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Let $u:[0,1]\to\mathbb{R}^n$ be a bounded Borel function. It is well-known that if, for any compact interval $I\subseteq [0,1]$, $$ \int_I|u-u_I|^2\le C|I|^{1+\alpha} $$ for some $C,\alpha>0$ (here $u_I:=\frac{1}{|I|}\int_I u$), then $u$ is in fact $\frac{\alpha}{2}$-Holder continuous: this was first proved by Campanato in 1963.

Q: Is it true that if $$ \int_I|u-u_I|^2\le |I|\omega(|I|) $$ for any $I$ then $u$ is continuous? Here $\omega$ denotes an arbitrary modulus of continuity. If this is false in general, can one characterize the $\omega$'s for which this is true?

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  • $\begingroup$ From what I remember, a Dini type condition suffices. $\endgroup$
    – Liviu Nicolaescu
    Mar 14, 2016 at 22:27
  • $\begingroup$ I think if you go through the proof of Campanato's result you end up using a condition of the form $$\sum_k\omega(2^{-k}r)\le C\omega(r), $$ where $2^{-k}$ can be replaced by $\rho^k$ for some $0<\rho<1$. I would guess the continuity can fail without any assumption on $\omega$ but I'm not sure the one above is necessary either. $\endgroup$
    – Teri
    Mar 14, 2016 at 22:52
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    $\begingroup$ I am afraid you misstated Campanato's theorem: take $u(x)=1$ for $x$ rational, and $u(x)=0$ otherwise. This is a bounded Borel function, is not it? $\endgroup$
    – Alexandre Eremenko
    Mar 15, 2016 at 2:19
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    $\begingroup$ @AlexandreEremenko: Certainly the claim is that there is a continuous function that agrees with $u$ a.e. $\endgroup$
    – Christian Remling
    Mar 15, 2016 at 2:28
  • $\begingroup$ @Christian Remling: Fine. But I expect an exact statement:-) $\endgroup$
    – Alexandre Eremenko
    Mar 16, 2016 at 1:18

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Since $u$ is assumed bounded, your condition is equivalent to $$ \frac{1}{|I|} \int_I |u-u_I|\, dx =o(1) $$ as $|I|\to 0$, uniformly in $I$, and this is the condition that defines VMO.

So you are asking if functions in $VMO\cap L^{\infty}$ are continuous, and this is known to be false. This classical paper by Sarason introduced VMO; Theorem 1(iv) there answers your question, modulo facts about the Hilbert transform. Here's a more direct reference.

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  • $\begingroup$ Nice! Can anything be said about the good moduli of continuity? $\endgroup$
    – Mizar
    Mar 15, 2016 at 9:16

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