Brezis and Nirenberg claim that function $f(x) = \sin(\log | \log |x||)$ belongs to VMO (vanishing mean oscillation). It is easier to verify that function $\log | \log |x||$ belongs to VMO. But how can I verify that the composite function $f(x) = \sin(\log | \log |x||)$ belongs to VMO?
Here the function $f$ is in $W^{1,n}(B_r(0)),\, r>0,\, 0\neq x \in \mathbb{R}^n\setminus \partial B_1(0)$.
The article I refer to is the following: https://link.springer.com/article/10.1007/BF01671566
The point $y\in\mathbb{R}^n$ is a Lebesgue point if:
$$ \lim_{r\to 0} \oint_{B_r(y)} |f(x)-f(y)|\; dx = 0 $$
The point $y\in\mathbb{R}^n$ is a VMO point (vanishing mean oscillation) if:
$$ \lim_{r\to 0} \oint_{B_r(y)} |f(x)-(f)_{y,r}|\; dx = 0 $$
How can I prove that 0 is not a Lebesgue point but a VMO point?
Here $\oint_U f=\frac{1}{|V|}\int_V f$ is the average integral of $f$ (I don't know the command in LateX to write an integral with a dash in the middle). And $(f)_{y,r} = \oint_{B_r(y)}$.
I tried to understand the references in the article itself. But there are a large number of propositions that justify the fact that a Lipschitz function composed with a VMO function also belongs to VMO. I have an intuition that the way to prove that the F function belongs to VMO may be more straightforward. But I still haven't succeeded. Any help in this regard would be appreciated.
So you need to show there exists $a_\delta \in \mathbb R$ such that $$ \frac1{2\delta}\int_{-\delta}^\delta |\sin(\log|\log|x||) - a_\delta| \, dx \to 0 $$ as $\delta \to 0$. I suggest choosing $a_\delta = \sin(\log|\log|\delta|)$. I think the trick is to realize that the intervals in which $\sin(\log(-\log(-x)))$ change sign get exponentially shorter and shorter. On the first interval, you can remove the outer absolute values, and do the usual integration by parts to see that it is very small. On the other intervals, just bound the integrand by $1$, but see that the sum of the lengths of the other intervals is very small compared to $\delta$.
This argument is going to be a bit fiddly around $\sin(\log|\log|\delta||) = \pm1$, because the sign changes will be close to $\delta$. But I think by slightly changing the value of $a_\delta$, you should be able to get around it.