Tame bounds for the quadratic-cubic trajectory remainder

First, multiplication by $a\in W^{1,\mathrm{\infty }}(\mathrm{\Omega })$ is bounded on $H^1$, with $\mathrm{\parallel }av{\mathrm{\parallel }}_{H^1}\le C\mathrm{\parallel }a{\mathrm{\parallel }}_{W^{1,\mathrm{\infty }}}\mathrm{\parallel }v{\mathrm{\parallel }}_{H^1}$ and $D_j(av)=a{D}_{j}v+(D_{j}a)v$. To prove this without assuming $a\in L^2$, start with smooth $v\in H^1$. Around the compact support of any test function, multiply $a$ and $v$ by a compactly supported smooth cutoff equal to one there. Both localized functions are bounded $H^1$ functions on a bounded neighborhood, so [F2] gives the weak product formula on that test support. The global coefficient bounds make the displayed derivative and value square integrable, proving the asserted global norm bound. For general $v$, approximate it in $H^1$ by smooth $H^1$ functions using [F2]; the same bound makes their products Cauchy in $H^1$. Their $L^2$ limit is $av$, so completeness and the weak-derivative norm identify the same formula and bound. Also every $H^1$ function supported in a compact $K⋐\mathrm{\Omega }$ lies in $H_0^1$: choose $\eta \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ equal to one near $K$, approximate the function by smooth $H^1$ functions, and multiply those approximants by $\eta$. The weak cutoff bound gives convergence in $H^1$ to $\eta v=v$, while the approximants are in $C_c^{\mathrm{\infty }}(\mathrm{\Omega })$.

Put $T_R(s)=s\chi (s\mathrm{/}R)$ and $S_{R}v=T_R(\mathrm{Re}v)+i{T}_R(\mathrm{Im}v)$ for positive integers $R$. The functions $T_R$ are smooth, fix zero and have bounded continuous derivatives. The map $S_R:H^1\to H^1$ is continuous. Indeed, along an $H^1$ convergent sequence take an almost-everywhere convergent subsequence; the derivative difference of a real component has the form $T_R^{\mathrm{\prime }}(v_k)(D_j{v}_k-D_{j}v)+(T_R^{\mathrm{\prime }}(v_k)-T_R^{\mathrm{\prime }}(v))D_{j}v$. The first term tends to zero in $L^2$ by its bounded coefficient, and dominated convergence gives the same for the second. The value difference is controlled by the Lipschitz bound. Applying this argument to every subsequence proves continuity of the full sequence. Since $T_R(0)=0$, its operation on $C_c^{\mathrm{\infty }}$ stays in $C_c^{\mathrm{\infty }}$; continuity and closure give $S_R(H_0^1)\subset H_0^1$. Each $S_{R}v$ has an essential bound depending only on $R$. If $u\in H_0^1\cap L^{\mathrm{\infty }}$, fix $R>\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}+1$ and apply $S_R$ to compactly supported smooth approximants to $u$. This produces $u_k\in C_c^{\mathrm{\infty }}$, uniformly bounded in $L^{\mathrm{\infty }}$, converging in $H^1$ to $u=S_{R}u$ and, after a subsequence, almost everywhere. This is also the bounded approximation used in [F1].

If $u\in H_0^1\cap L^{\mathrm{\infty }}$ and $v\in H^1\cap L^{\mathrm{\infty }}$, then $uv\in H_0^1$. Choose $u_k$ from step 1.2. Each $u_{k}v$ is compactly supported and in $H^1$ by [F2], so belongs to $H_0^1$ by step 1.1. Its value converges to $uv$ in $L^2$ by $\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }{u}_k-u{\mathrm{\parallel }}_2$. The derivative difference is $v(D_j{u}_k-D_{j}u)+(u_k-u)D_{j}v$; the first term converges in $L^2$, and the second does so by the uniform bound, almost-everywhere convergence and dominated convergence. Closure proves the claim. Conjugation preserves weak derivatives and $H_0^1$ by conjugating the test identities and approximants. Finally, any $a\in W^{1,\mathrm{\infty }}$ preserves $H_0^1$, without being in $L^2$: for ${\phi }_k\in C_c^{\mathrm{\infty }}$ converging to $u\in H_0^1$, each $a{\phi }_k$ is compactly supported $H^1$, and the multiplier estimate in step 1.1 gives convergence to $au$ in $H^1$. Thus all quadratic terms with the energy coefficient have zero-boundary closure.

