Is nonemptiness and boundedness of the subdifferential $f$ imply convexity $f$?

Question

Is inverse of following theorem correct? that's mean, if $\partial f(x)$ is nonempty and bounded, then $f$ is convex?

Let $f:E \longrightarrow (-\infty , \infty]$ be a proper function, and assume that $ x\in int(dom(f))$. Then $\partial f(x)$ is nonempty and bounded.

Answer 1

No. Take $$ f(x) = \begin{cases} 0 & \text{ if $x$ is integer}\\ +\infty & \text{ otherwise.} \end{cases} $$ Then $\partial f(x) = \{0\}$ for all $x\in dom(f)$.