The idea is to sharpen the a priori energy estimate from the [Local Well-Posedness Of The Euler Equations In Sobolev Spaces](/theorems/655). That proof bounds the commutator $\|[(1-\Delta)^{s/2}, u \cdot \nabla]u\|_{L^2}$ by $C\|u\|_{H^s}^2$ via the [Kato Ponce Commutator Estimate](/theorems/904) followed by the [Sobolev Embedding Theorem](/theorems/903) ($\|\nabla u\|_{L^\infty} \le C\|u\|_{H^s}$). The present proof instead retains $\|\nabla u\|_{L^\infty}$ without embedding, obtaining a linear-in-$\|u\|_{H^s}$ differential inequality that integrates via the [Gronwall Inequality](/theorems/872) to give exponential growth controlled by $\int_0^t \|\nabla u\|_{L^\infty}\,d\tau$. Write $\mathbb{P}$ for the Leray projector from the [Helmholtz Decomposition](/theorems/654). For $u \in H^s(\mathbb{R}^n; \mathbb{R}^n)$, the [integral formula](/page/Inhomogeneous%20Sobolev%20Space) gives \begin{align*} \|u\|_{H^s}^2 &= \int_{\mathbb{R}^n} (1+|\xi|^2)^{s}\,|\hat{u}_{\mathrm{rep}}(\xi)|^2\,d\mathcal{L}^n(\xi), \end{align*} where $\hat{u}_{\mathrm{rep}}$ is the $L^1_{\mathrm{loc}}$ representative of the [regular distribution](/page/Regular%20Distribution) $\hat{u}$ (applied component-wise to the vector field $u$). The Bessel potential operator $(1-\Delta)^{s/2}$ is defined on the Fourier side as multiplication by the scalar $(1+|\xi|^2)^{s/2}$, so $(1-\Delta)^{s/2}u$ has Fourier representative $(1+|\xi|^2)^{s/2}\hat{u}_{\mathrm{rep}}(\xi)$. Since $u \in H^s$, the [function](/page/Function) $(1+|\xi|^2)^{s/2}\hat{u}_{\mathrm{rep}} \in L^2$ by definition, and [Plancherel's theorem](/theorems/247) gives \begin{align*} \|(1-\Delta)^{s/2}u\|_{L^2}^2 &= \int_{\mathbb{R}^n} |(1+|\xi|^2)^{s/2}\hat{u}_{\mathrm{rep}}(\xi)|^2\,d\mathcal{L}^n(\xi) = \|u\|_{H^s}^2. \end{align*} This identity $\|u\|_{H^s} = \|(1-\Delta)^{s/2}u\|_{L^2}$ converts the $H^s$ norm into an $L^2$ norm, which can be differentiated using the chain rule. **Step 1: Pressure-free formulation.** Applying $\mathbb{P}$ to both sides of the Euler equation $\partial_t u + (u \cdot \nabla)u + \nabla p = 0$ and using $\mathbb{P}\nabla p = 0$ (since $\nabla p \in L^2_\sigma(\mathbb{R}^n)^\perp$ by the [Helmholtz Decomposition](/theorems/654)) and $\mathbb{P}u = u$ (since $u \in L^2_\sigma$): \begin{align*} \partial_t u &= -\mathbb{P}[(u \cdot \nabla)u]. \end{align*} **Step 2: Energy identity.** [claim:Energy Identity] The maximal solution $u \in C([0, T_+); H^s_\sigma(\mathbb{R}^n))$ satisfies \begin{align*} \frac{1}{2}\frac{d}{dt}\|u\|_{H^s}^2 &= -\int_{\mathbb{R}^n} (1-\Delta)^{s/2} u \cdot \bigl[(1-\Delta)^{s/2},\, u \cdot \nabla\bigr]u\,d\mathcal{L}^n, \end{align*} where $[(1-\Delta)^{s/2}, u \cdot \nabla]u := (1-\Delta)^{s/2}[(u \cdot \nabla)u] - (u \cdot \nabla)(1-\Delta)^{s/2}u$ is the commutator. [/claim] [proof] Set $w(t) := (1-\Delta)^{s/2}u(t) \in L^2(\mathbb{R}^n; \mathbb{R}^n)$. By the Plancherel identity derived above, $\|u(t)\|_{H^s}^2 = \|w(t)\|_{L^2}^2$. Since $(1-\Delta)^{s/2}$ acts only in the spatial variable, it commutes with $\partial_t$, giving $\partial_t w = (1-\Delta)^{s/2}(\partial_t u)$. The chain rule $\frac{d}{dt}\|w\|_{L^2}^2 = 2\int w \cdot \partial_t w\,d\mathcal{L}^n$ therefore gives \begin{align*} \frac{1}{2}\frac{d}{dt}\|u\|_{H^s}^2 &= \int_{\mathbb{R}^n} (1-\Delta)^{s/2} u \cdot (1-\Delta)^{s/2}(\partial_t u)\,d\mathcal{L}^n. \end{align*} Substituting $\partial_t u = -\mathbb{P}[(u \cdot \nabla)u]$ from Step 1: \begin{align*} &= -\int_{\mathbb{R}^n} (1-\Delta)^{s/2} u \cdot (1-\Delta)^{s/2}\bigl(\mathbb{P}[(u \cdot \nabla)u]\bigr)\,d\mathcal{L}^n. \end{align*} The Leray projector $\mathbb{P}$ can now be removed. Three facts are needed: (i) $\mathbb{P}$ is self-adjoint on $L^2(\mathbb{R}^n; \mathbb{R}^n)$ (from the [Helmholtz Decomposition](/theorems/654)); (ii) $\mathbb{P}$ commutes with $(1-\Delta)^{s/2}$ — both are Fourier multipliers, one acting component-wise as multiplication by $\delta_{ij} - \xi_i\xi_j/|\xi|^2$ and the other by the scalar $(1+|\xi|^2)^{s/2}$, and scalar multiplication commutes with matrix multiplication; (iii) $\mathbb{P}((1-\Delta)^{s/2}u) = (1-\Delta)^{s/2}u$ — since $u$ is divergence-free ($\xi \cdot \hat{u}_{\mathrm{rep}}(\xi) = 0$ for $\mathcal{L}^n$-a.e. $\xi$), and multiplying each component by the scalar $(1+|\xi|^2)^{s/2}$ preserves this condition, $(1-\Delta)^{s/2}u$ is also divergence-free, so $\mathbb{P}$ acts as the identity. Using self-adjointness of $\mathbb{P}$ to move it from the right factor to the left, then applying (iii): \begin{align*} &= -\int_{\mathbb{R}^n} (1-\Delta)^{s/2} u \cdot (1-\Delta)^{s/2}\bigl[(u \cdot \nabla)u\bigr]\,d\mathcal{L}^n. \end{align*} Now expand $(1-\Delta)^{s/2}[(u \cdot \nabla)u]$ using the commutator identity $AB = BA + [A,B]$ with $A = (1-\Delta)^{s/2}$ and $B = u \cdot \nabla$: \begin{align*} (1-\Delta)^{s/2}\bigl[(u \cdot \nabla)u\bigr] &= (u \cdot \nabla)(1-\Delta)^{s/2}u + \bigl[(1-\Delta)^{s/2},\, u \cdot \nabla\bigr]u. \end{align*} Substituting produces two [integrals](/page/Integral): \begin{align*} \frac{1}{2}\frac{d}{dt}\|u\|_{H^s}^2 &= -\underbrace{\int_{\mathbb{R}^n} (1-\Delta)^{s/2} u \cdot (u \cdot \nabla)(1-\Delta)^{s/2}u\,d\mathcal{L}^n}_{(I)} - \underbrace{\int_{\mathbb{R}^n} (1-\Delta)^{s/2} u \cdot \bigl[(1-\Delta)^{s/2}, u \cdot \nabla\bigr]u\,d\mathcal{L}^n}_{(II)}. \end{align*} It remains to show $(I) = 0$. Recalling $w = (1-\Delta)^{s/2}u$ and expanding component by component: \begin{align*} (I) &= \sum_{i=1}^n \sum_{j=1}^n \int_{\mathbb{R}^n} w_i\, u_j\, \partial_j w_i\,d\mathcal{L}^n. \end{align*} Using $w_i\,\partial_j w_i = \frac{1}{2}\partial_j(w_i^2)$ and summing over $i$: \begin{align*} (I) &= \frac{1}{2}\sum_{j=1}^n \int_{\mathbb{R}^n} u_j\,\partial_j(|w|^2)\,d\mathcal{L}^n = \frac{1}{2}\int_{\mathbb{R}^n} u \cdot \nabla(|w|^2)\,d\mathcal{L}^n. \end{align*} [Integration by parts](/theorems/210) moves $\nabla$ from $|w|^2$ to $u$ (justified since $u, w \in H^s \hookrightarrow L^\infty$ for $s > n/2$ by the [Sobolev Embedding Theorem](/theorems/903), so all terms are in $L^1(\mathbb{R}^n)$): \begin{align*} (I) &= -\frac{1}{2}\int_{\mathbb{R}^n} |w|^2\,(\nabla \cdot u)\,d\mathcal{L}^n = 0, \end{align*} since $\nabla \cdot u = 0$. Therefore only $(II)$ survives, giving the claimed identity. [/proof] **Step 3: Commutator bound.