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Is there a function $f:\mathbb R\to\mathbb R$ such that $f(x+y)=f(x)+f(y)$ which is not continuous? I have proved that if it's continuous in one point $a\in\mathbb R$ then it's continuous on all $\mathbb R$, but I didn't find such a function which is not continuous everywhere. Therefore I tried to prove that all function of this form is continuous at $x=0$ but with no success. I think that if such a function exist it would be of the form $f(x)=...$ if $x\in\mathbb Q$ and $f(x)=...$ if $x\in\mathbb R\backslash\mathbb Q$ but I didn't find it.

Martin Sleziak
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idm
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    Nope! Additive functions can be kind of crazy if you don't assume continuity. Look up Cauchy's functional equation. – Cameron Williams Jul 21 '15 at 06:24
  • google constructions of discontinuous 'additive functions' with Hamel Basis. – r9m Jul 21 '15 at 06:27
  • This is true if we suppose f measurable. – Euler88 ... Jul 21 '15 at 06:27
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    It is consistent with ZF that such a function is aways cointinuous – Hagen von Eitzen Jul 21 '15 at 07:47
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    You can find pointers to several related questions here. For example: http://math.stackexchange.com/questions/269434/non-continuous-function-satisfies-fxy-fxfy, http://math.stackexchange.com/questions/385586/do-there-exist-functions-satisfying-fxy-fxfy-that-arent-linear, http://math.stackexchange.com/questions/16175/on-sort-of-linear-functions, http://math.stackexchange.com/questions/487873/many-other-solutions-of-the-cauchys-functional-equation – Martin Sleziak Jul 21 '15 at 07:48

2 Answers2

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The field $\mathbb{R}$ is a vector space over $\mathbb{Q}$. Let $\{e_\alpha\}_{\alpha \in I}$ be a basis (note that it's uncountable!).

Then any $x \in \mathbb{R}$ can be uniquely written $x = \sum a_\alpha e_\alpha$ where all but finitely many of the $a_\alpha$ are $0$. Pick your favorite index $\beta \in I$ and define $f(\sum a_\alpha e_\alpha) = a_\beta$.

Then $f$ is linear, but it can't be continuous because it's a non-constant function from $\mathbb{R}$ to $\mathbb{Q}$.

hunter
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Yes, a discontinuous function satisfying this constraint exists.

For more, see here: Edwin Hewitt and Herbert S. Zuckerman: Remarks on the Functional Equation $f(x+y) = f(x)+f(y)$, Mathematics Magazine, Vol. 42, No. 3 (May, 1969), pp. 121-123 JSTOR, DOI: 10.2307/2689122

Martin Sleziak
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Danny
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