# Verify proof of $f(x)=e^x$ if $f(x+y)=f(x)f(y)$ and $f'(x)$ exists for all $x$

This is exercise 6.26.8 from Tom Apostol's Calculus I, I'd like to ask someone to verify my proof. I'd be also interested in alternative proofs:

If $f(x+y)=f(x)f(y)$ for all $x$ and $y$ and if $f(x)=1+xg(x)$, where $g(x) \to 1$ as $x \to 0$, prove that (a) $f'(x)$ exists for every $x$, and (b) $f(x)=e^x$.

(a) $$f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h} = \lim_{h \to 0}\frac{f(x)f(h) - f(x)}{h} = \lim_{h \to 0}f(x)\frac{1 + hg(h) - 1}{h} = \lim_{h \to 0}f(x)g(h) = f(x)$$

(b) $$\left(\frac{f(x)}{e^x}\right)' = \frac{f'(x)e^x - f(x)e^x}{e^{2x}} = \frac{f(x) - f(x)}{e^x} = 0 \implies f(x) = ke^x \; \text, \; k\in \mathbb R$$ $$k = ke^0 = f(0) = \lim_{x \to 0}1 + xg(x) = 1 \implies f(x) = e^x$$

• Yes, it's fine. – egreg Dec 15 '15 at 0:42

Yes, it's fine. You have less computations if you set $F(x)=f(x)e^{-x}$, so $$F'(x)=f'(x)e^{-x}-f(x)e^{-x}=0$$ and $F$ is constant.
An alternative proof could be by observing that $f(x)=0$ for some $x$ implies $f$ constant $0$, which contradicts the assumptions. So we know $f(x)\ne0$ for all $x$ and differentiability (part a) implies $f$ is continuous, so everywhere positive. Then $$F(x)=\log f(x)$$ is well defined and $$F'(x)=\frac{f'(x)}{f(x)}=1$$