Proving that an additive function $f$ is continuous if it is continuous at a single point 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.
 A: Fix $a\in \mathbb{R}.$
Then 
$\begin{align*}\displaystyle\lim_{x \rightarrow a} f(x) &= \displaystyle\lim_{x \rightarrow x_0} f(x - x_0 + a)\\ &= \displaystyle\lim_{x \rightarrow x_0} [f(x) - f(x_0) + f(a)]\\& = (\displaystyle\lim_{x \rightarrow x_0} f(x)) - f(x_0) + f(a)\\ & = f(x_0) -f(x_0) + f(a)\\ & = f(a).
\end{align*}$
It follows $f$ is continuous at $a.$
A: Suppose $x=x_0, f(x_0) + f(y) = f(x_0 + y)$
taking limit as $y$ tends to $0$, we get $\lim_{y\rightarrow 0} (f(x_0) + f(y)) = \lim_{y\rightarrow 0} f(x_0 + y)$
From the continuity at $x_0$, we know that RHS of the above equation is $f(x_0)$ which means that $\lim_{y\rightarrow 0} f(y) =0$
Next bit it simple. For any $x,y$ in the domain, $f(x+y)=f(x)+f(y)$ continuity can be established by checking whether $\lim_{y\rightarrow 0} f(x+y) =f(x)$ which is true since $f(x+y)=f(x)+f(y)$ and $\lim_{y\rightarrow 0}f(y)=0$. Hence $f(x)$ is continuous everywhere in the domain.
I'm new here so I don't know how  to input equations using latex. Pls bear with it.
A: Let's examine your situation. You have that $\lim_{x\rightarrow a} f(x) = f(a)$ for some $a\in \mathbb{R}$ and that for any $x,y\in \mathbb{R}, f(x)+f(y)=f(x+y)$. You want to prove that for any $c\in \mathbb{R}, \lim_{x\rightarrow c} f(x) = f(c)$. The key step here is to realize that
$$\lim_{x\rightarrow c} f(x) = \lim_{x\rightarrow a} f(x-a+c)$$
because $|(x-a+c)-c| = |x-a|$, so in plain english $x$ is close to $a$ if and only if $x-a+c$ is close to $c$. We can then complete the proof as follows:
$$\lim_{x\rightarrow a} f(x-a+c) = \lim_{x\rightarrow a} (f(x) + f(c-a)) = f(a) + f(c-a) = f(c)$$
A: 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.
