This is part of a homework assignment for a real analysis course taught out of "Baby Rudin." Just looking for a push in the right direction, not a full-blown solution. We are to suppose that $f(x)f(y)=f(x+y)$ for all real x and y, and that f is continuous and not zero. The first part of this question let me assume differentiability as well, and I was able to compose it with the natural log and take the derivative to prove that $f(x)=e^{cx}$ where c is a real constant. I'm having a little more trouble only assuming continuity; I'm currently trying to prove that f is differentiable at zero, and hence all real numbers. Is this an approach worth taking?
Real Analysis – Proving f(x)f(y)=f(x+y) Implies f(x)=e^{cx}
functional-equationsreal-analysis
Related Solutions
I assume that $A$ is supposed to be a Dedekind cut. Start by making a sketch:
+++++++++++++++++++++++++++++++)----------------------------------
A
The line as a whole represents $\Bbb Q$, and everything to the left of the parenthesis is in the cut $A$. Now look at the definition of $-A$:
$$-A=\{-(q+b):q\in\mathbb{Q}^+, b\in\mathbb{Q}\setminus A\}$$
It depends on two subsets of $\Bbb Q$, $\Bbb Q^+$, and $\Bbb Q\setminus A$, so we should figure out where those are in the picture. $\Bbb Q\setminus A$ is easy:
+++++++++++++++++++++++++++++++)(---------------------------------
A Q\A
it’s everything to the right of $A$. And we know just what $\Bbb Q^+$ is: it’s the set of positive rationals. What happens when you add a positive rational $q$ to every member of $\Bbb Q\setminus A$? You shift the set $\Bbb Q\setminus A$ to the right by $q$ units:
-------------------------------)(+++++++++++++++++++++++++++++++++
A Q\A
ooooooooooooooooooooooooooooooo)-----------(++++++++++++++++++++++
A q + Q\A
That last picture shows $\{q+b:b\in\Bbb Q\setminus A\}$ for a particular $q\in\Bbb Q^+$. Now what happens when you look at the negatives of these rational numbers, $\{-(q+b):b\in\Bbb Q\setminus A\}$? You simply flip the line $180$° around $0$ to get a picture more or less like this:
++++++++++++++++++++++)-----------(ooooooooooooooooooooooooooooooo
The plus signs mark the set $-(q+\Bbb Q\setminus A)$, and the o’s mark the set $\{-a:a\in A\}$. The gap in the middle has length $q$. $\{-a:a\in A\}$ is always in the same place, but the location of $-(q+\Bbb Q\setminus A)$ depends on $q$: when $q$ is large, it’s far to the left of $\{-a:a\in A\}$, and when $q$ is small, it’s very close to $\{-a:a\in A\}$.
The set $-A$ that you’re to prove is a Dedekind cut is just the union of all these sets $-(q+\Bbb Q\setminus A)$ as $q$ ranges over the positive rationals, so it’s the union of all possible sets like those marked with plus signs in the pictures below:
+++++++++++++++++)----------------(ooooooooooooooooooooooooooooooo
++++++++++++++++++++++)-----------(ooooooooooooooooooooooooooooooo
+++++++++++++++++++++++++)--------(ooooooooooooooooooooooooooooooo
++++++++++++++++++++++++++++++)---(ooooooooooooooooooooooooooooooo
What do you think that union will look like in a sketch of this kind? Won’t it look something like this?
+++++++++++++++++++++++++++++++++)(ooooooooooooooooooooooooooooooo
That looks an awful lot like a Dedekind cut. Now you just have to prove it. For instance, you have to show that there is some rational that is not in the set. From the picture it appears that any $-a$ with $a\in A$ should work, so you should try to prove that this is the case. (Remember, the $q$’s that are being added are all strictly positive.)
You need to show that if $p$ is rational and $p<r\in-A$, then $p\in -A$. The picture certainly makes that look plausible, and it’s not hard to show. If $r\in-A$, then $r=-(q+a)$ for some $q\in\Bbb Q^+$ and $a\in A$. Can you find another positive rational $q'$ such that $p=-(q'+a)$? The number $r-p$ may be useful.
Recall that if $f$ is invertible, then it is bijective. So, there exists a unique $y$ such that $y = f(x)$ on the domain of $g$, which is seen to be $(f(a),f(b))$ since we have an increasing function. By definition of the derivative:
$$g'(y) = \lim_{z \rightarrow f(x)} \frac{g(z) - g(f(x))}{z-f(x)} = \lim_{z \rightarrow f(x)} \frac{g(z) - x}{z-f(x)}$$
Now, as we tend $z$ closer and closer to $f(x)$, eventually it will have to belong to $(f(a),f(b))$, which means we can find another $x_z$ such that $f(x_z) = z$. We now choose $z$ sufficiently close, and take advantage of this fact. We then have:
$$g'(y) = \lim_{z \rightarrow f(x)} \frac{g(f(x_z)) - x}{f(x_z)-f(x)} = \lim_{z \rightarrow f(x)} \frac{x_z - x}{f(x_z)-f(x)}$$
We see that this final limit tends to $\frac{1}{f'(x)}$.
Best Answer
First note that $f(x) > 0$, for all $x \in \mathbb{R}$. This can be seen from the fact that $$f(x) = f\left(\dfrac{x}2 + \dfrac{x}2\right) = f \left(\dfrac{x}2\right)^2$$ Further, you can eliminate the case $f(x) = 0$, since this would mean $f \equiv 0$.
One way to go about is as follows.
$1$. Prove that $f(m) = f(1)^m$ for $m \in \mathbb{Z}^+$.
$2$. Now prove that $f(m) = f(1)^m$ for $m \in \mathbb{Z}$.
$3$. Now prove that $f(p/q) = f(1)^{p/q}$ for $p \in \mathbb{Z}$ and $q \in \mathbb{Z} \backslash \{0\}$.
$4$. Now make use of the fact that rationals are dense in $\mathbb{R}$ and hence you can find a sequence of rationals $r_n \in \mathbb{R}$ such that $r_n \to r \in \mathbb{R} \backslash \mathbb{Q}$. Now use continuity to conclude that $f(x) = f(1)^x$ for all $x \in \mathbb{R}$. You will see that you need only continuity at one point to conclude that $f(x) = f(1)^x$.