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Let $f:\mathbb{R}\to\mathbb{R}$ a convex decreasing function. Let $x_0 < x_1 < x_2$. Studying the behaviour of the difference quotient, it is clear that $$f(x_0)-f(x_2) \leq M (f(x_0)-f(x_1))$$ with $M=\frac{x_2-x_0}{x_1-x_0}>0$ .

Now take $F:\mathbb{R}^2\to\mathbb{R}$ convex and decreasing with respect to each variable. Let $x_0 < x_1 < x_2$ and $y_0 < y_1 < y_2$. I ask if a similar condition holds, say for example $$F(x_0,y_0)-F(x_2,y_2) \leq M (F(x_0,y_0)-F(x_1,y_1))$$ with $M=\max ( \frac{x_2-x_0}{x_1-x_0}, \frac{y_2-y_0}{y_1-y_0} )$ .

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1 Answer 1

If $f(x)$ is a function with many variables that is twice continuously differentiable, then its convexity is equivalent to the Hessian being positive semi-definite, $$x'H(f)(x)x \geq 0 \quad \forall \ x \in \textrm{dom}(f)$$.

In a two-variable case, the sign of the diagonals elements of the Hessian determines the convexity of $f$ in either $x_1$ or $x_2$ (holding the other variable constant); however, the cross-partial derivative is what determines whether $f$ is a convex function in its domain if the second derivatives with respect to $x_1$ and $x_2$ are both negative.

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