calculus - Prove if $f(x+y)=f(x)f(y)$ then $f(x)=a^x$

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that $f(x+y)=f(x)f(y), \ \forall x,y\in \mathbb{R}$. Prove: if $f \not \equiv 0$, then there exists constant $a$ such that $f(x)=a^x.$

I tried to deduce the result from this question and this question, but had hard time with it.

Any help appreciated.

Answer

First note that $f(0+0) = f(0)^2$, thus $f(0)$ is either $1$ or $0$. If it was $0$ then $f(x+0) = f(x)f(0) = 0$ and then $f\equiv 0$ which contradicts our hypothesis. It must be that $f(0) = 1$.

Let $a = f(1)$. Then $f(2) = a^2$. $f(3) = f(1)f(2) = a^3$ and inductively, $f(n) = a^n$ for all positive integer $n$.

Conversely, $f(1-1) = f(1)f(-1) = 1$, so $f(-1) = a^{-1}$ and now one can reason as before to conclude that $f(n) = a^n$ for any integer $n$.

Now to compute $f(p/q)$ where $p$ and $q$ are integers and $q$ is positive, we have that $a^p = f(p) = f(\underbrace{p/q + \ldots + p/q}_{q\text{ times}}) = f(p/q)^q$, thus $f(p/q) = \sqrt[q]{f(p)} = a^{p/q}$.

Now, we know that $f(x) = a^x$ for any rational number $x$. Since the set of rationals is dense in the set of reals, then by the continuity of $f$, it must be that $f(x) = a^x$ for any real number $x$.