# Can a smooth convex functions be non-differentiable?

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.

• This definition seems to implicitly assume that $f$ is differentiable, otherwise what does $\nabla f$ mean? – littleO Sep 6 '15 at 7:08
• Ah I see. For this definition, let's assume this definition holds for any sub-gradient. – Daniel Sep 6 '15 at 14:53
• That's an odd definition of smoothness... According to it $f(x) = x^2$ is not smooth. Where does this definition come from? – Rahul Sep 6 '15 at 15:00
• Updated it with parameter $\beta$. This is mostly used in convex analysis. For example see page 18 of ocobook.cs.princeton.edu/OCObook.pdf (mid-page equation) – Daniel Sep 6 '15 at 15:13
• Ill-stated question. Nothing to do. – dohmatob Sep 6 '15 at 21:28

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.