Let $A\ge B>0$ be real constants. I say that a function $f:[0,1]\rightarrow[0,1]$ satisfies the *$(A,B)$-condition* if for all $p\in [0,1]$, the expression
$$q(A-Bp-Bf(q))$$
is maximized (not necessarily uniquely) at $q=f(p)$. More precisely, this means that for all $p,q$ we have
$$f(p)(A-Bp-Bf(f(p)))\ge q(A-Bp-Bf(q))$$

(Of course, it's only the ratio $A/B$ that matters.)

I am interested in finding all functions from the unit interval to itself that satisfy the $(A,B)$-condition. I am particularly interested in the case $A=4,B=3$.

Some partial results (unless I've made mistakes, but I believe I've proved these):

**Theorem.** The only function satisfying the $(1,1)$-condition is
$$f(p)=\cases{1&if p $\neq 1$\cr 0&if $p=1$}$$

**Theorem.** Suppose that $f$ satisfies the $(4,3)$-condition. Then:

If $x>y$ and $y$ is in the range of $f$, then $f(x)>f(y)$.

There exists a $p$ such that $f(p)\le 1/3$.

$f$ does not take the value $0$.

For all $p$, we have $p+f(f(p))\le 4/3$

The range of $f$ is infinite.

Question 1. Are there any functions satisfying the $(4,3)$ condition? If so, what are they?

Question 2. Same question with $(4,3)$ replaced by $(A,B)$.

(Note: This earlier question is vaguely related to the current one, but probably not really terribly relevant.)