[proofplan] We differentiate $\|u(t)\|_{H^s}^2$ in time, substitute the Euler equation, and use self-adjointness of the Leray projector together with the divergence-free condition to remove $\mathbb{P}$. A commutator decomposition splits the resulting inner product into a transport term $T_1$ and a commutator term $T_2$. The transport term vanishes by integration by parts and the divergence-free constraint $\nabla \cdot u = 0$. To bound the commutator term, we pass to the Fourier side via Plancherel, apply a mean-value estimate on the Fourier multiplier symbol, then close with Cauchy--Schwarz and Young's convolution inequality. The condition $s > n/2 + 1$ ensures that a final $L^2$ weight is integrable, yielding $|T_2| \le C\|u\|_{H^s}^3$ and hence the quadratic energy estimate. [/proofplan] [step:Differentiate $\|u\|_{H^s}^2$ and remove the Leray projector using self-adjointness and divergence-freeness] **Notation.** Write $\langle \nabla \rangle^s$ for the Fourier multiplier with real-valued symbol $\langle \xi \rangle^s := (1 + |\xi|^2)^{s/2}$, acting componentwise on vector fields. The $H^s$ norm is defined by $\|u\|_{H^s(\mathbb{R}^n;\, \mathbb{R}^n)}^2 := \|\langle \nabla \rangle^s[u]\|_{L^2(\mathbb{R}^n;\, \mathbb{R}^n)}^2$, and the vector-valued $L^2$ inner product is $\langle v, w \rangle_{L^2(\mathbb{R}^n;\, \mathbb{R}^n)} := \sum_{k=1}^n \int_{\mathbb{R}^n} v_k(x)\, w_k(x)\, d\mathcal{L}^n(x)$. Differentiating in time and using $\partial_t u = -\mathbb{P}\bigl[(u \cdot \nabla)[u]\bigr]$: \begin{align*} \frac{1}{2}\frac{d}{dt}\|u\|_{H^s}^2 &= \bigl\langle \langle \nabla \rangle^s[u],\, \langle \nabla \rangle^s[\partial_t u] \bigr\rangle_{L^2(\mathbb{R}^n;\, \mathbb{R}^n)} \\ &= -\bigl\langle \langle \nabla \rangle^s[u],\, \langle \nabla \rangle^s\bigl[\mathbb{P}[(u \cdot \nabla)[u]]\bigr] \bigr\rangle_{L^2(\mathbb{R}^n;\, \mathbb{R}^n)}. \end{align*} We remove $\mathbb{P}$ using three facts: (i) $\mathbb{P}$ is self-adjoint on $L^2(\mathbb{R}^n;\, \mathbb{R}^n)$; (ii) $\mathbb{P} \circ \langle \nabla \rangle^s = \langle \nabla \rangle^s \circ \mathbb{P}$, since both are Fourier multipliers; (iii) $\mathbb{P}[u] = u$, since $\nabla \cdot u = 0$. Moving $\mathbb{P}$ to the left slot by (i), commuting past $\langle \nabla \rangle^s$ by (ii), and applying (iii): \begin{align*} \frac{1}{2}\frac{d}{dt}\|u\|_{H^s}^2 = -\bigl\langle \langle \nabla \rangle^s[u],\, \langle \nabla \rangle^s\bigl[(u \cdot \nabla)[u]\bigr] \bigr\rangle_{L^2(\mathbb{R}^n;\, \mathbb{R}^n)}. \tag{$\star$} \end{align*} [guided] We want to estimate the time evolution of the $H^s$ energy $\|u(t)\|_{H^s}^2$. The idea is to differentiate, substitute the equation, and simplify. **Setup.** Write $\langle \nabla \rangle^s$ for the Fourier multiplier with real-valued symbol $\langle \xi \rangle^s := (1 + |\xi|^2)^{s/2}$, acting componentwise on vector fields. The $H^s$ norm is $\|u\|_{H^s(\mathbb{R}^n;\, \mathbb{R}^n)}^2 := \|\langle \nabla \rangle^s[u]\|_{L^2(\mathbb{R}^n;\, \mathbb{R}^n)}^2$, and the vector-valued $L^2$ inner product is $\langle v, w \rangle_{L^2(\mathbb{R}^n;\, \mathbb{R}^n)} := \sum_{k=1}^n \int_{\mathbb{R}^n} v_k(x)\, w_k(x)\, d\mathcal{L}^n(x)$. Since $u \in C^1([0,T];\, H^s)$, we may differentiate the squared norm in time. Using the product rule for the $L^2$ inner product and the symmetry of $\langle \nabla \rangle^s$ (it has a real symbol, hence is self-adjoint on $L^2$): \begin{align*} \frac{1}{2}\frac{d}{dt}\|u\|_{H^s}^2 = \bigl\langle \langle \nabla \rangle^s[u],\, \langle \nabla \rangle^s[\partial_t u] \bigr\rangle_{L^2(\mathbb{R}^n;\, \mathbb{R}^n)}. \end{align*} Substituting the Euler equation $\partial_t u = -\mathbb{P}[(u \cdot \nabla)[u]]$: \begin{align*} \frac{1}{2}\frac{d}{dt}\|u\|_{H^s}^2 = -\bigl\langle \langle \nabla \rangle^s[u],\, \langle \nabla \rangle^s\bigl[\mathbb{P}[(u \cdot \nabla)[u]]\bigr] \bigr\rangle_{L^2(\mathbb{R}^n;\, \mathbb{R}^n)}. \end{align*} Now the Leray projector $\mathbb{P}$ is in the way. We remove it using three properties: 1. **Self-adjointness**: $\mathbb{P}$ is self-adjoint on $L^2(\mathbb{R}^n;\, \mathbb{R}^n)$ (its Fourier symbol $\delta_{jk} - \xi_j \xi_k / |\xi|^2$ is a real symmetric matrix for each $\xi$). So we can move $\mathbb{P}$ from the right slot to the left: $\langle f, \mathbb{P}[g] \rangle = \langle \mathbb{P}[f], g \rangle$. 