# Vanishing Mean Oscillation

The oscillation of function $f$ on a set $A$ is the diameter of $f(A)$. For real-valued functions it’s written as $\mathrm{osc}_A\,f=\sup_A f - \inf_A f$. The relation to uniform continuity is immediate: $f$ is uniformly continuous if and only if $\mathrm{osc}_I\,f$ is small ( $<\epsilon$) on all sufficiently short intervals ( $|I|<\delta$). Here I'm following the traditional definition of uniform continuity here, even though it’s wrong.

In statistics, the difference (maximum-minimum) is called the range of a set of data values, and is one of common measures of variation. It is a crude one, though, and is obviously influenced by outliers. There are better ones, such as standard deviation: subtract the mean, square each term, then average them, and take square root… you know the drill. What happens if we use the standard deviation in the definition of uniform continuity?

We get the space of functions of vanishing mean oscillation, known as VMO. To avoid technicalities, I will consider only the functions $f\colon \mathbb R\to\mathbb R$ that vanish outside of the interval $[-1,1]$. Like this one:

Given an interval $I\subset \mathbb R$, let $f_I$ denote the mean of $f$ on $I$, namely $\frac{1}{|I|}\int_I f$. By definition, $f$ is in VMO if and only if $|f - f_I|_I$ is small ( $<\epsilon$) on all sufficiently short intervals ( $|I|<\delta$).

Wait, what happened to squaring and taking a root afterwards? Turns out they are not really necessary: the above definition is equivalent to the one with standard deviation $\sqrt{(|f - f_I|^2)_I}$. In fact, VMO functions are integrable to any power $1\le p<\infty$, and any such $p$ can be used in the definition (a consequence of the John-Nirenberg lemma). Both squared and unsquared versions have their advantages:

• $|f - f_I|_I$ is finite for any locally integrable function; we don’t need to know in advance that $f$ is square integrable.
• $\sqrt{(|f - f_I|^2)_I}$ is better suited for precise computation, since it’s just $\sqrt{ (f^2)_I-f_I^2}$.

For example, taking $I=[-1,1]$ for the hat function, we get $f_I=1/2$ and $(f^2)_I=1/3$. So, $\sqrt{(|f - f_I|^2)_I} = 1/\sqrt{12}$, and this is as large as it gets. The supremum of mean oscillation over all intervals is called the BMO norm. The space BMO, naturally, includes all functions with finite BMO norm. VMO is a closed subspace of BMO — the closed span of continuous functions (uniformly continuous and vanishing at infinity, to be precise). In particular, VMO is separable while BMO is not.

The definition of VMO clearly rules out jump discontinuities such as $x/|x|$. Yet, it allows some discontinuous functions and even unbounded ones. The standard example of an unbounded function in VMO is the double logarithm:

If this function does not look unbounded, it’s because it reaches level $y=5$ only when $|x|\approx 10^{-64}$, which is well below the resolution of your screen. Even if you have Retina display.