# What does this monotonicity condition on the gradient correspond to?

We know that all the gradient of convex functions $$f: X \to \mathbb{R}, X$$ convex, are monotone maps, that is,

$$f$$ convex, continuously differentiable $$\implies (\nabla f(x) - \nabla f(x^\prime)^T(x-x^\prime) \geq 0$$

[Boyd Page 115 https://web.stanford.edu/~boyd/cvxbook/bv_cvxbook.pdf]

We can make this statement stronger,

$$f$$ strictly convex, cont. diff. $$\implies (\nabla f(x) - \nabla f(x^\prime)^T(x-x^\prime) > 0, \forall x$$ such that $$x \neq x^\prime$$

$$f$$ strongly convex, cont. diff.$$\implies (\nabla f(x) - \nabla f(x^\prime)^T(x-x^\prime) \geq c\| x-x^\prime\|, c>0, \forall x, x^\prime$$

$$f$$ ??? continuously diff.$$\implies (\nabla f(x) - \nabla f(x^\prime)^T(x-x^\prime) \geq -c\| x-x^\prime\|, c>0, \forall x, x^\prime$$
What condition must $$f$$ satisfy? Is this the case when $$f$$ is strongly concave?
The statement on the right-hand side in the last box (with $$||x-x'||^2$$) is equivalent to $$f + \frac c2 \,\|\cdot\|^2$$ being convex.
• $f \equiv 0$ or $f = c \, \|\cdot\|^2$ or (more generally) any smooth function whose Hessian is bounded from below (i.e., the smallest eigenvalue is bounded from below). – gerw May 11 '19 at 11:04