How to prove $$ \lVert uv\rVert_{\dot{B}^{\frac{N}{p}-1}_{p,1}}\leqslant C \lVert u\rVert_{\dot{B}^{\frac{N}{p}}_{p,1}} \lVert v\rVert_{\dot{B}^{\frac{N}{p}-1}_{p,1}}$$
when $N\geqslant2 $and$1\leqslant p<2N$.
I know it needs Bony decomposition, but I don’t know how to use the condition $N\geqslant2 $and$1\leqslant p<2N$. I could handle the paraproduct. But when dealing with the remainder, I have to assume $p\geqslant2 $and$p<2N$ instead of $N\geqslant2 $.
Any hints would be appreciated!
I shall write $\mathrm{P}_k$ for the homogeneous Littlewood-Paley projectors, and the paraproduct decomposition as
$$ uv = u\prec v + u\succ v + u\diamond v $$
Assume $1\le p<2$. Then $p'>p$ and Bernstein applies to estimate the $L^{p'}_x$ norm in terms of the $L^p_x$ norm, at a cost of derivatives. The resonant term $u\diamond v$ can be estimated as follows.
$$\begin{aligned} 2^{(\frac Np-1)k}\left\|\mathrm{P}_k(u\diamond v)\right\|_{L^p_x} &\lesssim 2^{(N-1)k}\left\|\mathrm{P}_k(u\diamond v)\right\|_{L^1_x} \\&\lesssim 2^{(N-1)k}\sum_{\substack{k_*\in\mathbb{Z}\,:\\k_*\ge k-O(1)}} \left\|\mathrm{P}_{k_*}u\right\|_{L^{p'}_x} \left\|\mathrm{P}_{\approx k_*}v\right\|_{L^p_x} \\&\lesssim 2^{(N-1)k}\sum_{\substack{k_*\in\mathbb{Z}\,:\\k_*\ge k-O(1)}} 2^{(\frac Np-\frac N{p'})k_*}\left\|\mathrm{P}_{k_*}u\right\|_{L^p_x} \left\|\mathrm{P}_{\approx k_*}v\right\|_{L^p_x} \\&\approx \sum_{\substack{k_*\in\mathbb{Z}\,:\\k_*\ge k-O(1)}} 2^{(N-1)(k-k_*)} \left(2^{\frac Npk_*}\left\|\mathrm{P}_{k_*}u\right\|_{L^p_x}\right) \left(2^{(\frac Np-1)k_*}\left\|\mathrm{P}_{\approx k_*}v\right\|_{L^p_x}\right) \;.\end{aligned} $$
Because $N-1>0$, we may apply Young's convolution inequality on $\mathbb{Z}$ to obtain
$$ \begin{aligned} \left\|u\diamond v\right\|_{\dot{B}^{\frac Np-1}_{p,1}} &\lesssim \sum_{k_*\in\mathbb{Z}} \left(2^{\frac Npk_*}\left\|\mathrm{P}_{k_*}u\right\|_{L^p_x}\right) \left(2^{(\frac Np-1)k_*}\left\|\mathrm{P}_{\approx k_*}v\right\|_{L^p_x}\right) \\&\lesssim \left\|u\right\|_{\dot{B}^{\frac Np}_{p,1}} \left\|v\right\|_{\dot{B}^{\frac Np}_{p,1}} \;.\end{aligned} $$
