Does infinitesimal variance imply continuity?

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?

• From what I remember, a Dini type condition suffices. – Liviu Nicolaescu Mar 14 '16 at 22:27
• 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. – Teri Mar 14 '16 at 22:52
• 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? – Alexandre Eremenko Mar 15 '16 at 2:19
• @AlexandreEremenko: Certainly the claim is that there is a continuous function that agrees with $u$ a.e. – Christian Remling Mar 15 '16 at 2:28
• @Christian Remling: Fine. But I expect an exact statement:-) – Alexandre Eremenko Mar 16 '16 at 1:18

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.