Given $f:\mathbb{R}\to\mathbb{R}$ with $f(x+y)=f(x)f(y), \forall x,y\in\mathbb{R}$. If $f$ continuous at $x=0$, and there exist $a$ such that $f(a)=0$, find $f(x)$.
I have read the definition of continuous function at $x=x_0$ in $E$, for all $\varepsilon>0$, there exist $\delta>0$, such that for all $x_1\in E$ with $\vert x_0-x_1\vert<\delta$, then satisfied $\vert f(x_0)-f(x_1)\vert<\varepsilon$.
In my problem, $f$ continuous at $x=0$, so for all $\varepsilon>0$, there exist $\delta>0$, such that for all $x_1\in \mathbb{R}$ with $\vert -x_1\vert<\delta$, then satisfied $$\vert f(0)-f(x_1)\vert\leq\varepsilon.$$
I don't know to find $f(x)$. How can I get $f(x)$? Thanks..
For any $x\in\mathbb{R}$, we see $$f(x)=f(a+(x-a))=f(a)f(x-a)=0\cdot f(x-a)=0$$ Therefore, $f$ is the constant $0$ function.