A strong solution of the projected periodic equation with all Sobolev norms continuous in time is $C^{\mathrm{\infty }}$

By hypothesis 3 and the divergence-free condition, $(u(t)\cdot \mathrm{\nabla })u(t)={\sum }_j{u}_j(t){\mathrm{\partial }}_{j}u(t)$ is well defined as a field; by [F1] and [F2] with $p$ arbitrarily large, $t\mapsto (u(t)\cdot \mathrm{\nabla })u(t)$ is continuous into $H^p$ for every $p>3\mathrm{/}2$. Hence $t\mapsto G(t):=\nu \mathrm{\Delta }u(t)-P((u(t)\cdot \mathrm{\nabla })u(t))$ is continuous into every real-order $H^p$ by [F5], the boundedness of $\mathrm{\Delta }:H^{p+2}\to H^p$, and inclusion from arbitrarily high orders. Since $u^{\mathrm{\prime }}=G$ by hypothesis 2, this also shows that $G$ is the derivative of $u$ in every $H^p$.

Choose $q>3\mathrm{/}2$. For each $x$, evaluation $E_x:H^q\to {\mathbb{R}}^3$, $E_{x}v=\tilde{v}(x)$, is bounded linear with $\mathrm{\mid }{E}_{x}v{\mathrm{\mid }}_2\le W_q\mathrm{\parallel }v{\mathrm{\parallel }}_{H^q}$ by the embedding. Applying $E_x$ to the $H^q$ difference quotient identifies the pointwise derivative ${\mathrm{\partial }}_{t}u(t,x)=G(t,x)$. More generally, $v\mapsto {\mathrm{\partial }}_x^{\alpha }\tilde{v}(x)$ is bounded on $H^q$ when $q>\mathrm{\mid }\alpha \mathrm{\mid }+3\mathrm{/}2$, so it also commutes with time differentiation. In particular ${\mathrm{\partial }}_{t}u=G$ is a continuous field whose spatial derivatives of every order are continuous, by [F1]; so $u$ is $C^1$ on $[0,T]\times {\mathbb{T}}^3$ and $u\in C([0,T],H^p)$ for every $p$.

Induction claim. We prove by induction on integers $r\ge 1$ that $u$ is $C^r$ into every $H^p$ and, for $0\le j\le r$, ${\mathrm{\partial }}_t^{j}u={\mathrm{\Phi }}_j$ with ${\mathrm{\Phi }}_j$ as in the Statement, continuous into every $H^p$. The zeroth identity is ${\mathrm{\partial }}_t^{0}u=u={\mathrm{\Phi }}_0$ by convention and hypothesis 1. For the base $r=1$, step 1.1 gives $u^{\mathrm{\prime }}=G={\mathrm{\Phi }}_1$ continuous into every $H^p$, so $u$ is $C^1$ there and both $j=0,1$ are covered.

Induction step. Suppose $u$ is $C^r$ into every $H^p$, with $r\ge 1$. For any target order $p$, choose $q>\max (p,3\mathrm{/}2)$. The product bound [F2] makes $(f,g)\mapsto (f\cdot \mathrm{\nabla })g$ a continuous bilinear map $H^{q+1}\times H^{q+1}\to H^q\subseteq H^p$. Applying the ordinary bilinear product rule [F3] repeatedly therefore shows that $t\mapsto (u\cdot \mathrm{\nabla })u$ is $C^r$ into $H^p$; its derivatives through order $r$ use only derivatives of $u$ already supplied by the induction hypothesis. Likewise $\mathrm{\Delta }:H^{p+2}\to H^p$ and $P:H^p\to H^p$ are bounded linear maps [F4,F5], so $G=\nu \mathrm{\Delta }u-P((u\cdot \mathrm{\nabla })u)$ is $C^r$ into $H^p$. Since $u\mathrm{\prime }=G$, this proves $u$ is $C^{r+1}$ into $H^p$, with derivative of order $r+1$ equal to the formal Leibniz expression. No differentiability of $u^{(r+1)}$ was assumed to reach this conclusion. As $p$ was arbitrary, induction proves the claim of step 2.2.

By step 2.2 with $r$ arbitrary, for every $r\ge 0$ and every $p$ the derivative ${\mathrm{\partial }}_t^{r}u$ is continuous into $H^p$; by [F1], every mixed derivative ${\mathrm{\partial }}_t^r{\mathrm{\partial }}_x^{\alpha }u$ is continuous on $[0,T]\times {\mathbb{T}}^3$. Hence $u\in C^{\mathrm{\infty }}([0,T]\times {\mathbb{T}}^3;{\mathbb{R}}^3)$, with one-sided time derivatives at the endpoints, and the differentiated identities of the equation hold by construction of the ${\mathrm{\Phi }}_j$.