Continuity of function where $f(x+y) = f(x)f(y) ~~\forall x, y \in \mathbb{R}$. Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a function which satisfies $f(x+y) = f(x)f(y)  ~~\forall x, y \in \mathbb{R}$ is continuous at $x=0$, then it is continuous at every point of $\mathbb{R}$.
So we know $\forall \epsilon > 0 ~~\exists \delta > 0$ such that $|x-0|<\delta \implies |f(x)-f(0)|<\epsilon$ and we want to show that given any $\epsilon > 0 ~~\exists \delta > 0$ such that $|x-y|<\delta \implies |f(x)-f(y)|<\epsilon$.
Now I see that the 'trick' that we can use is replacing $f(x)$ with $f(x)f(0)$ but I still cannot seem to finish the proof. Any advice?
 A: To show continuity at $x$ simply notice that:
$$\lvert f(x+h)-f(x)\rvert=\lvert f(x)f(h)-f(x)\rvert=\lvert f(x)\rvert\lvert f(h)-1\rvert$$
and notice that $f(0)=f(0+0)=f(0)f(0)$ so either $f(0)=0$ or $f(0)=1$. So either $f$ is identically $0$ and hence continuous or we use continuity at $x=0$.
A: Choose any $x$ and $\varepsilon>0$. You know that $f$ is continous at $t=0$, so there exists $\delta>0$, for which $|0-y|<\delta \Rightarrow |f(0)-f(y)|<\frac{\varepsilon}{|f(x)|} $. It's $\delta$ is good choice of delta in standard delta-epsilon proof, because for $y$, that $|x-y|<\delta$ you have($|t|<\delta$ $y=t+x$):
$$|f(x)-f(y)|=|f(x+0)+f(x+t)|=|f(x)f(0)-f(x)f(t)|<\\<|f(x)||f(0)-f(t)|<|f(x)|\cdot \frac{\varepsilon}{|f(x)|}=\varepsilon$$.
$\textbf{Edit:}$If $f(x)=0$ you can choose for example $\delta=1$, because in this case $|f(x)-f(y)|=|f(x)||f(0)-f(t)|=0$.
A: $f(x_0+h)=f(x_0)f(h)$, as $f(h)\to f(0)$ ($h\to 0$), then $f(x_0+h)=f(x_0)f(h)\to f(x_0)f(0)=f(x_0)$.
A: If $y$ is close to $x$, write $y = x + h$ for some small $h$. Then
\begin{align*}
|f(y) - f(x)| &= |f(x + h) - f(x + 0)| \\
&= |f(x)| \cdot |f(h) - f(0)|
\end{align*}
The first part of this product is fixed, and the second part is small. Can you finish from here?
