## About this post

The Hahn-Banach theorem has been a central tool for functional analysis and therefore enjoys a wide variety, many of which have a numerous uses in other fields of mathematics. Therefore it’s not possible to cover all of them. In this post we are covering two ‘abstract enough’ results, which are sometimes called the dominated extension theorem. Both of them will be discussed in real vector space where topology is not endowed. This allows us to discuss any topological vector space.

Another interesting thing is, we will be using axiom of choice, or whatever equivalence you may like, for example Zorn’s lemma or well-ordering principle. Before everything, we need to examine more properties of vector spaces.

## Vector space

It’s obvious that every complex vector space is also a real vector space. Suppose $X$ is a complex vector space, and we shall give the definition of real-linear and complex-linear functionals.

An addictive functional $\Lambda$ on $X$ is called

real-linear(complex-linear) if $\Lambda(\alpha x)=\alpha\Lambda(x)$ for every $x \in X$ and for every real (complex) scalar $\alpha$.

For *-linear functionals, we have two important but easy theorems.

If $u$ is the real part of a complex-linear functional $f$ on $X$, then $u$ is real-linear and

*Proof.* For complex $f(x)=u(x)+iv(x)$, it suffices to denote $v(x)$ correctly. But

we see $\Im(f(x)=v(x)=-\Re(if(x))$. Therefore

but $\Re(f(ix))=u(ix)$, we get

To show that $u(x)$ is real-linear, note that

Therefore $u(x)+u(y)=u(x+y)$. Similar process can be applied to real scalar $\alpha$. $\square$

Conversely, we are able to generate a complex-linear functional by a real one.

If $u$ is a real-linear functional, then $f(x)=u(x)-iu(ix)$ is a complex-linear functional

*Proof.* Direct computation. $\square$

Suppose now $X$ is a complex topological vector space, we see a complex-linear functional on $X$ is continuous if and only if its real part is continuous. Every continuous real-linear $u: X \to \mathbb{R}$ is the real part of a unique complex-linear continuous functional $f$.

### Sublinear, seminorm

Sublinear functional is ‘almost’ linear but also ‘almost’ a norm. Explicitly, we say $p: X \to \mathbb{R}$ a sublinear functional when it satisfies

for all $t \geq 0$. As one can see, if $X$ is normable, then $p(x)=\lVert x \rVert$ is a sublinear functional. One should not be confused with semilinear functional, where inequality is not involved. Another thing worth noting is that $p$ is not restricted to be nonnegative.

A seminorm on a vector space $X$ is a real-valued function $p$ on $X$ such that

for all $x,y \in X$ and scalar $\alpha$.

Obviously a seminorm is also a sublinear functional. For the connection between norm and seminorm, one shall note that *$p$ is a norm if and only if it satisfies $p(x) \neq 0$ if $x \neq 0$.*

## Dominated extension theorems

Are the results will be covered in this post. Generally speaking, we are able to extend a functional defined on a subspace to the whole space as long as it’s dominated by a sublinear functional. This is similar to the dominated convergence theorem, which states that if a convergent sequence of measurable functions are dominated by another function, then the convergence holds under the integral operator.

(Hahn-Banach)Suppose

- $M$ is a subspace of a real vector space $X$,
- $f: M \to \mathbb{R}$ is linear and $f(x) \leq p(x)$ on $M$ where $p$ is a sublinear functional on $X$
Then there exists a linear $\Lambda: X \to \mathbb{R}$ such that

for all $x \in M$ and

for all $x \in X$.

### Step 1 - Extending $f$ by one dimension

With that being said, if $f(x)$ is dominated by a sublinear functional, then we are able to extend this functional to the whole space with a relatively proper range.

*Proof.* If $M=X$ we have nothing to do. So suppose now $M$ is a nontrivial proper subspace of $X$. Choose $x_1 \in X-M$ and define

It’s easy to verify that $M_1$ satisfies all axioms of vector space (warning again: no topology is endowed). Now we will be using the properties of sublinear functionals.

Since

for all $x,y \in M$, we have

Let

By definition, we naturally get

and

Define $f_1$ on $M_1$ by

So when $x +tx_1 \in M$, we have $t=0$, and therefore $f_1=f$.

To show that $f_1 \leq p$ on $M_1$, note that for $t>0$, we have

which implies

Similarly,

and therefore

Hence $f_1 \leq p$.

### Step 2 - An application of Zorn’s lemma

#### Side note: Why Zorn’s lemma

It seems that we can never stop using step 1 to extend $M$ to a larger space, but we have to extend. (If $X$ is a finite dimensional space, then this is merely a linear algebra problem.) This meets exactly what William Timothy Gowers said in his blog post:

If you are building a mathematical object in stages and find that (i) you have not finished even after infinitely many stages, and (ii) there seems to be nothing to stop you continuing to build, then Zorn’s lemma may well be able to help you.

