TITLE: A question about specific function with this property $f(x+y)=f(x)f(y)$
QUESTION [5 upvotes]: $$\forall x,y \in \mathbb{R} :f(x+y)=f(x)f(y) ,f(0)\neq0$$ Exponential function $f(x)=a^x $ works fine here .
My question :Is there any $\bf\color{red} {\text{other function}}$  exept of exponential functions work for this property  ?
$\bf\text{I know like this question asked before}$ ,but My question is about uniqueness type of functions which work with this property.
On the other hand is there a proof that only exponential function work for that property ?

REPLY [2 votes]: It's very easy to show that $f(0) = 1$ and if $f(1)= b$ then by induction $f(n)= b^n$ for any natural $n$ and that $f(\frac 1n) = \sqrt[n]b$, and so for all $q \in \mathbb Q$ that $f(q) = b^q$  ($b^{\frac nm} = (\sqrt[m]{b})^n$).
And so as $\mathbb Q $ is dense in $\mathbb R$ it follows that if $f$ if continuous, that $f(x) = b^x$.
But $f$ is not continuous it does not.  For any irrational $k$ we could set $f(k)= c$ by any value and we could have and $f(qk) = f(k)^q$ for all rational $q$ but as $qk$ is not rational no contradiction need arise. ($f(r + qk) = b^rc^q \not \in \mathbb Q$ (unless $q=0$) for all $r,q \in \mathbb Q$.)
But we will have for any $w \in \mathbb R$ that for and $q_i\in \mathbb R$ that $f(\sum q_i*w^i) = \prod f(w^i)^{q_i}$.