proof of Hahn-Banach theorem


Consider the family of all possible extensionsPlanetmathPlanetmathPlanetmath of f, i.e. the set of all pairings (F,H) where H is a vector subspace of X containing U and F is a linear map F:HK such that F(u)=f(u) for all uU and |F(u)|p(u) for all uH. is naturally endowed with an partial orderMathworldPlanetmath relationMathworldPlanetmathPlanetmath: given (F1,H1),(F2,H2) we say that (F1,H1)(F2,H2) iff F2 is an extension of F1 that is H1H2 and F2(u)=F1(u) for all uH1. We want to apply Zorn’s Lemma to so we are going to prove that every chain in has an upper bound.

Let (Fi,Hi) be the elements of a chain in . Define H=iHi. Clearly H is a vector subspace of V and contains U. Define F:HK by “merging” all Fi’s as follows. Given uH there exists i such that uHi: define F(u)=Fi(u). This is a good definition since if both Hi and Hj contain u then Fi(u)=Fj(u) in fact either (Fi,Hi)(Fj,Hj) or (Fj,Hj)(Fi,Hi). Notice that the map F is linear, in fact given any two vectors u,vH there exists i such that u,vHi and hence F(αu+βv)=Fi(αu+βv)=αFi(u)+βFi(v)=αF(u)+βF(v). The so constructed pair (F,H) is hence an upper bound for the chain (Fi,Hi) because F is an extension of every Fi.

Zorn’s Lemma then assures that there exists a maximal elementMathworldPlanetmath (F,H). To completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof we will only need to prove that H=V.

Suppose by contradictionMathworldPlanetmathPlanetmath that there exists vVH. Then consider the vector spaceMathworldPlanetmath H=H+Kv={u+tv:uH,tK} (H is the vector space generated by H and v). Choose

λ=supxH{F(x)-p(x-v)}.

We notice that given any x,yH it holds

F(x)-F(y)=F(x-y)p(x-y)=p(x-v+v-y)p(x-v)+p(y-v)

i.e.

F(x)-p(x-v)F(y)+p(y-v);

in particular we find that λ<+ and for all yH it holds

F(y)-p(y-v)λF(y)+p(y-v).

Define F:HK as follows:

F(u+tv)=F(u)+tλ.

Clearly F is a linear functionalMathworldPlanetmathPlanetmath. We have

|F(u+tv)|=|F(u)+tλ|=|t||F(u/t)+λ|

and by letting y=-u/t by the previous estimates on λ we obtain

F(u/t)+λF(u/t)+F(-u/t)+p(-u/t-v)=p(u/t+v)

and

F(u/t)+λF(u/t)+F(-u/t)-p(-u/t-v)=-p(u/t+v)

which together give

|F(u/t)+λ|p(u/t+v)

and hence

|F(u+tv)||t|p(u/t+v)=p(u+tv).

So we have proved that (F,H) and (F,H)>(F,H) which is a contradiction.

Title proof of Hahn-Banach theorem
Canonical name ProofOfHahnBanachTheorem
Date of creation 2013-03-22 13:31:58
Last modified on 2013-03-22 13:31:58
Owner paolini (1187)
Last modified by paolini (1187)
Numerical id 8
Author paolini (1187)
Entry type Proof
Classification msc 46B20