Let $X$ be a Fréchet space and let $Y\subseteq X$ such that $\overline{\operatorname{span}(Y)}=X$. It seems intuitive to me that $\operatorname{int}\big(\overline{\operatorname{co}(Y)}\big)$ is a non-empty open subset of $X$. But how to show this?
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1$\begingroup$ What happens if $X=\ell^2$ and $Y$ consists of the vectors $2^ne_n$, where the $e_n$ are an orthonormal basis for $X$? It seems to me that, if there are counterexamples, they'd look something like this. (Iosif's answer, which appeared as I was typing mine, indicates that I got it backward; $2^{-n}$ instead of $2^n$.) $\endgroup$– Andreas BlassCommented Jun 9, 2020 at 15:17
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The answer is no. E.g., let $X=\ell^2$ and $$Y=\{x\in\ell^2\colon\sum_n nx_n^2\le1\}.$$ Then $Y$ is a closed convex set, spanning $X$, but the interior of $Y$ is empty.
Details: $Y$ is convex because the function $\mathbb R\ni u\mapsto u^2$ is convex. To show that $Y$ is closed one may use the Fatou lemma.
If the interior of $Y$ were not empty, then, by the symmetry of $Y$ (that is, by the property $-Y=Y$), $0$ would be in the interior of $Y$. So, for some real $h>0$ and all natural $n$, we would have $he_n\in Y$, where $(e_1,e_2,\dots)$ is the standard basis of $\ell^2$. But $he_n\notin Y$ for $n>1/h^2$, a contradiction.
answered Jun 9, 2020 at 15:16
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$\begingroup$ what would be a reasonable condition on $Y$ (or $X$ for that matter) such that this doesn't happen? $\endgroup$– ABIMCommented Jun 9, 2020 at 15:21
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1$\begingroup$ @Zorn'sLama : I suspect $X$ would have to be finite dimensional for this to hold for all $Y$, but don't know at the moment whether this is true. $\endgroup$ Commented Jun 9, 2020 at 15:22
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$\begingroup$ If for example, $co(Y)$ is dense in $Ball(0,1)$ in $X$, then everything should work no? $\endgroup$– ABIMCommented Jun 9, 2020 at 15:23
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1$\begingroup$ @Zorn'sLama : Concerning your last comment: yes, of course. $\endgroup$ Commented Jun 9, 2020 at 15:25
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2$\begingroup$ @Zorn'sLama: Your previous comment is just saying something trivial: if $\operatorname{co}(Y)$ is dense in the ball then its closure contains the ball, so of course the interior of the closure is nonempty. $\endgroup$ Commented Jun 9, 2020 at 15:45