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?


1 Answer 1


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?

  • $\begingroup$ Thanks! You are probably talking about the Uniform Boundedness Theorem? Yes, I can see how to apply it here to show that $\sup_{x\in S}\|J_x\|<\infty$. $\endgroup$
    – Algebra
    Aug 24, 2018 at 21:57
  • $\begingroup$ But I am not sure about $\|J_x\|=\|x\|$: I see $\|J_x\|=\inf\{c\geq 0\mid \|J_x(f)\|\leq c\|f\| \ \forall f\in X^*\}=\inf\{c\geq 0 \mid \|f(x)\|\leq c\|f\| \ \forall f\in X^*\}$. And I also know that $\|f(x)\|\leq \|f\| \|x\|$, so $\|J_x\|\leq \|x\|$. How do I get equality? $\endgroup$
    – Algebra
    Aug 24, 2018 at 22:01
  • $\begingroup$ @mathstackuser I think you can consider a function $f$ satisfying $f(cx/\|x\|) = c$. $\endgroup$
    – angryavian
    Aug 24, 2018 at 23:01
  • $\begingroup$ Indeed, by the Hahn Banach theorem, for all $x$ there is $f\in X^*$ such that $\|f\|=1$ and $J_x(f)=f(x)=\|x\|$. $\endgroup$
    – Lorenzo Q
    Aug 25, 2018 at 13:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .