# Continuous choice of Hahn-Banach extensions

QUESTION RETRACTED - My original argument was fundamentally mistaken (mixing up lower and upper semi-continuity). Sorry (and thanks for the useful comments)

I need, and (unless I am seriously mistaken) can prove, the following:

Let $E \subseteq F$ be an (isometric) inclusion of Banach spaces, and let $E^*_1$, $F^*_1$ denote the closed unit balls of their respective duals. Then the restriction map $F^*_1 \to E^*_1$ admits a weak$^\*$-continuous section from $E^*_1$ to $F^*_1$.

On the unit sphere this section gives you norm-preserving extensions.

This must be known - can anyone provide a reference? (am not a Banach space theorist)

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Note for other people reading: Phil Brooker's answer was to an earlier version of this question – Yemon Choi Nov 28 '11 at 21:46
So having extensions that are not norm preserving (or a weak$^*$ continuous selection into a larger ball) is not useful for you? – Bill Johnson Nov 29 '11 at 1:06

If I understand the claims of the OP correctly, I don't think that such a section can actually exist (if there is a misunderstanding on my part, I will happily retract this answer!).

Upon reading the question, I immediately thought of topological vector space versions of the Michael continuous selection theorem (for instance, Theorem 1.2 of Michael's paper Three mapping theorems, Proc. Amer. Math. Soc. 15 (1964), 410-415.) and Mori Zippin's later work on operator extension problems, which uses continuous selections.

In one of Zippin's papers in particular, namely The embedding of Banach spaces into spaces with structure, Illinois J. Math. 34(3) (1990), 586-606, we see that the existence of a $w^\ast$-continuous selection $E_1^\ast\longrightarrow F^\ast$ is equivalent to $E$ being almost complemented in $F$; by almost complemented we mean that there exists a number $\lambda>0$ such that for every compact, Hausdorff $K$ and operator $T: E\longrightarrow C(K)$, there is a continuous linear extention $\overline{T}: F\longrightarrow C(K)$ of $T$ such that $\Vert \overline{T}\Vert \leq \lambda\Vert T\Vert$.

So the opposition to the claim of the OP comes from the existence of Banach spaces admitting subspaces that are not almost complemented. From the definition of almost complemntedness, the most easily illustrated examples are found by taking compact Hausdorff spaces $K$ and $L$ such that $C(L)$ contains an uncomplemented subspace isomorphic to $C(K)$, and considering the operator $T$ to be the identity operator on the the uncomplemented copy of $C(K)$. One example of such spaces $K$ and $L$ is obtained by taking $K=\gamma\mathbb{N}$ (one-point compactification) and $L=\beta \mathbb{N}$ (Stone-Cech compactification). Another example is to take $K=L = \omega^\omega+1$ (the set of ordinals not exceeding $\omega^\omega$, equipped with its natural compact, Hausdorff order topology).

Whilst writing my answer I noticed that Andreas Thom posted an answer mentioning the Bartle-Graves theorem, which also crossed my mind when thinking of the answer to this question. Conincidentally, only weeks ago I was wondering whether Bartle-Graves can be done $w^\ast$-continuously for the adjoint of an isometric inclusion map, and now I can see that the answer is no... so I, for one, am very grateful for this question!

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Is it true that if $F/E$ is finite dimensional then $E$ is almost complemented in $F$? (Sorry, just trying to understand the definitions...) If so, I may be OK, it's true I went a bit too quickly over the transfinite induction for the general case, but for my purposes all I need is the finite dimensional case. – Itaï BEN YAACOV Nov 28 '11 at 15:08
@Itai: Sure-- pick a basis for $F/E$ of size $n$ say, and then the Open Mapping Theorem shows that $F \cong E \oplus \mathbb R^n$ where you give the direct sum the $\max$ norm, and $\mathbb R^n$, well, any norm. So pick your extension $\overline T$ by setting it to be $T$ on $E$ and $0$ on $\mathbb R^n$. You only need $\|\overline T\| \leq \lambda \|T\|$ and that's ensured by $F$ being isomorphic to $E \oplus \mathbb R^n$. – Matthew Daws Nov 28 '11 at 15:17
And that last statement would be saying that $C(K)$ is 1-injective in classical terminology, which happens only when $K$ is Stonean by a result of Kelley, Nachbin and someone else IIRC – Yemon Choi Nov 28 '11 at 21:45
(Clarification of my last comment: the 1-injectivity would be equivalent to requiring norm-preserving extensions for every $E$ and $F$, not just those where $F/E$ is finite-dimensional) – Yemon Choi Nov 28 '11 at 21:47
@Yemon, In his Memoirs, Lindenstrauss proved that to get 1-injectivity it is enough to consider the cases where $F/E$ is one dimensional. – Bill Johnson Nov 28 '11 at 22:35

After Phil's answer and the ensuing discussion, the remaining question can be formulated as:

Estimate $\lambda$ s.t. if $F\subset E$ and $E/F$ is finite dimensional, then every norm one linear operator $T$ from $F$ into a $C(K)$ space can be extended to an operator from $E$ into the $C(K)$ space which has norm at most $\lambda$.

I want to point out that $\lambda$ cannot be less than two. I think that it is known that $\lambda$ can be anything larger than two, but I did not find a reference after a quick search. I called Morry Zippin's attention to the problem; maybe he will comment. Or maybe this was proved after he worked on almost complementation; perhaps by Jesus Castillo and colleagues. I more or less thought through that $\lambda$ can be anything larger than three.

Before saying why $\lambda$ cannot be less than two, let's further reformulate the problem. The second dual of every $C(K)$ is $1$-injective, so the problem reduces to the case where $F$ is a $C(K)$ space and $T$ is the identity operator on $F$. That is, we need to estimate how well complemented a $C(K)$ must be in a superspace in which the $C(K)$ space has finite codimension. In fact, the superspace of $C(K)$ can be assumed to be in $C(K)^{**}$, but that is not particularly useful in getting the lower bound on $\lambda$.

Consider the $C(K)$ space $c$, the space of convergent sequences under the sup norm. (I'll treat the real case, but the complex case is only slightly more complicated.) The superspace $E$ is the span in $\ell_\infty$ of $c$ and the sequence $x$ defined by $x(n)= (-1)^n$. Let $P_n$ be the projection onto the span of the first $n$ unit vectors in $\ell_\infty$ and let $P$ be any projection from $E$ onto $c$. Write $Px = a\cdot 1 + y$ with $y$ in $c_0$ and where $1$ is the constant one function. Now for every $n$, the norm of $x-2P_n x$ is one, and $P(x- 2P_n x)=a 1 + y - 2P_n x$. Since $y$ is in $c_0$, letting $n\to \infty$ gives $\|P\| \ge |a|+2$. The choice $a=0$, $y=0$ of course gives a projection of norm two.

EDIT Nov. 29, 2011: Zippin's article in Handbook of the Geometry of Banach Spaces, vol. 2, contains information about almost complementation (differently named) including many open problems.

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