We say a function $f\in L^1_{loc}(\mathbb{R})$ is in $\mathrm{BMO}(\mathbb{R})$ if
$$\|f\|_{\mathrm{BMO}}=\sup_{I}\frac{1}{|I|}\int\limits_I |f(y)-f_I|\, dy<\infty$$ for all intervals $I\subset\mathbb{R},$ where $f_I$ is the average value of $f$: $$f_I=\frac{1}{|I|}\int_I f(y)\, dy.$$
$f\in \mathrm{VMO}(\mathbb{R})$ if $f \in\mathrm{BMO}$ and $$\lim_{|I|\rightarrow 0}\frac{1}{|I|}\int_I|f(y)-f_I|\, dy\rightarrow 0. $$
For more details, see Bounded mean oscillation.
Some facts that maybe useful
- UC $\cap$ BMO $\subset$ VMO, where UC means uniformly continuous functions in $\mathbb{R}$.
- VMO is the closure of UC $\cap$ BMO in BMO.
My question is
1.If $f\in $ VMO, is $f^2\in$ VMO? (If yes, next question is trivial; if not, please see next question)
2.If $\omega\in A_\infty$,A_p weight and $\log \omega\in$ VMO, is $(\log \omega)^2 \in$ VMO?
My try:
Divide $f=f_1+f_2$, where $f_1\in \mathrm{UC}\cap \mathrm{BMO},\ \|f_2\|_{BMO}<\epsilon$. But I can't go on next. I am not sure it is correct, if not, could you give me a counterexample? Thanks very much.