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Let $f: \mathbb R\rightarrow \mathbb R$ a derivable function but not zero, such that $f'(0) = 2$ and $$ f(x+y)= f(x)\cdot \ f(y)$$ for all $x$ and $y$ belongs $\mathbb R$. Find $f$.

My first answer is $f(x) = e^{2x}$, and I proved that there are not more functions like $f(x) = a^{bx}$ by Existence-Unity Theorem (ODE), but I don't know if I finished.

What do you think about this sketch of proof's idea?


I'll be asking more things.

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Differentiate $f(x+y)=f(x)\cdot f(y)$ with respect to $y$ to obtain $$f'(x+y)=f(x)\cdot f'(y).$$ Now by letting $y=0$ and noting that $f'(0)=2,$ we obtain $$f'(x)=2f(x).$$

The solutions to the preceding equation are of the form $f(x)=Ce^{2x}$ for some constant $C$. Using the fact that $f'(0)=2$, we find that $f'(0)=2C=2$, so that $C=1$. Hence $f(x)=e^{2x}.$ It is easy to check that this does indeed satisfy the original functional equation.

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$f(0)=(f(0))^2 \Rightarrow f(0) \in \{0,1\}$. If $f(0)=0$, then $f(x) =0$ for every $x$, therefore $f(0)=1$.

For every $x \in \mathbb{R}$ one has $f(x)f(-x)=f(0)=1$ and $f(x)=(f(x/2))^2$, i.e. $f(x)>0$ for every $x \in \mathbb{R}$.

By induction one has $f(nx)=(f(x))^n$ for every $n \in \mathbb{N}$. Since $f(x)f(-x)=f(0)=1$, one has $f(-nx)=(f(nx))^{-1}=(f(x))^{-n}$ for every $n \in \mathbb{N}$. Hence $f(nx)=(f(x))^n$ for every $n \in \mathbb{Z}$.

For every $n \in \mathbb{Z}\setminus\{0\}$ one also has $(f(x/n))^n=f(x)$, so $f(x/n)=(f(x))^{1/n}$. Setting $a=f(1)$ one has, for every $m/n \in \mathbb{Q}$: $$ f(m/n)=(f(m))^{1/n}=[(f(1))^m]^{1/n}=a^{m/n}. $$ Since $\mathbb{Q}$ is dense in $\mathbb{R}$, given $x \in \mathbb{R}$, there is a sequence $(x_k) \subset \mathbb{Q}$ such that $x_k \to x$ as $k \to \infty$, and by continuity one has $$ f(x)=\lim_kf(x_k)=\lim_k a^{x_k}=a^x. $$ One has $f'(x)=a^x\ln a$ for every $x$, and $2=f'(0)=\ln a$, i.e. $a=e^2$. Thus $f(x)=e^{2x}$ for every $x \in \mathbb{R}$.

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