Proof of Weak Hahn-Banach Theorem

Question

I am going through Differential and Riemannian Manifolds by Serge Lang. In Proposition 2.3 it is stated a weak version of the Hahn-Banach theorem. It is stated as follows

Let $\mathbf{E}$ be a Banach Space and $x\neq0$ an element of $\mathbf{E}$. Then there exists a linear continuous map $\lambda:\mathbf{E}\rightarrow \mathbb{R}$ such that $\lambda(x)\neq 0$

Now, the proof is not given in the text, nevertheless a small explanation is given, it is the following:

One constructs $\lambda$ by Zorn's lemma, supposing that $\lambda$ is defined on some subspace, and having a bounded norm. One then extends $\lambda$ to the subspace generated by one additional element, without increasing the norm.

From here I take that $\lambda$ is first supposed to exist, which makes no sense to me. Maybe I am not understanding the explanation at all, so I ask if anyone could give some insight or make a mora detailed explanation of this.

Thanks in advance!

Answer 1

I'll try to give a brief proof sketch.

You first define $\lambda: \mathbb{R}x \to \mathbb{R}: c x \mapsto c$. Clearly $\lambda(x) = 1 \neq 0$. This gives you a functional on the subspace $\mathbb{R}x$ of $E$ on which you want to apply Hahn-Banach.

Consider $\mathcal{P}$ which is the set of all couples $(W, \theta)$ such that

• $W$ is a subspace of $V$
• $\mathbb{R}x \subseteq W$
• $\Vert\theta \Vert \le \Vert \lambda \Vert$
• $\theta: W \to \mathbb{R}$ is a linear functional
• $\theta\vert_{\mathbb{R}x} = \lambda$, i.e. $\theta$ extends $\lambda$.

Partially order this in the obvious way.

Applying Zorn's lemma, there is a maximal $(W, \theta)$ among those pairs.

We want to show that $W= V$, because then we have found our functional defined on all of $V$. Suppose it is not, then $W \subsetneq V$ and we can pick $y \notin W$.

The idea and the hard step in the proof is then to show that we can find a further extension $$\mu: W + \mathbb{R}y \to \mathbb{R}$$

such that $\mu$ extends $\theta$. This is what the author means with the one-dimensional extension.

The existence of such an extension will give a contradiction to the maximality of $(W, \theta)$, such that we must have had $W=V$ to begin with.

Answer 2

The idea is the following. Let $x \in E$. Then you want to show that there exists a continuous linear functional $f \in E^*$ s.t. $f(x) \neq 0$.

Consider the 1-dimensional linear subspace $S = \langle x \rangle$ of $E$ generated by $x$. Then we can define a linear functional on $S$ via $f(\lambda x) = \lambda$. Then we have $f$ continuous and linear with $f(x) = 1$. Now choose some $y \not\in S$ and extend $f$ to $S \oplus \langle y \rangle$ which has the same norm i.e. construct a continuous linear functional $f_1$ on $S \oplus \langle y \rangle$ s.t. $f_1 \vert_S \equiv f$ and $\Vert f_1 \Vert = \Vert f \Vert$. Then $\Vert f_1 \Vert < \infty$ and is hence continuous. Then continue this process and apply Zorn's lemma to get a continuous linear functional on all of $E$. Showing that such a functional exists is the non-trivial part of the theorem.