2. **Commutativity with $\langle \nabla \rangle^s$**: Both $\mathbb{P}$ and $\langle \nabla \rangle^s$ are Fourier multipliers, so they commute. After moving $\mathbb{P}$ to the left slot, it sits next to $\langle \nabla \rangle^s$, and we commute it through: $\mathbb{P} \circ \langle \nabla \rangle^s = \langle \nabla \rangle^s \circ \mathbb{P}$. 3. **Projection annihilation**: Since $\nabla \cdot u = 0$, the field $u$ is divergence-free, so $\mathbb{P}[u] = u$. Applying (i), then (ii), then (iii) to the left slot: \begin{align*} \frac{1}{2}\frac{d}{dt}\|u\|_{H^s}^2 &= -\bigl\langle \mathbb{P}\bigl[\langle \nabla \rangle^s[u]\bigr],\, \langle \nabla \rangle^s\bigl[(u \cdot \nabla)[u]\bigr] \bigr\rangle_{L^2(\mathbb{R}^n;\, \mathbb{R}^n)} \\ &= -\bigl\langle \langle \nabla \rangle^s\bigl[\mathbb{P}[u]\bigr],\, \langle \nabla \rangle^s\bigl[(u \cdot \nabla)[u]\bigr] \bigr\rangle_{L^2(\mathbb{R}^n;\, \mathbb{R}^n)} \\ &= -\bigl\langle \langle \nabla \rangle^s[u],\, \langle \nabla \rangle^s\bigl[(u \cdot \nabla)[u]\bigr] \bigr\rangle_{L^2(\mathbb{R}^n;\, \mathbb{R}^n)}. \tag{$\star$} \end{align*} The Leray projector has been completely eliminated. This is possible only because $u$ is divergence-free; without this constraint, $\mathbb{P}[u] \neq u$ and the pressure term would remain. [/guided] [/step] [step:Decompose via the commutator $[\langle \nabla \rangle^s, u \cdot \nabla]$ into a transport term $T_1$ and a commutator term $T_2$] Apply the operator identity $\langle \nabla \rangle^s \circ (u \cdot \nabla) = (u \cdot \nabla) \circ \langle \nabla \rangle^s + [\langle \nabla \rangle^s, u \cdot \nabla]$ to $u$: \begin{align*} \langle \nabla \rangle^s\bigl[(u \cdot \nabla)[u]\bigr] = (u \cdot \nabla)\bigl[\langle \nabla \rangle^s[u]\bigr] + \bigl[\langle \nabla \rangle^s,\, u \cdot \nabla\bigr][u]. \end{align*} Substituting into $(\star)$ gives $\frac{1}{2}\frac{d}{dt}\|u\|_{H^s}^2 = -T_1 - T_2$, where \begin{align*} T_1 &:= \bigl\langle \langle \nabla \rangle^s[u],\; (u \cdot \nabla)\bigl[\langle \nabla \rangle^s[u]\bigr] \bigr\rangle_{L^2(\mathbb{R}^n;\, \mathbb{R}^n)}, \\ T_2 &:= \bigl\langle \langle \nabla \rangle^s[u],\; \bigl[\langle \nabla \rangle^s,\, u \cdot \nabla\bigr][u] \bigr\rangle_{L^2(\mathbb{R}^n;\, \mathbb{R}^n)}. \end{align*} [guided] The expression $(\star)$ involves $\langle \nabla \rangle^s$ applied to the nonlinear term $(u \cdot \nabla)[u]$. The difficulty is that $\langle \nabla \rangle^s$ does not commute with the transport operator $u \cdot \nabla$ (the latter has variable coefficients). The standard strategy is to separate the part that does commute (the transport term) from the remainder (the commutator). Write the operator identity: \begin{align*} \langle \nabla \rangle^s \circ (u \cdot \nabla) = (u \cdot \nabla) \circ \langle \nabla \rangle^s + \bigl[\langle \nabla \rangle^s,\, u \cdot \nabla\bigr], \end{align*} where the commutator is $[\langle \nabla \rangle^s, u \cdot \nabla] := \langle \nabla \rangle^s \circ (u \cdot \nabla) - (u \cdot \nabla) \circ \langle \nabla \rangle^s$. Applying both sides to $u$: \begin{align*} \langle \nabla \rangle^s\bigl[(u \cdot \nabla)[u]\bigr] = (u \cdot \nabla)\bigl[\langle \nabla \rangle^s[u]\bigr] + \bigl[\langle \nabla \rangle^s,\, u \cdot \nabla\bigr][u]. \end{align*} Substituting into $(\star)$: \begin{align*} \frac{1}{2}\frac{d}{dt}\|u\|_{H^s}^2 = -T_1 - T_2, \end{align*} where \begin{align*} T_1 &:= \bigl\langle \langle \nabla \rangle^s[u],\; (u \cdot \nabla)\bigl[\langle \nabla \rangle^s[u]\bigr] \bigr\rangle_{L^2(\mathbb{R}^n;\, \mathbb{R}^n)}, \\ T_2 &:= \bigl\langle \langle \nabla \rangle^s[u],\; \bigl[\langle \nabla \rangle^s,\, u \cdot \nabla\bigr][u] \bigr\rangle_{L^2(\mathbb{R}^n;\, \mathbb{R}^n)}. \end{align*} Why this decomposition? The transport term $T_1$ is an $L^2$ inner product of $v := \langle \nabla \rangle^s[u]$ against its own transport $(u \cdot \nabla)[v]$. Because $u$ is divergence-free, this term will vanish by integration by parts -- it is an antisymmetric bilinear form. The commutator term $T_2$ measures how much $\langle \nabla \rangle^s$ fails to commute with transport; it will be bounded using Fourier analysis. The whole proof rests on $T_1 = 0$ (energy conservation of transport) and $|T_2| \lesssim \|u\|_{H^s}^3$ (the commutator is lower order). [/guided] [/step] [step:Show $T_1 = 0$ via integration by parts and the divergence-free condition] Set $v := \langle \nabla \rangle^s[u] \in L^2(\mathbb{R}^n;\, \mathbb{R}^n)$. Expanding the vector $L^2$ inner product and the transport operator: \begin{align*} T_1 = \sum_{k=1}^n \int_{\mathbb{R}^n} v_k(x) \sum_{j=1}^n u_j(x)\, \partial_j v_k(x)\, d\mathcal{L}^n(x). \end{align*} Using the product rule $v_k\, \partial_j v_k = \frac{1}{2}\partial_j(v_k^2)$ and writing $|v(x)|^2 := \sum_{k=1}^n v_k(x)^2$: \begin{align*} T_1 = \frac{1}{2} \sum_{j=1}^n \int_{\mathbb{R}^n} u_j(x)\, \partial_j |v(x)|^2\, d\mathcal{L}^n(x). \end{align*} Integrating by parts in the $x_j$ variable (the boundary terms vanish since $u \in H^s(\mathbb{R}^n;\, \mathbb{R}^n) \subset L^2(\mathbb{R}^n;\, \mathbb{R}^n)$ and $|v|^2 \in L^1(\mathbb{R}^n)$, so the product $u_j |v|^2$ decays at infinity): \begin{align*} T_1 = -\frac{1}{2} \sum_{j=1}^n \int_{\mathbb{R}^n} (\partial_j u_j)(x)\, |v(x)|^2\, d\mathcal{L}^n(x) = -\frac{1}{2} \int_{\mathbb{R}^n} (\nabla \cdot u)(x)\, |v(x)|^2\, d\mathcal{L}^n(x) = 0, \end{align*} where the final equality uses $\nabla \cdot u = 0$. [guided] The transport term $T_1$ has a special structure: it pairs the vector field $v = \langle \nabla \rangle^s[u]$ against its own transport by $u$. We will show that this always vanishes for divergence-free transport velocities -- this is the key cancellation that makes the Euler energy estimate possible. Set $v := \langle \nabla \rangle^s[u] \in L^2(\mathbb{R}^n;\, \mathbb{R}^n)$. Expanding the vector $L^2$ inner product $\langle v, (u \cdot \nabla)[v] \rangle_{L^2(\mathbb{R}^n;\, \mathbb{R}^n)} = \sum_{k=1}^n \langle v_k, (u \cdot \nabla)[v_k] \rangle_{L^2(\mathbb{R}^n;\, \mathbb{R})}$ and writing $(u \cdot \nabla)[v_k] = \sum_{j=1}^n u_j\, \partial_j v_k$: \begin{align*} T_1 = \sum_{k=1}^n \int_{\mathbb{R}^n} v_k(x) \sum_{j=1}^n u_j(x)\, \partial_j v_k(x)\, d\mathcal{L}^n(x). \end{align*} The crucial observation is the algebraic identity $v_k\, \partial_j v_k = \frac{1}{2}\partial_j(v_k^2)$. Summing over $k$ and writing $|v(x)|^2 := \sum_{k=1}^n v_k(x)^2$: \begin{align*} T_1 = \frac{1}{2} \sum_{j=1}^n \int_{\mathbb{R}^n} u_j(x)\, \partial_j |v(x)|^2\, d\mathcal{L}^n(x). \end{align*} Now integrate by parts in the $x_j$ variable. The boundary terms vanish: since $u \in H^s(\mathbb{R}^n;\, \mathbb{R}^n) \subset L^2(\mathbb{R}^n;\, \mathbb{R}^n)$ (as $s > 0$) and $v = \langle \nabla \rangle^s[u] \in L^2(\mathbb{R}^n;\, \mathbb{R}^n)$ by definition, the product $u_j \cdot |v|^2$ decays sufficiently at infinity (in the sense that the integration-by-parts identity holds for $H^s$ functions on $\mathbb{R}^n$). Transferring the derivative from $|v|^2$ to $u_j$: \begin{align*} T_1 = -\frac{1}{2} \sum_{j=1}^n \int_{\mathbb{R}^n} (\partial_j u_j)(x)\, |v(x)|^2\, d\mathcal{L}^n(x) = -\frac{1}{2} \int_{\mathbb{R}^n} (\nabla \cdot u)(x)\, |v(x)|^2\, d\mathcal{L}^n(x). \end{align*} Since $\nabla \cdot u = 0$ by hypothesis, we conclude $T_1 = 0$. Why does this matter? Without the divergence-free condition, the transport term $T_1$ would contribute a term proportional to $\|\nabla \cdot u\|_{L^\infty} \|u\|_{H^s}^2$, which cannot be controlled purely in terms of $\|u\|_{H^s}$. The constraint $\nabla \cdot u = 0$ is what allows the $H^s$ energy estimate to close as a quadratic ODE rather than something worse. [/guided] [/step] [step:Bound $|T_2|$ by passing to Fourier space via Plancherel and applying a mean-value estimate on the commutator symbol] **Pass to Fourier space.** Expand the vector inner product componentwise: $T_2 = \sum_{k=1}^n \langle \langle \nabla \rangle^s[u_k],\, ([\langle \nabla \rangle^s, u \cdot \nabla][u])_k \rangle_{L^2(\mathbb{R}^n;\, \mathbb{R})}$. By [Plancherel's Theorem](/theorems/247), applied to each scalar $L^2$ inner product (the hypotheses are satisfied since both entries lie in $L^2(\mathbb{R}^n;\, \mathbb{R})$): \begin{align*} T_2 = \sum_{k=1}^n \bigl\langle \widehat{\langle \nabla \rangle^s[u_k]},\; \widehat{\bigl([\langle \nabla \rangle^s, u \cdot \nabla][u]\bigr)_k} \bigr\rangle_{L^2(\mathbb{R}^n;\, \mathbb{C})}. \end{align*} **Compute the Fourier transforms.