# proof of Hahn-Banach theorem

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 $|F(u)|\leq p(u)$ 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 $(F_{1},H_{1})\leq(F_{2},H_{2})$ 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 $(F_{i},H_{i})\leq(F_{j},H_{j})$ or $(F_{j},H_{j})\leq(F_{i},H_{i})$. 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 $H^{\prime}=H+Kv=\{u+tv\colon u\in H,\quad t\in K\}$ ($H^{\prime}$ is the vector space generated by $H$ and $v$). Choose

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

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

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

i.e.

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

in particular we find that $\lambda<+\infty$ and for all $y\in H$ it holds

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

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

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

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

 $\lvert F^{\prime}(u+tv)\rvert=\lvert F(u)+t\lambda\rvert=\lvert t\rvert\,% \lvert F(u/t)+\lambda\rvert$

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

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

and

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

which together give

 $\lvert F(u/t)+\lambda\rvert\leq p(u/t+v)$

and hence

 $\lvert F^{\prime}(u+tv)\rvert\leq\lvert t\rvert p(u/t+v)=p(u+tv).$

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

Title proof of Hahn-Banach theorem ProofOfHahnBanachTheorem 2013-03-22 13:31:58 2013-03-22 13:31:58 paolini (1187) paolini (1187) 8 paolini (1187) Proof msc 46B20