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I am trying to prove that if I have $f:\mathbb{R} \to \mathbb{R}$ satisfying $\forall x,y\in\mathbb{R},f(x+y) = f(x) + f(y)$. Which is assumed continuous at $0$, that $f$ is continuous on $\mathbb{R}$

I am fairly sure it is, for that property seems to be a property of polynomials, and we know polynomials are continuous where defined(all reals)

Any ideas for rigor?

$|x-a| \lt \delta \implies |f(x+a)-f(x)-f(a)| \lt \epsilon$

No idea where to go from here, this is my first time doing $\epsilon-\delta$ stuff.

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  • $\begingroup$ The only polynomials satisfying that property are those of the form $p(x) = ax$. $\endgroup$ – Santiago Canez Apr 29 '14 at 22:53
  • $\begingroup$ Cauchy's Functional Equation, See: cofault.com/2010/01/hunt-for-addictive-monster.html $\endgroup$ – Alan Apr 29 '14 at 23:12
  • $\begingroup$ Proving that an additive function $f$ is continuous if it is continuous at a single point. Pointers to a few more facts about such functions can be found here. $\endgroup$ – Martin Sleziak Nov 2 '14 at 10:00
  • $\begingroup$ @MartinSleziak I don't think I realised at the time, but this is(the satisfied property) homomorphism isn't it. Thank you for the links $\endgroup$ – user142198 Nov 2 '14 at 10:10
  • $\begingroup$ Yes, you are right. The condition is equivalent to saying that $f$ is a group homomorphism from $(\mathbb R,+)$ to $(\mathbb R,+)$. (I stressed the word group homomorphism, since $\mathbb R$ is not only a group, but also a ring, field, poset and many other things.) $\endgroup$ – Martin Sleziak Nov 2 '14 at 10:23
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For any $x$, $$f(x+h)-f(x)=f(h)$$

Since $f$ is continuous at $x=0$, $f(h)\to 0$ as $h\to 0$.

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Putting $x=y=0$, $f(0) = 0$.

Since $f$ is assumed to be continuous at $0$, $\lim_{x \to 0} f(x) = f(0) = 0$.

Finally, as Pedro Tamaroff wrote, since $f(x+h)-f(x) = f(h)$, $\lim_{h \to 0} f(x+h)-f(x) = \lim_{h \to 0} f(h) =0$, so $f$ is continuous at all $x$.

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