Find all pairs of functions $ f,g : \mathbb{R}\rightarrow \mathbb{R}$ such that

(a) if $ x < y$, then $ f(x) < f(y)$;

(b) for all $ x,y \in \mathbb{R}$, $ f(xy) = g(y)f(x) + f(y)$.

My work :

Let $ P(x,y) : f(xy) = g(y)f(x) + f(y)$

$x<y \leftrightarrow f(x) < f(y)$

$ P(x,1) : f(1)(1-g(1)) = f(1)$

$g(1) = 1 \rightarrow f(xy) = f(x) + f(y)$

so $f(1) = f(1) + f(1) \rightarrow f(1) = 0$

$ P(x,1) :$

$f(1) = 0 \rightarrow f(x) = g(1)f(x)f(1)$

so $f(x)(1-g(1)) = 0 \rightarrow g(1) = 1$

Please suggest how to proceed.


The answer is, $(f,g)$ satisfies $g=\text{sgn}(x)\exp({b\log|x|})$ and $f(x)=a(g(x)-1)$ for some real numbers $a>0$ and $b>0$.

It is easy to see that all such $(f,g)$ satisfy the conditions. We will show that they are the only possibilities.

The proof is the following:

Take $y=1$, we get $$f(x)=g(1)f(x)+f(1).$$ Since $f$ is not constant, $f(1)=0$, $g(1)=1$.

Take $y=0$, $$ f(0)=g(0)f(x)+f(0). $$ Hence $g(0)=0$.

Take $x=0$ $$ f(0)=g(y)f(0)+f(y), $$ Hence $$f(x)=a(g(x)-1),$$ here we assume $f(0)=-a.$ Of course $a>-f(1)=0$ by assumption, and $g$ is monotonic increasing as $f$ is.

Use this equation to the original one, we get $$ g(xy)=g(x)g(y). $$ Take $x=y=-1$, note that $g$ is monotonic, hence $g(-1)=1$. Take $y=-1$, we know that $g$ is an odd function. It is enough to show that $g=\exp({b\log x})$ for $x>0$. This is actually a standard Cauchy functional equation problem.

Take $G(x)=\log (g(e^x))$. Note that it is well defined because $g(y)>0$ if $y>0$. Also note that $G$ is monotonic increasing and $$ G(x+y)=G(x)+G(y). $$ Hence $G(x)=bx$ for some $b>0$ and we finished the proof.

  • $\begingroup$ However your answer and mine are same. Did you aware $g=sgn(x)\exp (b\log|x|)=x^b$? $\endgroup$ – Takahiro Waki Jan 18 '17 at 8:23
  • $\begingroup$ If your b is even, what happens? Then nothing is written, though. $\endgroup$ – Takahiro Waki Jan 18 '17 at 9:16
  • $\begingroup$ @TakahiroWaki In my answer, b can be any positive real number. $\endgroup$ – Chen Jiang Jan 18 '17 at 9:20
  • $\begingroup$ So, b of your answer is never even as you say. $\endgroup$ – Takahiro Waki Jan 18 '17 at 9:26
  • 1
    $\begingroup$ @TakahiroWaki, it could be even, as it is well-known that, an even number is a real number. $\endgroup$ – Chen Jiang Jan 18 '17 at 9:27

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