As in the title, I'd like to know if it's true that:
$$| u^j|_{B^{-1}_{2,\infty}} \leq C 2^{-j\sigma} | u^j |_{B^{\sigma-1}_{2,\infty}} $$ if $\hat{u^j} \subset 2^j B$ ($B$ being a ball, $| \cdot |_{B^{\sigma}_{pq}}$ a Besov space) and let $\sigma > 1$
If $\hat{u^j} \subset 2^j C$ for $C$ being an annulus, then it's true by the following computation:
$$| u^j |_{B^{s_1}_{p,\infty}} = 2^{js_1} | u^j |_p = 2^{js_1} 2^{js_2} 2^{-js_2} | u^j |_p = 2^{-js_2} | u^j |_{B^{s_1+s_2}_{p,\infty}}$$
however, if $u^j$ is in the ball, then we can't just equate Besov norm to the $p$-norm multiplied by $2^{js}$, but we have to do $$| u^j |_{B^{\sigma}_{p\infty}} = \sup_{k < j + 2} 2^{k\sigma} | \Delta_k u^j |_p $$,
which in my special case causes an issue with $2^{-k}$. In the absence of the lower bound for $k$ I'm not able to deal with it properly. In particular, if I try similar "multiply by 1" trick:
$$\sup_k 2^{-k} 2^{j\sigma} 2^{-j\sigma} | \Delta_k u^j | = 2^{-j\sigma} \sup_{k<j+2} (2^{j\sigma} 2^{-k}) | \Delta_k u^j | \leq 2^{-j\sigma} \sup_{k < j+2} 2^{k\sigma -k } | \Delta_k u^j |$$
By Prop 2.33 in the Bahouri-Chemin-Danchin I have:
$$| u^j |_{B^{-1}_{p,\infty}} \leq C \sup_k 2^{-k} | S_k u^j |_p $$
with $S_k u = \sum_{l \leq k} = \Delta_l u$, but it doesn't seem to help much.
If it matters, my function $u^j$ has FT supported in a ball because it is a product of two other functions which are localized in the same annulus.