Strongly bounded set is weakly bounded I have the following two definitions:
Let $S$ be a subset of a Banach space $X$.

*

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


*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.
 A: 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?
A: 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.

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.
