A limit of Galerkin solutions is a Sobolev and then a classical solution of the symmetric hyperbolic system

(The limit is in $H^s$ with the uniform bound (a).) Fix $t$. For every finite $F\subseteq {\mathbb{Z}}^3$, [L7] gives ${\sum }_{k\in F}(1+\mathrm{\mid }k{\mathrm{\mid }}^2{)}^s\mathrm{\mid }\widehat{u}(t)(k){\mathrm{\mid }}^2={\lim }_j{\sum }_{k\in F}(1+\mathrm{\mid }k{\mathrm{\mid }}^2{)}^s\mathrm{\mid }\widehat{u_{N_j}(t)}(k){\mathrm{\mid }}^2\le 4^s{R}_1^2$; taking the supremum over finite $F$ gives $N_s(u(t){)}^2\le 4^s{R}_1^2$, so $u(t)\in H^s$ and $\mathrm{\parallel }u(t){\mathrm{\parallel }}_{H^s}\le C_s{R}_1$ by the norm equivalence, the bound being uniform in $t$. Values in $K_1$ and continuity into $H^{s-1}$ follow from [L1]. To establish strong measurability into $H^s$, for each $M$ the finite Fourier sum $t\mapsto P_{M}u(t)$ is continuous into $H^s$, since its finitely many coefficients are continuous by the $H^{s-1}$ continuity. Moreover $P_{M}u(t)\to u(t)$ in $H^s$ for every $t$, by the now finite weighted Fourier sum. Explicitly choose compact intervals $J_M$ increasing to $(-T,T)$. On each $J_M$, uniform continuity of $P_{M}u$ permits a finite partition into Borel intervals and a finite-valued simple function $h_M$ whose values are sampled from $P_{M}u$ and whose $H^s$ error is at most $1\mathrm{/}M$; put $h_M=0$ off $J_M$. For every fixed $t$, eventually $t\in J_M$ and $\mathrm{\parallel }{h}_M(t)-u(t){\mathrm{\parallel }}_{H^s}\le 1\mathrm{/}M+\mathrm{\parallel }{P}_{M}u(t)-u(t){\mathrm{\parallel }}_{H^s}\to 0$. Thus $u$ is a pointwise limit of measurable simple functions into $H^s$, hence strongly measurable, and its uniform norm bound gives the claimed $L^{\mathrm{\infty }}{H}^s$ membership. This is (a).

(A convergent subsequence of the time derivatives.) Put $g_j:={\mathrm{\partial }}_t{u}_{N_j}$. By [L4] the family is bounded in $H^{s-1}$ and Lipschitz in $H^{s-2}$; by [L5] there is a strictly increasing $j_n$ with $g_{j_n}\to g$ uniformly on every compact subinterval of $(-T,T)$ in $H^{s-2}$, for some $g\in C^0((-T,T);H^{s-2})$. Discarding the complementary terms, rename the subsequence by $N_j$: still $u_{N_j}\to u$ on compact subintervals in $H^{s-1}$ and now $g_j\to g$ on compact subintervals in $H^{s-2}$.

(The time derivative (b).) (i) Let $\psi \in C_c^{\mathrm{\infty }}((-T,T))$ and $v\in H^{s-2}$; put $J^{\mathrm{\prime }}:=\mathrm{supp}\psi$. By [L6], ${\int }_{J^{\mathrm{\prime }}}{\psi }^{\mathrm{\prime }}(t)⟨u_{N_j}(t),v⟩dt=-{\int }_{J^{\mathrm{\prime }}}\psi (t)⟨g_j(t),v⟩dt$ for every $j$; by step 1.2 the limits ${\int }_{J^{\mathrm{\prime }}}{\psi }^{\mathrm{\prime }}⟨u,v⟩dt$ and ${\int }_{J^{\mathrm{\prime }}}\psi ⟨g,v⟩dt$ are attained uniformly on $J^{\mathrm{\prime }}$, so $\int {\psi }^{\mathrm{\prime }}⟨u,v⟩dt=-\int \psi ⟨g,v⟩dt$. Hence the $H^{s-2}$-valued distributional derivative ${\mathrm{\partial }}_{t}u$ equals the class of $g$. (ii) Fix $t$; by step 1.2, $\widehat{g_j(t)}(k)\to \widehat{g}(t)(k)$ for every $k$ [L7], so for every finite $K\subseteq {\mathbb{Z}}^3$ the finite sums ${\sum }_{k\in K}(1+\mathrm{\mid }k{\mathrm{\mid }}^2{)}^{s-1}\mathrm{\mid }\widehat{g_j(t)}(k){\mathrm{\mid }}^2$ are at most $c_{s-1}^{-1}\mathrm{\parallel }{g}_j(t){\mathrm{\parallel }}_{H^{s-1}}^2\le c_{s-1}^{-1}{C}_{R_1}^2$ by [L4]; passing to the limit and taking the supremum over finite $K$ gives $N_{s-1}(g(t){)}^2\le c_{s-1}^{-1}{C}_{R_1}^2$, so $g(t)\in H^{s-1}$ with $\mathrm{\parallel }g(t){\mathrm{\parallel }}_{H^{s-1}}\le C_{s-1}^{1\mathrm{/}2}{c}_{s-1}^{-1\mathrm{/}2}{C}_{R_1}$ by [L7]. For each $M$, $t\mapsto P_{M}g(t)$ is continuous into $H^{s-1}$ because $g\in C^0{H}^{s-2}$ and the projection has finite rank, and $P_{M}g(t)\to g(t)$ in $H^{s-1}$ for each $t$. Applying the compact-interval simple-function construction of step 1.1 at order $s-1$ proves strong measurability of $g$ into $H^{s-1}$. Thus ${\mathrm{\partial }}_{t}u$ is represented by the function $g\in L^{\mathrm{\infty }}((-T,T);H^{s-1})$, uniformly bounded as asserted. This is (b).

