3
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ Yes, it's fine. $\endgroup$ – egreg Dec 15 '15 at 0:42
2
$\begingroup$

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 $$ F'(x)=\frac{f'(x)}{f(x)}=1 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.