** [claim:Sharper Commutator Bound] For $s > n/2 + 1$ and $u \in H^s_\sigma(\mathbb{R}^n)$, \begin{align*} \|[(1-\Delta)^{s/2}, u \cdot \nabla]u\|_{L^2} &\le C_s\,\|\nabla u\|_{L^\infty}\,\|u\|_{H^s}. \end{align*} [/claim] [proof] The commutator decomposes as $[(1-\Delta)^{s/2}, u \cdot \nabla]u = \sum_{j=1}^n [(1-\Delta)^{s/2}, u_j]\,\partial_j u$. By the [Kato Ponce Commutator Estimate](/theorems/904) applied to each summand with $f = u_j \in H^s \cap W^{1,\infty}$ and $g = \partial_j u \in H^{s-1} \cap L^\infty$: \begin{align*} \|[(1-\Delta)^{s/2}, 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). \end{align*} Both terms on the right are bounded by $C_s\,\|\nabla u\|_{L^\infty}\,\|u\|_{H^s}$: the first directly by $\|\partial_j u\|_{H^{s-1}} \le \|u\|_{H^s}$, and the second by $\|\partial_j u\|_{L^\infty} \le \|\nabla u\|_{L^\infty}$. Summing over $j = 1, \ldots, n$ and absorbing the factor $n$ into $C_s$ gives the claim. [/proof] **Step 4: Gronwall closure.** Combining Steps 2 and 3 with the [Cauchy-Schwarz Inequality](/theorems/432): \begin{align*} \frac{1}{2}\frac{d}{dt}\|u\|_{H^s}^2 &\le \|u\|_{H^s}\,\|[(1-\Delta)^{s/2}, u \cdot \nabla]u\|_{L^2} \le C_s\,\|\nabla u\|_{L^\infty}\,\|u\|_{H^s}^2. \end{align*} Dividing both sides by $\|u\|_{H^s}$ (which is positive for $u_0 \neq 0$; if $u_0 = 0$ the result is trivial) yields the differential inequality \begin{align*} \frac{d}{dt}\|u\|_{H^s} &\le C_s\,\|\nabla u\|_{L^\infty}\,\|u\|_{H^s}. \end{align*} This is of the form $\phi'(t) \le \beta(t)\,\phi(t)$ with $\phi(t) := \|u(t)\|_{H^s}$ and $\beta(t) := C_s\|\nabla u(t)\|_{L^\infty}$. The [Gronwall Inequality](/theorems/872) gives \begin{align*} \|u(t)\|_{H^s} &\le \|u_0\|_{H^s}\exp\!\Bigl(C_s\int_0^t \|\nabla u(\tau)\|_{L^\infty}\,d\tau\Bigr), \end{align*} which is the claimed estimate. **Step 5: Blow-up alternative.** Suppose for contradiction that $T_+ < \infty$ and $M := \int_0^{T_+}\|\nabla u(\tau)\|_{L^\infty}\,d\tau < \infty$. The estimate from Step 4 gives $\sup_{t \in [0, T_+)}\|u(t)\|_{H^s} \le \|u_0\|_{H^s}\,e^{C_s M} < \infty$. For any [sequence](/page/Sequence) $t_k \nearrow T_+$, the uniform $H^s$ bound and the equation $\partial_t u = -\mathbb{P}[(u \cdot \nabla)u]$ together with the [Algebra Property Of Inhomogeneous Sobolev Spaces](/theorems/468) give $\|\partial_t u\|_{H^{s-1}} \le C\|u\|_{H^s}^2 \le C\|u_0\|_{H^s}^2 e^{2C_s M}$, so $u$ is Cauchy in $C([0, T_+]; H^{s-1})$ and converges to a [limit](/page/Limit) $u(T_+) \in H^{s-1}$. The uniform $H^s$ bound and weak compactness of bounded [sets](/page/Set) in the [Hilbert space](/page/Hilbert%20Space) $H^s$ then give $u(T_+) \in H^s_\sigma$, and the [Local Well-Posedness Of The Euler Equations In Sobolev Spaces](/theorems/655) restarts the solution from $u(T_+)$, extending $u$ past $T_+$. This contradicts the maximality of $T_+$.