Cubic and quintic time bounds from L4 Linfinity control

By [F6], $H^1(\mathrm{\Omega })=W^{1,2}(\mathrm{\Omega };\mathbb{C})$ is Banach under the already assumed Axiom of Choice. The test functions form a complex linear subspace. Their norm closure $H_0^1(\mathrm{\Omega })$ is also linear: for $u,v$ in that closure and $a,b\in \mathbb{C}$, approximate $u,v$ by test functions with errors less than $\epsilon \mathrm{/}(1+\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid })$ and use the norm triangle inequality to approximate $au+bv$. The closure is closed: if $w$ is in its ambient closure, choose an element $u\in H_0^1$ within $\epsilon \mathrm{/}2$ of $w$, then a test function within $\epsilon \mathrm{/}2$ of $u$; hence $w$ itself belongs to the test-function closure. Its norm is the restriction of the $H^1$ norm, so the closed-subspace result in [F6] makes $H_0^1(\mathrm{\Omega })$ Banach. This verifies the target-space hypothesis of the time spaces and the Banach-valued inputs in [F2].

Choose the cutoff $\chi$ in [F5] and set $T_R(s)=s\chi (s\mathrm{/}R)$ for positive integers $R$. It is smooth, globally Lipschitz, fixes zero and agrees with $s$ on $[-R,R]$. Apply it to the real and imaginary parts of $w\in H_0^1$ and call the resulting complex function $w_R$. By [F4], $w_R\in H^1$. It lies in $H_0^1$: apply the same smooth zero-fixing operation to compactly supported smooth approximants of $w$. Their images are compactly supported and smooth. For convergence in $H^1$, use the Lipschitz bound for the functions and the chain rule for derivatives; after taking an almost-everywhere convergent subsequence, subtract the derivatives as $T_R^{\mathrm{\prime }}(q_k)(D{q}_k-Dq)+(T_R^{\mathrm{\prime }}(q_k)-T_R^{\mathrm{\prime }}(q))Dq$ and use boundedness and continuity of $T_R^{\mathrm{\prime }}$ and dominated convergence. The same argument, applied to any convergent sequence, proves that $w\mapsto w_R$ is continuous in $H^1$. Here the subsequence argument gives convergence of the whole sequence because every subsequence admits a further convergent subsequence.

The functions $w_R$ have a spatial essential bound depending only on $R$. On any uniformly bounded sequence converging in $H^1$, the polynomial map $g$ is continuous into $H^1$: take an almost-everywhere convergent subsequence, use the polynomial value and weak-derivative formulas in [F1], subtract the derivative terms by adding the coefficient at the approximating function times the limiting derivative, and apply dominated convergence to the bounded coefficient differences times that limiting derivative. The other derivative terms tend to zero in $L^2$ by their uniform coefficient bounds. The subsequence criterion again gives full sequential convergence. Consequently $t\mapsto g(v_R(t))$ is a continuous $H_0^1$-valued path, where $v_R(t)$ is the componentwise truncation of $v(t)$; [F1] supplies zero-boundary preservation.

Each continuous path on the compact interval is a uniform limit of finite-range measurable time-step functions, obtained from sufficiently fine finite partitions. Thus each path $g(v_R)$ is strongly measurable. For almost every $t$, the mixed-norm hypothesis gives $\mathrm{\parallel }v(t){\mathrm{\parallel }}_{\mathrm{\infty }}<\mathrm{\infty }$ and the continuous and mixed representatives agree as spatial classes; for all sufficiently large integers $R$, $v_R(t)=v(t)$ in $H^1$, so $g(v_R(t))=g(v(t))$. Diagonal approximation of the countably many strongly measurable paths, or equivalently choosing simple approximants with errors at most $1\mathrm{/}R$ outside null sets, shows that their pointwise limit is strongly measurable. This uses only the measurable mixed section norm, not strong measurability into $L^{\mathrm{\infty }}$. The same reasoning applies to $u$, and differences of the resulting strongly measurable maps are strongly measurable by subtracting simple approximants.

Write $A_v(t)=\mathrm{\parallel }v(t){\mathrm{\parallel }}_{\mathrm{\infty }}$, $B_v={\sup }_t\mathrm{\parallel }v(t){\mathrm{\parallel }}_{H^1}$ and $a_v=\mathrm{\parallel }{A}_v{\mathrm{\parallel }}_{L^4(0,T)}$. By [F1], $\mathrm{\parallel }g(v(t)){\mathrm{\parallel }}_{H^1}\le C_m\mathrm{\mid }\lambda \mathrm{\mid }{A}_v(t{)}^{m-1}{B}_v$ for almost every $t$. Apply [F3] to $A_v^{m-1}$ and $1$ with reciprocal exponents $(m-1)\mathrm{/}4$ and $(5-m)\mathrm{/}4$, interpreting reciprocal exponent zero as $L^{\mathrm{\infty }}$. This gives ${\int }_0^T{A}_v^{m-1}dt\le T^{(5-m)\mathrm{/}4}{a}_v^{m-1}$, including the $m=5$ endpoint. Thus the nonlinear norm is integrable; [F2] gives Bochner integrability and the first displayed bound since $a_v,B_v\le \mathrm{\parallel }v{\mathrm{\parallel }}_{X_T}$.

Put $h=v-u$ and use the difference estimate in [F1]. Its first part is bounded in time by $C_m\mathrm{\mid }\lambda \mathrm{\mid }{B}_h(A_u^{m-1}+A_v^{m-1})$, whose integral is controlled as in step 4.1. For the second part use $C_m\mathrm{\mid }\lambda \mathrm{\mid }(B_u+B_v)(A_u^{m-2}+A_v^{m-2})A_h$. Repeated Hölder gives ${\int }_0^T{A}_w^{m-2}{A}_{h}dt\le T^{(5-m)\mathrm{/}4}{a}_w^{m-2}{a}_h$: the reciprocal exponents are $(m-2)\mathrm{/}4$, $1\mathrm{/}4$, and $(5-m)\mathrm{/}4$, with the last factor interpreted as $L^{\mathrm{\infty }}$ for $m=5$. Since every $a$ and $B$ is bounded by the corresponding $X_T$ norm, and $x^{m-2}y+x{y}^{m-2}\le x^{m-1}+y^{m-1}$ for nonnegative $x,y$, the second displayed bound follows after enlarging $C_m$. The integrable bound and step 3.1 also give the asserted $L^1$ membership of the difference.