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Let E be a normed space, and let $T$:E * $\rightarrow$ E * be a nonlinear operator.

Suppose that :

1) $T$ is continuous from (E *, ||.||) to itself (i.e., it is norm-continuous).

and

2) $T$ is continuous from (E *, w *) to itself (i.e., it is weakly-star continuous).

Then does it follow that

3) $T$ is continuous from (E *, w) to itself (i.e., it is weakly continuous) ?


I guess the case E = ℓ∞ would be particularly interesting.

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Ady, is your interest in $\ell_\infty$ because it is a Grothendieck space whose dual unit ball is not weak* sequentially compact? –  Philip Brooker Jan 23 '10 at 0:51
    
Very true. :-) Indeed. Mainly because it is Grothendieck (and with Dunford-Pettis). I just cannot see a counterexample here. –  Ady Jan 23 '10 at 1:02
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1 Answer

I think Ady's question has a negative answer.

Recall that the Mazur map, $T$, from the unit ball of $\ell_2$ onto the unit ball of $\ell_1$, is defined by $T(\sum a_i e_i)= \sum a_i^2 b_i e_i$, where $b_i$ is the sign of $a_i$. It is a uniform homeomorphism in the norm topologies and obviously is coordinatewise continuous, which means that it is weak ($=$ weak$*$ in $\ell_2$) to weak$^*$ continuous (since on the unit ball of $\ell_p$ with $p$ finite the weak* topology is the topology of coordinatewise convergence). $T$ is not weak to weak continuous since the unit vector basis converges weakly to zero in $\ell_2$ but not in $\ell_1$.

The problem is to extend $T$ to all of $\ell_2$. I think that the easiest way to do this is to show that there is a retraction $R$ from $\ell_2$ onto its unit ball which is both norm to norm continuous and weak to weak continuous. Define $R$ on the complement of the unit ball by

$$ R(\sum a_i e_i) = \sum_{i=1}^n a_i e_i + t e_{n+1}, $$ where $\sum_{i=1}^n a_i^2 \le 1 < \sum_{i=1}^{n+1} a_i^2$ and $t$ is chosen to have the same sign as $a_{n+1}$ and to make the image vector have norm one. $R$ is obviously continuous (even Lipschitz) in the norm topology and is continuous in the topology of coordinatewise convergence, hence is weak to weak continuous (since in all of $\ell_2$, the weak topology is stronger than the topology of coordinatewise convergence).

To get a counterexample that maps one dual space to itself, work with the space $\ell_2 \oplus \ell_1$.

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Thank you very much ! It seems that your construction heavily relies on Schur. Therefore, may I ask: do you think that the answer is negative for all nonreflexive spaces, including the quasi-reflexive ones ? E.g., what about E = J, the James space ? –  Ady Jan 13 '10 at 16:21
    
Any non reflexive space with a basis has a normalized basis that does not converge weakly to zero, but I don't know a replacement for the Mazur map even for J. My guess is that the answer is negative for all non reflexive dual spaces, but my guesses have a probability of only around 1/2 of being correct. –  Bill Johnson Jan 13 '10 at 16:49
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