Strongly bounded set is weakly bounded

Question

I have the following two definitions:

Let $S$ be a subset of a Banach space $X$.

1. We say that $S$ is weakly bounded if $l\in X^*$, the dual space of $X$, then $\sup\{|l(s)|:s\in S\}<\infty$.

2. We say that $S$ is strongly bounded if $\sup\{||s||:s\in S\}<\infty$.

I need to prove those definitions are equivalent. I have already proved that 2 implies 1, but I'm stuck with the other direction. I suspected that I have to use Hahn-Banach's theorem, or maybe some of their corollaries, but I don't know how to proced.

Thanks to all of you.

Answer 1

Assume that $1$ holds. For $s\in S$ define $\phi_s:X^*\to\mathbb{C}$ by $\phi_s(f)=f(s)$. Then $\phi_s$ is a bounded linear functional, since $|\phi_s(f)|=|f(s)|\leq\|s\|\cdot\|f\|$. Now since $$\sup_{s\in S}|\phi_s(f)|<\infty$$ for all $f\in X^*$ by our assumption, we can apply the principle of uniform boundedness and conclude that $\sup_{s\in S}\|\phi_s\|<\infty$.

If you show that $\|\phi_s\|=\|s\|$, you are done (note that it is immediate that $\|\phi_s\|\leq\|s\|$). The other inequality follows from Hahn-Banach. Can you do that?

Answer 2

The fact that X is Banach it's not necessary. Suppose that X is a normed space. Recall that the map

$j: X\to X^{**}$, $s\mapsto \hat{s}$

is an isometry.

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Per your definition of weakly boundedness the set of operators $\{ \hat{s}\}_{s\in S}\subseteq X^{**}$ is punctually bounded.
Apply now the Uniform boundedness principle and get that $\Vert s\Vert\leq K$ for all $s\in S$.
Apply now the fact that $j$ is an isometry and conclude that $S$ is bounded in the norm topology of $X$.
Hope it helps.