5

I'm trying to find an example of when the extension of a functional in the Hahn-Banach theorem is not necessarily unique. I'm looking at the space of continuous functions on $[0,1]$ and I'm trying to find a subspace and a functional defined on the subspace that has two different norm preserving extensions.

So far I've thought about the subspace of polynomials, but then the only linear functionals I can think of are things like $f(p) = p(0)$ or $f(p) = \|p(x)\|_{\infty}$ but I can't think of any non-unique way to extend these.

Martin Sleziak
  • 56,060
Wooster
  • 3,885
  • Polynomials are dense in the continuous functions so there are unique extensions of this bounded linear functional (the second one you give is not linear) to the continuous functions (you can work this out explicitly or see it as a special case of a proposition from topology which tells us that continuous functions into a Hausdorff space, here $\mathbf R$ or $\mathbf C$, are determined by their values on a dense subspace of the domain). Maybe try out on a 1-dimensional subspace instead. – a... Jan 13 '15 at 12:35
  • I meant to say there is a unique extension. – a... Jan 13 '15 at 12:41
  • Maybe the result mentioned here can, in some situations, help to find out whether there is only one extension. – Martin Sleziak Jan 13 '15 at 12:56
  • To give a specific example, any sequence which is not almost convergent can be assigned different values for different Banach limits. Banach limits are simply extensions of $\lim$ to the space $\ell_\infty$ which are shift-invariant. See also here. – Martin Sleziak Jan 13 '15 at 13:00
  • It seems that the example I have suggested in the above comment is unnecessarily complicated. – Martin Sleziak Jan 13 '15 at 13:27

2 Answers2

5

What about the subspace $$S=\{f\in C(0,1); f(0)=f(1)\}$$ and the functional $T\colon f \mapsto f(0)$ defined on $S$.

The you have two extensions $T_0\colon f\mapsto f(0)$ and $T_1\colon f\mapsto f(1)$.

For these functionals we have $\|T\|=\|T_1\|=\|T_2\|=1$.

In fact, for every $a\in[0,1]$ we have an extension $T_a\colon f\mapsto af(0)+(1-a)f(1)$ with $\|T_a\|=1$.

Martin Sleziak
  • 56,060
2

The following is true:

If $X$ is a Banach space, then each Hahn-Banach (norm preserving) extension is unique iff the unit ball of $X^{\ast}$ is strictly convex.

In particular, this would be true for a Hilbert space. Furthermore, there is an interesting theorem of Phelps (See this) which relates this property for a fixed subspace to a best approximation property of the annihilator of that subspace in $X^{\ast}$.

Prahlad Vaidyanathan
  • 32,330
  • 1
    An interesting result. Here are some posts from this site where this was mentioned: http://math.stackexchange.com/questions/769619/when-is-the-closed-unit-ball-b-in-the-dual-space-strictly-convex, http://math.stackexchange.com/questions/827816/uniqueness-of-hahn-banach-extension-for-convex-dual-spaces, http://math.stackexchange.com/questions/332350/hilbert-spaces-and-unique-extensions-of-linear-functions – Martin Sleziak Jan 13 '15 at 14:38