** The multiplier gives $\widehat{\langle \nabla \rangle^s[u_k]}(\eta) = \langle \eta \rangle^s\, \widehat{u_k}(\eta)$. The $k$-th component of $(u \cdot \nabla)[u]$ is $\sum_{j=1}^n u_j\, \partial_j u_k$, and by the convolution theorem for products of $L^2$ functions: \begin{align*} \widehat{u_j\, \partial_j u_k}(\eta) = (2\pi)^{-n/2} \int_{\mathbb{R}^n} \widehat{u_j}(\eta - \xi)\, (i\xi_j)\, \widehat{u_k}(\xi)\, d\mathcal{L}^n(\xi). \end{align*} Applying $\langle \nabla \rangle^s$ to $(u \cdot \nabla)[u]$ places $\langle \eta \rangle^s$ outside the convolution integral. Applying $(u \cdot \nabla)$ to $\langle \nabla \rangle^s[u]$ instead places $\langle \xi \rangle^s$ on $\widehat{u_k}(\xi)$ inside. The commutator is the difference: \begin{align*} \widehat{\bigl([\langle \nabla \rangle^s, u \cdot \nabla][u]\bigr)_k}(\eta) = (2\pi)^{-n/2} \sum_{j=1}^n \int_{\mathbb{R}^n} \bigl(\langle \eta \rangle^s - \langle \xi \rangle^s\bigr)\, \widehat{u_j}(\eta - \xi)\, (i\xi_j)\, \widehat{u_k}(\xi)\, d\mathcal{L}^n(\xi). \end{align*} **Take absolute values.** Writing $|\hat{u}(\zeta)|^2 := \sum_{k=1}^n |\widehat{u_k}(\zeta)|^2$ and applying the Cauchy--Schwarz inequality in $\mathbb{R}^n$ to the $j$-sum ($|\sum_{j=1}^n \widehat{u_j}(\eta - \xi)\, \xi_j| \le |\hat{u}(\eta - \xi)|\, |\xi|$) and to the $k$-sum in the outer inner product: \begin{align*} |T_2| \le (2\pi)^{-n/2} \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} \langle \eta \rangle^s\, |\hat{u}(\eta)| \cdot \bigl|\langle \eta \rangle^s - \langle \xi \rangle^s\bigr| \cdot |\hat{u}(\eta - \xi)| \cdot |\xi| \cdot |\hat{u}(\xi)|\, d\mathcal{L}^n(\xi)\, d\mathcal{L}^n(\eta). \end{align*} **Mean-value estimate.** For $s \ge 1$, the function $\zeta \mapsto \langle \zeta \rangle^s$ satisfies the mean-value inequality \begin{align*} \bigl|\langle \eta \rangle^s - \langle \xi \rangle^s\bigr| \lesssim |\eta - \xi|\, \bigl(\langle \xi \rangle^{s-1} + \langle \eta - \xi \rangle^{s-1}\bigr), \end{align*} where the implicit constant depends only on $s$. Substituting and using $|\xi|\, \langle \xi \rangle^{s-1} \le \langle \xi \rangle^s$ and $|\eta - \xi|\, \langle \eta - \xi \rangle^{s-1} \le \langle \eta - \xi \rangle^s$, the double integral splits as $|T_2| \lesssim I_1 + I_2$ with \begin{align*} I_1 &:= \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} \langle \eta \rangle^s\, |\hat{u}(\eta)| \cdot |\eta - \xi|\, |\hat{u}(\eta - \xi)| \cdot \langle \xi \rangle^s\, |\hat{u}(\xi)|\, d\mathcal{L}^n(\xi)\, d\mathcal{L}^n(\eta), \end{align*} and $I_2$ defined identically but with the roles of $\langle \cdot \rangle^s |\hat{u}|$ and $|\cdot|\, |\hat{u}|$ interchanged between the $\xi$ and $(\eta - \xi)$ factors. To see that the same estimates apply to $I_2$, substitute $\zeta = \eta - \xi$ (so $\xi = \eta - \zeta$, with Jacobian one and no change of domain): under this substitution, the factor $|\eta - \xi|\, |\hat{u}(\eta - \xi)|$ becomes $|\zeta|\, |\hat{u}(\zeta)|$ and the factor $\langle \xi \rangle^s |\hat{u}(\xi)|$ becomes $\langle \eta - \zeta \rangle^s |\hat{u}(\eta - \zeta)|$, so $I_2$ takes precisely the form of $I_1$ with the roles of the $\langle \cdot \rangle^s$-weighted and $|\cdot|$-weighted functions exchanged; the Cauchy--Schwarz and Young's inequality steps are therefore identical. [guided] We now face the commutator term $T_2$, which we must bound in terms of $\|u\|_{H^s}$. The strategy is: (1) pass to Fourier space where the commutator has an explicit kernel, (2) estimate the kernel using a mean-value bound on $\langle \cdot \rangle^s$, and (3) reduce to standard convolution estimates. **Passing to Fourier space.** Expand the vector inner product: $T_2 = \sum_{k=1}^n \langle \langle \nabla \rangle^s[u_k], ([\langle \nabla \rangle^s, u \cdot \nabla][u])_k \rangle_{L^2(\mathbb{R}^n;\, \mathbb{R})}$. Both $\langle \nabla \rangle^s[u_k]$ and $([\langle \nabla \rangle^s, u \cdot \nabla][u])_k$ lie in $L^2(\mathbb{R}^n;\, \mathbb{R})$ (the first by definition of $H^s$; the second because the commutator maps $H^s$ to $L^2$ for Lipschitz $u$). By [Plancherel's Theorem](/theorems/247), which preserves the $L^2$ inner product: \begin{align*} T_2 = \sum_{k=1}^n \bigl\langle \widehat{\langle \nabla \rangle^s[u_k]},\; \widehat{\bigl([\langle \nabla \rangle^s, u \cdot \nabla][u]\bigr)_k} \bigr\rangle_{L^2(\mathbb{R}^n;\, \mathbb{C})}. \end{align*} **Computing the Fourier transforms.