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

  • $f$ is positive, 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?


  • $\begingroup$ We may assume that $f$ is differentiable. $\endgroup$ – mike Oct 12 '16 at 18:58
  • $\begingroup$ But $x+g(x)y$ may like outside this interval, if it is finite, then how LHS is defined? $\endgroup$ – Fedor Petrov Oct 12 '16 at 19:46
  • $\begingroup$ $f$ is guaranteed to be defined on the entire interval of $x+g(x)y$, with $x,y$ in some initial interval. $\endgroup$ – mike Oct 12 '16 at 21:29

There are other solutions which is similar to your last one. They are: $f(x)=\ln(x+\frac{b}{a})$, and $g(x)=ax+b$ for $a\neq 0$, $b$ constants. I think this answers your question.

And for the general solution, if one assumes both functions are differentiable. There would be no other solutions except what we listed. This can be shown by combining the equation after taking derivative for $y$ (and evaluating at $y=0$), with the equation after taking derivative for $x$.

The precise equations are:$$g(x)f'(x)=g(0)f'(0)=C,$$and $$f'(x+g(x)y)\cdot(1+g'(x)y)=f'(x).$$ Assume $C\neq 0$. For the second equation above, let $t=x+g(x)y$, then $$f'(t)\cdot\left(1+\frac{t-x}{g(x)}g'(x)\right)=f'(x).$$From here, multiplying both sides with $g(x)g(t)$, and use the first equation above, you will get $$g(x)-g(t)=(x-t)g'(x),$$ which means that $g$ is linear!

In the case when $C=0$, the two situations are obtained already by you!

| cite | improve this answer | |
  • $\begingroup$ Thanks. That answers my question (though not what I wanted - :( ). $\endgroup$ – mike Oct 12 '16 at 21:20
  • $\begingroup$ Changguang: can you explain in a little more detail the proof that these are the only solution (assuming all functions are differentiable)? Thanks. $\endgroup$ – mike Oct 12 '16 at 22:06

Assuming $f$ and $g$ are differentiable and et $a=f(0), b=f'(0), c=g(0), d=g'(0)$. Here are the details you ask.

Suppose neither $f$ not $g$ is a constant (you already know if not that).

Differentiate w.r.t. $x$ to get $(1+g'(x)y)f'(x+g(x)y)=f'(x)$ then replace $x=0$: $$f'(cy)(1+dy)=b \Longrightarrow f(cy)=\frac{cb}d\log(1+dy)+a \Longrightarrow f(y)=\frac{cb}d\log\left(1+\frac{d}cy\right)+a.$$

Differentiate the original equation w.r.t. $y$ to get $g(x)f'(x+g(x)y)=cf'(cy)$. Replace $y=0$: $$g(x)f'(x)=cb \Longrightarrow g(x)\cdot \frac1{c+dx}=1 \Longrightarrow g(x)=dx+c.$$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.