12
$\begingroup$

I was thinking the following problem while reading some functional analysis notes.

Is it possible to characterize the Hahn-Banach extensions (meaning, extensions with the same norm) of a functional in a Banach space $E$ to the double continuous dual $E^{**}$?

My intuition tells me that there aren't many. This is due to two facts:

Theorem 1 (Goldstine) The unit ball $B_E$ of $E$ is $w^*$-dense in the unit ball of $E^{**}$, $B_{E^{**}}$

Theorem 2 There is a canonical (H-B) extension of a functional $\varphi \in E^*$ given by the natural map $E^* \to E^{***}$. Even more so, this map splits.

Are there examples of spaces and functionals that admit many H-B extensions to its double dual?

Just for reference, the first example one would think is $c_0$ which has $c_0^{**}=\ell^\infty$. But in this case, it is well known that the H-B extension is always unique.

Edit: I'll extend the results I know a little in order to attain a possible answer.

Theorem 3 (Phelps) Given a closed subspace $Y$ of a Banach space $X$, every functional on $Y$ has a unique norm-preserving extension if and only if the distance from a functional $f \in X^*$ to $Y^\perp$ is attained uniquely, in the sense that there exists a unique $g \in Y^\perp$ such that $$ d(f,Y^\perp) = \|f-g\|$$

Form this is an easy exercise that the Hahn-Banach extension is always unique if and only if the dual space is strictly convex.

Using theorem 3 and theorem 2, it is also possible to observe the following:

The triple dual, $E^{***}$ splits $$ E^{***} = E^* \oplus E^\perp$$ Then every functional on $E$ admits unique H-B extension to $E^{**}$ if and only if the norm on $E^{***}$ satisfies $$\|f\|_{E^{***}} = \|f|_E\|_{E^*} + \| f - f_E \|_{E^\perp} $$

Is this always the case? Thanks in advance.

$\endgroup$
  • $\begingroup$ I think I'm misunderstanding your question - how are you extending a point in $E$ to something in $E**$? Is it just the evaluation map? $\endgroup$ – B. Mehta Apr 12 '18 at 0:44
  • $\begingroup$ @B.Mehta Yes, I'm identifying $E$ with it's image in $E^{**}$ through the evaluation map. $\endgroup$ – sjvega Apr 12 '18 at 1:31
  • $\begingroup$ And given $\varphi \in E^{**}$, it is an extension of $f \in E$ if they have the same norm? $\endgroup$ – B. Mehta Apr 12 '18 at 1:33
  • $\begingroup$ @B.Mehta I'm using the term "Hahn-Banach" extension to mean that it is an extension with the same norm, as is written on the question. $\endgroup$ – sjvega Apr 12 '18 at 1:38
  • $\begingroup$ Yes, but I'm unclear how you're defining extension. $\endgroup$ – B. Mehta Apr 12 '18 at 1:39
2
$\begingroup$

$C(K)$ for $K$ compact, Hausdorff, infinite satisfies the requirements. From arXiv:math/9605213 Remark 7, we see

For $C(K)$ spaces, the property of being nicely smooth is equivalent to reflexivity.

Further, it is well known that for $C(K)$ spaces, $K$ finite is equivalent to reflexivity.

Finally, Remark 4 in the same paper states that

Hahn-Banach smooth spaces are nicely smooth

where Hahn-Banach smooth spaces are exactly those for which every functional on $X$ Hahn-Banach extends uniquely to $X^{**}$. The paper however does not justify this result, though this paper attributes it to Godefroy:

Godefroy, G., Nicely smooth Banach spaces, in “Texas Functional Analysis Seminar 1984–1985”, (Austin, Tex.), 117 – 124, Longhorn Notes, Univ. Texas Press, Austin, TX, 1985

I can't find an online version of this, nor can I prove myself that Hahn-Banach smooth spaces are nicely smooth (admittedly, I haven't tried much). Many other papers seem to cite this result however, so it seems reasonably reliable.

$\endgroup$
  • $\begingroup$ I'm aware of the result you mention (I believe is a theorem of Phelps). $\endgroup$ – sjvega Apr 12 '18 at 18:53
  • 1
    $\begingroup$ I believe the space $C([0,1])$ satisfies the requirements, I will post more details when I'm able. $\endgroup$ – B. Mehta Apr 13 '18 at 22:14
  • $\begingroup$ @sjvega I have updated. $\endgroup$ – B. Mehta Apr 15 '18 at 0:56
0
$\begingroup$

It took me some time to come back to this question but I've found an answer that somewhat satisfies me.

Proposition: Let $E$ be a Banach space and denote as $J_E:E \to E^{**}$ its inclusion in its double dual. Then given a norm one functional $\varphi \in E^*$ the following are equivalent.

  • $J_{E^*}(\varphi)$ is the unique functional of norm one in $E^{***}$ that extends $\varphi$ from $J_E(E)$ to $E^{**}$.

  • The identity map $(B_{E^*},w^*) \to (B_{E^*},w)$ of the unit ball of $E^*$ with the weak-$*$ topology to itself with the weak topology is continuous at $\varphi$ .

An immediate corollary is

Corollary: Let $E$ be a Banach space. The following are equivalent.

  • Every functional in $E$ has a unique norm-preserving extension to $E^{**}$.
  • The $w$ and $w^*$ topologies coincide in the unit ball of $E^*$.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.