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?



A hyperplane separation question
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 $C^1([0,1])$.


