Let $f:\mathbb R^n\times\mathbb R^n\rightarrow \mathbb R$ be a twice continuously differentiable function.
The cross-derivative of $f$ is denoted by $D_{xy}f$. I want to check whether the following condition is equivalent to the positive definiteness of $D_{xy}f$: $$f(x',y')-f(x',y)>f(x,y')-f(x,y),\forall x, x', y, y'.$$ i.e., $D_{xy}>0$ if and only if $f(x',y')-f(x',y)>f(x,y')-f(x,y),\forall x, x', y, y'.$
To show this, I tried to connect the cross derivative and the original functional form. I referred to this and this and tried to use the monotonicity.
Using the monotonicity of $D_xf(x,y)$ (from the positive definiteness of the derivative with respect to the second term $y$), we have $$(z'-z)^T(D_xf(x,z')-D_xf(x,z))>0~ \forall x,z,~\textrm{and}~z'\tag{1}.$$
If I could take the same step one more time, it looks like I could get something like $$(z'-z)^T(f(z',z')-f(z',z)-(f(z,z')-f(z,z)))(z'-z)>0~ \forall z,~\textrm{and}~z',\tag{2}$$ which means $f(z',z')-f(z',z)>f(z,z')-f(z,z).$ However, as $D_xf(x',z')-D_xf(x',z)$ is not a matrix, I cannot use the monotonicity which is from the positive definiteness.
How can I show the equivalence of the two conditions? or how can I show at least one direction of the equivalence?
The original formulation of the question seems wrong. Here is a counter-example. Let $n=1$, and consider $$f:{\mathbb R}\times {\mathbb R}\to {\mathbb R}; \ (x, y)\to xy.$$ Then $D_{xy}f$ is the matrix $(1)$, which is positive-definite. Now consider $x'=-1, x=0, y'=1, y=0$. We have $f(x',y')=-1$, and the other terms are zero (since $x=0$ and $y=0$), so
$$ f(x',y')-f(x',y)>f(x,y')-f(x,y) $$ does not hold.
Of course, the problem is that even for a positive-definite matrix $A$, we only have $x^TAx\geq 0$, not $y^TAx\geq 0$ for all $x, y$.
I do also want to point out that $D_{x,y}f$ is not necessarily symmetric, so the definiteness defined by $$ w^T (D_{x,y}f) w\geq 0,\quad w\in {\mathbb R}^n $$ is actually in terms of the symmetrization of $D_{x,y}f$. Just consider the example of $f:{\mathbb R}^2\times {\mathbb R}^2\to {\mathbb R}; ((x_1,x_2),(y_1,y_2))\to x_1y_2 + 2x_1y_2$, where $D_{xy}f$ is represented by the matrix $\begin{pmatrix} 0 & 2\\1 & 0\end{pmatrix}$.
With these in mind, the following seems the only meaningful version, to me.
If $D_{xy}f$ is positive definite then $f(x+z,y+\lambda z)-f(x+z,y)>f(x,y+\lambda z)-f(x,y),\ \forall z\neq 0, \lambda>0$.
This can be proved by the mean value theorem. We have \begin{align} &\quad\ (f(x+z,y+\lambda z)-f(x+z,y))-(f(x,y+\lambda z)-f(x,y))\\ &=D_x|_c(f(\cdot, y+\lambda z)-f(\cdot,y)) z\\ &=(D_xf(c,y+\lambda z)-D_xf(c,y))z\\ &=\lambda z^T \big(D_{xy}f(c,d)\big) z>0, \end{align} where $c$ and $d$ lie on the line segments connecting $x$ to $x+z$, and $y$ to $y+\lambda z$ respectively.
In terms of positive semi-definiteness, we actually have the following iff statement.
$D_{xy}f$ is positive semi-definite iff $f(x+z,y+\lambda z)-f(x+z,y)\geq f(x,y+\lambda z)-f(x,y),\ \forall z, \lambda\geq 0.$
For the $\Rightarrow$ direction, just repeat the above argument with semi-definiteness and $\geq$.
For the $\Leftarrow$ direction, let $\lambda = 1$, and let $z=tw\in {\mathbb R}^n$ and $w\in {\mathbb R}^n$. The above mean value identity becomes $$t^2w^T\big(D_{xy}f(c, d)\big)w=(f(x+tw,y+tw)-f(x+tw,y))-(f(x,y+tw)-f(x,y))\geq 0. $$ Dividing by $t^2$, and taking limit as $t\to 0$, we get $$ w^T \big(D_{xy}f(x, y)\big) w\geq 0. $$