Series continuity at one point - prove continuity at all points I'm struggling with an assignment on function continuity. I'm asked to show two things:

Part A) Let $f:R\rightarrow R$ be a function where $f(x+y) = f(x)f(y)$ for all $x,y \in R$
I'm asked to prove that if $f$ is continuous at $x_0=0$ then $f$ is
  continuous at all points $x_0 \in R$

I'm rather at a loss here. For the first part (A), we see that
$$||f(x_0)-f(x)||=||f(x_0 + (x-x_0))-f(x_0)||=||f(x_0)f(x-x_0)-f(x_0)||$$
Continuity at $x_0$ implies that for any $\epsilon>0$ there exists a $\delta>0$ such that when
$$||x-x_0||<\delta \rightarrow ||f(x)-f(x_0)||<\epsilon$$
I know this holds for $x_0=0$ but I don't see how this helps to prove general continuity?

Part B) Let $g:R \rightarrow R$ with $g(xy)=g(x)+g(y)$. And I'm similarly to prove that if $g$ is contious at $x_0=1$ then it is
  continous for all $x_0>0$

I suspect this second part will be elucidated, if I manage to understand the first, so I'll let this rest for now. Noting only that
$$||g(x)-g(x_0)||=||g\big(x_0\frac{x}{x_0}\big)-g(x_0)||=||g(x_0)+g(\frac{x}{x_0})-g(x_0)||=||g(\frac{x}{x_0})||$$
But again I'm not sure how this leads to continuity at all $x_0>0$.
 A: You are close. Note that for both questions, $f \equiv 0$ and $g \equiv 0$ are possible solutions, in which then they are clearly continuous. If $f \not\equiv 0$, then $f(0) = 1$. Fix $\epsilon > 0$, and choose $\delta$ such that $|x| < \delta \Rightarrow |f(x) - 1| < \frac{\epsilon}{|f(x_0)|}$. We then have $|x - x_0| < \delta \Rightarrow |f(x - x_0) - 1| < \frac{\epsilon}{|f(x_0)|}$. Following your working:
\begin{align*}
|f(x) - f(x_0)| = |f(x_0)f(x - x_0) - f(x_0)| = |f(x_0)||f(x - x_0) - 1| < \epsilon
\end{align*}
For part b), you have $g(1) = 0$ if $g \not\equiv 0$. The working is very similar, so you can try it out yourself.
A: By putting $x=0=y$ we have
$$f(x+y)=f(x)f(y) \implies f(0)=f^2(0) \implies f(0)=0 ~or~ 1.$$
By putting $y=0$, we get $f(x)=f(x) f(0)= f(x)$, thus $f(0)=1$.
Next, the right limit of $f(x)$ as $h\rightarrow 0$ $$f(0^+)=\lim_{h \rightarrow 0} f(0+h)=\lim_{h \rightarrow 0} f(0)f(h)=\lim_{h \rightarrow 0} f(h) ~~~~~~~(1)$$
Smilarly the left limit
$$f(0^-)=\lim_{h \rightarrow 0} f(0-h)=\lim_{h \rightarrow 0} f(0)f(-h)=\lim_{h \rightarrow 0} f(-h) ~~~~~~~(2)$$
Given that $f(x)$ is cntinuous at $x=0$, then
$$\lim_{h \rightarrow 0} f(h) = \lim_{h \rightarrow 0} f(-h)=f(0)=1~~~~~~~(3)$$
Let us show the continuity at any real number  $x=a$
$$f(a^+)=\lim_{h \rightarrow 0} f(a+h) = f(a)\lim_{h \rightarrow 0} f(h)=f(a)~~~from~~(3)$$
Dimilarly the left limit
$$f(a^-)=\lim_{h \rightarrow 0} f(a-h) = f(a)\lim_{h \rightarrow 0} f(-h)=f(a)~~~from~~(3)$$
This shows that the function is continuous at any real non zero value of $x$ provided iy is
continuous at $x=0$.
