# Weak-* continuous functionals with bounded level sets

Let $$X$$ be an infinite-dimensional Banach space, $$X^*$$ its topological dual and $$f:X^*\to\mathbb{R}$$ some weak-* continuous functional (not necessarily linear).

Is it possible for $$f^{-1}(a)$$ to be nonempty and bounded?

Attempt: I thought perhaps the operator norm $$\|\cdot\|_{op}$$ on $$X^*$$ would be an example, but it is actually not even weak-* continuous since weak-* open sets in infinite dimensions are not bounded. Linear functionals clearly don't work either.

Follow-up: If the weak-* continuous functionals cannot have nonempty bounded level sets, then a natural follow-up question would be:

Is it possible for $$f^{-1}(a)$$ to have a connected component which is unbounded?

• Do you want to have this for one $a$ or for all $a$? – gerw Mar 14 at 19:35

Yes, there is such a function. Let $$X=\ell^2$$ with its standard norm, and identify $$X^*$$ with $$X$$ via the mapping $$y\mapsto \langle -,y\rangle$$. Denote the standard orthonormal basis by $$e_n$$. Define $$f:X^*\to\mathbb C$$ by $$f(y)=\sum_{n\in\mathbb N} |\langle 2^{-n}e_n,y\rangle|.$$
First, that this sum is convergent for any fixed $$y$$: $$|\langle e_n,y\rangle|\le\|y\|$$, so the sum of nonnegative terms is bounded above by $$\sum_{n\in\mathbb N} 2^{-n}\|y\|<\infty$$.
If $$f$$ were weak-star continuous, it would provide an example of such a function. $$f(0)=0$$, so the preimage of $$0$$ is nonempty. Assume that $$f(y)=0$$. Then $$|\langle e_n,y\rangle|=0$$ for each $$e_n$$, so $$\langle e_n,y\rangle=0$$, so $$y$$ is $$0$$ on a (Schauder) basis. So $$y$$ must be the zero functional, implying that $$f^{-1}(0)=\{0\}$$, a bounded set.
Now we need only show the weak-star continuity. By separability of $$\ell^2$$, it is enough to check the convergence of limits. Assume that $$\{y_j\}_{j\in\mathbb N}\xrightarrow{w*} y$$. We need to show that $$f(y_j)\rightarrow f(y)$$. Note that by the Banach-Steinhaus theorem, weak-star convergent sequences must be bounded. Pick $$R$$ such that $$\|y_j\| for each $$j$$.
Now let $$\epsilon>0$$. Pick $$N$$ large so that $$\sum_{n>N} \frac{1}{2^n}R<\frac{\epsilon}{4}$$. Pick $$K_1,\ldots,K_N$$ large (from the weak-star convergence of $$\{y_j\}$$) so that $$\forall j>K_i,\quad |\langle e_i,(y-y_j)\rangle|<\frac{\epsilon}{4N}$$ and let $$M=\text{max}\{N,K_1,\ldots,K_N\}$$. Then for all $$j>M$$, \begin{align*} |f(y_j)-f(y)|&=\bigg|\sum_{n\in\mathbb N}|\langle 2^{-n}e_n,y_j\rangle|-\sum_{n\in\mathbb N}|\langle 2^{-n}e_n,y\rangle|\bigg|\\ &\le\bigg|\sum_{n\le N}\big[|\langle 2^{-n}e_n,y_j\rangle|-|\langle 2^{-n}e_n,y\rangle|\big]\bigg|+\bigg|\sum_{n>N}\big[|\langle 2^{-n}e_n,y_j\rangle|-|\langle 2^{-n}e_n,y\rangle|\big]\bigg|\\ &\le \bigg|\sum_{n\le N}\frac{\epsilon}{2N}\bigg|+\bigg|\sum_{n>N} 2^{-n}\cdot 2R\bigg|\\ &< \frac{\epsilon}{2}+\frac{\epsilon}{2}\\ &<\epsilon \end{align*} so the function is weak-star continuous.