The functional equation $f(x+x) = f(x)f(x)$ 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.
 A: Your claim is false. For example
$$f(x)= \begin{cases} e^x & x\ge 0 \\ e^{-x} & x<0 \end{cases}$$
satisfies the functional equation, is continuous and is not an exponential function.
A: 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. 
A: Summary: There is an uncountably infinite family of $\mathcal{C}^\infty$ functions on $(0, +\infty)$ satisfying this equation, and these functions are also continuous at $x = 0$ if we demand $f(0) = 1$.  If we further demand that the function be differentiable at $x = 0$, $f(x)$ must be an exponential:  $f(x) = e^{cx}$ for some constant $c$.

Let $g(x): [1,2] \to \mathbb{R}$ be any function satisfying $g(2) = 2g(1)$.  Define a function $h(x): [0, +\infty) \to \mathbb{R}$ recursively:
$$
h(x) = \begin{cases} g(x) & 1 < x \leq 2 \\ \frac{1}{2} h(2x) & 0 < x \leq 1 \\ 2 h(x/2)  & x > 2 \\ 0 & x = 0
\end{cases}
$$
For example, by this definition 
$$
h(3) = 2 h(3/2) = 2 g(3/2)
$$
and 
$$
h(3/16) = \frac{1}{2} h(3/8) = \frac{1}{4} h(3/4) = \frac{1}{8} h(3/2) = \frac{1}{8} g(3/2).
$$
The graph of this function would be self-similar:

By construction, $h(2x) = 2 h(x)$ for all $x$.  But we can then define $f(x) = e^{h(x)}$;  and therefore $f(2x) = e^{h(2x)} = e^{2 h(x)} = (e^{h(x)})^2 = f(x)^2$.  Moreover, we have 
$$
\lim_{x \to 0_+} h(x) = 0 \quad \Rightarrow \quad \lim_{x \to 0_+} f(x) = 1
$$
and so the function is continuous at $x = 0$.
If you want the function to satisfy any particular derivative conditions, you can also demand that $g^{(n)}(1) = 2^{n-1} g^{(n)}(2)$ for any and all values of $n$ you desire.  In particular, you can choose $g$ to be a $\mathcal{C}^{\infty}$ function for which all of the derivatives match at $x = 1$ and $x = 2$, and $f(x)$ will be a $\mathcal{C}^{\infty}$ function as well.
The exponential family of functions you found correspond to choosing $g(x)$ to be a linear function on $[1,2]$.  A similar construction can be made to extend $h(x)$ (and thereby $f(x)$) to the negative real numbers as well;  the $g(x)$ we choose to construct $f(x)$ on this domain need not be the same function as the one we used for the positive reals.  @mfl's answer corresponds to taking $g(x) = x$ for the positive reals and $g(x) = -x$ for the negative reals.

EDIT: After further thought, I'm convinced that even if $g(x)$ is a $\mathcal{C}^n$ function on its domain, the resulting function $h(x)$ will not be $\mathcal{C}^n$ at $x = 0$ unless $n = 0$.  This can be seen by noting that we have
$$
h^{(n)}(x) = \begin{cases} g^{(n)}(x) & 1 < x \leq 2 \\ 2^{n-1} h^{(n)}(2x) & 0 < x \leq 1 \\ 2^{1-n} h^{(n)}(x/2)  & x > 2 \\ 0 & x = 0
\end{cases}
$$
For $n > 1$, consider the sequence of values $x_m = a/2^m$, with $1<a<2$. It can be seen that $h^{(n)}(x_m) = 2^{m(n-1)} g^{(n)}(a)$.   This sequence of values will diverge as $x_m \to 0$, and so the $n$th derivative of $h$ cannot approach a finite value as $x \to 0$.  The only exception here is if $g^{(n)}(a) = 0$ for all $a$, i.e., $g(x)$ is a linear function.  In this case, the resulting $f(x)$ is an exponential.
For $n = 1$, meanwhile, $h'(x)$ will take on all possible values of $g'(x)$ in every interval containing the origin;  and so $h'(x)$ cannot have a well-behaved limit as $x \to 0$ either.  The only exception here is if $g'(x)$ is a constant over its domain, in which case we get $f(x)$ as an exponential function as well.
