proof of Hahn-Banach theorem

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

Let $(F_i,H_i)$ be the elements of a chain in $ℱ$. Define $H={\bigcup }_i{H}_i$. Clearly $H$ is a vector subspace of $V$ and contains $U$. Define $F:H\to K$ 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 $(F,H)\in ℱ$. 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 $H^{\prime }=H+Kv=\{u+tv:u\in H,t\in K\}$ ($H^{\prime }$ is the vector space generated by $H$ and $v$). Choose

$$\lambda =\underset{x\in H}{sup}\{F(x)-p(x-v)\}.$$

We notice that given any $x,y\in H$ it holds

$$F(x)-F(y)=F(x-y)\le p(x-y)=p(x-v+v-y)\le p(x-v)+p(y-v)$$

i.e.

$$F(x)-p(x-v)\le F(y)+p(y-v);$$

in particular we find that and for all $y\in H$ it holds

$$F(y)-p(y-v)\le \lambda \le F(y)+p(y-v).$$

Define $F^{\prime }:H^{\prime }\to K$ as follows:

$$F^{\prime }(u+tv)=F(u)+t\lambda .$$

Clearly $F^{\prime }$ is a linear functional. We have

$$|F^{\prime }(u+tv)|=|F(u)+t\lambda |=|t||F(u/t)+\lambda |$$

and by letting $y=-u/t$ by the previous estimates on $\lambda$ we obtain

$$F(u/t)+\lambda \le F(u/t)+F(-u/t)+p(-u/t-v)=p(u/t+v)$$

and

$$F(u/t)+\lambda \ge F(u/t)+F(-u/t)-p(-u/t-v)=-p(u/t+v)$$

which together give

$$|F(u/t)+\lambda |\le p(u/t+v)$$

and hence

$$|F^{\prime }(u+tv)|\le |t|p(u/t+v)=p(u+tv).$$

So we have proved that $(F^{\prime },H^{\prime })\in ℱ$ and $(F^{\prime },H^{\prime })>(F,H)$ 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