It is quite well known that for any polyhedral set $K$ and any linear mapping $A$, the set $AK$ is polyhedral and hence closed. I am more curious about the converse problem, namely:


Suppose $K$ is a closed convex cone and for any linear mapping $A$, the set $AK$ is closed. Is $K$ a polyhedral cone?


For closed convex set, counterexample is easy to construct, e.g., $\{(x,y):y\geq x^2\}$. But for closed convex cone, I didn't find any easy counterexample.



A converse question about the polyhedrality under linear mapping
I think we can argue as in <https://mathoverflow.net/a/423284/32507> to answer the question in the affirmative.


Let $\mathcal R\_K(x)$ be the radial cone of $K$ at $x$ (as defined in the other answer). Further, for fixed $x \in K$, let $A$ be the projection onto the orthogonal complement of $\operatorname{span}(x)$. By assumption, $A K$ is closed. Due to
$$
\mathcal R\_K(x) = K + \operatorname{span}(x) = A K + \operatorname{span}(x)$$
(and since $A K$ and $\operatorname{span}(x)$ are orthogonal), the radial cone $\mathcal R\_K(x)$ is closed for each $x \in K$.


Hence, by Proposition 2 of [Duality of linear conic problems by Shapiro and Nemirovski](http://www.optimization-online.org/DB_FILE/2003/12/793.pdf)
we get that $K$ is a polyhedral cone.


