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It is known that, if a function $f$ from a planar domain $D$ to a Banach space $A$ is weakly analytic [i.e. $l(f)$ is analytic for every $l$ in $A^*$], then $f$ is strongly analytic [i.e. $\lim_{h \to 0} h^{-1}[f(z+h)-f(z)]$ exists in norm for every $z$ in $D$].

Now the question is, if above $f$ is assumed to be weakly continuous [i.e.$l(f)$ is continuous for every $l$ in $A^*$], then is it true that $f$ will be strongly continuous.[i.e. $\lim_{h \to 0} [f(z+h)-f(z)] = 0$ in norm for every $z$ in $D$.]

Rudy the Reindeer
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Timon
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2 Answers2

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Regarding the clarified question with finite dimensional domain. Let $X$ be an infinite dimensional separable and reflexive Banach space. Its unit ball $B$ is weakly compact and metrizable. It is also convex.

So by Hahn–Mazurkiewicz theorem there exists a continuous function $f\colon [0,1]\to B$ that is onto $B$. This function is not norm continuous as $B$ is not norm compact.

Rabee Tourky
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  • can you give me some reference for the Hahn-Mazurkiewicz theorem? Does not there exist any function in explicit form? – Timon Sep 20 '12 at 16:06
  • there is definitely no function in explicit form. Basically, one shows that the set $B$ is the image of the cantor set. try finding an explicit form for a path filling curve from $[0,1]$ onto $[0,1]^2$. Now try finding one for the unit ball of $L_2$. I'll add, the answer to your question is that weak continuity does not imply strong continuity even if the domain is finite dimensional; and we can find such an example for every metrizable domain in which we embed $[0,1]$ as a closed set. – Rabee Tourky Sep 20 '12 at 16:09
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Let $X$ be an infinite dimensional Banach space. Let $T_w$ denote the weak topology and let $T$ denote the norm topology. Then $\mathrm{id}: (X, T_w) \to (X, T)$ is not continuous but $\mathrm{id}: (X, T_w) \to (X, T_w)$ is.

Note that every map $f: X \to Y$ that is weakly continuous where weakly means $f: (X, T_w) \to Y$ is also strongly continuous, $f: (X, T) \to Y$, since the norm (or strong) topology contains more open sets than the weak topology (by definition).

Rudy the Reindeer
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  • Well in my actual question domain of the function was a planar domain with usual euclidean norm. If the range is again complex plane then we know for finite dimension Banach space weak topology and strong topology are same. – Timon Sep 20 '12 at 15:05
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    @timon Such important information ought to be included in the body of your question. –  Sep 20 '12 at 15:12
  • Yes, to what Byron said. Thanks, Byron. – Rudy the Reindeer Sep 20 '12 at 15:14
  • @timon Sure, $X$ has to be infinite dimensional. That's quite vital for the argument to be true. Sorry for not writing the most important thing. In any case: can you please make your question a bit clearer? Like: edit in all the information. Thanks! – Rudy the Reindeer Sep 20 '12 at 15:18
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    sorry matt, it was my mistake that i forgot to include those information.Next time i will be careful.This time i am editing now. – Timon Sep 20 '12 at 15:37
  • @timon No problem : ) Could you point me to a definition of "planar domain"? I googled it but could not find a definition. – Rudy the Reindeer Sep 20 '12 at 15:48
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    @matt planar domain means a open connected subset of the complex plane. – Timon Sep 20 '12 at 16:08
  • @timon Nice, thanks! – Rudy the Reindeer Sep 20 '12 at 16:09