Monotonicity of a multivariate function

Question

Let $f:\mathbb R^n\times \mathbb R^n\rightarrow \mathbb R$. Assume $f(x,y)$ is twise continuously differentiable and the derivative with respect to both arguments (i.e., $x$ and $y$) are monotone in the sense that the following holds true for any $x,x'\in\mathbb R^n$ such that $x\neq x'$: $$(x'-x)(D_xf(x,x')-D_xf(x,x))>0~\textrm{and}$$ $$(x'-x)(D_yf(x',x)-D_yf(x,x))>0~\textrm.$$ Suppose now we have $$f(x,x')-f(x,x)>0.$$ Can we prove that the following statement is true: $f(x',x')-f(x',x)\geq0?$

Answer 1

Yes, that's true. Let $p=x'-x.$ Define

\begin{align*} I_1&=f(x',x)-f(x,x)=\int_0^1 p\cdot D_xf(x+tp,x) dt\\ I_2&=f(x,x')-f(x,x)=\int_0^1 p\cdot D_yf(x,x+tp) dt\\ I_3&=f(x',x')-f(x,x)=\int_0^1 [p\cdot D_xf(x+tp,x+tp)+p\cdot D_yf(x+tp,x+tp) ]dt \end{align*}

The monotonicity conditions give \begin{align*} (-tp)\cdot (D_xf(x+tp,x) - D_xf(x+tp,x+tp))&>0\\ (-tp)\cdot (D_yf(x,x+tp) - D_yf(x+tp,x+tp))&>0 \end{align*} and hence $I_1+I_2<I_3.$ Rearranging and cancelling $f(x,x)$ terms, $f(x',x')-f(x',x)>f(x,x')-f(x,x),$ which is positive by assumption.