1
$\begingroup$

This question already has an answer here:

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a function such that $f(x+y)=f(x)+f(y)$. If $f$ is continuous at zero how can I prove that is continuous in $\mathbb{R}$.

$\endgroup$

marked as duplicate by Adriano, Daniel Fischer real-analysis Oct 20 '14 at 19:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @Daniel: D'oh! :-) In my defense, the Hebrew "vav" is used, amongst other things, to express "u" sounds. :-P $\endgroup$ – Asaf Karagila Oct 20 '14 at 19:56
  • $\begingroup$ By sliding around a neighborhood of zero. In other words if $\delta$ is very close to zero, then by continuity, so is $f(\delta)$, and so is $f(x + \delta) = f(x) + f(\delta)$. Now just formalize that argument. $\endgroup$ – user4894 Oct 20 '14 at 19:59
  • $\begingroup$ See also here: math.stackexchange.com/questions/93816/… (And as was already pointed out, you can find more related posts here: math.stackexchange.com/questions/423492/…) $\endgroup$ – Martin Sleziak Feb 15 '16 at 13:51
2
$\begingroup$
  1. Show $f(0)=0$
  2. For any $x\in\mathbb{R}$, $|f(x+h)-f(x)|=|f(h)|$
$\endgroup$
1
$\begingroup$

Prove it is Lipschitz. (Actually you can prove it is of the form $f(x) = cx$ for some constant $c$.) See also this thread.

$\endgroup$
1
$\begingroup$

Hint: Use the definition, prove that $f(0)=0$ and use $$f(a+\varepsilon)-f(a)=f(\varepsilon)\,.$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.