For bounded $H^1$ functions define $A_u=\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}$ and $B_u=\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}$. The weak product formula gives $\mathrm{\parallel }uv{\mathrm{\parallel }}_{H^1}\le C(A_u{B}_v+A_v{B}_u)$. With $d=z-q$, the identities $z^2-q^2=dz+qd$ and $\mathrm{\mid }z{\mathrm{\mid }}^2-\mathrm{\mid }q{\mathrm{\mid }}^2=d\bar{z}+q\bar{d}$ therefore give, for either quadratic expression $Q$, $\mathrm{\parallel }Q(z){\mathrm{\parallel }}_{H^1}\le C{A}_z{B}_z$ and $\mathrm{\parallel }Q(z)-Q(q){\mathrm{\parallel }}_{H^1}\le C((A_z+A_q)B_d+(B_z+B_q)A_d)$. Apply the multiplier estimate to $2w\mathrm{\mid }z{\mathrm{\mid }}^2+\bar{w}{z}^2$ to obtain the same bounds times $C{M}_1$. This calculation uses only the value and first derivative bounds of $w$, not its global $L^2$ norm. The cubic term and its pointwise-in-time difference estimate are already supplied by [F1].

To justify energy measurability, $t\mapsto S_{R}z(t)$ is continuous in $H_0^1$ by step 1.2 and is uniformly spatially bounded for fixed $R$. On uniformly bounded $H^1$ sets, the maps $v\mapsto v^2$ and $v\mapsto \mathrm{\mid }v{\mathrm{\mid }}^2$ are continuous into $H^1$: for a convergent sequence, subtract the product derivative formulas by adding the approximating coefficient times the limiting derivative; the terms with derivative differences converge by the uniform bound, and the other terms converge by an almost-everywhere subsequence and dominated convergence. The subsequence criterion gives convergence of the whole sequence. Since $w$ is norm-continuous into $W^{1,\mathrm{\infty }}$, the two truncated quadratic expressions with $w$ are continuous $H_0^1$ paths by the multiplier bound and step 2.1. Each continuous path on a compact interval admits uniform finite-valued time-step approximations. For almost every $t$, compatibility of the scalar mixed representative gives $\mathrm{\parallel }z(t){\mathrm{\parallel }}_{\mathrm{\infty }}<\mathrm{\infty }$, so $S_{R}z(t)=z(t)$ for every sufficiently large integer $R$. Thus the quadratic expression is a pointwise almost-everywhere limit of these strongly measurable paths, and hence strongly measurable; one may choose each path's finite-valued approximant with uniform error at most $1\mathrm{/}R$. The cubic term is strongly measurable by [F1]. Differences of the resulting maps are strongly measurable as well. No strong measurability into spatial $L^{\mathrm{\infty }}$ is needed.

