Can a smooth convex functions be non-differentiable?

Question

Consider the definition of the $\beta$-smoothness (for some constant $\beta$): $$ \|\left. \nabla f \right|_{ y } - \left. \nabla f \right|_{ x } \| \leq \beta \| x - y \| $$ And convexity: $$ f(x) \geq f(y)+ \left. \nabla f \right|_{ y } . (x - y), \forall x, y $$ Can a smooth convex function be non-differentiable at some points on its domain? (and why?)

In the definitions, $\left. \nabla f \right|_{y}$ is subgradient of the function $f$ at point $y$, if it is not differentiable at this point; so the definitions of convexity and smoothness hold even for non-differentiable functions.

Answer 1

A convex function $f$ is non-differentiable at a point $x$ iff the subgradient $\nabla f|_x$ has more than one vector. I presume the inequality $\|\nabla f|_x - \nabla f|_y\| \le \beta \|x - y \|$ means $\|v - w\| \le \beta \|x - y\|$ for all $v \in \nabla f|_x$ and $w \in \nabla f|_y$. In particular, for $y = x$ you see that $\nabla f|_x$ must contain only one vector for this to be true.

Answer 2

I presume that the definition of smooth-ness you are using is something specific to convex functions. Its not the standard definition of a function that a function is continuously differentiable and has derivatives of all orders C∞;

https://en.wikipedia.org/wiki/Smoothness although I note that you are using a particular definition here.

I note that discrete analogues involving subgradients and the like are also used in characterizing (not necessarily diff-erentiable, in the traditional sense) 'pseudo convex functions' using more discrete notion such as the "dini 'derivative" see https://en.wikipedia.org/wiki/Dini_derivative