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I have the functional equation $$ f(x+g(x)y)=f(x)+f(g(0)y)-f(0) $$ where

  • $f$ is monotone increasing and continuous
  • $g$ is continuous and positive
  • The domain of both functions is a closed interval that includes 0.

The obvious solutions are:

  • $g$ constant and $f$ linear
  • $f$ constant and $g$ arbitrary.

There is also another solution:

  • $g(x)=x+1$, and $f(x)=ln(x+1)$.

The question is whether there are other solutions? In particular, I am interested in the question: if $g(0)\neq 1$ is $f$ necessarily linear?

Thanks.

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Let $g(0)=0$ for this example.

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

Since $f(x)$ is injective, this can only hold for $0=g(x)y$, except we have the problem that $y$ can be anything. Thus the only solution for this case is $g(x)=0$, which gives $f(x)$ be any monotone increasing continuous function defined at $x=0$.

We may then choose $f(x)$ to be non-linear, a contradiction to your claim.

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  • $\begingroup$ You are totally right. I forgot to mention that $g$ are $h$ are both positive. I will fix in the question. Thanks. $\endgroup$ – mike Oct 12 '16 at 17:46
  • $\begingroup$ Sorry, $h$ is not present in this problem. $g$ is positive. $\endgroup$ – mike Oct 12 '16 at 17:48

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