In my Analysis class, we started to prove a theorem that said:

Let a > 1. So there is a unique increasing function $f:(0,\infty)\to\mathbb{R}$ so that:

1. $f(a) = 1$
2. $f(xy) = f(x) + f(y)\quad\forall x, y > 0$

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 $\mathbb{R}$.

$f(x) = f(x\cdot1) = f(x)+f(1)$, so $f(1)=0=f(x\cdot x^{-1})=f(x)+f(x^{-1})$

Then $f(x^{-1}) = -f(x)$.

We affirm that $f(x^n) = n f(x)\quad\forall x>0, n \in \mathbb{N}$ (and then we proved by induction).

Let x > 0 and $n\in \mathbb{N}^*$. 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. $m\le nf(x) \le m+1$

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.

In my Analysis class, we started to prove a theorem that said:

Let a > 1. So there is a unique increasing function $f:(0,\infty)\to\mathbb{R}$ so that:

1. $f(a) = 1$
2. $f(xy) = f(x) + f(y)\quad\forall x, y > 0$

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 $\mathbb{R}$.

$f(x) = f(x\cdot1) = f(x)+f(1)$, so $f(1)=0=f(x\cdot x^{-1})=f(x)+f(x^{-1})$

Then $f(x^{-1}) = -f(x)$.

We affirm that $f(x^n) = n f(x)\quad\forall x>0, n \in \mathbb{N}$ (and then we proved by induction).

Let x > 0 and $n\in \mathbb{N}^*$. So exists $m \in \mathbb{Z}$ so that $a^m \le x^n \le a^{m+1}$

So $f(a^m) \le f(x^n) \le f(a^{m+1})$, i.e. $m\le nf(x) \le m+1$

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.

In my Analysis class, we started to prove a theorem that said:

Let a > 1. So there is a unique increasing function $f:(0,\infty)\to\mathbb{R}$ so that:

 

1. $f(a) = 1$
2. $f(xy) = f(x) + f(y)\quad\forall x, y > 0$

 

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 $\mathbb{R}$.

 

$f(x) = f(x\cdot1) = f(x)+f(1)$, so $f(1)=0=f(x\cdot x^{-1})=f(x)+f(x^{-1})$

 

Then $f(x^{-1}) = -f(x)$.

 

We affirm that $f(x^n) = n f(x)\quad\forall x>0, n \in \mathbb{N}$ (and then we proved by induction).

 

Let x > 0 and $n\in \mathbb{N}^*$. 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. $m\le nf(x) \le m+1$

 

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.

In my Analysis class, we started to prove a theorem that said:

Let a > 1. So there is a unique increasing function $f:(0,\infty)\to\mathbb{R}$ so that:

1. $f(a) = 1$
2. $f(xy) = f(x) + f(y)\quad\forall x, y > 0$

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 $\mathbb{R}$.

$f(x) = f(x\cdot1) = f(x)+f(1)$, so $f(1)=0=f(x\cdot x^{-1})=f(x)+f(x^{-1})$

Then $f(x^{-1}) = -f(x)$.

We affirm that $f(x^n) = n f(x)\quad\forall x>0, n \in \mathbb{N}$ (and then we proved by induction).

Let x > 0 and $n\in \mathbb{N}^*$. 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. $m\le nf(x) \le m+1$

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.