neq \emptyset$, so there is $x_1\in S$ with $f(x_1)=0$.

Since $S$ is connected, from the intermediate value theorem it follows that there is $x_2\in S$ with $f(x_2)=1-\delta$.

Lorenzo Q
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