A Recursive Maximization Problem 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.)
 A: 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.
