real analysis - If $f(xy)=f(x)f(y)$, then $f(x)=x^b$.

I'm trying to prove that if $g:\mathbb R^+\to \mathbb R$, defined by $g(uv)=g(u)+g(v)$, for all $u, v \in \mathbb R^+$, $g$ continuous, then $g(x)=b\cdot \ln x$

In order to prove this, I'm trying to prove that if there is a function $f:\mathbb R \to \mathbb R$ such that $f(xy)=f(x)f(y)$, then $f(x)=x^b, b\in \mathbb R$.

If I prove the last statement, I prove the question. First, I was trying this by induction on $\mathbb N$, what I found very difficult to prove.

I need help. Thanks!

Answer

Your approach has problems because, when, say, $b=1/2$, $x^b$ is not defined on all of $\mathbb R$.

The safer approach is to prove that if $h:\mathbb R\to\mathbb R$ is continuous, and for all $x,y\in\mathbb R$, $h(x)+h(y)=h(x+y)$, then there is a $b$ such that $h(x)=bx$ for all $x$.

Setting $h(x)=g(e^x)$ above, from this result it would follow since $$h(x+y)=g(e^{x}e^{y})=g(e^x)+g(e^u)=h(x)+h(y)$$

So for $z>0$, $g(z)=h(\log z)=b\log z$.

To prove this statement about $h$, we proceed by first proving it for natural numbers $x$, then for positive rational numbers $x$, and then for all rationals, and then finally for all real $x$.

We first show $h(0)=0$. Then we show that $h(-x)=-h(x)$ for all $x$. Both of these follow relatively directly from the condition on $h$.

Then we show by induction that if $n$ is a natural number and $x$ any real, then $h(nx)=nh(x)$.

In particular, if $x=1$, this means that $h(n)=nh(1)$. Let $b=h(1)$. So $h(n)=bn$ for all natural numbers $n$.

Then we show that if $x=m/n$ is a positive rational number, then $$bm=h(m)=h(nx)=nh(x)$$ so $h(x)=b\frac{m}{n}=bx$ for $x$ positive rational, as well.

Now, if $f(x)=bx$, then $f(-x)=-f(x)=-bx=b(-x)$. So this means that $f(x)=bx$ for all rationals $x$.

But since the we assumed $h$ was continuous, this means that $h(x)=bx$ for all real number numbers. (This is why you need continuity.)

[I've left out some of the steps - the induction argument, the $h(0)=0$ and $h(-x)=-h(x)$ steps.]