(Note: this is a simplified version of my previous question, which was not answered).

I am seeking the solution for the functional equation


where $f,g,h$ are continuous.

Clearly, $f$ constant and $g=1$ is a solution.

If $h(y)=1$ for all $y$ then there is also the solution:

  • $f(x)=ce^{\lambda x}$, $g(y)=e^{\lambda y}$.

My question is: for the case that $h(y)$ is not identically $1$, is there any other solution but $f$ constant and $g=1$?

  • 2
    $\begingroup$ $f(x) = x+1$, $g(y)= h(y)=y+1$ is also a solution. $\endgroup$
    – cr001
    Oct 1, 2015 at 9:36
  • $\begingroup$ You are right. Thanks! $\endgroup$
    – mike
    Oct 1, 2015 at 9:40
  • $\begingroup$ More generally, we can find solutions $h(y)=y+1, g(x)=(x+1)^k, f(x)=cg(x)$ for any constants $c,k$. $\endgroup$
    – Tintarn
    Oct 1, 2015 at 9:44
  • $\begingroup$ Another family of solutions satisfies $h(y)=1-y, g(x)=(1-x)^k, f(x)=cg(x)$ for any constants $c,k$. In fact, it is not hard to prove that if $g$ is injective then $h$ must be of the form $h(y)=1+cy$ for some constant $c$. $\endgroup$
    – Tintarn
    Oct 1, 2015 at 9:52

2 Answers 2


It seems like you are satisfied with the fact that there are indeed other solutions than the ones given in your original post... Well, for me it seems interesting to find not only some, but all solutions to this equation.

So I will collect the work given in the comments above and write a somewhat partial answer:

First, denote $P(x,y)$ the assertion that $f(h(y)x+y)=g(y)f(x)$. Denote $f(0)=c$. $P(0,x)$ implies $g(x)=cf(x)$. So in case $c=0$ $f$ must be identically zero which is a solution whatever $g,h$. Now, assume $c \ne 0$. Then the functional equation simplifies to $P'(x,y): g(h(y)x+y)=g(x)g(y)$. Now, clearly the RHS is symmetric in $x,y$ so comparing $P'(x,y)$ and $P'(y,x)$ we conclude $$g(h(y)x+y)=g(h(x)y+x)$$ Assuming that $g$ is injective (which would be obvious if it's monotone!) we conclude $$h(y)x+y=h(x)y+x$$ and hence with $y=1$ this can be rewritten as $h(x)=kx+1$ for $k=h(1)-1$. Now, some special cases: If $k=0$ we have $h(x)=1$ and the equation $g(x+y)=g(x)g(y)$ which famously solves for $g(x)=a^x$ and hence in this case we have the family of solutions: $$h(x)=1, g(x)=a^x, f(x)=c \cdot a^x$$ If $k=1$ we have $h(x)=x+1$ and the equation $g(xy+x+y)=g(x)g(y)$. Substituting $m(x)=g(x-1)$ we conclude $m((x+1)(y+1))=m(x+1)m(y+1)$ i.e. $m$ is multiplicative and hence famously $m(x)=x^t$ and hence the family of solutions: $$h(x)=1+x, g(x)=(x+1)^t, f(x)=c(x+1)^t$$ If $k=-1$ we have $h(x)=1-x$ and the equation $g(-xy+x+y)=g(x)g(y)$. Substituting $m(x)=g(-x)$ we obtain $m(xy-x-y)=m(-x)m(-y)$ i.e. $m(xy+x+y)=m(x)m(y)$ and now, similar to above we conclude $m(x)=(x+1)^t$ and hence another family of solutions: $$h(x)=1-x, g(x)=(1-x)^t, f(x)=c(1-x)^t$$

Now, it remains to solve the other cases for $k$ (which probably won't give such nice solutions) and - if we don't assume monotonicity - the case where $g$ is not injective.

EDIT: In fact, the case of general $k$ is also not hard to solve. Note that we want to find solutions to the equation $g(kxy+x+y)=g(x)g(y)$ for some $k \ne 0$. Then, substituting $m(x)=g(\frac{x-1}{k})$ we obtain the equivalent equation $m((kx+1)(ky+1))=m(kx+1)m(ky+1)$ i.e. $m(x)=x^t$ similar to the cases above. Hence $g(x)=(kx+1)^t$ which is indeed a solution. So the general solution in the case that $g$ is injective can be written as

$h(x)=1, g(x)=a^x, f(x)=c \cdot a^x$ (the case $k=0$)

$h(x)=1+kx, g(x)=(kx+1)^t, f(x)=c \cdot (kx+1)^t$ (the case $k \ne 0$).

  • $\begingroup$ Thanks. The original equation was $f(h(y)x+y)=g(y)f(x)$ not $g(h(y)x+y)=g(y)f(x)$. I am now checking if this makes a difference, but just FYI $\endgroup$
    – mike
    Oct 1, 2015 at 10:12
  • $\begingroup$ @mike: Thanks. It was just a spelling mistake. I edited. I also added the complete proof for the case that g is injective which gives two families of solutions. $\endgroup$
    – Tintarn
    Oct 1, 2015 at 11:10

Let me try with the case $f$, $g$, $h$ linear function.

One has $LHS = f((h_1y+h_2)x+y) = f_1((h_1y+h_2)x+y) + f_2$ and $RHS = (g_1y+g_2)(f_1x+f_2)$.

Identify two sides, we get $g_2=h_2=1$, $g_1=h_1$, $f_1=g_1f_2$.

For example, we have $f(x) = 2x+2$, $g(x) = x+1$, $h(x) = x+1$.

  • $\begingroup$ But $x=0$ implies $f(y)=cg(y)$ where $c=f(0)$ so $f$ must be a multiple of $g$, right? But $f(x)=x+2$ is not a multiple of $g(x)=x+1$, so you must have made some mistake... $\endgroup$
    – Tintarn
    Oct 1, 2015 at 9:46
  • $\begingroup$ You are right. It is my mistake. Thanks :) $\endgroup$
    – GAVD
    Oct 1, 2015 at 9:50
  • $\begingroup$ Thanks!! That settles this question. My original question was on the equation $f(h(y)x+y)=g(y)f(x)+f(y)$, with $f$ monotone increasing, and $f,g,h$ continuous. Is there any solution but linear $f$, when $h\neq 1$. Any ideas? $\endgroup$
    – mike
    Oct 1, 2015 at 9:54

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