proof of Riesz’ Lemma


Let’s consider xE-S and let r=d(x,S). Recall that r is the distance between x and S: d(x,S)=inf{d(x,s) such that sS}. Now, r>0 because S is closed. Next, we consider bS such that

x-b<rα

This vector b exists: as 0<α<1 then

rα>r

But then the definition of infimum implies there is bS such that

x-b<rα

Now, define

xα=x-bx-b

Trivially,

xα=1

Notice that xαE-S, because if xαS then x-bS, and so (x-b)+b=xS, an absurd. Plus, for every sS we have

s-xα=s-x-bx-b=1x-bx-bs+b-xrx-b

because

x-bs+bS

But

rx-b>αrr=α

QED.

Title proof of Riesz’ Lemma
Canonical name ProofOfRieszLemma
Date of creation 2013-03-22 14:56:14
Last modified on 2013-03-22 14:56:14
Owner gumau (3545)
Last modified by gumau (3545)
Numerical id 6
Author gumau (3545)
Entry type Proof
Classification msc 54E35
Classification msc 15A03