In my Analysis class, we started to prove a theorem that said:
Let a > 1. So there is a unique increasing function $f:(0,infty)tomathbb{R}$ so that:
- $f(a) = 1$
- $f(xy) = f(x) + f(y)quadforall x, y > 0$
First we supposed its existence. Then through 2 it follows that f is a group homomorphism between the multiplicative group $(0,infty)$ and the additive group $mathbb{R}$.
$f(x) = f(xcdot1) = f(x)+f(1)$, so $f(1)=0=f(xcdot x^{-1})=f(x)+f(x^{-1})$
Then $f(x^{-1}) = -f(x)$.
We affirm that $f(x^n) = n f(x)quadforall x>0, n in mathbb{N}$ (and then we proved by induction).
Let x > 0 and $nin mathbb{N}^*$. So exists $m in mathbb{Z}$ so that $a^m le x^n le a^{m+1}$
So $f(a^m) le f(x^n) le a^{m=1}$, i.e. $mle nf(x) le m+1$
And finally $m/n le f(x) le (m+1)/n$
Let $A_x = { frac{m}{n} : m in mathbb{Z},: n in mathbb{N},: a^m le x^n}$
So $f(x) = sup A_x$
He said that this means f is unique, but I can see no reason why. I've omitted some lemmas and parts of the proof, but kept all the results. It's quite clear f is $log_ax$, but why it is unique?
Edit: I can't get LaTeX to work. It all seems fine while editing, but wrong in the question page.
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