The following exercise is from "Modern Fourier Analysis" by Loukas Grafakos in page 129 Exercise 7.1.5.
Let $a > 1$, let $B$ be a ball (or a cube) in $\mathbb{R}^n$, and let $aB$ be a concentric ball whose radius is $a$ times the radius of $B$. Show that there is a dimensional constant $C_n$ such that for all $f$ in BMO we have $$ |f_{aB}-f_B|\le C_n \log(a+1)\|f\|_{BMO}. $$
When i write formally i found that $|f_{aB}-f_B|\le a^n \|f\|_{BMO}$ and then i could not continue. Are there any hints?
By your computation the result holds if $1<a\leq 2$. Now for $a>2$ pick $k$ to be the integer such that $2^kB\subset aB\subset 2^{k+1}B$ and 'telescope' $$ |f_{aB}-f_B|\leq |f_{aB}-f_{2^kB}|+ \sum_{i=1}^k |f_{2^iB}-f_{2^{i-1}B}|. $$ Each of these terms can be estimated by the case $a\leq 2$ above to get $$ |f_{aB}-f_B| \leq C_n(1+k)\| f\|_{BMO}. $$ Now just notice that $k+1 \leq \frac{2}{\log(2)} \log(a)$ since $a>2$ and $2^k\leq a$.