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Can you guys help me with this question? I've been thinking over it all day long.

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a Lebesgue measurable function, and there exist two constants $a,b$ $\in \mathbb{R} $ $s.t.$ for $\forall$ $l,n$ $\in \mathbb{Z}$ that are all not equal to zero, $la+nb \neq 0$ and $ f(x) \doteq f(x+a), f(x) \doteq f(x+b)$. Prove that there exists a constant c $\in \mathbb{R} s.t. f \doteq c$.

Dotequal means "equality holds a.e. by Lebesgue measure" here.

Hodge Chen
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  • Related – TheSilverDoe Jul 07 '22 at 14:50
  • Proof idea: first compose $f$ with a squashing function (i.e., an injective continuous function $f :\Bbb{R}\to (-1,1)$) to make $f$ bounded. Then smooth $f$ out by convolution and show that it satisfies the same condition. This reduces the question to the case where $f$ is continuous. This case should not be too hard. – PhoemueX Jul 09 '22 at 15:30

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