Does $\frac{ d v(x,y) }{\mathbb d |x-y|}$ denote the change in $v$ with respect to a change in the difference between $x$ and $y$? I have a function $v(x,y)$, where $x,y \in \mathbb{R}$.
I would like to turn the following statement into math: “the function $v(x,y)$ is increasing at a decreasing rate as the difference between $x$ and $y$ is increasing.”
Is there a simple way to express this mathematically, such as
$\frac{ d v(x,y) }{\mathbb d |x-y|} >0$, e.t.c?
 A: Fix $x$ and let $y=x+u$. Then as $u$ increases $v(x,x+u)$ also increases. Assuming that $f_x(u) = v(x, x+u)$ and $g_x(u)=v(x+u,x)$  are twice differentiable, then the assumptions can be written as
$$\frac{dv(x,x+u)}{du}>0, \qquad \frac{d^2 v(x,x+u)}{du^2}<0,$$
for all $x\in \mathbb{R}.$
Similarly,
$$\frac{dv(x+u,x)}{du}>0, \qquad \frac{d^2 v(x+u,x)}{du^2}<0,$$
for all $x\in \mathbb{R}.$
In general without differentiability assumption, one can say
$$v(x,x+u)>v(x,x+r)\quad  \text{if} \quad  u>r,$$ and $$v(x,x+u)-v(x,x+r) < v(x,x+r)-v(x,x+s)$ \quad \text{if} \quad u-r > r-s >0. $$
or
$$v(x,x+u)+v(x,x+s) < 2\,v(x,x+r) \quad \text{if} \quad u+s > 2\,r. $$
Other cases also can be easily obtained.
A: One way of expressing your statement mathematically (assuming $v$ is twice differentiable on all of $\mathbb{R}^2$) is the following:

$\underbrace{\forall x,y\in \mathbb{R},}_\text{for all points $(x,y)$ in $\mathbb{R}^2$} \overbrace{\forall d\in Q(x,y),}^\text{for all directions in $Q(x,y)$}  $
$\left( \underbrace{(\nabla v)^Td>0}_\text{directional derivative of $v(x,y)$ is positive}  \text{and} \overbrace{\nabla((\nabla v)^Td)^Td<0}^\text{2nd directional derivative of $v(x,y)$ is negative}\right)$
where
$\underbrace{Q(x,y)=\{d\in\mathbb{R}^2:D_d(|x-y|)>0\}}_\text{set of directions (away from point $(x,y)$) for which the directional derivative of $|x-y|$ is positive}$.

In plain English, this expression, which holds for all points $(x,y)$ in the plane, assures that for each direction $d\in Q(x,y)$ away from a point $(x,y)$ ($d\in Q(x,y)$ means that $d$ is a direction in which the difference between $x$ and $y$ increases as one moves an infinitesimally small amount away from point $(x,y)$ in direction $d$) the directional derivative of $v$ at point $(x,y)$ and in direction $d$ is positive and the second directional derivative (the directional derivative in direction $d$ of the first directional derivative) is negative.
Because the function $|x-y|$ is not differentiable (i.e., the gradient does not exist) for all points $(x,y)$ satisfying $y=x$, we denoted the directional derivative in direction $d$ of $|x-y|$ at point $(x,y)$ by $D_d(|x-y|)$ instead of a dot product of a gradient and direction.
If some part of this answer is unclear, please comment below and I will make the necessary edits.
