I have a subset $K\subset X^\ast$ of the dual of a Banach space $X$. (In fact $X$ is $C^1(M)$ for some smooth compact manifold $M$.) I hope that there exists $x\in X$ such that every $k\in K$ satisfies $k(x)>0$. I know that the convex hull of $K$ does not contain the origin, and I know that $K$ is compact. Is that enough?
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asked Oct 24, 2019 at 0:44
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$\begingroup$ Now that I see the subtlety in the question ($K$ is compact but we don't know its convex hull is compact) I'm going to delete my answer. I suspect the answer is no, but I have to think. $\endgroup$– Nik WeaverCommented Oct 24, 2019 at 1:07
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$\begingroup$ "Convex hull does not contain the origin". Actually, you need more: the weak* closure of the convex hull does not contain the origin. "$K$ is compact" Actually you need less: $K$ is bounded, consequently, the weak* closure of the convex hull of $K$ is weak* compact. $\endgroup$– Gerald EdgarCommented Oct 24, 2019 at 1:07
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$\begingroup$ I know that for every finite set $S\subset K$ there exists $x$ such that for all $k\in S$ $k(x)>0$. $\endgroup$– Tom GoodwillieCommented Oct 24, 2019 at 2:17
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$\begingroup$ The answer is certainly yes if $K$ consists of positive functionals $\varphi$ only (where positive means that $\langle \varphi, f \rangle \ge 0$ for each $f\ge 0$). For more general $K$ I have severe doubts whether this is true, but I also couldn't find a counterexample yet. $\endgroup$– Jochen GlueckCommented Oct 24, 2019 at 5:52
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Here is a counterexample:
Let $M$ be the one-dimensional unit circle, so we can identify $C^1(M)$ with \begin{align*} X = \{f \in C^1([0,1]): \, f(0) = f(1), \; f'(0) = f'(1)\}. \end{align*}
For each $z \in [0,1)$ let $d_z \in X^*$ be given by $\langle d_z,f\rangle = f'(z)$ for each $f \in X$. Then the set \begin{align*} K := \{d_z: \, z \in [0,1)\} \end{align*} is weak${}^*$-compact (since $M$ is compact), but $K$ cannot be separated from $0$ by an element of $X$. Indeed, let $f \in X$. If $f$ is constant, then every functional in $K$ vanishes on $f$. If $f$ is not constant then, due to the periodic nature of $f$, there exist $z_1,z_2 \in [0,1)$ such that $f'(z_1) < 0$ and $f'(z_2) > 0$. Hence, a convex combination of $d_{z_1}$ and $d_{z_2}$ vanishes on $f$.
Remark (Edited after a comment by Tom Goodwillie). The counterexample above uses that the manifold $M$ is closed. The comment by Tim Goodwillie below explains how this example can be adjusted to obtain a counterexample on $C^1([0,1])$.
answered Oct 24, 2019 at 6:43
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$\begingroup$ Thank you! And your Remark is very interesting. $\endgroup$ Commented Oct 24, 2019 at 14:43
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1$\begingroup$ Oh, here is an example for the case of [0,1]. For 0≤𝑥≤1 let 𝑑𝑥∈𝑋∗ be defined by <𝑑𝑥,𝑓>=𝑓′(𝑥). Also define 𝑒∈𝑋∗ by <𝑒,𝑓>=𝑓(0)−𝑓(1). There is no 𝑓 on which all of these are strictly positive, but if you delete some of them then there is such an 𝑓. $\endgroup$ Commented Oct 24, 2019 at 23:47
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$\begingroup$ @TomGoodwillie: Thank you very much for your comment! I changed the remark at the end of my post accordingly and referred to your comment for a counterexample on $C^1([0,1])$. $\endgroup$ Commented Oct 25, 2019 at 8:05