TITLE: The functional equation $f(x+x) = f(x)f(x)$
QUESTION [5 upvotes]: Consider the exponential functional equation on the diagonal
\begin{align}
f(x+x) = f(x)f(x)\,, \quad f(0) = 1\,.
\end{align}
What are the solutions for this problem? Either for all $x $ or for non-negative $x $.
EDIT: Apparently the following is wrong:  I guess the exponential functions $x \mapsto e^{cx}$ are the only solutions if we assume measurability, continuity or some other regularity property.

REPLY [2 votes]: Be $f(x):=\sum\limits_{k=0}^\infty a_k x^k$ and $a_0:=1$. 
With $f(2x)=f(x)^2$ follows $\sum\limits_{v=0}^k a_v a_{k-v}=2^k a_k$.
This means, that $a_1$ can be any value. 
We get the exponential function $e^{a_1 x}$. $\enspace$(as $\pi r8$ and Michael Seifert have mentioned below)
Short hint: $\enspace\sum\limits_{v=0}^k \frac{(a_1)^v}{v!} \frac{(a_1)^{k-v}}{(k-v)!}=a_1^k\frac{(1+1)^k}{k!}=2^k \frac{(a_1)^k}{k!}$  
This shows that functions which can be represented by Taylor series, can only be exponentially, if they solve the claim above.