ProofOfHahnBanachTheorem 2003-03-25 11:10:12 2004-07-29 18:25:58 Proof Hahn-Banach theorem 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here 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 \[ \lambda = \sup_{x\in H}\{ F(x)-p(x-v)\}. \] We notice that given any $x,y\in H$ it holds \[ F(x)-F(y) = F(x-y) \le p(x-y) = p (x-v+v-y) \le p(x-v) + p(y-v) \] i.e. \[ F(x)-p(x-v) \le F(y) + p(y-v); \] in particular we find that $\lambda < +\infty$ and for all $y\in H$ it holds \[ F(y)-p(y-v) \le \lambda \le F(y)+p(y-v). \] Define $F'\colon H'\to K$ as follows: \[ F'(u+tv) = F(u) + t\lambda. \] Clearly $F'$ is a linear functional. We have \[ \lvert F'(u+tv)\rvert = \lvert F(u) + t\lambda \rvert = \lvert t\rvert \, \lvert F(u/t) + \lambda \rvert \] and by letting $y=-u/t$ by the previous estimates on $\lambda$ we obtain \[ F(u/t) + \lambda \le F(u/t) + F(-u/t) + p(-u/t-v ) = p(u/t+v) \] and \[ F(u/t) + \lambda \ge F(u/t) + F(-u/t) - p(-u/t -v) = -p(u/t+v) \] which together give \[ \lvert F(u/t) + \lambda \rvert \le p(u/t+v) \] and hence \[ \lvert F'(u+tv) \rvert \le \lvert t\rvert p(u/t+v) = p(u+tv). \] So we have proved that $(F',H')\in\mathcal F$ and $(F',H')> (F,H)$ which is a contradiction.