TITLE: Proving that an additive function $f$ is continuous if it is continuous at a single point
QUESTION [21 upvotes]: Suppose that $f$ is continuous at $x_0$ and $f$ satisfies $f(x)+f(y)=f(x+y)$. Then how can we prove that $f$ is continuous at $x$ for all $x$? I seems to have problem doing anything with it. Thanks in advance.

REPLY [2 votes]: Preliminaries: $f(0)=f(0+0)=f(0)+f(0)\implies 2f(0)=f(0)\implies$
$$f(0)=0$$
$f(0)=f(x-x)=f(x+(-x))=f(x)+f(-x)\implies f(x)+f(-x)=0\implies$
$$ f(-x)=-f(x)$$
$\displaystyle{\lim_{x\to x_0}(f(x-x))}=\displaystyle{\lim_{x\to x_0}(f(x)-f(x))}=\displaystyle{\lim_{x\to x_0}(f(x))-\lim_{x\to x_0}(f(x))}=f(x_0)-f(x_0)=0$
$$\mbox{By theorem: If }f~\mbox{and }g~\mbox{are continuous, then }f-g~\mbox{is continuous}.$$
$$\therefore f~\mbox{is continuous at }0$$
Proof:
Choose any $c\in \mathbb{R}$ and let $h=x-c$. (as $x$ approaches $c$, $h$ approaches $0$)
$$x\to c\iff h\to 0$$
\begin{align*}
\lim_{x\to c}(f(x))&=\lim_{h\to 0}(f(h+c))~~~~~~~~~~~~~~(\mbox{Because }x=h+c)\\
&=\lim_{h\to 0}(f(h)+f(c))\\
&=\lim_{h\to 0}(f(h))+\lim_{h\to 0}(f(c))\\
&=f(0)+f(c)~~~~~~~~~~~~~~~~~~(f~\mbox{continuous at 0 and }f(c)~\mbox{constant})\\
&=0+f(c)~~~~~~~~~~~~~~~~~~~~~~~(f(0)=0)\\
&=f(c)
\end{align*}
Because
$\displaystyle{\lim_{x\to c}(f(x))=f(c)}$ ($f$ continuous at c) and $c$ is arbitrary, then $f$ is continuous for all real numbers.