If $f(x+y) = f(x)f(y)$ and $f(x)$ is monotone, prove $f(x)$ differentiable. If $f(x)$ is monotone and $f(x)$ is not a constant, $\forall x,y \in (-\infty,+\infty)$ $f(x+y) = f(x)f(y)$, prove that:
(1)$f(0) = 1$;
(2)$f(x)>0$;
(3)$f(x)$ is continuous;
(4)$f(x)$ is differentiable;
I can prove (1) (2) but got stuck on (3) and (4), as
$$
f'(x) = \lim_{\Delta x \to 0}\frac{ f(x+\Delta x)-f(x)}{\Delta x}=\lim_{\Delta x \to 0}\frac{f(x)[f(\Delta x)-1]}{\Delta x} = f(x)f'(0)
$$
I Guess if it's differentiable at one point $x=0$, it will be differentiable at any x, but how to prove $x=0$ is differentiable? Thank you!
 A: Claim 1. If $f$ is continuous at $0$, then $f$ is continuous everywhere.
Proof. Note that $$f(x + h) - f(x) = f(x)f(h) - f(x) = f(x)[f(h) - f(0)]$$
for all $x, h \in \Bbb R$. Thus, if $f$ is continuous at $0$, then taking limit $h \to 0$ above show that $f$ is continuous at $x$. $\Box$
Claim 2. $f$ is continuous at $0$. Thus, $f$ is continuous everywhere.
Proof. Here's where we use monotonicity for the first time.
Since $f$ is monotonic, the limits $\lim_{h \to 0^+} f(h)$ and $\lim_{h \to 0^-} f(h)$ exist.
It suffices to show that both are equal $f(0)$, i.e., equal to $1$.
Since we already know the existence of the limit, we can use the fact that $$\lim_{h \to 0^+}f(h) = \lim_{n \to \infty} f\left(\frac{1}{n}\right).$$
Note that $$f(1) = f\left(\underbrace{\frac{1}{n} + \cdots + \frac{1}{n}}_{n \text{ times}}\right) = \left[f\left(\frac{1}{n}\right)\right]^n.$$
Thus, $$f\left(\frac{1}{n}\right) = f(1)^{1/n}.$$
Note that since $f(1) > 0$, we see that the above limit is $1$ as $n \to \infty$. This shows us that $$\lim_{h \to 0^+} f(h) = 1,$$
which is (half of) what we want.
To see the same for the other limit, note that $$f(x)f(-x) = f(0) = 1$$ for all $x \in \Bbb R$. $\Box$
Claim 3. $f(x) = a^x$ for all $x \in \Bbb R$, where $a := f(1)$. In particular, $f$ is differentiable everywhere.
Proof. By a similar trick as before, we see that $f(n) = f(1)^n$ for all positive integers $n$. By using $f(n)f(-n) = 1$, we see that $f(n) = f(1)^n$ for all integers $n$. (We already know it for $n = 0$.)
Now, suppose we have integers $m, n$ with $n > 0$. Then, we have
$$\left[f\left(\frac{m}{n}\right)\right]^n = f(m) = f(1)^m$$
and thus,
$$f\left(\frac{m}{n}\right) = f(1)^{m/n}.$$
This shows that $f(x) = a^x$ for all $x \in \Bbb Q$.
Now, we use continuity to conclude the same for all $x \in \Bbb R$. Let $x \in \Bbb R$ be arbitrary. Then, there exists a sequence $(x_n)_n$ of rationals such that $x_n \to x$. By continuity of $f$, we have
$$f(x) = \lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} a^{x_n} = a^x,$$
where the last equality follows by continuity of the exponential function. $\Box$
A: To prove continuity: first you prove continuity at x=0 (I prove it from the right without loss of generality).
Iterating the property of $f$, you clearly have $f(1)=f(\frac{1}{n})^n$.
Thus $f(\frac{1}{n})=f(1)^{1/n}$ that converges to $1$ for $n\rightarrow+\infty$.
Using the monotonicity of $f$, it is then clear that $f(x)\in[1,f(1/n)]\quad\forall x\in[0,1/n]$, thus it converges to 0. As I said, same thing for convergence from the left.
To show that $f$ is continuous in $x_0$, just write $f(x)=f(x-x_0)f(x_0)$ and continuity in $x=x_0$ follows from the continuity of $f$ in $x=0$.
For the differentiability the same techniques should work.
Edit:
The other answer is the most clever thing to do of course. If instead you want to verify the differentiability in $x=0$ by hand, let's suppose w.l.o.g. that $f$ is increasing in $x$. As you said we want to find the limit
$\lim_{x \to 0}\frac{f(x)-1}{x}$.
Analogously to what written before, $\lim_{n\to\infty}\frac{f(1/n)-1}{1/n}=\lim_{n\to\infty}\frac{f(1)^{1/n}-1}{1/n}=\log(f(1))$ by standard calculus.
Then, if $x\in(1/n,1/(n+1)]$, by monotonicity of both $x$ and $f(x)$,
$\frac{f(1/(n+1))-1}{1/n}\leq \frac{f(x)-1}{x}\leq \frac{f(1/n)-1}{1/(n+1)}$,
and the two quantities on the two sides go to $\log(f(1))$.
