Function obeying some constraints

Question

Let $X:=(0,\infty)$. I am looking for a function $f:X^2\to X$ such that:

1. $\inf(f(X^2))=0$.
2. $x\leq f(x,y)\forall(x,y)\in X^2$.
3. $y\geq \frac{1}{f(x,y)}\forall(x,y)\in X^2$.
4. $\lim_{\varepsilon\to 0}f(x+\varepsilon,y+\varepsilon)=f(x,y)\forall(x,y)\in X^2$.
5. $f(x,\frac{1}{\max(\{x-y,y\})})\leq x\forall x\geq y\in X.$

Does such a function exist? Or are the constraints somehow mutually exclusive? I cannot figure it out. I've been playing around with various such functions such as $f(x,y)=x/y$ or $f(x,y)=x\coth(1/y)$ but each such examples fails to satisfy one of the constraints.

The pursuit of such a function arises while trying to define a norm.

Answer 1

It seems $f(x,y)=\max(x,\frac1y)$ is something that works. But it's 3:00 AM here, so I may have made a silly mistake.