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 $f(0)=0$

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:

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


Similarly, differentiating by $y$ we get:


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?


  • $\begingroup$ A continuous, monotonic function may be non differentiable. Take for example $f(x)=x$ for $x\le 0$ and $f(x)=2x$ for $x>0$. $\endgroup$ – ajotatxe Sep 29 '15 at 22:02
  • $\begingroup$ Surely, a continuous monotonic function need not be differentiable. However, it is differentiable almost everywhere. Now. suppose that our $f$ is not differentiable at some point $x_0$. Then, by (*) for any $y$, it is not differentiable at all points $h(y)x_0+y$, Since $h(y)$ is continuous, these points form an interval. In contradiction. $\endgroup$ – mike Sep 30 '15 at 18:51

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