An introduction to the Hahn-Banach extension theorem: Part V

In this post we prove the general Hahn-Banach Extension Theorem for continuous linear functionals on real normed spaces. (As we saw in Part IV, the result for complex normed spaces then follows from this.)

Here we give the usual proof using Zorn's Lemma. In the next post we will give a second proof using transfinite induction, similar to the proof we gave for separable normed spaces.

Theorem (Hahn-Banach Extension Theorem for linear functionals on normed spaces over $\R$) 

Let $(E,\jnorm)$ be a normed space over $\R,$ and let $F$ be a linear subspace of $E$. Let $\psi \in F^*$. Then there exists $\phi \in E^*$ with $\|\phi\|_\op=\|\psi\|_\op$ and such that $\phi$ is an extension of $\psi$.

Proof using Zorn's Lemma

Let set $S$ to be the set of all ordered pairs $(Y,\theta)$ of the following form:

• $Y$ is a linear subspace of $E$ with $F \subseteq Y$ and
• $\theta \in Y^*$, $\theta|_F=\psi$, and $\|\theta\|_\op = \|\psi\|_\op.$

Certainly  $(F,\psi) \in S$, and so $S \neq \emptyset$.

We need to show that there exists $\phi \in E^*$ such that $(E,\phi) \in S.$ To show this, we introduce a suitable partial order $\lesssim$ on $S$, and then apply Zorn's Lemma.

We define $\lesssim$ as follows. Let $(Y_1,\theta_1). (Y_2,\theta_2) \in S$. Then

\[(Y_1,\theta_1) \lesssim (Y_2,\theta_2) \Leftrightarrow Y_1 \subseteq Y_2 \text{ and } \theta_2|_{Y_1}=\theta_1\,.\]

A routine set of checks shows that $\lesssim$ really is a partial order on $S$: we leave these checks to the reader!

Now let $C$ be a non-empty chain in $(S,\lesssim)$. We show that $C$ has an upper bound in $S$.

[Since $S \neq \emptyset$, we do not need to consider the empty chain.]

In fact we prove a bit more, and this will be something that we can also use in the second proof (in our next post).

Set

\[
Y_0=\bigcup_{(Y,\theta) \in C} Y\,.
\]

We  attempt to define $\theta_0:Y_0 \to \F$ as follows.

Let $x \in Y_0$. Then there exists $(Y,\theta) \in C$ with $x \in Y$. We wish to define $\theta_0(x)=\theta(x)$.

We claim:

1. $\theta_0$ is well defined on $Y_0$;
2. $(Y_0,\theta_0) \in S$;
3. $(Y_0,\theta_0)$ is an upper bound for $C$.

[We will refer back to this claim in our second proof in the next post.]

1. To check that $\theta_0$ is well defined, let $x \in Y_0$, and suppose that $(E_1,\theta_1)$ and $(E_2,\theta_2)$ are in $C$ with $x \in E_1$ and $x \in E_2$. We show that $\theta_1(x)=\theta_2(x)$.

Since $C$ is a chain in $S$, by symmetry we may assume that $(E_1,\theta_1)\lesssim (E_2,\theta_2).$ But then $\theta_2|_{E_1} = \theta_1$, and so $\theta_1(x)=\theta_2(x)$. this proves the first part of our claim.

2. None of this is hard, but it takes a while to write it out in full!

It is relatively standard (and routine) that, with respect to the usual partial order (by inclusion), the union of a non-empty chain of linear subspaces of $E$ is a linear subspace of $E$.

[The reader may check the details, which are similar to the linearity check below.] 

From this, and the definition of $S$, it is clear that $Y_0$ is a linear subspace of $E$, and that $F \subseteq Y_0$.

We next check that $\theta_0$ is a linear map from $S_0$ to $\F$. This is again relatively routine, but let's look at the details this time.

Let $x_1,x_2 \in S_0$ and let $\alpha_1,\alpha_2 \in \F$. Then there exist $(E_1,\theta_1)$ and $(E_2,\theta_2)$ in $C$ such that $x_1 \in E_1$ and $x_2 \in E_2$.

