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Let $f:\mathbb{R}^{+}\to \mathbb{R}$ such that $f$ is differentiable and has the property $f(xy) =f(x) +f(y)$.Show that $f(x) =f′(1) \ln(x)$.

Rafiyah Noor
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  • Welcome to MathSE. Please, show your attempts for to solve the problem. –  Dec 07 '20 at 11:02
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    $g\colon \Bbb R\to\Bbb R$, $x\mapsto f(e^x)$ is additive and continuous (you do not need that $f$ is diff'able). By Cauchy functional equation, $g$ is of the form $g(x)=ax$ – Hagen von Eitzen Dec 07 '20 at 11:12
  • This has been asked and answered before: https://math.stackexchange.com/q/1548921, https://math.stackexchange.com/q/1090301, https://math.stackexchange.com/q/43964 and some more: https://math.stackexchange.com/questions/linked/43964. – Martin R Dec 07 '20 at 12:23

1 Answers1

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Let $$(*) \quad f(xy)=f(x)+f(y)$$

for $x,y >0.$

If we differentiate $(*)$ with respect to $x$, we get

$$yf'(xy)=f'(x).$$

If we differentiate $(*)$ with respect to $y$, we get

$$xf'(xy)=f'(y).$$

These equations give

$$xf'(x)=yf'(y)$$

for all $x,y >0.$

With $y=1$ we derive

$$xf'(x)=f'(1),$$

hence $f'(x)= \frac{f'(1)}{x}.$ Therefore

$$f(x)= f'(1) \ln x+c.$$

Now show that $f(1)=0$, to see that $c=0.$

Fred
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