** The Fourier multiplier $\langle \nabla \rangle^s$ acts by multiplication on the Fourier side: $\widehat{\langle \nabla \rangle^s[u_k]}(\eta) = \langle \eta \rangle^s\, \widehat{u_k}(\eta)$. For the nonlinear term, the $k$-th component of $(u \cdot \nabla)[u]$ is $\sum_{j=1}^n u_j\, \partial_j u_k$. By the convolution theorem (the product $u_j \cdot \partial_j u_k$ maps to a convolution of their Fourier transforms): \begin{align*} \widehat{u_j\, \partial_j u_k}(\eta) = (2\pi)^{-n/2} \int_{\mathbb{R}^n} \widehat{u_j}(\eta - \xi)\, (i\xi_j)\, \widehat{u_k}(\xi)\, d\mathcal{L}^n(\xi), \end{align*} where the factor $i\xi_j$ comes from $\widehat{\partial_j u_k}(\xi) = i\xi_j\, \widehat{u_k}(\xi)$. Now, applying $\langle \nabla \rangle^s$ to the entire expression places $\langle \eta \rangle^s$ outside: \begin{align*} \widehat{\langle \nabla \rangle^s[(u \cdot \nabla)[u]]_k}(\eta) = (2\pi)^{-n/2} \langle \eta \rangle^s \sum_{j=1}^n \int_{\mathbb{R}^n} \widehat{u_j}(\eta - \xi)\, (i\xi_j)\, \widehat{u_k}(\xi)\, d\mathcal{L}^n(\xi). \end{align*} If instead we first apply $\langle \nabla \rangle^s$ to $u$ and then transport, the weight $\langle \xi \rangle^s$ appears on $\widehat{u_k}(\xi)$ inside the integral. The commutator is the difference of these two operations: \begin{align*} \widehat{\bigl([\langle \nabla \rangle^s, u \cdot \nabla][u]\bigr)_k}(\eta) = (2\pi)^{-n/2} \sum_{j=1}^n \int_{\mathbb{R}^n} \bigl(\langle \eta \rangle^s - \langle \xi \rangle^s\bigr)\, \widehat{u_j}(\eta - \xi)\, (i\xi_j)\, \widehat{u_k}(\xi)\, d\mathcal{L}^n(\xi). \end{align*} The factor $\langle \eta \rangle^s - \langle \xi \rangle^s$ is the commutator kernel in frequency space. When $\eta \approx \xi$ (low-frequency interaction), this factor is small -- the commutator penalizes only high-frequency mismatches. **Taking absolute values.** Write $|\hat{u}(\zeta)|^2 := \sum_{k=1}^n |\widehat{u_k}(\zeta)|^2$ for the pointwise norm of the vector-valued Fourier transform. Apply the Cauchy--Schwarz inequality in $\mathbb{R}^n$ to the $j$-sum: $|\sum_{j=1}^n \widehat{u_j}(\eta - \xi)\, \xi_j| \le |\hat{u}(\eta - \xi)|\, |\xi|$, and similarly sum over $k$ in the outer $L^2$ inner product. The result: \begin{align*} |T_2| \le (2\pi)^{-n/2} \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} \langle \eta \rangle^s\, |\hat{u}(\eta)| \cdot \bigl|\langle \eta \rangle^s - \langle \xi \rangle^s\bigr| \cdot |\hat{u}(\eta - \xi)| \cdot |\xi| \cdot |\hat{u}(\xi)|\, d\mathcal{L}^n(\xi)\, d\mathcal{L}^n(\eta). \end{align*} **Mean-value estimate on the symbol.** For $s \ge 1$, the function $\zeta \mapsto \langle \zeta \rangle^s = (1 + |\zeta|^2)^{s/2}$ is $C^1$ and its gradient satisfies $|\nabla_\zeta \langle \zeta \rangle^s| = s|\zeta|\, \langle \zeta \rangle^{s-2} \le s\, \langle \zeta \rangle^{s-1}$. By the mean-value theorem applied along the segment from $\xi$ to $\eta$, together with the Peetre inequality $\langle \zeta \rangle^{s-1} \lesssim \langle \xi \rangle^{s-1} + \langle \eta - \xi \rangle^{s-1}$ for points $\zeta$ on that segment: \begin{align*} \bigl|\langle \eta \rangle^s - \langle \xi \rangle^s\bigr| \lesssim |\eta - \xi|\, \bigl(\langle \xi \rangle^{s-1} + \langle \eta - \xi \rangle^{s-1}\bigr), \end{align*} where the implicit constant depends only on $s$. **Splitting.