All solutions to a set of integral equations

Question

I would like a better understanding of the set of pairs $(f_1,f_2)$ of functions $[0,1] \times [0,1] \to [0,1]$ which satisfy the following conditions:

1. For all $y \in [0,1]$, $f_1(x,y) \geq f_1(x',y) \Leftrightarrow x>x'$. For all $x \in [0,1]$, $f_2(x,y) \geq f_2(x,y') \Leftrightarrow y>y'$.

2. $f_1(x,y)+f_2(x,y) = 1$, for all $(x,y) \in [0,1] \times [0,1]$

3. $f_1(x,y)=f_2(y,x)$ for all $(x,y) \in [0,1] \times [0,1]$

4. Due to monotonicity (Condition 1), the functions $f_1$ and $f_2$ are integrable. Now define $$p_1(x,0) = xf_1(x,0) - \int_0^x f_1(t,0) dt,$$ $$p_2(0,y) = yf_2(0,y) - \int_0^y f_2(0,t) dt,$$ $$p_1(x,y) = p_1(x,0) + xf_1(x,y) - \int_0^x f_1(t,y) dt,$$ $$p_2(x,y) = p_2(0,y) + yf_2(x,y) - \int_0^y f_2(x,t) dt.$$

Then for all $(x,y) \in [0,1] \times [0,1]$, $p_1(x,y)+p_2(x,y)=0$.

Example: Let $\alpha \geq 0$. For $x \geq y$, let $$f_1(x,y) = (1/2)+(1/2)(x^\alpha-y^\alpha)$$ $$f_2(x,y) = (1/2)-(1/2)(x^\alpha-y^\alpha)$$ For $x < y$, let $$f_1(x,y) = (1/2)-(1/2)(y^\alpha-x^\alpha)$$ $$f_2(x,y) = (1/2)+(1/2)(y^\alpha-x^\alpha)$$ Then $(f_1,f_2)$ satisfy the conditions.

Answer 1

This is not as clear as it might be. I give at the end a more general set of solutions to what I think are your intended conditions. I wonder if these are all the solutions, however I am not sure that I fully understand what you want.

Three issues (in decreasing order of importance)

1. I am guessing that (maybe) you wanted to have $$p_1(x,y) = p_1(x,0) + yf_1(x,y) - \int_0^y f_1(x,u) du$$ This is valid when you put $y=0$ whereas what you currently have written would give the generally invalid $$p_1(x,0) = p_1(x,0) + xf_1(x,0) - \int_0^x f_1(t,0) dt.$$

2. Your example at the end has cases but there is no difference between them since $$\frac{1}{2}+\frac{1}{2}(x^\alpha-y^\alpha)=\frac{1}{2}-\frac{1}{2}(y^\alpha-x^\alpha)$$ (similarly for $f_2$). Nothing wrong with that (if that was your intention), but confusing.

3. In your first condition you need to use $\ge$ or $\gt$ in both places or (else change $ \Leftrightarrow$ to just $\Leftarrow$). As I say, least important. But do you want $f_1(x,y)$ strictly increasing in $x$ or merely non-decreasing?

Here are a few suggestions:

Since your conditions give $f_1(x,x)=\frac{1}{2}$ and $f_2(x,y)=1-f_1(x,y)$, define $f_1(x,y)=\frac{1}{2}+g(x,y)$ and $f_2(x,y)=\frac{1}{2}-g(x,y).$ We should note that $g$ is a function $[0,1] \times [0,1] \to [-\frac{1}{2},\frac{1}{2}]$.

Then condition 2 is automatic, condition 3 says $g(y,x)=-g(x,y)$ and the $f_1$ part of condition 1 (corrected) says that for fixed $v$ and $u \gt u'$ we have $$g(u,v) \gt g(u',v).$$ Then, using condition 3, we then have $$-g(v,u) \gt -g(v,u')$$ meaning that the $f_2$ part is automatic and also that $$g(v,u) \lt g(v,u').$$

Your could rewrite your definition $$p_1(x,0) = xf_1(x,0) - \int_0^x f_1(t,0) dt=\int_0^x \big(f_1(x,0)-f_1(t,0)\big) dt$$ so $$p_1(x,0)=\int_0^x \big(g(x,0)-g(t,0)\big) dt.$$ Then with my suggest revision one would have in general

$$p_1(x,y)=\int_0^x \big(g(x,0)-g(t,0)\big) dt+\int_0^y \big(g(x,y)-g(x,u)\big) du$$

$$p_2(x,y)=\int_0^y -\big(g(0,y)-g(0,u)\big) du+\int_0^x -\big(g(x,y)-g(t,y)\big) dt$$ $$=\int_0^y \big(g(y,0)-g(u,0)\big) du+\int_0^x \big(g(y,x)-g(y,t)\big) dt$$

I'm not so sure about this last part but, if it is correct, then I suppose generalizing your example to $$g(x,y)=q(x)-q(y)$$ where $q:[0,1] \to [0,\frac{1}{2}]$ is increasing would meet your conditions as

$$p_1(x,y)=\int_0^x \big(q(x)-q(t)\big) dt+\int_0^y \big(q(u)-q(y)\big) du=xq(x)-yq(y)-\int_x^yq(v)dv$$

$$p_2(x,y)=\int_0^y \big(q(y)-q(u)\big) du+\int_0^x \big(q(t)-q(x)\big) dt=yq(y)-xq(x)+\int_x^yq(v)dv.$$