I am looking for the solution to the following two variable functional equation:

(*) $f(h(y)\cdot x+y)= g(y)f(x)+f(y)$


  • $h$ 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 $f(0)=0$

Clearly, $f$ linear and $g(y)=h(y)$ is a solution. The question is: is this the only solution for the case that $h$ is not identically $1$?

(For $h(y)=1$ for all $y$, I believe that it is known that the only three possible solutions are:

  1. $f$ is linear,
  2. $f(x)=c(a^{x}-1)$ for some $a,c >0$,
  3. $f(x)=c(1-a^{-x})$ for some $a,c >0$. )

I tried asking this question first in the Stack-Exchange Mathematics site, but did not get an answer. Hoping maybe this forum can help.

Directions I attempted:

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, I think 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 I am not sure how to complete the proof.

Direction 2:

Suppose that $f$ is continuously differential (I think it is possible to prove that this must be so in this case (since $f$ is monotonic, and the structure of the functional equation). Then, differentiating (*) by $x$ we get


Similarly, differentiating by $y$ we get:


So, now we can get rid of the term $f'(h(y)x+y)$ and remain with 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. Again, not sure how to complete the proof.

Any ideas?



1 Answer 1


It turns out that $f$ linear is not the only solution, even for the case that $h$ is not identically 1.


$h(y)=y+1, g(y)=(y+1)^2, f(x)=x^2+2x$.


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