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Some related facts I already know:

1) In a Banach space $X$, weakly bounded sets are strongly bounded and vice-versa (Thm 3.18 - "Functional Analysis", Rudin);

2) From 1, it follows that my question is equivalent to proving that weak$^*$ bounded sets are bounded with respect to the weak topology of the dual $X'$.

3) If X is reflexive, then 2 is easy to show and then my question is true. But reflexiveness is really necessary?

Yiorgos S. Smyrlis
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FMC
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  • Perhaps this would be also inteeresting: http://math.stackexchange.com/questions/99987/what-facts-about-the-weak-topology-fail-in-spaces-that-arent-banach – Poppy Jan 30 '14 at 11:04

1 Answers1

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Reflexivity is not necessary. Since $X$ is a Banach space, the Banach-Steinhaus theorem asserts that a family of continuous linear functionals on $X$ that is pointwise bounded (weak* bounded) is equicontinuous (norm-bounded).

Daniel Fischer
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  • Hmm..I'm not sure that weak* bounded implies that the family of linear functionals is pointwise bounded. I can only control their bound on a finite number of points of X, not all of them! – FMC Jan 30 '14 at 11:10
  • The definition of pointwise bounded is "for each $x\in X$, there is a $K_x$ such that $\lvert \lambda(x)\rvert \leqslant K_x$ for all $\lambda\in M$". You only consider finitely many points at once [namely one single point at a time]. – Daniel Fischer Jan 30 '14 at 11:18
  • Sure! Now I see! Thanks!! :) – FMC Jan 30 '14 at 11:33