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.$\implies (\nabla f(x) - \nabla f(x^\prime)^T(x-x^\prime) \geq c\| x-x^\prime\|, c>0, \forall x, x^\prime$
What about?
$f$ ??? 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.