(The limit solves the equation distributionally (c), first part.) Let $\psi \in C^{\mathrm{\infty }}((-T,T)\times {\mathbb{T}}_{2\pi }^3;{\mathbb{R}}^m)$ have compact support in $t$. Testing the Galerkin equation [L2] with $\psi$ and using self-adjointness of $P_{N_j}$ gives $\int \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\int {\psi }^{\mathsf{T}}[A^0(u_{N_j}){\mathrm{\partial }}_t{u}_{N_j}+A^i(u_{N_j}){\mathrm{\partial }}_i{u}_{N_j}+B(u_{N_j})]=\int \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\int (\psi -P_{N_j}\psi {)}^{\mathsf{T}}[\cdots ]$, and the right-hand side tends to $0$ because $\psi -P_{N_j}\psi \to 0$ in $L^2$ [L2] while the bracket is bounded in $L^{\mathrm{\infty }}{L}^2$ by [L3] and [L4]. On the left, all factors converge on the compact $t$-support of $\psi$: $A^0(u_{N_j})\to A^0(u)$, $A^i(u_{N_j})\to A^i(u)$ and $B(u_{N_j})\to B(u)$ uniformly by [L1]; ${\mathrm{\partial }}_i{u}_{N_j}\to {\mathrm{\partial }}_{i}u$ uniformly by [L1]; and ${\mathrm{\partial }}_t{u}_{N_j}=g_j\to g={\mathrm{\partial }}_{t}u$ uniformly in $H^{s-2}\subseteq L^2$ by step 1.2 and step 2.1; hence the limit is $\int \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\int {\psi }^{\mathsf{T}}[A^0(u){\mathrm{\partial }}_{t}u+A^i(u){\mathrm{\partial }}_{i}u+B(u)]=0$ for every test $\psi$, which is the distributional form of $(\star )$.

(Classical regularity and initial value.) From the distributional equation of step 3.1, ${\mathrm{\partial }}_{t}u=-(A^0(t,\cdot ,u){)}^{-1}[A^i(t,\cdot ,u){\mathrm{\partial }}_{i}u+B(t,\cdot ,u)]$ as $H^{s-2}$-valued distributions; the right-hand side agrees with the continuous function $g={\mathrm{\partial }}_{t}u\in C^0((-T,T);H^{s-2})$ as a distribution, so the classical identity ${\mathrm{\partial }}_{t}u(t)=-(A^0(t,\cdot ,u(t)){)}^{-1}[A^i(t,\cdot ,u(t)){\mathrm{\partial }}_{i}u(t)+B(t,\cdot ,u(t))]$ holds for every $t$, both sides being continuous in $t$. The right-hand side is continuous in $(t,x)$: indeed $u\in C^0{H}^{s-1}$ gives ${\mathrm{\partial }}_{i}u\in C^0{H}^{s-2}\subseteq C^0$ with $s-2>3\mathrm{/}2$; the inverse coefficient is continuous by [L3]; hence ${\mathrm{\partial }}_{t}u$ has a continuous representative in $(t,x)$. Passing to the limit in $u_{N_j}(t)-u_{N_j}(t^{\mathrm{\prime }})={\int }_{t^{\mathrm{\prime }}}^t{g}_j(\tau )d\tau$ using the uniform $H^{s-2}$ convergence on the segment gives $u(t)-u(t^{\mathrm{\prime }})={\int }_{t^{\mathrm{\prime }}}^{t}g(\tau )d\tau$; the continuous-curve fundamental theorem [L6] therefore makes $t\mapsto u(t)$ $C^1$ into $H^{s-2}$ with continuous derivative and the first spatial derivatives are continuous, so $u\in C^1((-T,T)\times {\mathbb{T}}_{2\pi }^3)$ and $(\star )$ holds pointwise. Finally $u_{N_j}(0)=P_{N_j}{u}_0\to u_0$ in $H^{s-1}$ and $u_{N_j}(0)\to u(0)$ by the convergence of step 1.2, so $u(0)=u_0$.