Real Analysis Continuous Function Problem Show that the only continuous function on $(-1,+1)$, which is not identically
zero and satisfies the equation $f(x + y) = f(x)f(y)$ for all $x,y \in \mathbb{R}$, is the exponential function $f(x) = a^x$ with $a = f(1) > 0$.
 A: Assume $f(x_0)=0$ for some $x_0\in(-1,1)$. Then from $0=f(x_0)=f(x_0/2+x_0/2)=f(x_0/2)^2$ we conclude $f(x_0/2)=0$. By induction, $f(2^{-n}x_0)=0$ and by continuity $f(0)=0$. Then $f(x)=f(x+0)=f(x)f(0)=0$ for all $x\in(-1,1)$. As $f$ is not supposed to be the zero function, we conclude $f(x)\ne 0$ for all $x\in (-1,1)$. Together with $f(x)=f(x/2)^2\ge0$ we conclude $f(x)>0$ for all $x\in(-1,1)$.
Specifically  $f(0)=f(0+0)=f(0)^2$ implies $f(0)=1$.
Pick $x_1\in (0,1)$ and let $a=\exp(\frac1{x_1}\ln f(x_1)))>0$. Let 
$$ A=\{\,x\in(-1,1)\mid f(x)=a^x\,\}.$$
We have $x_1\in A$ and $0\in A$. 
If $x,y,z\in(-1,1)$ with $x+y=z$ and two of $x,y,z$ are in $A$ then all are in $A$: For example if $x,y\in A$, then $f(z)=f(x)f(y)=a^xa^y=a^{x+y}=a^z$; and if $x,z\in A$, then $f(y)=\frac{f(z)}{f(x)}=\frac {a^z}{a^x}=a^{z-x}=a^y$.
Especially, $x\in A$ implies $-x\in A$ and then $nx\in A$ for all $n\in\mathbb Z$ with $nx\in(-1,1)$. Also $x\in A$ implies $\frac12x\in A$ and by induction $2^{-n}x\in A$ for all $n\in\mathbb N$. We conclude that $\frac m{2^n}x_1\in A$ for all $n\in \mathbb N$, $m\in\mathbb Z$ with $-1<\frac{m}{2^n}x_1<1$. Then $A$ is dense in $(-1,1)$ and by continuity of $f$, it is all of $(-1,1)$.
Note that this implies $\lim_{x\to 1}f(x)=a$.
A: This answer is basically in the same spirit as Hagen von Eitzen's answer but probably in a more "algorithmic" way. From his answer we know that $f(0)=1$.
Step 1: For all $n\in\mathbb N$, we have:
$$
f(2)=f(1)^2\\
\vdots\\
f(n)=f(1)^n
$$
So for all positive integers $n$ , we have $f(n)=a^n$ with $a=f(1)$.
Step 2: We know that $f(nx)=f(x)^n$ for all positive integers $n$. Therefore $f(\frac 1n)^n=a$. Also:
$$
f(\frac mn)=f(\frac 1n)^m=a^{\frac mn}
$$
So for all positive rationals $\frac mn$ , we have $f(\frac mn)=a^{\frac mn}$ with $a=f(1)$.
Step 3: For an arbitrary positive real number $r$, there is a sequence of $\{a_n\}$ of rational numbers such that $\lim_{n\to\infty}a_n=r$. Now $f$ is supposed to be continuous and we have:
$$
f(r)=f(\lim_{n\to\infty}a_n)=\lim_{n\to\infty}f(a_n)=\lim_{n\to\infty}a^{a_n}=a^r.
$$
So for all positive real numbers $r$ , we have $f(r)=a^{r}$.
Step 4: Use $f(-x)=\frac 1{f(x)}$ to prove the result for all $r\in\mathbb R$. 

Usually, when we deal with a functional equation, we can use these steps to find the function. Frist we find the value of function for some obvious choice of inputs (like $x=0$ here). Then we try to find the function for all integers and then all rationals. Finally, if the function is assumed to be continuous, we can use a sequence of rationals to find the function at all irrational number. That's why I called this approach "algorithmic". No need to say that there is no universal solution applicable to all functional equations. 
