TITLE: Does infinitesimal variance imply continuity?
QUESTION [8 upvotes]: 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?

REPLY [8 votes]: 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.