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Let $X$ be a locally convex space, $T$ a balanced convex compact set in $X$, and $f:X\to\mathbb{C}$ a linear functional which is (not necessarily continuous on $X$, but) continuous on $T$. It is not difficult to show that then for each $\varepsilon>0$ there exists a linear continuous functional $f^\varepsilon:X\to\mathbb{C}$ such that $$ \sup_{x\in T}|f(x)-f^\varepsilon(x)|<\varepsilon. $$

Question:

Is the same true if in addition we require that $f$ belongs to the polar $B^\circ$ of a barrel $B$ in $X$, and we want $f^\varepsilon$ to belong to $B^\circ$ as well?

I.e. we take a linear functional $f:X\to\mathbb{C}$ which is continuous on $T$, and bounded by 1 on some barrel $B\subseteq X$, $$ \sup_{x\in B}|f(x)|\le 1, $$ and for a given $\varepsilon>0$ we want to find a linear continuous functional $f^\varepsilon:X\to\mathbb{C}$ such that $$ \sup_{x\in B}|f^\varepsilon(x)|\le 1, $$ and $$ \sup_{x\in T}|f(x)-f^\varepsilon(x)|<\varepsilon. $$ Is it possible?

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  • $\begingroup$ In the "not difficult to show" statement, compactness of $T$ is not needed, is it? You need the theorem of bipolars and Banach-Alaoglu. I don't know the answer to your question, technically it becomes a problem how to estimate $(A+B)\cap C$ (which appears quite often in the locally convex theory). $\endgroup$ Nov 22, 2021 at 8:09
  • $\begingroup$ @Jochen, I expected you to answer. I actually was thinking about totally bounded $T$. I wrote "compact" because I don't know, perhaps when we add the barrel $B$ in our considerations, the compactness of $T$ becomes important. For me compact $T$ will be enough. $\endgroup$ Nov 22, 2021 at 8:52
  • $\begingroup$ I can explain why I need this: if this is true, then this gives a criterion of quasicompleteness for stereotype spaces. $\endgroup$ Nov 22, 2021 at 8:58

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