TITLE: Prove that if $f(x+y)=f(x)f(y)$ and f is continuous in $ x=0$, then it is continuous in all its domain
QUESTION [5 upvotes]: Prove that, for $f:\Bbb{R} \rightarrow \Bbb{R}$,   if $f(x+y)=f(x)f(y)$ and f is continuous at $x=0,$ then it is continuous in all $\Bbb{R}$.
I haven't figured out how to prove this. What would you suggest?
I've already tried finding and expression for $f(0)$ which is either $0$ or $1$.

REPLY [10 votes]: Actually, we do not need to know the value of $f(0)$. For any $a \in \mathbb{R}$,
\begin{align*}
\lim_{x \rightarrow a} f(x) & = \lim_{x \rightarrow a} f\big((x-a)+a\big) = \lim_{x \rightarrow a} f(x-a)f(a) \\
& = f\Big(\lim_{x \rightarrow a} (x-a)\Big)f(a) \quad \text{because } \lim_{x \rightarrow a} (x-a) = 0 \text{ and } f \text{ is continuous at } 0 \\
& =f(0)f(a) = f(0+a) = f(a).
\end{align*}
Hence $f$ is continuous at $a$.

REPLY [5 votes]: Case $1$: If $f(0) = 1$, then we have
$$\lim_{y \to 0} \;[f(x + y) - f(x) ] = \lim_{y \to 0} \; [f(x)f(y) - f(x)] = \lim_{y \to 0} \;f(x)[f(y) - 1]= f(x)(f(0) - 1) = 0.$$
Case $2$: If $f(0) = 0$, then we are done because the problem told us $f$ was continuous.