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

In this post we establish (in Lemma 2 below) the "key step" needed for the Hahn-Banach theorem (as described in the previous post), where we extend our linear functional by one real dimension.

First we note an elementary fact about the operator norm for continuous linear functionals on real normed spaces.

Lemma 1 (Operator norm for real linear functionals)

Let $(E,\jnorm)$ be a normed space over $\R$, and let $\phi \in E^*.$ Then

$(1)\Jdisplay\|\phi\|_\op = \sup \{\phi(x): x \in E, \|x\| \leq 1\,\}\,.$

Comment

• This looks very like the definition of the operator norm! However the subtle difference is that we work with $\phi(x)$ instead of $|\phi(x)|$ here. Our actual definition of the operator norm is that
  $$\|\phi\|_\op = \sup \{|\phi(x)|: x \in E, \|x\| \leq 1\,\}\,.$$
• Note that this would make no sense, as it stands, for complex normed spaces (though we could work with the real part of $\phi(x)$ in that case, and we would again obtain a true result).
• Let us refer to the left hand side and right hand side of $(1)$ as $\LHS$ and $\RHS$ respectively.

Proof of Lemma 1

It is clear that $\RHS\leq\LHS$ (since $\phi(x) \leq |\phi(x)|$).

For the reverse inequality, let $x \in E$ with $\|x\|\leq 1.$ Then $\phi(x) \leq \RHS.$ 

But we also have $\|{-x}\|=\|x\| \leq 1,$ and so we also have have $-\phi(x)=\phi(-x) \leq \RHS.$

Since $|\phi(x)|$ is $\phi(x)$ or $-\phi(x)$, we have $|\phi(x)|\leq \RHS.$

This holds for all such $x$. Thus (taking the $\sup$ of $|\phi(x)|$ over such $x$), we see that $\LHS \leq \RHS$, and hence equality holds. $\quad\square$

Armed with Lemma 1, we can now prove the key lemma we need for the Hahn-Banach theorem.

Lemma 2 (Hahn-Banach key lemma) 

Let $(E,\|\cdot\|)$ be a normed space over $\R$, let $F$ be a linear subspace of $E$, and suppose that $F$ has codimension $1$ in $E.$

Let $\psi \in F^*$. Then there exists $\phi \in E^*$ with $\|\phi\|_\op=\|\psi\|_\op$ such that $\phi$ is an extension of $\psi$.

Comments

• As usual, the norm we use on $F$ is the restriction to $F$ of the norm on $E$.
• There are several equivalent definitions of codimension $1$ we could work with. We will use the one that says that $F$ has a $1$-dimensional algebraic complement in $E$.
• Once we have this result, an easy induction gives the result for subspaces of finite codimension.

Proof of Lemma 2

Since $F$ has codimension $1$ in $E$, there exists $y \in E\setminus F$ such that 

$$E = F \oplus \{\alpha y:\alpha \in \R\,\}\,.$$

[In fact this will be true for all $y \in E\setminus F$, but we only need one.]

Given any constant $c \in \R$, we can define a linear extension $\phi_c$ of $\psi$ to $E$ as follows: for all $x \in F$ and $\alpha \in \R$, we set

$$\phi_c(x+\alpha y) = \psi(x) + \alpha c\,.$$

[This is the unique extension which takes the value $c$ at $y$.]

It is not hard to show that each of these extensions $\phi_c$ is continuous, but we need to show that there is at least one real number $c$ satisfying $\|\phi_c\|_\op = \|\psi\|_\op.$

Set $M= \|\psi\|_\op.$ We want to have $\|\phi_c\|_\op \leq M$ (so that equality holds).

By Lemma 1 (and the usual scaling trick), it is necessary and sufficient for $c$ to satisfy, for all $x \in F$ and all $\alpha \in \R$,

$$\psi(x) + \alpha c \leq  M \|x + \alpha y\|\,.$$

It now turns out to be decisive for this to hold when $\alpha=\pm 1$ (because the general case then follows, again by a scaling trick using $|\alpha|$). So what we need is, for all $x_1$ and $x_2$ in $F$,

$$\psi(x_1) + c \leq M \|x_1 + y\|$$

and

$$\psi(x_2)- c \leq M \|x_2 - y\|\,.$$

Rearranging, we see that we need (for all $x_1$ and $x_2$ in $F$)

$$\psi(x_2)-M\|x_2-y\| \leq c \leq M \|x_1+y\| - \psi(x_1)\,.$$

Why should there be any reason to believe that we can actually find such a real number $c$?

Well, this is what I regard as the "Hahn-Banach trick"!

We do know that, for all $x \in F$, we have $\psi(x) \leq M \|x\|$. So, in particular, for all $x_1$ and $x_2$ in $F$, we have

$$\psi(x_1) + \psi(x_2) = \psi(x_1+x_2) \leq M \|x_1 + x_2\| = M \|(x_1+y)+(x_2-y)\| \leq M \|x_1+y\| + M\|x_2-y\|\,.$$

Rearranging gives us

$$\psi(x_2)-M\|x_2-y\| \leq M \|x_1+y\| - \psi(x_1)\,.$$

Fixing $x_1$, we see that the set $\{\psi(x_2)-M\|x_2-y\|:x_2 \in F\}$ is bounded above, and

$$\sup\{\psi(x_2)-M\|x_2-y\|:x_2 \in F\} \leq M \|x_1+y\| - \psi(x_1)\,.$$

Then, we can let $x_1$ vary, and we see that the set $\{M \|x_1+y\| - \psi(x_1): x_1 \in F\,\}$ is bounded below, and

$$\sup\{\psi(x_2)-M\|x_2-y\|:x_2 \in F\} \leq \inf\{M \|x_1+y\| - \psi(x_1): x_1 \in F\,\}\,.$$

And now we can choose any $c$ between these two numbers (inclusive), and set $\phi=\phi_c$.

[This $c$ is unique if and only if the two numbers are the same.]

The result follows.$\quad\square$

Comments

• This argument still looks like a bit of magic for me. However, I have come across a number of similar arguments elsewhere, so I think of this type of argument as the Hahn-Banach trick.
• It works because we can express $x_1+x_2$ as $(x_1+y)+(x_2-y)$, and then use what we know about $\psi$ on $F$ to obtain exactly the estimates we need. But although I have seen some authors try to express a geometric reason why this was bound to work, I have never yet been entirely convinced. I am going to have another think about it!
• The proof above can be modified easily to work with "sublinear functions" instead of norms, but I'll come back to that in a future post.
• After another think: the key inequality we need above is that, for all $x_1$ and $x_2$ in $F$, we have
  $$\psi(x_2)-M\|x_2-y\| \leq M \|x_1+y\| - \psi(x_1)\,.$$
  This means that we really can, in some sense, reduce the problem to the case of extending from two dimensions to three dimensions. Of course we have already proven this inequality true above based on the fact that $\psi(x_1+x_2) \leq M \|x_1+x_2\|.$ But if it is more obviously bound to be true in the 2d to 3d case, that might make me happier! For now I'll continue to think that this works for "Hahn-Banach reasons".