Assume that f is twice differentiable, that is, its Hessian or second derivative $\nabla^2f$ exists at each point in dom$f$, which is open. Then f is convex if and only if dom$f$ is convex and its Hessian is positive semidefinite: for all x ∈ dom$f$, $\nabla^2f\geq 0$.
Is it possible that a function can be convex but not 2 twice differentiable?
On
Convexity doesn't even imply continuity(e.g one can construct a convex function on a closed domain which is nowhere continuous on the boundary).
However, Alexandrov's Theorem states that a convex function is $\mathcal C^2$ almost-everywhere. Now, that's alot of differentiability (depending on what you want to do with it)!
What about $f(x)=\|x\|$? (Euclidean norm)