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~
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For more, see https://math.stackexchange.com/questions/423492/overview-of-basic-facts-about-cauchy-functional-equation – Chris Culter Sep 26 '17 at 08:06
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Also this one https://math.stackexchange.com/questions/385586/do-there-exist-functions-satisfying-fxy-fxfy-that-arent-linear – velut luna Sep 26 '17 at 08:12
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Here's a MathJax tutorial :) – Shaun Sep 26 '17 at 08:19
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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 $f(r)$.)
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Thanks I just got this part from another post actually, so now I am left the second part of my problem above. – Homaniac Sep 26 '17 at 09:02