# 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 $F:H\to K$ such that $F(u)=f(u)$ for all $u\in U$ and $|F(u)|\le p(u)$ for all $u\in H$.
$\mathcal{F}$ 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 $F:H\to K$ 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^{} ${H}^{\prime}=H+Kv=\{u+tv:u\in H,t\in K\}$ (${H}^{\prime}$ 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 $({F}^{\prime},{H}^{\prime})\in \mathcal{F}$ 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 |