The forward direction is immediate from the Sobolev embedding. The converse replaces $\|\nabla u\|_{L^\infty}$ in the [Lipschitz Continuation Criterion For The Euler Equations](/theorems/656) by a term involving $\|\omega\|_{L^\infty}$ plus a logarithmic correction, using the Biot–Savart law and the [Logarithmic Estimate For Singular Integrals](/theorems/657). The resulting differential inequality is then integrated using a nonlinear Gronwall argument. **Step 1: Forward direction.** Since $s > n/2 + 1$, the [Sobolev Embedding Into Continuous Functions](/theorems/226) gives $H^{s-1}(\mathbb{R}^n) \hookrightarrow L^\infty(\mathbb{R}^n)$. Therefore $\|\omega(t)\|_{L^\infty} \le C\|u(t)\|_{H^s}$, and $\limsup_{t \to T^-}\|u(t)\|_{H^s} < \infty$ implies $\int_0^T\|\omega\|_{L^\infty}\,d\tau < \infty$. **Step 2: Biot–Savart and logarithmic estimate.** By the Biot–Savart law, $|\nabla u(x)| \le C(|\omega(x)| + |T\omega(x)|)$ where $T$ is a Calderón–Zygmund operator. Since $\omega = \nabla \times u$ and $s - 1 > n/2$, the Sobolev embedding $H^{s-1} \hookrightarrow C^{0,\alpha}$ holds for some $\alpha \in (0, \min\{s - 1 - n/2, 1\})$. Applying the [Logarithmic Estimate For Singular Integrals](/theorems/657) to $T\omega$ and using that $\omega = \nabla \times u$ gives \begin{align*} \|T\omega\|_{L^\infty} &\le C\bigl(\|u_0\|_{L^2} + \|\omega\|_{L^\infty}[1 + \log^+(\|u\|_{H^s}/\|\omega\|_{L^\infty})]\bigr), \end{align*} where conservation of kinetic energy was used to replace $\|\omega\|_{L^2}$ by $\|u_0\|_{L^2}$. **Step 3: Differential inequality.** Combining Step 2 with the [Lipschitz Continuation Criterion For The Euler Equations](/theorems/656) yields an ODE for $y(t) := \log(1 + \|u(t)\|_{H^s})$ of the form \begin{align*} \frac{dy}{dt} &\le a(t) + b(t)\,y(t), \end{align*} where $b(t) = C\|\omega(t)\|_{L^\infty}$ and $a(t) = C(\|u_0\|_{L^2} + \|\omega(t)\|_{L^\infty})$. **Step 4: Nonlinear Gronwall and conclusion.** Integrating the linear ODE comparison gives \begin{align*} \sup_{0 \le t \le T}(1 + \|u(t)\|_{H^s}) &\le C\bigl(\|u_0\|_{H^s},\, T,\, M(T)\bigr) < \infty, \end{align*} where $M(T) := \int_0^T\|\omega(\tau)\|_{L^\infty}\,d\tau < \infty$ by assumption. Since the $H^s$ norm remains bounded, the solution extends beyond $T$.