A smooth function satisfying these functional constraints

Question

I am looking for any function on a square $$f:[-1,1]\times [-1,1] \rightarrow [0,1]$$ with the following properties:

1. The function $f$ is as smooth as possible, e.g. differentiable almost everywhere.
2. The function $f$ is symmetric: $f(a,b) = f(b,a)$
3. Thinking of the values of $f$ as probabilities, negating both arguments "inverts" the result: $f(a,b) + f(-a,-b) = 1$.
4. Increasing an argument increases the result: $f(a + \Delta, b) \geq f(a,b)$
5. The function $f$ attains its maximum when both arguments do: $f(1,1) = 1$.
6. The function is a linear interpolation when either argument is zero: $f(0,a) = \frac{1+a}{2}$

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Is there a simple instance of a function with this form?

What sorts of functions have this behavior?

How do you think about finding functions of this form?

Answer 1

The simplest function I can come up with is the following:

$$f(a,b)= \begin{cases} 0 & a + b \leq -1 \\ \frac{1+a+b}{2} & -1 < a + b < 1 \\ 1 & a + b \geq 1 \end{cases}$$

Here is a 3D graph of $z=f(x,y)$. Of course, it is not differentiable on the two lines where the function definition changes.

Graphically, you can imagine making it smoother by "dulling" the edges. You just need the function to pass through this skeleton, obey the appropriate symmetries (namely about the point $(0,0,0.5)$ for condition $3$ and about the plane passing through $x=y$ and the $z$-axis for condition $2$), and obey your non-decreasing condition.

Answer 2

Here is another piecewise linear solution:

Write $$f(x,y)={1\over2}\bigl((1+g(x,y)\bigr)$$ whereby the function $g$ is defined by $$g(x,y):=\cases{x+y\quad&$(-x\leq y\leq 0)$ \cr x\quad&$(0\leq y\leq x)$ \cr}$$ in the sector $x\geq|y|$ and has the required symmetries. Here is a picture of the graph of $f$: