1

Let $f: (0, +\infty) \to\mathbb R$ be a differentiable function such that $f(xy)=f(x)f(y)$ for all $x,y \in (0, +\infty)$.

Show that if $f(1)=1$, then there exists a constant $\alpha$ such that $f(x)=x^\alpha$ for all $x \in (0, +\infty)$.

So far: $$f(x^\alpha)=f(\underbrace{x\cdot x\cdots x}_{\alpha\text{ times}})=\underbrace{f(x)\cdot f(x)\cdots f(x)}_{\alpha\text{ times}}=\alpha\cdot f(x)$$ Hence, $f(x^\alpha)=\alpha\cdot f(x)$.

From here I am not quite sure, but I believe I need to differentiate both sides, then move all terms to one side equaling zero and solve?

user113009
  • 37

1 Answers1

1

First note that $f(y) > 0$(Why?). Now let $g(x) = \ln(f(a^x))$, where $a>0$. We then have $$g(x+y) = \ln(f(a^{x+y})) = \ln(f(a^xa^y)) = \ln(f(a^x) f(a^y)) = g(x) + g(y)$$ This is the Cauchy function equation and if $g(x)$ is continuous, the solution is $$g(x) = cx$$ Hence, we have $$f(a^x) = e^{cx} \implies f(x) = x^t$$ Note that in the above proof, we only relied on the continuity of $f$.

Below are some similar problems based on Cauchy function equation:

Is there a name for function with the exponential property $f(x+y)=f(x) \cdot f(y)$?

Classifying Functions of the form $f(x+y)=f(x)f(y)$

If $f\colon \mathbb{R} \to \mathbb{R}$ is such that $f (x + y) = f (x) f (y)$ and continuous at $0$, then continuous everywhere

continuous functions on $\mathbb R$ such that $g(x+y)=g(x)g(y)$

What can we say about functions satisfying $f(a + b) = f(a)f(b) $ for all $a,b\in \mathbb{R}$?

Community
  • 1
  • But my problem states that $f(xy)=f(x)f(y)$ for all $x,y \in (0, +\infty)$. Not that $f(x+y)=f(x)f(y)$ for all $x,y \in (0, +\infty)$. Sorry, I edited my comment I had the + sign in the wrong place. Also, where is the use of $f(1)=1$? – user113009 Dec 07 '13 at 23:21
  • @user113009 Where did I use $f(xy) = f(x) + f(y)$. I only used $f(xy) = f(x)f(y)$. Kindly read the proof completely. –  Dec 07 '13 at 23:22
  • @user113009 Where did I use $f(x+y) = f(x)f(y)$? Why don't you read my answer completely from the first line and see if you follow? –  Dec 07 '13 at 23:44
  • Sorry, I am trying to understand why the $g(x+y)=...=g(x)+g(y)$. Is it just to get that g(x)=cx? Also, I am not seeing where f(1)=1 was used. Was it used to get $f(a^x) = e^{cx} \implies f(x) = x^t$ – user113009 Dec 07 '13 at 23:49
  • @user113009 $f(1)=1$, was used in the very first step, to conclude that $f(x) \neq 0$. Note that if $f(1) = b \neq 1$, we would then get $f(x\cdot 1) = f(x) \cdot f(1) = bf(x) \implies f(x) = 0$, since $b \neq 1$. Hence, for non-trivial $f(x)$, we need $f(1) = 1$. –  Dec 07 '13 at 23:56
  • @user113009 Which step is unclear in $g(x+y) = \cdots = g(x) + g(y)$? –  Dec 07 '13 at 23:56
  • First, thanks for the clarification on the $f(1)=1$ use. All the steps are clear but, I do not understand why we need it unless it is to obtain that $g(x)=cx$. – user113009 Dec 07 '13 at 23:59
  • @user113009 Yes, we need it to obtain $g(x) = cx$. The reason we want to obtain $g(x) = cx$ is that, we can then obtain $f(x) = x^t$ as you wanted. –  Dec 08 '13 at 00:03
  • Thank you for the clarification. I wanted to make sure I fully understood the solution. :-) – user113009 Dec 08 '13 at 00:05
  • @user113009 No problem. Feel free to ask. –  Dec 08 '13 at 00:06