It is known that a non-concave smooth function must be locally convex somewhere.
Let me drop smoothness.
Let $f:\mathbb R^2\to\mathbb R$ be non-concave.
We say $f$ is locally pseudo-convex if there exists $x,y\in\mathbb R^2$ such that $f(x)=f(y)$ and $f(x)\geq f(\overline{xy})$. $f(\overline{xy})$ is the image of the segment $\overline{xy}$.
Can we say that $f$ must be pseudo-convex somewhere?
The only counter-example that I can think of must be generated by Weierstrass function or similar ones. However, a Weierstrass function seems to be pseudo-convex looking at the graph.
