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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?

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  • $\begingroup$ Do you want to have this for one $a$ or for all $a$? $\endgroup$ – gerw Mar 14 at 19:35
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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 $\|y_j\|<R$ 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.

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