I am looking for the solution to the following two variable functional equation:
(*) $f(h(y)\cdot x+y)= g(y)f(x)+f(y)$
where:
- $h$ is some given continuous function,
- $f, g,$ unknown functions on some interval $[0,\alpha]$ for some $\alpha>0$,
- $f$ is continuous and monotone increasing, with $f(0)=0$
The question is for what functions $f$ can this functional equation hold?
If $h(y)=1$ for all $y$, then (I believe that) it is known that the only three possible solutions are:
- $f$ is linear,
- $f(x)=c(e^{\lambda x}-1)$ for some $c,\lambda >0$,
- $f(x)=c(1-e^{-\lambda x})$ for some $c,\lambda >0$.
The question is what happens when $h(y)$ is not identically $1$? I suspect that in this case $f$ must be linear, but am seeking a proof.
Directions I explored:
Direction 1: Fixing $y$, we get a one variable functional equation: $f(h_y x+y)=g_y f(x)+f_y$, (where $h_y=h(y), g_y=g(y), f_y=f(y)$). The solution to this functional equation is:
$f(x)=(x+a)^bp(x)+c$, where:
- $b=\frac{\ln g_y}{\ln h_y}, a = \frac{y}{h_y-1}, c=\frac{f_y}{1-g_y}$
- $p(x)$ is an arbitrary periodic function such that $p(h_yx+y)=p(x)$.
This is true for any $y$. Looking at these solutions it seems that the only way that such solutions can ``fit together'' is that the exponent $b=1$. If so, $g_y=h_y$ and $f$ is linear. However, the periodic function $p(x)$ messes things up, and the proof looks messy.
Direction 2:
Suppose that $f$ is continuously differential (I think it is possible to prove that this must be so, since $f$ is monotone). Then, differentiating (*) by $x$ we get
$h(y)f'(h(y)x+y)=g(y)f'(x)+f(y)$.
Similarly, differentiating by $y$ we get:
$(xh'(y)+1)f'(h(y)x+y)=g'(y)f(x)+f'(y)$.
So, now we can get rid of the term $f'(h(y)x+y)$ and remain an equation containing only functions of either $y$ or $x$ but not both. Now this equation holds for infinitely many $x$'s and $y$'s, and somehow the only solution should be that $f$ is linear.
This again seems messy.
Any ideas?
Thanks.