Consider $f_1(x)=-1/\log(|x|)$ and $f_2(x)=1/\sqrt{-\log(|x|)}$ around $x=0$. It can be shown, just by contradiction, that the previous functions do not verify any (positive) Hölder condition: $$ |f(x)-f(y)|\leq C|x-y|^\alpha $$ with $C>0$ and $\alpha\in(0,1)$. When $\alpha\in\mathbb{R}$ the Besov spaces (with $p=q=\infty$) can be considered as the "natural extension" of the Hölder spaces (roughly speaking the set of functions that verifies a $\alpha$-Hölder condition).
The question is, any of you know how to prove (or a hint) that $f_1$ and $f_2$ are in "negative" Besov space (that is $\alpha$-Hölder with $\alpha<0$ if it makes sense...)? They are in "different spaces" as $x\log(x)$ and $\sqrt{x}$?
Thanks guys!
Write $f_1=F_1'$ and $f_2=F_2'$. Then check $F_j\in B^{1+\alpha}_{\infty,\infty}$. The latter holds even with $\alpha=0$, because both $F_j$ are Lipschitz - they are antiderivatives of bounded functions. Hence, $f_1,f_2\in B^0_{\infty,\infty}$. The derivative operator reduces $\alpha$ by $1$ and leaves $p,q$ unchanged.
By the same reasoning, $L^\infty$ functions on a bounded domain are in $B^0_{\infty,\infty}$. See this MO answer.