Let $f:\mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}$ be a strictly convex function in both its arguments. Suppose $f$ is differentiable in both its arguments. Does the following hold? $$\langle \nabla_x f(x_1,y_1)-\nabla_x f(x_2,y_2),x_1-x_2 \rangle >0, \forall x_1,x_2,y_1,y_2\,,$$ where $\nabla_xf$ denotes the gradient of $f(x,y)$ with respect to $x$? In other words, are the partial gradients of a strictly convex function strictly monotone?

  • $\begingroup$ Note that there is no reason to assume that a strictly convex function is differentiable. $\endgroup$ – Michael Grant May 31 '18 at 23:28
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    $\begingroup$ I added the assumption that it is differentiable in $x$ and $y$. $\endgroup$ – pulosky May 31 '18 at 23:33

There is no reason to expect such an inequality with arbitrary $y_1, y_2$ in it. For example, let $n=1$, $$f(x, y) = (x+y)^2 + y^2 $$ so that $$ f_x(x, y) = 2(x+y) $$ The question becomes: does $x_1 > x_2$ imply $$ 2(x_1+y_1) > 2(x_2 + y_2) $$ for arbitrary $y_1, y_2$? This shows just how absurd the purported inequality is.


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