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

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?

Answer

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 $f$.

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