Convexity (for a differentiable function $f$) is equivalent to the condition that the following holds $$ \langle \nabla f(x) - \nabla f(y),\, x-y \rangle \geq 0. $$ If $f$ is non-differentiable, is convexity equivalent to the same condition replaced with subgradients? It's clear that if $f$ is convex, then the condition holds by summing up subgradient inequalities, but for the other direction, does the subgradient $\partial f$ have to be a cyclically monotone map (see here), or is $$ \langle \partial f(x) - \partial f(y),\, x-y \rangle \geq 0 $$ enough?
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How do you define 'subgradient' for non-convex functions? Many non-convex functions do not have any subgradients, e.g. $x \mapsto -x^2$ on $\mathbb{R}$. – gerw Nov 09 '20 at 13:57
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1Only convex functions can have subgradients at arbitrary points. You can plot a simple non-convex function on a piece of paper and see why that is true – iarbel84 Nov 09 '20 at 15:16