Let $f $ be such that $f(0) = f'(0) = 1$ and for any $x,y$ real numbers, $f(x+y) = f(x)f(y)$. Prove that $f(x) = \exp(x)$.

Question

Suggestion: See that $f'(x) = f'(0)f(x)$, using the definition for derivative with limits when $h \rightarrow 0$

I would like to know if my proof is fine or if I'm missing something. Thank you!

Proof $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$ By hypothesis, $$f'(x) = \lim_{h \to 0} \frac{f(x)f(h)-f(x)} {h} = \lim_{h \to 0} \frac{f(x)(f(h) - 1)}{h} = f(x)\lim_{h \to 0} \frac{f(h) - 1}{h} = f(x)\lim_{h \to 0} \frac{f(h) - f(0)}{h - 0} $$ Then, $$f'(x) = f'(0)f(x)$$ $$\therefore f'(x) = f(x)$$

We know that $exp(x)$ is the only function that satisfies that $f(x) = f'(x)$. $$\therefore f(x) = \exp(x)$$

Answer 1

Yes, it is correct, save for your last statement. We know that $e^x$ is the only function that satisfies $f(x)=f'(x)$ and $f(0)=1$. Otherwise for any $c$ the function $ce^x$ also satisfies the same equation, where $c=f(0)$.

Answer 2

Your argument is correct.

Let me just explain why if $f'(x)=f(x)$, and $f(0)=1$, then $f(x)=\mathrm{e}^x$.

We have: $$ f'(x)=f(x)\quad\Longrightarrow\quad \mathrm{e}^{-x}\big(f'(x)-f(x)\big)=0\quad\Longrightarrow\quad \big(\mathrm{e}^{-x}f(x)\big)'=0 $$ and hence there exists a $c\in\mathbb R$, such that $$ \mathrm{e}^{-x}f(x)=c \quad\Longleftrightarrow\quad f(x)=c\mathrm{e}^{x} $$ and since $f(0)=1$, then $c=1$, and thus $$ f(x)=\mathrm{e}^{x}. $$