Boundedness of subspace of a normed space

Question

I am dealing with the following easy problem, and I am sure when I'll see a solution I'll go "ahhh", but until now I couldn't figure it out:

Let $X$ be a normed space and $S \subseteq X$. Show that if $\{f(x):x \in S\}$ is bounded for every continuous linear functional $f \in X^*$ then the set $S$ is bounded.

So, we have that for every $f: X \rightarrow \mathbb{R}$ it holds that $\|f(S)\| \leq K$ for some $K$. Also, since $f$ is continuous on a normed space then is it bounded, and thus we get $\|f(S)\| \leq \|f\| \|S\| \leq C \|S\|$ for some $C$. From here it seems I sould be close to finish, but I don't see a conclusion. Any hint?

Answer 1

I am pretty sure I figured it out.

For every $s \in S$, we define an operator $$ \begin{matrix} T_s: &X^* &\rightarrow &\mathbb{R} \\ &f &\mapsto &f(s). \end{matrix} $$ We have that $T_s \in X^{**}$ and $$ \|T_s\|=\sup_{\|f\|=1}\|T_s(f)\|=\sup_{\|f\|=1}\|f(s)\| \leq K_s. $$ Therefore the family $\{T_s\}_{s\in S}$ is pointwise bounded, and thus bu the Uniform Boundedness Principle there exists a uniform bound $K>0$. Also, from the natural embedding $X \to X^{**}$ we have that $\|T_s\|=\|s\|_X$, and thus we can conclude that $$ \sup\{\|x\|:x \in S \} \leq K. $$