Does the closed convex hull of a compact set in the interior of a convex cone is still contained in the interior of the cone?

Question

Let $C$ be a convex cone in a Banach space $X$ with nonempty interior. The set $A\subset {\rm Int}C$ is a compact subset, where ${\rm Int}C$ means the interior of $C$. Denote the closed convex hull of $A$ as $\overline{{\rm Co}(A)}$. Do we have $\overline{{\rm Co}(A)}\subset {\rm Int}C$?

Answer 1

Consider the function $$f(x)=\inf_{x\notin C}\|x-y\|.$$ This function is continuous. Since $A\subset C$ is compact and $C$ is open we get $$f(a)\ge \delta >0,\ \ a\in A.$$ Therefore for any $a\in A$ the closed ball $B(a,\delta)=\{x\in X\,:\, \|x-a\|\le \delta\}$ is contained in $C.$ Let ${\rm conv}(A)$ denote the convex hull of $A.$ This set consists of elements of the form $$y=\sum_{k=1}^n \lambda_k a_k,\qquad n\in \mathbb{N},\ \lambda_k\ge 0, \ a_k\in A,\ \sum_{k=1}^n \lambda_k=1.\qquad (*)$$ We claim that $f(y)\ge \delta$ for $y\in {\rm conv}(A).$ Fix $y\in {\rm conv }A$ of the form $(*).$ Then \begin{multline*} C\supset \lambda_1B(a_1,\delta)+\ldots + \lambda_nB(a_n,\delta) = [\lambda_1 a_1+\lambda_1\delta\, B(0,1)]+\ldots + [\lambda_n a_n+\lambda_n\delta\, B(0,1)] \\ =y +\delta[\lambda_1B(0,1)+\ldots +\lambda_nB(0,1)]=y+\delta B(0,1)=B(y,\delta). \end{multline*} On the way we have used a simple fact that $aB(0,1)+bB(0,1)=(a+b)B(0,1)$ for $a,b\ge 0.$

Thus $f(y)\ge \delta$ for $y\in {\rm conv}(A).$ Since the function $f$ is continuous the inequality $f(y)\ge \delta$ holds for $y\in \overline{{\rm conv}(A)}.$ Hence $\overline{{\rm conv}(A)}\subset C.$