For $u,v\in \mathcal{S}(\mathbb R^d)$, $s,s_1,s_2\in\mathbb R$,
$$
\|uv\|_{H^s}\lesssim_{d,s,s
_1}\|u\|_{H^{s_1}}\|v\|_{H^{s_2}}.
$$
where $s_1+s_2=s+d/2>0$.
My attempts: We work in a direct estimate in Fourier side except introducing Bony's decomposition. Indeed, $$ |(\hat{u}\ast \hat{v})(\xi)|\le \left(\int_{|\eta|\le |\xi-\eta|}+\int_{|\eta|\ge |\xi-\eta|} \right)|\hat{u}(\eta)\hat{v}(\xi-\eta)| \,\mathrm{d}\eta. $$ For the first term we estimate as $$ \begin{aligned} & [\int_{|\eta|\le |\xi-\eta|}|\hat{u}(\eta)\hat{v}(\xi-\eta)| \,\mathrm{d}\eta]^2\\ \le&[\int_{|\eta|\le |\xi-\eta|}\langle \eta \rangle^{t} |\hat{u}(\eta)|\cdot \langle \eta \rangle^{-t} |\hat{v}(\xi-\eta)| \,\mathrm{d}\eta]^2 \\ \le& \|u\|_{H^t}^2 \int_{|\eta|\le |\xi-\eta|} \langle \eta \rangle^{-2t} |\hat{v}(\xi-\eta)|^2 \,\mathrm{d}\eta=\|u\|_{H^t}^2 \int_{|\eta|\ge |\xi-\eta|} \langle \xi-\eta \rangle^{-2t} |\hat{v}(\eta)|^2 \,\mathrm{d}\eta\,. \end{aligned} $$ Thus, $$ \|uv\|_{H^s}^2= \int_{\mathbb R^d}\langle \xi \rangle^{2s}|(\hat{u}\ast \hat{v})(\xi)|^2 \,\mathrm{d}\xi\le J_1+J_2, $$ where $$ \begin{aligned} J_1:&=\|u\|_{H^t}^2 \int_{\mathbb R^d}\langle \xi \rangle^{2s}\mathrm{d}\xi\int_{|\xi-\eta|\le |\eta|} \langle \xi-\eta \rangle^{-2t} |\hat{v}(\eta)|^2 \,\mathrm{d}\eta \\ &=\|u\|_{H^t}^2 \int_{\mathbb R^d}|\hat{v}(\eta)|^2\,\mathrm{d}\eta\int_{|\xi-\eta|\le |\eta|} \langle \xi-\eta \rangle^{-2t} \langle \xi \rangle^{2s}\,\mathrm{d}\xi. \end{aligned} $$ The last integral can be estimated as $$ \int_{B(\eta,|\eta|)} \langle \eta-\xi \rangle^{-2t} \langle \xi \rangle^{2s}\,\mathrm{d}\xi\lesssim \langle \eta \rangle^{d+2s-2t}(\forall s,t\in\mathbb R). $$ Thus we obtain $$ J_1\lesssim \|u\|_{H^t}^2 \|v\|_{H^{\frac{d}{2}+s-t}}^2. $$ If we adapt the step for $J_2$ to obtain the same estimate, the proof is complete.
However I could not find any step in the proof requires the condition $s_1+s_2>0$. Is there anything wrong in the proof? Or the proof is right but it can't be generalized to the case when $u,v$ are in negative Sobolev spaces, so that they are not necessarily to be functions?
24.03.29 Updated: The problem comes from the last integral, for estimate of which requires $s>-d/2$ and $t<d/2$. We spilt the ball into three areas: $$ A_1:=B(0,|\eta|/2)\cap B(\eta,|\eta|), A_2:=B(\eta,|\eta|/2),A_3:=B(\eta,|\eta|)\setminus(A_1\cup A_2). $$ In $A_1$ we have $|\eta|/2\le|\eta-\xi|\le|\eta|$, hence $$ \int_{A_1} \langle \eta-\xi \rangle^{-2t} \langle \xi \rangle^{2s}\,\mathrm{d}\xi\sim \langle \eta \rangle^{-2t}\int_{A_1} \langle \xi \rangle^{2s}\,\mathrm{d}\xi\le \langle \eta \rangle^{-2t} I(s,|\eta|/2), $$ where $$I(s,R):=\int_{B(0,R)} \langle \xi \rangle^{2s} \,\mathrm{d}\xi. $$ Similarly, $$ \int_{A_2} \langle \eta-\xi \rangle^{-2t} \langle \xi \rangle^{2s}\,\mathrm{d}\xi\sim\langle \eta \rangle^{2s} I(-t,|\eta|/2), $$ and $$ \int_{A_3} \langle \eta-\xi \rangle^{-2t} \langle \xi \rangle^{2s}\,\mathrm{d}\xi\sim \langle \eta \rangle^{2s-2t}|A_3|\sim \langle \eta \rangle^{d+2s-2t}. $$ We are on the stage to deal with $I(s,R)$ for all $s\in\mathbb{R}$ and $R>0$. However the estimate $$ I(s,R)\lesssim \langle R \rangle^{d+2s} $$ only holds for $s>-d/2$. Hence the proof collapsed if $s\le -d/2$ or $t\ge d/2$. Conclusively, the proof above $$ \|uv\|_{H^s}\lesssim_{d,s,s _1}\|u\|_{H^{s_1}}\|v\|_{H^{s_2}} $$ only applies when $s_1+s_2>0$, $s\le s(s_1,s_2)$ where $$ s(s_1,s_2)=\begin{cases} s_1+s_2-d/2, \min\{s_1,s_2\}<d/2,\\ \max\{s_1,s_2\}-\epsilon, \min\{s_1,s_2\}\ge d/2. \end{cases} $$