Grf $\not=$ Cl(epigraph of f) $\cap$ Cl(subgraph of f) can be concave/convex?

Question

For $f\colon\mathbb{R}^p\to\mathbb{R}$, define $\operatorname{Gr}f=\{(x,y) \in\mathbb{R}^p\times\mathbb{R}: f(x)=y\}$. Can a function $f\colon\mathbb{R}^p\to\mathbb{R}$ such that $\operatorname{Gr}f\neq\operatorname{Cl}(\text{epigraph of }f) \cap\operatorname{Cl}(\text{subgraph of }f)$ be concave/convex? (Cl: closure)

Answer 1

No.

Note that $\operatorname{Cl}(\operatorname{epigraph}f))=\operatorname{epigraph}f$ when $f$ is continuous, since it is the inverse image of the nonnegative reals under the continuous map $(x,y)\mapsto y-f(x)$. Similarly for subgraph. But convex/concave functions are continuous.