** Substitute into the double integral. The two terms from the mean-value estimate give two contributions. For the first, use $|\xi|\, \langle \xi \rangle^{s-1} \le \langle \xi \rangle^s$ (since $|\xi| \le \langle \xi \rangle$). For the second, use $|\eta - \xi|\, \langle \eta - \xi \rangle^{s-1} \le \langle \eta - \xi \rangle^s$. This yields $|T_2| \lesssim I_1 + I_2$ where \begin{align*} I_1 &:= \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} \langle \eta \rangle^s\, |\hat{u}(\eta)| \cdot |\eta - \xi|\, |\hat{u}(\eta - \xi)| \cdot \langle \xi \rangle^s\, |\hat{u}(\xi)|\, d\mathcal{L}^n(\xi)\, d\mathcal{L}^n(\eta), \end{align*} and $I_2$ is the same expression with the roles of $\langle \cdot \rangle^s |\hat{u}|$ and $|\cdot|\, |\hat{u}|$ interchanged between the $\xi$ and $(\eta - \xi)$ slots. Concretely, in $I_1$ the "high" weight $\langle \xi \rangle^s$ sits on $|\hat{u}(\xi)|$ and the "low" weight $|\eta - \xi|$ sits on $|\hat{u}(\eta - \xi)|$, while in $I_2$ these are swapped. [/guided] [/step] [step:Estimate $I_1$ and $I_2$ via Cauchy--Schwarz and Young's convolution inequality] Define \begin{align*} F: \mathbb{R}^n \to [0, \infty), \quad F(\eta) &:= \langle \eta \rangle^s\, |\hat{u}(\eta)|, \\ H: \mathbb{R}^n \to [0, \infty), \quad H(\xi) &:= \langle \xi \rangle^s\, |\hat{u}(\xi)|, \\ g: \mathbb{R}^n \to [0, \infty), \quad g(\zeta) &:= |\zeta|\, |\hat{u}(\zeta)|. \end{align*} Then $I_1 = \int_{\mathbb{R}^n} F(\eta)\, (g * H)(\eta)\, d\mathcal{L}^n(\eta)$, which by the Cauchy--Schwarz inequality in $L^2(\mathbb{R}^n;\, \mathbb{R})$ satisfies \begin{align*} I_1 \le \|F\|_{L^2(\mathbb{R}^n)}\, \|g * H\|_{L^2(\mathbb{R}^n)}. \end{align*} By [Young's Convolution Inequality](/theorems/463) with exponents $(1, 2, 2)$ (i.e., $\|g * H\|_{L^2} \le \|g\|_{L^1}\, \|H\|_{L^2}$, valid since $1 + 1/2 = 1/1 + 1/2$): \begin{align*} I_1 \le \|F\|_{L^2}\, \|g\|_{L^1}\, \|H\|_{L^2}. \end{align*} Compute each factor: \begin{align*} \|F\|_{L^2}^2 &= \int_{\mathbb{R}^n} \langle \eta \rangle^{2s}\, |\hat{u}(\eta)|^2\, d\mathcal{L}^n(\eta) = \|u\|_{H^s}^2, \\ \|H\|_{L^2}^2 &= \int_{\mathbb{R}^n} \langle \xi \rangle^{2s}\, |\hat{u}(\xi)|^2\, d\mathcal{L}^n(\xi) = \|u\|_{H^s}^2, \\ \|g\|_{L^1} &= \int_{\mathbb{R}^n} |\zeta|\, |\hat{u}(\zeta)|\, d\mathcal{L}^n(\zeta) = \|\,|\cdot|\, \hat{u}\,\|_{L^1(\mathbb{R}^n)}. \end{align*} Hence $I_1 \le \|u\|_{H^s}^2\, \|\,|\cdot|\, \hat{u}\,\|_{L^1}$. The term $I_2$ admits the same bound by the symmetric roles of the factors, so \begin{align*} |T_2| \lesssim \|u\|_{H^s}^2\, \|\,|\cdot|\, \hat{u}\,\|_{L^1(\mathbb{R}^n)}. \end{align*} [guided] We now estimate the double integrals $I_1$ and $I_2$ using standard tools from harmonic analysis. The idea is to recognize $I_1$ as a convolution integral, then apply Cauchy--Schwarz and Young's inequality to separate the factors. **Rewrite as a convolution.** Define the non-negative functions \begin{align*} F: \mathbb{R}^n \to [0, \infty), \quad F(\eta) &:= \langle \eta \rangle^s\, |\hat{u}(\eta)|, \\ H: \mathbb{R}^n \to [0, \infty), \quad H(\xi) &:= \langle \xi \rangle^s\, |\hat{u}(\xi)|, \\ g: \mathbb{R}^n \to [0, \infty), \quad g(\zeta) &:= |\zeta|\, |\hat{u}(\zeta)|. \end{align*} Then $I_1 = \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} F(\eta)\, g(\eta - \xi)\, H(\xi)\, d\mathcal{L}^n(\xi)\, d\mathcal{L}^n(\eta)$. By Fubini's theorem (the integrand is non-negative, so Tonelli applies without integrability hypotheses), the inner integral is the convolution $(g * H)(\eta) = \int_{\mathbb{R}^n} g(\eta - \xi)\, H(\xi)\, d\mathcal{L}^n(\xi)$. Thus: \begin{align*} I_1 = \int_{\mathbb{R}^n} F(\eta)\, (g * H)(\eta)\, d\mathcal{L}^n(\eta). \end{align*} **Cauchy--Schwarz.** Apply the Cauchy--Schwarz inequality in $L^2(\mathbb{R}^n;\, \mathbb{R})$ to the pair $(F, g * H)$: \begin{align*} I_1 \le \|F\|_{L^2(\mathbb{R}^n)}\, \|g * H\|_{L^2(\mathbb{R}^n)}. \end{align*} **Young's convolution inequality.** We apply [Young's Convolution Inequality](/theorems/463) with exponents $p = 1$, $q = 2$, $r = 2$: for $g \in L^1$ and $H \in L^2$, $\|g * H\|_{L^2} \le \|g\|_{L^1}\, \|H\|_{L^2}$. The exponent condition $1 + 1/r = 1/p + 1/q$, i.e., $1 + 1/2 = 1 + 1/2$, is satisfied. Thus: \begin{align*} I_1 \le \|F\|_{L^2}\, \|g\|_{L^1}\, \|H\|_{L^2}. \end{align*} **Compute each norm.