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In the texts on functional analysis I'm reading right now, the Hahn-Banach theorem is used to prove, among others, those statements (all spaces are over $\mathbb{R}$ or $\mathbb{C}$):

Lemma 1: Let $Y\subset X$ be a subspace of a normed space $X$. Then, for each $y'\in Y'$ (the space of all continuous linear functions from $Y$ into the ground field) there is an $x'\in X'$ with $x' = y'$ on $Y$ and $\|x'\|_{X'} = \|y'\|_{Y'}$.

Lemma 2: Let $Y$ be a closed subspace of a normed space $X$ and let $x_0\in X\setminus Y$. Then, there is a $x'\in X'$ with $x'=0$ on $Y$, $\|x'\|_{X'} = 1$ and $x'(x_0) = \operatorname{dist}(x_0, Y)$.

In the proofs of these 2 lemmas, invoking Hahn-Banach is a central step.

But couldn't we simply prove these 2 lemmas algebraically without Hahn-Banach? For example Lemma 1:

Extend a basis of $Y$ to a basis of $X$, so that one can write $X=Y\oplus C$. Then, extend $y'$ by $0$ to $X$ (and call the new linear form, by abuse of notation, also $y'$). The question is now if the extended linear form is continuous and if the norm is right. But obviously

$$ \sup_{a\in Y, \|a\|=1} \|y'(a)\| = \sup_{a\in X, \|a\|=1} \|y'(a)\| $$ and hence $\|y'\|_{Y'} = \|y'\|_{X'}$, so everything is as desired.

What do we need Hahn-Banach for?

Sh4pe
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  • The problem is with your obviously. The only obvious thing is that $\sup_Y \leq \sup_X$. You probably think that $Y$ and $C$ are orthogonal and that you can do Pythagoras, or something like that. – Julien Feb 19 '13 at 17:21
  • Since every $x\in X$ can be written uniquely as $x=y+c$ for $y\in Y$ and $c\in C$, we have $|y'(x)|=|y'(y)|\le \operatorname{sup}_Y$, so $\operatorname{sup}_X\le \operatorname{sup}_Y$. – Sh4pe Feb 19 '13 at 17:24
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    yes, but what is the relation between $|x|$ and $|y|$? Your approach is fundamentally hopeless, believe me. – Julien Feb 19 '13 at 17:25
  • Of course - didn't see that until now. That might prove a problem... :) – Sh4pe Feb 19 '13 at 17:27

1 Answers1

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Hahn-Banach is an essential ingredient in order to carry over extension statements familiar from linear algebra. The point is that not every closed inclusion of normed vector spaces is a split monomorphism. You claim that we can just apply the forgetful functor to vector spaces, and choose a splitting, but this morphism cannot be lifted along the forgetful functor, i.e. it is not continuous. Your proof is wrong where you compute the norm. It would be correct if $X = Y \cup C$, but remember what $X = Y \oplus C$ means ...

Martin Brandenburg
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