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[parent] proof of Hahn-Banach theorem (Proof)

Consider the family of all possible extensions of $ f$, i.e. the set $ \mathcal F$ of all pairings $ (F,H)$ where $ H$ is a vector subspace of $ X$ containing $ U$ and $ F$ is a linear map $ F\colon H \to K$ such that $ F(u)=f(u)$ for all $ u\in U$ and $ \vert F(u)\vert \le p(u)$ for all $ u\in H$. $ \mathcal F$ is naturally endowed with an partial order relation: given $ (F_1,H_1),(F_2,H_2)\in \mathcal F$ we say that $ (F_1,H_1)\le (F_2,H_2)$ iff $ F_2$ is an extension of $ F_1$ that is $ H_1\subset H_2$ and $ F_2(u)=F_1(u)$ for all $ u\in H_1$. We want to apply Zorn's Lemma to $ \mathcal F$ so we are going to prove that every chain in $ \mathcal F$ has an upper bound.

Let $ (F_i,H_i)$ be the elements of a chain in $ \mathcal F$. Define $ H=\bigcup_i H_i$. Clearly $ H$ is a vector subspace of $ V$ and contains $ U$. Define $ F\colon H \to K$ by “merging” all $ F_i$'s as follows. Given $ u\in H$ there exists $ i$ such that $ u\in H_i$: define $ F(u)=F_i(u)$. This is a good definition since if both $ H_i$ and $ H_j$ contain $ u$ then $ F_i(u)=F_j(u)$ in fact either $ (F_i,H_i)\le (F_j,H_j)$ or $ (F_j,H_j)\le (F_i,H_i)$. Notice that the map $ F$ is linear, in fact given any two vectors $ u,v\in H$ there exists $ i$ such that $ u,v\in H_i$ and hence $ F(\alpha u + \beta v) = F_i( \alpha u + \beta v) = \alpha F_i(u) + \beta F_i(v) = \alpha F(u) + \beta F(v)$. The so constructed pair $ (F,H)$ is hence an upper bound for the chain $ (F_i,H_i)$ because $ F$ is an extension of every $ F_i$.

Zorn's Lemma then assures that there exists a maximal element $ (F,H)\in \mathcal F$. To complete the proof we will only need to prove that $ H=V$.

Suppose by contradiction that there exists $ v\in V\setminus H$. Then consider the vector space $ H'= H+ Kv=\{ u+tv\colon u\in H,\quad t\in K\}$ ($ H'$ is the vector space generated by $ H$ and $ v$). Choose

$\displaystyle \lambda = \sup_{x\in H}\{ F(x)-p(x-v)\}. $
We notice that given any $ x,y\in H$ it holds
$\displaystyle F(x)-F(y) = F(x-y) \le p(x-y) = p (x-v+v-y) \le p(x-v) + p(y-v) $
i.e.
$\displaystyle F(x)-p(x-v) \le F(y) + p(y-v); $
in particular we find that $ \lambda < +\infty$ and for all $ y\in H$ it holds
$\displaystyle F(y)-p(y-v) \le \lambda \le F(y)+p(y-v). $

Define $ F'\colon H'\to K$ as follows:

$\displaystyle F'(u+tv) = F(u) + t\lambda. $

Clearly $ F'$ is a linear functional. We have

$\displaystyle \lvert F'(u+tv)\rvert = \lvert F(u) + t\lambda \rvert = \lvert t\rvert \, \lvert F(u/t) + \lambda \rvert $
and by letting $ y=-u/t$ by the previous estimates on $ \lambda$ we obtain
$\displaystyle F(u/t) + \lambda \le F(u/t) + F(-u/t) + p(-u/t-v ) = p(u/t+v) $
and
$\displaystyle F(u/t) + \lambda \ge F(u/t) + F(-u/t) - p(-u/t -v) = -p(u/t+v) $
which together give
$\displaystyle \lvert F(u/t) + \lambda \rvert \le p(u/t+v) $
and hence
$\displaystyle \lvert F'(u+tv) \rvert \le \lvert t\rvert p(u/t+v) = p(u+tv). $

So we have proved that $ (F',H')\in\mathcal F$ and $ (F',H')> (F,H)$ which is a contradiction.



"proof of Hahn-Banach theorem" is owned by paolini.
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Cross-references: estimates, linear functional, generated by, vector space, contradiction, complete, maximal element, vectors, map, contains, upper bound, chain, Zorn's lemma, iff, relation, partial order, linear map, vector subspace, pairings, extensions

This is version 5 of proof of Hahn-Banach theorem, born on 2003-03-25, modified 2004-07-29.
Object id is 4127, canonical name is ProofOfHahnBanachTheorem.
Accessed 4197 times total.

Classification:
AMS MSC: 46B20 (Functional analysis :: Normed linear spaces and Banach spaces; Banach lattices :: Geometry and structure of normed linear spaces)
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