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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.

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As I understand the question, the answers below don't address the poster's confusion. The poster apparently understands the proof that $f(x) = \sup A_x$ already, but doesn't see why this implies uniqueness of $f$, probably due to not appreciating what "unique" means. So to clarify: saying that such an $f$ is unique means that if $f$ and $g$ are both increasing functions that satisfy 1. and 2., then $f=g$. The proof you outline implies this since it shows that for each $x$, $f(x)$ and $g(x)$ are both equal to $\sup A_x$. –  Mark Meckes Apr 27 '10 at 13:01
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Also, the LaTeX looks fine to me. Try clicking the "(Re)process math" link on the right. –  Mark Meckes Apr 27 '10 at 13:02
    
If the proof shows that for a function $f(x)$ one has to have $f(x)=\sup A_x$ at each point $x$, then $f(x)$ is defined uniquely, so no need to introduce another function $g(x)$. But one can do, without changing the proof. –  Wadim Zudilin Apr 27 '10 at 14:10
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The reason to introduce another function $g$ is merely to clarify what "unique" means. For the last few years I've been teaching students in the first or second proof-based math classes, and I've found that "unique" is one of the words mathematicians are most likely to take for granted and don't necessarily even try explain the meaning of, but that many students don't understand. –  Mark Meckes Apr 27 '10 at 15:37

1 Answer 1

up vote 3 down vote accepted

Suppose that at some point $x$ the function $f(x)$ assumes a different value $f(x)\ne\sup A_x$.

If $f(x)>\sup A_x$, take a rational number $m/n$ in the interval $(\sup A_x,f(x))$. It does not belong to $A_x$ (otherwise it's at most $\sup A_x$), so $a^m>x^n$ implying $m>nf(x)$, hence $f(x)$ is less than $m/n$, a contradiction.

If $f(x)<\sup A_x$, then take any $m/n\in A_x$ which is greater than $f(x)$ (if all such $m/n\in A_x$ are less or equal than $f(x)$, then $\sup A_x$ is less or equal than $f(x)$ as well by the definition of the supremum). In this case $m/n\in A_x$ implies $a^m\le x^n$, so that $m\le nf(x)$ or $f(x)\ge m/n$, again a contradiction.

Therefore, the only choice for the function $f(x)$ to satisfy (1), (2) and the increasing property is to assume the value $\sup A_x$ at the point $x$.

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Now it's clear, and special thanks to Mark Meckes who understood my problem. –  Silva May 1 '10 at 22:10

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