— How to use Zorn’s lemma

And we will show that, as W. T. Gowers said,

If the resulting partial order satisfies the chain condition and if a maximal element must be a structure of the kind one is trying to build, then the proof is complete.

To apply Zorn’s lemma, we need to construct a partially ordered set. Let $\mathscr{P}$ be the collection of all ordered pairs $(M’,f’)$ where $M’$ is a subspace of $X$ containing $M$ and $f’$ is a linear functional on $M’$ that extends $f$ and satisfies $f’ \leq p$ on $M’$. For example we have

The partial order $\leq$ is defined as follows. By $(M’,f’) \leq (M’’,f’’)$, we mean $M’ \subset M’’$ and $f’ = f’’$ on $M’$. Obviously this is a partial order (you should be able to check this).

Suppose now $\mathcal{F}$ is a chain (totally ordered subset of $\mathscr{P}$). We claim that $\mathcal{F}$ has an upper bound (which is required by Zorn’s lemma). Let

and

whenever $(M’,f’) \in \mathcal{F}$ and $y \in M’$. It’s easy to verify that $(M_0,f_0)$ is the upper bound we are looking for. But $\mathcal{F}$ is arbitrary, therefore by Zorn’s lemma, there exists a maximal element $(M^\ast,f^\ast)$ in $\mathscr{P}$. If $M^* \neq X$, according to step 1, we are able to extend $M^\ast$, which contradicts the maximality of $M^\ast$. And $\Lambda$ is defined to be $f^\ast$. By the linearity of $\Lambda$, we see

The theorem is proved. $\square$

### How this proof is constructed

This is a classic application of Zorn’s lemma (well-ordering principle, or Hausdorff maximality theorem). First, we showed that we are able to extend $M$ and $f$. But since we do not know the dimension or other properties of $X$, it’s not easy to control the extension which finally ‘converges’ to $(X,\Lambda)$. However, Zorn’s lemma saved us from this random exploration: Whatever happens, the maximal element is there, and take it to finish the proof.

## Generalizing onto $\mathbb{C}$

Since inequality is appeared in the theorem above, we need more careful validation.

(Bohnenblust-Sobczyk-Soukhomlinoff)Suppose $M$ is a subspace of a vector space $X$, $p$ is a seminorm on $X$, and $f$ is a linear functional on $M$ such thatfor all $x \in M$. Then $f$ extends to a linear functional $\Lambda$ on $X$ satisfying

for all $x \in X$.

*Proof.* If the scalar field is $\mathbb{R}$, then we are done, since $p(-x)=p(x)$ in this case (can you see why?). So we assume the scalar field is $\mathbb{C}$.

Put $u = \Re f$. By dominated extension theorem, there is some real-linear functional $U$ such that $U(x)=u$ on $M$ and $U \leq p$ on $X$. And here we have

where $\Lambda(x)=f(x)$ on $M$.

To show that $|\Lambda(x)| \leq p(x)$ for $x \neq 0$, by taking $\alpha=\frac{|\Lambda(x)|}{\Lambda(x)}$, we have

since $|\alpha|=1$ and $p(\alpha{x})=|\alpha|p(x)=p(x)$. $\square$

## Extending Hahn-Banach theorem under linear transform

To end this post, we state a beautiful and useful extension of the Hahn-Banach theorem, which is done by R. P. Agnew and A. P. Morse.

(Agnew-Morse)Let $X$ denote a real vector space and $\mathcal{A}$ be a collection of linear maps $A_\alpha: X \to X$ that commute, or namelyfor all $A_\alpha,A_\beta \in \mathcal{A}$. Let $p$ be a sublinear functional such that

for all $A_\alpha \in \mathcal{A}$. Let $Y$ be a subspace of $X$ on which a linear functional $f$ is defined such that

- $f(y) \leq p(y)$ for all $y \in Y$.
- For each mapping $A$ and $y \in Y$, we have $Ay \in Y$.
- Under the hypothesis of 2, we have $f(Ay)=f(y)$.
Then $f$ can be extended to $X$ by $\Lambda$ so that $-p(-x) \leq \Lambda(x) \leq p(x)$ for all $x \in X$, and

To prove this theorem, we need to construct a sublinear functional that dominates $f$. For the whole proof, see *Functional Analysis* by Peter Lax.

## References / Further Readings

- Walter Rudin,
*Functional Analysis*. - Peter Lax,
*Functional Analysis*. - William Timothy Gowers,
*How to use Zorn’s lemma*.