** By Plancherel and the definition of $H^s$: \begin{align*} \|F\|_{L^2}^2 = \int_{\mathbb{R}^n} \langle \eta \rangle^{2s}\, |\hat{u}(\eta)|^2\, d\mathcal{L}^n(\eta) = \|u\|_{H^s}^2, \qquad \|H\|_{L^2}^2 = \|u\|_{H^s}^2. \end{align*} For the $L^1$ norm: $\|g\|_{L^1} = \int_{\mathbb{R}^n} |\zeta|\, |\hat{u}(\zeta)|\, d\mathcal{L}^n(\zeta) = \|\,|\cdot|\, \hat{u}\,\|_{L^1(\mathbb{R}^n)}$. This is the $L^1$ norm of the Fourier transform of $\nabla u$ (since $\widehat{\nabla u}(\zeta) = i\zeta\, \hat{u}(\zeta)$ and $|i\zeta| = |\zeta|$), i.e., $\|g\|_{L^1} = \|\widehat{\nabla u}\|_{L^1}$. Hence $I_1 \le \|u\|_{H^s}^2\, \|\widehat{\nabla u}\|_{L^1}$. The term $I_2$ has the same structure with $F$ and $g$ swapped, and the same estimates apply. Therefore: \begin{align*} |T_2| \lesssim \|u\|_{H^s}^2\, \|\widehat{\nabla u}\|_{L^1(\mathbb{R}^n)}. \end{align*} It remains to control $\|\widehat{\nabla u}\|_{L^1}$ in terms of $\|u\|_{H^s}$. This is where the Sobolev regularity condition $s > n/2 + 1$ will be consumed. [/guided] [/step] [step:Close the estimate using $s > n/2 + 1$ to control $\|\widehat{\nabla u}\|_{L^1}$ by $\|u\|_{H^s}$] Insert the weight $\langle \zeta \rangle^{s-1}$ and apply the Cauchy--Schwarz inequality in $L^2(\mathbb{R}^n;\, \mathbb{R})$: \begin{align*} \|\widehat{\nabla u}\|_{L^1(\mathbb{R}^n)} &= \int_{\mathbb{R}^n} |\zeta|\, |\hat{u}(\zeta)|\, d\mathcal{L}^n(\zeta) \\ &= \int_{\mathbb{R}^n} \langle \zeta \rangle^{-(s-1)} \cdot \langle \zeta \rangle^{s-1}\, |\zeta|\, |\hat{u}(\zeta)|\, d\mathcal{L}^n(\zeta) \\ &\le \bigl\|\langle \cdot \rangle^{-(s-1)}\bigr\|_{L^2(\mathbb{R}^n)} \cdot \bigl\|\langle \cdot \rangle^{s-1}\, |\cdot|\, \hat{u}\bigr\|_{L^2(\mathbb{R}^n)}. \end{align*} **Second factor.** Since $\langle \zeta \rangle^{s-1}\, |\zeta| \le \langle \zeta \rangle^s$ (because $|\zeta| \le \langle \zeta \rangle$): \begin{align*} \bigl\|\langle \cdot \rangle^{s-1}\, |\cdot|\, \hat{u}\bigr\|_{L^2} \le \bigl\|\langle \cdot \rangle^s\, \hat{u}\bigr\|_{L^2} = \|u\|_{H^s}. \end{align*} **First factor.** The squared $L^2$ norm is $\int_{\mathbb{R}^n} \langle \zeta \rangle^{-2(s-1)}\, d\mathcal{L}^n(\zeta)$. This integral is finite if and only if $2(s-1) > n$, i.e., $s > n/2 + 1$, which holds by hypothesis. Denote this finite constant by $C_1 = C_1(n,s) := \|\langle \cdot \rangle^{-(s-1)}\|_{L^2(\mathbb{R}^n)}$. Combining: \begin{align*} \|\widehat{\nabla u}\|_{L^1} \le C_1\, \|u\|_{H^s}, \end{align*} and hence \begin{align*} |T_2| \le C_2\, \|u\|_{H^s}^3, \end{align*} where $C_2 = C_2(n,s) > 0$ absorbs $C_1$ and the implicit constants from the mean-value and convolution estimates. [guided] We need to show that $\|\widehat{\nabla u}\|_{L^1(\mathbb{R}^n)} \lesssim \|u\|_{H^s}$. The $L^1$ norm of a Fourier transform is not controlled by the $L^2$ norm alone (by Cauchy--Schwarz, $\|\hat{f}\|_{L^1} \le \|\hat{f}\|_{L^2} \cdot \|\mathbb{1}\|_{L^2}$, but $\|\mathbb{1}\|_{L^2(\mathbb{R}^n)} = \infty$). The idea is to introduce a decaying weight that makes the integral converge, at the cost of requiring more regularity from $u$. **Insert a weight.** Write the integrand as a product of a decaying factor and a growing factor: \begin{align*} |\zeta|\, |\hat{u}(\zeta)| = \langle \zeta \rangle^{-(s-1)} \cdot \langle \zeta \rangle^{s-1}\, |\zeta|\, |\hat{u}(\zeta)|. \end{align*} Apply the Cauchy--Schwarz inequality in $L^2(\mathbb{R}^n;\, \mathbb{R})$ to these two factors: \begin{align*} \|\widehat{\nabla u}\|_{L^1} \le \bigl\|\langle \cdot \rangle^{-(s-1)}\bigr\|_{L^2(\mathbb{R}^n)} \cdot \bigl\|\langle \cdot \rangle^{s-1}\, |\cdot|\, \hat{u}\bigr\|_{L^2(\mathbb{R}^n)}. \end{align*} **Bound the second factor.** Since $|\zeta| \le (1 + |\zeta|^2)^{1/2} = \langle \zeta \rangle$, we have $\langle \zeta \rangle^{s-1}\, |\zeta| \le \langle \zeta \rangle^s$, so \begin{align*} \bigl\|\langle \cdot \rangle^{s-1}\, |\cdot|\, \hat{u}\bigr\|_{L^2} \le \bigl\|\langle \cdot \rangle^s\, \hat{u}\bigr\|_{L^2} = \|u\|_{H^s}. \end{align*} **Bound the first factor -- this is where $s > n/2 + 1$ is consumed.** We need \begin{align*} C_1^2 := \int_{\mathbb{R}^n} \langle \zeta \rangle^{-2(s-1)}\, d\mathcal{L}^n(\zeta) < \infty. \end{align*} Since $\langle \zeta \rangle \ge |\zeta|$ for all $\zeta$, the integrand is bounded by $(1 + |\zeta|^2)^{-(s-1)}$, which behaves like $|\zeta|^{-2(s-1)}$ for large $|\zeta|$. In polar coordinates, the integral converges if and only if $\int_1^\infty r^{n-1} \cdot r^{-2(s-1)}\, d\mathcal{L}^1(r) < \infty$, i.e., $2(s-1) - n + 1 > 1$, equivalently $2(s-1) > n$, i.e., $s > n/2 + 1$. This is precisely our hypothesis. **Combining.