Suppose we have an annulus $A \subset \mathbb R^n$, which is the set $\{x|0<r \leqslant\|x\| \leqslant R\}$, $\alpha \in \mathbb R$ and $\{ u_j\}_{j\geqslant-1} $ be a seqence of smooth functions such that the Fourier transformations $\mathcal{F}(u_j) $ is supported in $2^jA$, and $\|u_j\|_{L^\infty}\leqslant C2^{-j\alpha}$ for all $j$, where $C$ is a constant independent of $j$. How to show the series$$\sum_{j\geqslant-1}u_j$$ converges in the Besov space $B^\alpha_{\infty,\infty}$?
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$\begingroup$ You can't. (This is essentially the same phenomenon that $L^\infty$ functions cannot always be approximated by functions with compact support.) // However, you can prove convergence in $B^{\alpha-\epsilon}_{\infty,\infty}$ for any $\epsilon > 0$. $\endgroup$– Willie WongCommented Dec 9, 2021 at 14:10
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$\begingroup$ @WillieWong Then how to understand the statement of lemma A.3 in the paper arxiv.org/abs/1210.2684 ? $\endgroup$– InuyashaCommented Dec 9, 2021 at 16:17
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1$\begingroup$ The series can converge in a weaker space (as I mentioned in my previous comment) AND the limit can belong to the stronger space, without the convergence being in the stronger space. (Basic example, let $f_j = \chi_{[j,j+1)}$, then $\sum_{j\geq 0} f_j$ converges pointwise to $\chi_{[0,\infty)}$, the limit is in $L^\infty$, but the convergence is not in $L^\infty$.) $\endgroup$– Willie WongCommented Dec 9, 2021 at 16:46
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