Identifying the exponential function $f(x)=e^x$ from its functional equation

Question

Prove that if $f(x+y)=f(x)f(y)$ for all $x,y$ and $f(x)=1+xg(x)$ where $\lim_{x\to 0}g(x)=1$, then:

a) $\exists f'(x)$ $\forall x$
b) $f(x)=e^x$

I would really appreciate your help.

Answer 1

Hint: \begin{align*} \frac{f(x + h) - f(x)}{h} &= f(x) \frac{f(h) - 1}{h} \\ &= f(x) \frac{h g(h)}{h} \end{align*}

Can you finish from here?

Answer 2

To prove the uniqueness of the function after arriving at $f^{'}(x)=f(x) $:

Let $t(x)$ be a function that satisfies the aforementioned equation, and $k(x)$ be another solution. Let $u(x)=\dfrac{k(x)}{t(x)}\implies u'(x)=\dfrac{k'(x)t(x)-t'(x)k(x)}{(t(x))^2}=0$

Thus by mean value theorem $u(x)$ is a constant, say $c$. But $t(x)\rightarrow 1$ and $k(x)\rightarrow 1$ as $x\rightarrow 0$. Thus $c=1$. And there exists a unique solution.

To prove that the function is $e^{x}$ you will have to specify the definition of $e^x$, which you haven't.