Let $X$ be a normed vector space over $\mathbb R$, not necessarily Banach. Let $X'$ denote the dual of $X$, that is, the set of all bounded, linear functionals on $X$: $$X'\equiv\{f:X\to\mathbb R\,|\,\text{$f$ is linear and bounded}\}.$$ Suppose that $A'\subseteq X'$ is a topologically bounded set in the weak* topology on $X'$. This means that for any $x\in X$, there exists some $M_x>0$ such that $|f(x)|\leq M_x$ for all $f\in A'$.

Question: Is $A'$ necessarily bounded in the norm? That is, does there exist some $M>0$ such that $\|f\|\leq M$ for any $f\in A'$ (where $\|\cdot\|$ is the operator norm on $X'$)?

The answer is affirmative if $X$ is a Banach space, as can be shown easily using the uniform boundedness principle. Is this also true is $X$ is not a Banach space? If not, do there exist some easy counterexamples?


It's not true for an arbitrary normed space:

Take $X$ to be the space of sequences with finite support with the $\ell_1$-norm.

For each $m$, define the continuous linear functional, $x_m^*$, on $X$ via $$x^*_m\bigl((a_n)\bigr)=m a_m.$$ Then $\{ x_m^* :m\in\Bbb N\}$ is weak*-bounded (since any $x\in X$ has finite support) but not norm-bounded (since for each $m$, $x_m^*(e_m)=m$, where $e_m$ is the standard $m$'th unit vector).

  • $\begingroup$ In the meanwhile, I found this counterexample, too. Thank you. $\endgroup$ – triple_sec Jan 10 '15 at 17:42
  • 1
    $\begingroup$ Moreover, weak*-compact subsets of $X'$ need not be norm bounded, either (unless $X$ is a Banach space). In your counterexample, it is easy to see that $x_m^*$ converges to the zero functional in the weak*-topology. This implies that $\{x_m^*\,|\,m\in\mathbb N\}\cup\{0\}$ is a weak*-compact subset of $X'$, but this set is a fortiori not norm-bounded. $\endgroup$ – triple_sec Jan 10 '15 at 18:07
  • $\begingroup$ Another addendum: On the other hand, if $A'$ is a norm-bounded and weak*-closed subset of $X'$, then it follows from Alaoglu's theorem that $A'$ is weak*-compact, even if $X$ is not a Banach space. $\endgroup$ – triple_sec Jan 10 '15 at 18:13

Take as $X:= C^{\infty}_0(\mathbb{R})$ the space of smooth compactly supported functions, and endow it with the supremum norm. Then it is not banach, because you could have cauchy sequences of compactly supported functions that converges to a non compactly supported function. take f a smooth compactly supported function whose support is $[0;1]$ , then f is uniformly continuous, so there is $\delta_k<\frac{1}{k}$ such that $|f(x)-f(y)|< \frac{1}{k} |x-y|$ whenever $|x-y|<\delta_k$. Define now $f_n(x)=f(\frac{1}{n}x)$ then the support of $f_n$ is going to contain $[0;n]$, and for any $n_1,n_2$ sufficiently large so that $|\frac{1}{n_1}-\frac{1}{n_2}|< \frac{1}{k}$ you will have: $\forall x |f_{n_1}(x)-f_{n_2}(x)|=|f(\frac{1}{n_1}x)-f(\frac{1}{n_2}x)|<\frac{1}{k^2}$ therefore it is a cauchy sequence but, it doesn't converge in $C^{\infty}_0$.

Now take in the dual the subset of functionals that make the evaluation at the point $n$ and multiply by $n$ namely, $F_n(f):= nf(n)$. Then for any $f$ the set $\{ F_n(f), n=1,...\}$ is bounded because $f$ is compactly supported. But on the other hand $||F_n||=n$ as you can verify by evaluating $F_n$ on a function that achieves its maximum at point $n$.


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