TITLE: Proving a function is continuous everywhere
QUESTION [6 upvotes]: Suppose $f:\mathbb{R} \rightarrow \mathbb{R}$ continuous at 0, and $f(a+b) = f(a) + f(b) \forall a,b \epsilon \mathbb{R}$. Show that f is in fact continuous everywhere.
Let $\epsilon > 0$.
There exists $\delta_1>0$ such that $|x - 0| < \delta_1 \rightarrow |f(x) - f(0)|<\epsilon$. We want to show that, for c other than 0, there exists $\delta>0$ such that $|x-c|<\delta \rightarrow |f(x) - f(c)|<\epsilon$. We note that $|f(x) - f(c)| = |f(x-c)|$. 
Let $y = x-c$. Then, we want to prove $|y| < \delta \rightarrow |f(y)| < \epsilon$.
But, we set $\delta = \delta_1$.
Thus, $|f(y)| = |f(x-c)| < \epsilon.$
Thus $f(x)$ is continuous everywhere.
Is this proof correct? If not, I would appreciate guidance towards the right direction.

REPLY [3 votes]: It is correct. You might also like this one :

If $f(a) + f(b) = f(a+b)$ for all $a,b \in \mathbb R$, and if for some $\eta >0$ and $M$ positive integer we have $|f(x)|< M$  for all $|x| \leq \eta$, then $f$ is continuous everywhere.

Proof : Suppose this is the case. Then, I claim $f$ is continuous at $0$. Indeed, let $\epsilon > 0$, WLOG $\epsilon$ is rational, say $\epsilon = \frac pq$ with $p,q$ 
 positive integers (otherwise take a rational between $\epsilon$ and zero and use that in place of $\epsilon$ in the below argument), note that if $|x| \leq \frac{\eta \epsilon}{\ M}$ then $$
|x| \leq \frac{p\eta}{qM} \implies \frac{qM|x|}{p} \leq \eta \implies \left|f\left(\frac {qMx}{p}\right)\right| \leq M \tag{*}
$$
Now note that $f\left(\frac{qMx}{p}\right) = \frac {qM}{p}f(x)$ since 
$$
f(qMx) = \underbrace{f\left(\frac{qMx}{p}\right) + \ldots + f\left(\frac{qMx}{p}\right)}_{p \textrm{ times}} = pf\left(\frac{qMx}{p}\right)
$$
and
$$
f(qMx) = \underbrace{f(x)+ \ldots + f(x)}_{qM \textrm{ times}}
 = qMf(x)$$
combining these results gives the claim. Now, from $(^*)$ we get $|f(x)| \leq \epsilon$ , as desired.
In general, for linear functions from $\mathbb R^d \to \mathbb R^m$, continuity and boundedness around zero(or any arbitrary fixed point) are equivalent, and also equivalent to continuity everywhere.