Does there exist a such function from a subset of $\mathbb{R}^2$ to $\mathbb{R}$ satisfying these conditions? Let $D$ be the set $\{ (x, y) \in \mathbb{R}^2 : x \geq y \geq 0 \}$.

Does there exist a function $f : D \to \mathbb{R}$ satisfying all of the following conditions?

For all $x, x_1, x_2, y, y_1, y_2 \in \mathbb{R}$,

*

*Suppose $x_1 - y_1 = x_2 - y_2$. If $x_1 < x_2$, then $f(x_1, y_1) > f(x_2, y_2)$.

*Suppose $\frac{x_1}{y_1} = \frac{x_2}{y_2}$. If $x_1 < x_2$ then $f(x_1, y_1) < f(x_2, y_2)$.

*If $y_1 < y_2$, then $f(x, y_1) > f(x, y_2)$.

*$f(x, x) = 0$.

*$f$ is continuous.

*$f(x, y) \geq 0$.

And finally, a somewhat optional item (which I still would like):


*Fix $y$. Then $\lim_{x \to \infty} f(x, y) = \infty$.


I've been trying to come up with such a function but haven't yet been able to do so. I want to use the function $f$ as a quantitative description of how close $x$ is to $y$ for some particular use I have in mind.
EDIT:
Let me thank user psl2Z for answering the original question in a comment. I am still interested in this slight modification of the above:
For 1, I should have stated that $x_1 - y_1 = x_2 - y_2 > 0$.
 A: After some more thought, I am trying out the following function:
$f(x, y) = x \cdot \log(x/y)$
This satisfies conditions 2 through 7, and I ran some expirements in Python for 1, for which it passed all the inputs I gave it.

EDIT:
The answer to the modified question is yes. For the function $f$ above, also define $f(0, 0)$ to be $0$. This function is continuous at $0$ because $x \geq y$ and the leading term of $x$ make $\lim_{(x,y) \to (0,0)} f(x,y) = 0$.
As @psl2Z suggested in the comment below, the meaning of my modified 1 is the following:

For all $a > 0$, the function $g$, defined by $g(t) = f(a+t, t)$, is strictly decreasing for $t \geq 0$.

We have (for $t \neq 0$) that $g(t) = (t + a)\log(1+at^{-1})$, and so
$$
\begin{align}
g'(t) &= (t + a) \frac{-at^{-2}}{1+at^{-1}} + \log(1 + at^{-1}) \\
&= \frac{-a}{t} + \log(1 + a/t).
\end{align}$$
But $a > 0$ and $t > 0$ imply that $a/t > 0$.
Notice that $g'(t)$ is negative because for all $x > 0$ we have
$$\log(1 + x) < x.$$
Therefore $g(t)$ is decreasing.
