recover a bivariate function from a system

Question

Suppose the bivariate function $f:[0,1]^2\to \mathbb R$ satisfies

(i) $0\leq f(x,y)\leq 1$, for all $x,y\in[0,1]$

(ii) $f(x,y)+f(y,x)=1$ for all $x,y \in[0,1]$

(iii) $\int_0^1 f(x,y)dy=\frac{1}{2}$, for all $x\in[0,1]$.

For example, $f(x,y)=\frac{1}{2}(1-a(x+y)+2axy)$, for $a\in[-1,1]$ satisfies (i) and (iii) but violates (ii).

Is is possible to give a solution to this system?

Thanks.

Answer 1

Let $h,g$ be two continuous functions, you can look for

$$f(x,y)=\frac{1}{2}+h(g(x)-g(y))-h(g(y)-g(x))$$

for example, you can choose: $h(z)=\frac{1}{4}\sin(\frac{z}{2\pi})$ and $g(z)=z$ will give you

$$f(x,y)=\frac{1+\sin(\frac{x-y}{2\pi})}{2}$$ that satisfies all $(i),(ii),(iii)$.

Obvious generalization:

Let $h$ be $1$ periodic function and such that $h(z)=-h(-z)$ then, $$f(x,y)=\frac{1+\frac{h(x-y)}{||h||_{\infty}}}{2}$$ is a candidate