Again (since $C$ is a chain in $S$), by symmetry, we may assume that $(E_1,\theta_1)\lesssim (E_2,\theta_2).$ But then $x_1$, $x_2$ and $\alpha_1 x_1 + \alpha_2 x_2$ are all in the linear subspace $E_2$, and we have, since $\theta_2$ is linear,

\[\theta_0(\alpha_1 x_1 + \alpha_2 x_2) = \theta_2(\alpha_1 x_1 + \alpha_2 x_2) = \alpha_1 \theta_2(x_1) + \alpha_2 \theta_2(x_2) = \alpha_1 \theta_0(x_1) + \alpha_2 \theta_0(x_2)\,.\] 

Thus $\theta_0$ is a linear map from $Y_0$ to $\mathbb{F}$.

We next check that $\theta_0$ is continuous, and satisfies $\|\theta_0\|_\op \leq \|\psi\|_\op$. This is easy from the definition of $\theta_0$ and $S$.

Let $x \in Y_0$. Then there exists $(Y,\theta) \in C$ with $x \in Y$. By definition of $\theta_0$, we have $\theta_0(x)=\theta(x)$. 

By the definition of $S$, we know that $\|\theta\|_\op = \|\psi\|_\op$. Thus 

\[|\theta_0(x)|=|\theta(x)| \leq \|\psi\|_\op \|x\|\,.\]

This shows that $\theta_0 \in Y_0^*$, and $\|\theta_0\|_\op \leq \|\psi\|_\op$.

Next we check that $\theta_0|_F=\psi$. This is again easy. Since $C$ is non-empty, take any $(Y,\\theta) \in C$.

Let $x \in F$. Then, by definition of $S$, we have $x \in Y$ and $\theta(x)=\psi(x)$. Also, by definition of $\theta_0$, $\theta_0(x)=\theta(x)=\psi(x)$. This holds for all $x\in F$, and so we have $\theta_0|_F=\psi$. 

From this we also obtain $\|\theta_0\|_\op \geq \|\psi\|_\op$, and so equality holds. 

This (finally!) completes the proof of 2.: $(Y_0,\theta_0) \in S$.

3. Given the work above, it is now almost immediate that $(S_0,\theta_0)$ is an upper bound for $C$. The only comment that might be worth making is the fact that, for all $(E,\theta)\in C$, $\theta_0|_E=\theta$ by the definition of $\theta_0$.

We now know that every chain (including the empty chain) in the non-empty partially ordered set $S$ has an upper bound. So by Zorn's Lemma, $S$ has a maximal element $(Y_\max,\theta_\max)$.

We claim that $Y_\max=E$. If not, then, there exists $y \in E \setminus Y_\max$, and, by the Hahn-Banach key lemma, $\theta_\max$ can be extended to a bounded linear functional on $Y_\max \oplus \lin\{y\}$ with the same operator norm. But this would contradict the maximality of  $(Y_\max,\theta_\max)$ in $S$.

[Again, the reader seeing this for the first time may wish to fill in a few extra details.]

Thus we must have $Y_\max = E$, and we may take $\phi = \theta_\max \in E^*$. This $\phi$ has the desired properties.$\quad\quad\square$

In the next post (the final post in this series, I think) we use the claim we proved above along with transfinite induction to prove the theorem again, in a similar way to the proof of the separable case from an earlier post.

All that we really need for this is, for some ordinal $\gamma$, vectors $v_\alpha$ $(\alpha<\gamma)$ in $E$ such that $F+\lin\{v_\alpha:\alpha<\gamma\}$ is dense in $E$. On the other hand, assuming the axiom of choice, $E$ itself can be well-ordered, and we can actually find an ordinal $\gamma$ and vectors $v_\alpha$ $(\alpha<\gamma)$ such that $\{v_\alpha:\alpha<\gamma\}=E$. Or we might prefer to arrange that $\{v_\alpha:\alpha<\gamma\}$ is a Hamel basis for $E$.

Without some form of the axiom of choice, if $E$ is not separable, we would not usually expect to find such vectors $v_\alpha$ explicitly, unless there is something special about $E$ and/or the subspace $F$.