Functional equation $f(h(y)x+y)=g(y)f(x)$

Question

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

I am seeking the solution for the functional equation

$f(h(y)x+y)=g(y)f(x)$

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 $g=1$?

Answer 1

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 $k \ne 0$).

Answer 2

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