A Recursive Maximization Problem

Question

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:

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

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

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

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

5. 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.)

Answer 1

Here is the sketch of how one can work this problem out: first, we can assume that $a=\frac{A}{B}>1,$ and the problem is equivalent to $$f(p)(a-p-f(f(p)))\ge q(a-p-f(q)).$$

Replacing $q\to f(q),$ we can rewrite the inequality as follows: $$(f(p)-f(q))(a-p)\ge f(p)f(f(p))-f(q)f(f(q)).$$ Switch $q$ and $p$ and add the results up to end up with $$(f(p)-f(q))(q-p)\ge 0.$$ The last inequality implies that $f$ is decreasing function. To simplify life, let me assume that $a> 2,$ but everything below can be adjusted to work for general case.

Since $a>2> p+f(f(p)),$ then for any $q\ge f(p)$ we must have $$a- p-f(f(p))> a-p-f(q)$$ or $f(q)\ge f(f(p))$ which is impossible since $f$ should decrease unless $f$ is constant on $[f(p),1].$ Now if $f(p)<f(q)$ for some $p\ne q$ we have $$(f(p)-f(q))(a-p)\ge f(p)f(f(p))-f(q)f(f(q))=f(f(p))(f(p)-f(q)),$$ we must have $a-p<f(f(p))$ which is impossible.

The general case could be handled in the same manner since you can always choose $p$ close to zero to satisfy $a>f(f(p))+p$ to obtain immediately that $f$ is $1$ near zero.