[proof]
Compute
\begin{align*}
\frac{1}{2}\frac{d}{dt}\|u\|_{H^s}^2 &= -\int_{\mathbb{R}^n} \langle\nabla\rangle^s u \cdot \langle\nabla\rangle^s \mathbb{P}[(u \cdot \nabla)u]\,d\mathcal{L}^n.
\end{align*}
Since $\mathbb{P}$ is self-adjoint on $L^2$ and $\mathbb{P}\langle\nabla\rangle^s u = \langle\nabla\rangle^s u$ (because $\langle\nabla\rangle^s$ commutes with $\mathbb{P}$ as both are Fourier multipliers, and $u \in L^2_\sigma$), this equals $-\int \langle\nabla\rangle^s u \cdot \langle\nabla\rangle^s[(u \cdot \nabla)u]\,d\mathcal{L}^n$. Writing
\begin{align*}
\langle\nabla\rangle^s[(u \cdot \nabla)u] &= (u \cdot \nabla)\langle\nabla\rangle^s u + [\langle\nabla\rangle^s, u \cdot \nabla]u
\end{align*}
and using the Divergence Free Cancellation claim to eliminate the first term:
\begin{align*}
\frac{1}{2}\frac{d}{dt}\|u\|_{H^s}^2 &= -\int_{\mathbb{R}^n} \langle\nabla\rangle^s u \cdot [\langle\nabla\rangle^s, u \cdot \nabla]u\,d\mathcal{L}^n.
\end{align*}
By the [Cauchy-Schwarz Inequality](/theorems/432), the right-hand side is bounded by $\|u\|_{H^s}\,\|[\langle\nabla\rangle^s, u \cdot \nabla]u\|_{L^2}$. The commutator equals $\sum_{j=1}^n [\langle\nabla\rangle^s, u_j]\,\partial_j u$. Applying the [Kato Ponce Commutator Estimate](/theorems/904) to each term:
\begin{align*}
\|[\langle\nabla\rangle^s, u_j]\,\partial_j u\|_{L^2} &\le C_s\bigl(\|\nabla u_j\|_{L^\infty}\,\|\partial_j u\|_{H^{s-1}} + \|u_j\|_{H^s}\,\|\partial_j u\|_{L^\infty}\bigr) \le C_s\,\|\nabla u\|_{L^\infty}\,\|u\|_{H^s}.
\end{align*}
Summing over $j$ and using the [Sobolev Embedding Into Continuous Functions](/theorems/226) ($\|\nabla u\|_{L^\infty} \le C\|u\|_{H^s}$ since $s - 1 > n/2$) gives $\|[\langle\nabla\rangle^s, u \cdot \nabla]u\|_{L^2} \le C_s\,\|u\|_{H^s}^2$. Therefore $\frac{1}{2}\frac{d}{dt}\|u\|_{H^s}^2 \le C_s\,\|u\|_{H^s}^3$, giving $\frac{d}{dt}\|u\|_{H^s} \le C_s\,\|u\|_{H^s}^2$.
[/proof]