What is a function satisfying these constraints?

Question

I'm looking for a monotonically increasing function $f\colon (0,1)\times(0,1)\mapsto(0,1)$ satisfying:

1. $f(x,y)=1-f(1-y,1-x)$
2. If $x >\frac{1}{2}$, $f(x, y)>y$
3. $f(\frac{1}{2}, y)=y$
4. If $x <\frac{1}{2}$, $f(x, y)<y$

So far, I've figured out that if $x+y=1$, then $f(x,y)=\frac{1}{2}$. Therefore by monotonicity, $f(x,y)\ge\frac{1}{2}$ in the upper half, and $f(x,y)\le\frac{1}{2}$ in the lower half. 1. and 3. imply that if $y=\frac{1}{2}$, then $f(x,y)=x$. Also, the lower left quadrant is less then $\frac{1}{2}$, and the upper right quadrant is greater than $\frac{1}{2}$. I conjecture the function might be symmetric about $x=y$.

$\frac{x+y}{2}$ satisfies 1., but fails 3. because $\frac{\frac{1}{2}+y}{2}\ne y$.

$x+y-\frac{1}{2}$ goes out of the image for values near the edges.

Any ideas?

Answer 1

I can do it piecewise. For the lower-left quarter: $$f(x, y) = 2xy$$ For the upper-right quarter: $$f(x, y) = 1 - 2(1-x)(1-y)$$ For the other two quarters: $$f(x, y) = x + y - \frac12$$