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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.

sjvega
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  • 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? – B. Mehta Apr 12 '18 at 00:44
  • @B.Mehta Yes, I'm identifying $E$ with it's image in $E^{**}$ through the evaluation map. – sjvega Apr 12 '18 at 01:31
  • And given $\varphi \in E^{**}$, it is an extension of $f \in E$ if they have the same norm? – B. Mehta Apr 12 '18 at 01:33
  • @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. – sjvega Apr 12 '18 at 01:38
  • Yes, but I'm unclear how you're defining extension. – B. Mehta Apr 12 '18 at 01:39
  • @B.Mehta Given any functional $\phi:E \to \mathbb{C}$, it is possible to extend it to $\tilde{\phi}:E^{**} \to \mathbb{C}$ through the Hahn-Banach theorem. What I'm asking is, "How many possible extensions are that have the same norm?" and "Which functionals on $E^{**}$ come from extensions of functionals of the same norm on $E$?" – sjvega Apr 12 '18 at 01:43
  • Ah! I see, I thought we were extending points **in** $E$ but you're asking about functionals **on** $E$. My mistake. – B. Mehta Apr 12 '18 at 01:45
  • When $E=E^{**}$, I know of the following partial characterisation: the extension is unique if and only if the unit ball of $E^*$ is strictly convex. I'm not sure what happens when $E$ is not reflexive, however... – Theoretical Economist Apr 12 '18 at 15:17
  • @TheoreticalEconomist if $E=E^{**}$ (that is, $E$ is reflexive) then every functional on $E$ is the same as a functional on $E^{**}$, no extension is necessary. I'm aware of the fact that every continous functional on a **subspace** of $E$ extends uniquely (with norm preservation) if and only if $E^*$ is strictly convex. – sjvega Apr 12 '18 at 17:32
  • Sorry, I had misread your post. I was thinking of subspaces of $E$. My mistake. – Theoretical Economist Apr 12 '18 at 17:33

2 Answers2

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$C(K)$ 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.

B. Mehta
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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 $E^*$.
sjvega
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