Directional Distance 
Let $f:\mathbb R^n\to\mathbb R$ be a strictly increasing function in all
arguments. $A=\{x:f(x)<0\}$ is a non-empty open set.
$D_{\mathbf 1}(x,A)=\inf\{d:x−(d,...,d)∈A\}$.
Question: how to prove that $D_1(x,A)$ is also strictly increasing in all arguments whenever $D_1(x,A)>0$?

The "direction distance" or "vector distance" in  the direction of vector $v$ is generally defined as follows:
$D_{v}(x,A)=\inf\{d:x−dv∈A\}$
I tried Google Scholar search but all the papers about "directional distance" are economics papers. I wonder if there are any math papers or books on this topic; any help is appreciated.
 A: Here is a solution in the special case that $f$ is linear:
In this case, $f$ has the form $$ f(x) = a_1 x_1 + a_2 x_2 + \cdots + a_n x_n $$
The assumption that $f$ is increasing in all arguments means that $a_i > 0$ for all $i$ (just note that $\frac{\partial f}{\partial x_i} = a_i$).
Let $P = \{x ~|~ f(x) = 0\}$, which is an ($n-1$)-dimensional plane. Then your set $A$ is simply the space underneath $P$. Also $D_1(x,A)$ is the same as the value of $d$ for which $x - (d,d,\dots,d)$ intersects $P$. In other words, it is the value of $d$ such that $f(x_1-d, x_2-d, \dots, x_n-d) = 0$.
You can see (from the expression for $f$) that $f(x_1-d, \dots, x_n-d) = f(x) - d(a_1+a_2+\cdots+a_n)$, and so the condition that this equals zero means that $f(x) = d(a_1+a_2+\cdots+a_n)$. Now you can just solve for $d$ to get
$$ D_1(x,A) = d = \frac{f(x)}{a_1+a_2+\cdots+a_n} = \frac{a_1 x_1 + a_2 x_2 + \cdots + a_nx_n}{a_1 + a_2 + \cdots + a_n} $$
As noted above, all $a_i$'s are positive. Now take partial derivatives to see that $D_1(x,A)$ is increasing in all arguments.
