Let $S\subset X$ be a subset of a normed linear space such that $\sup_{x\in S} |f(x)|<\infty$ for all $f\in X^*$, the continuous dual of $X$. Prove that the set $S$ is bounded.
By definition $S$ is bounded if $d(S)<M$ for some $0<M\in\mathbb{R}$, where $d(S)$ denotes the diameter. I already proved that this is equivalent to $\sup_{x\in S}\|x\|<\infty$ in the previuos exercise, so probably I should use that here, but I don't know how.
I also thought about using Hahn-Banach: I know that for every $a\in S$ there is $f_a\in X^*$ such that $f_a(a)=\|a\|$ and $\|f_a\|=1$, but I don't get to use the second condition. What I can show is
$$\|a\|=f_a(a)\leq\sup_{x\in S}|f_a(x)|<\infty.$$
But know I don't know what happen if I take the supremum on both sides. How do I see that the right side stays bounded?
For all $x\in S$ define the evaluation functional $J_x\in X^{**}$ given by $$J_x(f)=f(x),\qquad \forall f\in X^* $$ We have $\|J_x\|_{X^{**}}=\|x\|$, so that $S$ is bounded iff $$\sup_{x\in S}\|J_x\|_{X^{**}}=\sup_{x\in S}\|x\|<\infty $$ Here the $\sup$ ranges over the operator norms of a family of functionals $\left\{J_x\right\}_{x\in S}$.
Can you remember a general result in functional analysis which allows you to deduce (under certain assumptions) that a family of functionals (or operators) is bounded in the operator norm?