$g(x+y)=g(x)g(y)$ for all $x,y \in\mathbb{R}$. If $g$ is continuous at $0$, prove that $g$ is continuous on $\mathbb{R}$

Question

$g(x+y)=g(x)g(y)$ for all $x,y \in\mathbb{R}$.

If $g$ is continuous at $0$, prove that $g$ is continuous on $\mathbb{R}$.

Answer 1

For every $x\in\mathbb{R}$, we have $$ g(x)=g(x+0)=g(x)g(0), $$ so that $g(x)(1-g(0))=0$. So, either $g(x)=0$ for all $x$, in which case $g$ is certainly continuous on $\mathbb{R}$, or $g(0)=1$.

Suppose $g(0)=1$. Fix $x\in\mathbb{R}$, and let $\epsilon>0$ be given. Note that for any $\delta>0$, we have $$ g(x+\delta)=g(x)g(\delta)\qquad\text{and}\qquad g(x-\delta)=g(x)g(-\delta). $$ But, we can make $g(\pm\delta)$ close to 1 by choosing $\delta$ small, by continuity of $g$ at 0. Do you see how to use this idea to finish the problem?

Answer 2

\begin{align} g(0) & = 1 \\ \lim_{h \to 0} g(x + h) & = \lim_{h\to 0} g(x)g(h) = g(x)g(0) = g(x) \end{align}