** With $C_1 = C_1(n,s) := \|\langle \cdot \rangle^{-(s-1)}\|_{L^2(\mathbb{R}^n)} < \infty$: \begin{align*} \|\widehat{\nabla u}\|_{L^1} \le C_1\, \|u\|_{H^s}. \end{align*} Substituting into the bound $|T_2| \lesssim \|u\|_{H^s}^2\, \|\widehat{\nabla u}\|_{L^1}$ from the previous step: \begin{align*} |T_2| \le C_2\, \|u\|_{H^s}^3, \end{align*} where $C_2 = C_2(n,s) > 0$ absorbs $C_1$ and all implicit constants. This is the critical cubic bound: the commutator contributes at most one power of $\|u\|_{H^s}$ beyond the quadratic term $\|u\|_{H^s}^2$. [/guided] [/step] [step:Combine $T_1 = 0$ and $|T_2| \le C\|u\|_{H^s}^3$ to obtain the quadratic energy estimate and the ODE blowup bound] From the decomposition $\frac{1}{2}\frac{d}{dt}\|u\|_{H^s}^2 = -T_1 - T_2$ with $T_1 = 0$: \begin{align*} \frac{1}{2}\frac{d}{dt}\|u\|_{H^s}^2 = -T_2, \qquad |T_2| \le C_2\, \|u\|_{H^s}^3. \end{align*} Hence $\bigl|\frac{d}{dt}\|u\|_{H^s}^2\bigr| \le 2C_2\, \|u\|_{H^s}^3$. Writing $\frac{d}{dt}\|u\|_{H^s}^2 = 2\|u\|_{H^s}\, \frac{d}{dt}\|u\|_{H^s}$ (valid when $\|u\|_{H^s} > 0$) and dividing by $2\|u\|_{H^s}$: \begin{align*} \frac{d}{dt}\|u\|_{H^s} \le C_2\, \|u\|_{H^s}^2. \end{align*} Setting $C := C_2 = C(n,s)$ gives the claimed differential inequality. By ODE comparison with the Bernoulli equation $\dot{y} = Cy^2$ (whose solution is $y(t) = y_0 / (1 - Cy_0 t)$), we obtain \begin{align*} \|u(t)\|_{H^s(\mathbb{R}^n;\, \mathbb{R}^n)} \le \frac{\|u_0\|_{H^s(\mathbb{R}^n;\, \mathbb{R}^n)}}{1 - C\,\|u_0\|_{H^s(\mathbb{R}^n;\, \mathbb{R}^n)}\, t} \end{align*} for all $t < (C\,\|u_0\|_{H^s})^{-1}$, where $u_0 := u(0, \cdot)$. [guided] We now assemble the estimate from the two preceding steps and derive the ODE comparison bound stated in the theorem. **Recovering the differential inequality for the norm.** From the commutator decomposition we have $\frac{1}{2}\frac{d}{dt}\|u\|_{H^s}^2 = -T_1 - T_2$. Since $T_1 = 0$ (step 3) and $|T_2| \le C_2\|u\|_{H^s}^3$ (steps 4--5), we obtain \begin{align*} \Bigl|\frac{d}{dt}\|u\|_{H^s}^2\Bigr| = 2|T_2| \le 2C_2\,\|u\|_{H^s}^3. \end{align*} We want to pass from an inequality on $\frac{d}{dt}\|u\|_{H^s}^2$ to one on $\frac{d}{dt}\|u\|_{H^s}$. When $\|u\|_{H^s} > 0$, the chain rule gives \begin{align*} \frac{d}{dt}\|u\|_{H^s}^2 = 2\,\|u\|_{H^s}\,\frac{d}{dt}\|u\|_{H^s}. \end{align*} Dividing both sides by $2\|u\|_{H^s} > 0$: \begin{align*} \frac{d}{dt}\|u\|_{H^s} \le C_2\,\|u\|_{H^s}^2. \end{align*} Setting $C := C_2 = C(n,s) > 0$, this is the claimed differential inequality. (If $\|u(t_0)\|_{H^s} = 0$ at some $t_0$, then $u(t_0) \equiv 0$ and by the Euler equation $\partial_t u = 0$ at $t_0$, so the zero solution persists and the inequality holds with equality. We therefore lose no generality by assuming $\|u\|_{H^s} > 0$ throughout.) **ODE comparison argument.** Consider the scalar Bernoulli ODE $\dot{y} = Cy^2$ with initial data $y(0) = y_0 := \|u_0\|_{H^s} \ge 0$. Separating variables and integrating: $\int_{y_0}^{y(t)} \frac{d\sigma}{\sigma^2} = C\int_0^t d\mathcal{L}^1(\tau)$, giving $-1/y(t) + 1/y_0 = Ct$, i.e., the unique solution is \begin{align*} y(t) = \frac{y_0}{1 - Cy_0 t}, \qquad t \in \bigl[0,\, (Cy_0)^{-1}\bigr). \end{align*} Since $\frac{d}{dt}\|u\|_{H^s} \le Cy^2$ and $y$ satisfies $\dot{y} = Cy^2$ with $y(0) = \|u\|_{H^s(0)}$, the comparison principle for scalar ODEs (valid because the right-hand side $Cy^2$ is locally Lipschitz and the inequality is one-sided) yields $\|u(t)\|_{H^s} \le y(t)$ for all $t \in [0, (Cy_0)^{-1})$. Substituting $y_0 = \|u_0\|_{H^s}$: \begin{align*} \|u(t)\|_{H^s(\mathbb{R}^n;\,\mathbb{R}^n)} \le \frac{\|u_0\|_{H^s(\mathbb{R}^n;\,\mathbb{R}^n)}}{1 - C\,\|u_0\|_{H^s(\mathbb{R}^n;\,\mathbb{R}^n)}\,t}. \end{align*} **Blowup time interpretation.** The denominator $1 - Cy_0 t$ vanishes as $t \to (Cy_0)^{-1}$, and the bound on $\|u(t)\|_{H^s}$ diverges to $+\infty$. This does not prove that the actual solution blows up — the Euler equations could remain smooth — but it establishes that the $H^s$ energy estimate alone cannot rule out blowup before time $T^* = (C\|u_0\|_{H^s})^{-1}$. The estimate is therefore sharp in that it provides a lower bound $T^*$ on the guaranteed lifespan: the solution must exist on $[0, T^*)$ but may or may not continue beyond it. [/guided] [/step]