0
$\begingroup$

Say we have $f(x+y) = f(x) + f(y) \quad \forall x,y \in \mathbb R$ and $f$ is continuous at one point at least. I wish to show there must be some $c$ such that $f(x)=cx$ for all $x$. Think I can do so by first showing $f$ is continuous everywhere I'm not sure how then let $f(q) = 1$ somehow and show that $f(q) = cq$ where $q$ is rational. But then the aim is to show for all real $x$ so I am not sure~

$\endgroup$

marked as duplicate by Chris Culter, 5xum, Shaun, Matthew Leingang, José Carlos Santos Sep 26 '17 at 9:03

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.

1
$\begingroup$

Hint:

Let $x_0$ be the point where the function is continous. By continuity we have

$$\lim_{x\to x_0}f(x)=f(x_0)$$ or

$$\lim_{x\to x_0}f(x)-f(x_0)=\lim_{x\to x_0}f(x-x_0)=0.$$

Then

$$\lim_{x\to x_1}f(x)=\lim_{x\to x_0}f(x-x_0+x_1)=\lim_{x\to x_0}f(x-x_0)+f(x_1)=f(x_1)$$ and the function is continuous everywhere.

Then you can indeed show that for all rationals $f(q)=cq$, and by squeezing this extends to all rationals.

$$q_0<r<q_1\implies f(q_0)=cq_0<f(r)<f(q_1)=cq_1.$$

(You will need to fiddle a little with neighborhoods of $r$ and $f(r)$.)

$\endgroup$
  • $\begingroup$ Thanks I just got this part from another post actually, so now I am left the second part of my problem above. $\endgroup$ – Homaniac Sep 26 '17 at 9:02

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