In the order-two setting [F6] gives $H^2\subset H^1\cap L^{\mathrm{\infty }}$ and its global algebra bound. Every quadratic-cubic summand in $F$ has at least one factor $z\in H_0^1$, while every remaining factor, including $w$ and its conjugate, lies in $H^1\cap L^{\mathrm{\infty }}$. Repeatedly apply step 2.1 to obtain actual $H_0^1$ membership, with no zero trace assumption on $w$. Repeated multiplication in the exterior $H^2$ algebra gives $\mathrm{\parallel }F(w,z){\mathrm{\parallel }}_{H^2}\le C_{\mathrm{\Omega }}\mathrm{\mid }\lambda \mathrm{\mid }(\mathrm{\parallel }w{\mathrm{\parallel }}_{H^2}\mathrm{\parallel }z{\mathrm{\parallel }}_{H^2}^2+\mathrm{\parallel }z{\mathrm{\parallel }}_{H^2}^3)$. Telescoping the factors in $z^2-q^2$, $\mathrm{\mid }z{\mathrm{\mid }}^2-\mathrm{\mid }q{\mathrm{\mid }}^2$, and $z^2\bar{z}-q^2\bar{q}$ gives the pointwise difference bound $C_{\mathrm{\Omega }}\mathrm{\mid }\lambda \mathrm{\mid }[\mathrm{\parallel }w{\mathrm{\parallel }}_{H^2}(\mathrm{\parallel }z{\mathrm{\parallel }}_{H^2}+\mathrm{\parallel }q{\mathrm{\parallel }}_{H^2})+\mathrm{\parallel }z{\mathrm{\parallel }}_{H^2}^2+\mathrm{\parallel }q{\mathrm{\parallel }}_{H^2}^2]\mathrm{\parallel }z-q{\mathrm{\parallel }}_{H^2}$. Continuous multiplication and conjugation in this algebra show that $t\mapsto F(w(t),z(t))$ is continuous in $H^2$, and its values are in the closed subspace $V_2$, hence the path is continuous into that Banach space.

For the energy bounds write $a_z=\mathrm{\parallel }\mathrm{\parallel }z(t){\mathrm{\parallel }}_{\mathrm{\infty }}{\mathrm{\parallel }}_{L_t^4}$ and $b_z={\sup }_t\mathrm{\parallel }z(t){\mathrm{\parallel }}_{H^1}$. Holder gives ${\int }_0^T\mathrm{\parallel }z(t){\mathrm{\parallel }}_{\mathrm{\infty }}dt\le T^{3\mathrm{/}4}{a}_z$. Thus step 2.2 bounds the quadratic growth by $C\mathrm{\mid }\lambda \mathrm{\mid }{M}_1{T}^{3\mathrm{/}4}{a}_z{b}_z$. Its quadratic difference is bounded after integration by $C\mathrm{\mid }\lambda \mathrm{\mid }{M}_1{T}^{3\mathrm{/}4}((a_z+a_q)b_d+(b_z+b_q)a_d)$. Every $a$ and $b$ is bounded by the corresponding $X_1$ norm. The $m=3$ case of [F1] supplies the remaining cubic estimates with $T^{1\mathrm{/}2}$. These yield both displayed energy inequalities. In view of step 3.1 the norms are integrable norms of strongly measurable $H_0^1$ maps, so they are the claimed Bochner $L^1$ quantities by [F5].

Integrating the pointwise bounds in step 3.2 over the finite interval gives the order-two estimates with factor $T$. The continuous path there is strongly measurable by uniform time-step approximation and has bounded $V_2$ norm, so Bochner integrability follows from [F3] and [F5]. To identify its obstacle trace, choose $R$ with $\mathrm{\Theta }⋐B_R$ and let $G=\mathrm{\Omega }\cap B_R$, the bounded smooth truncation in [F6]. Restrict any $H_0^1(\mathrm{\Omega })$ approximating sequence from $C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ to $G$. Each approximant has zero obstacle trace and convergence in $H^1(G)$; boundedness of the trace in [F6] gives zero obstacle trace for the limit. For any $u\in H^2(\mathrm{\Omega })$, restriction gives $u{\mathrm{\mid }}_G\in H^2(G)$. Apply the bounded-domain extension theorem of [F6] with $k=p=2$ to make $G$ an $H^2$ extension domain, and then the higher-order embedding there with $m=0$, $\alpha =1\mathrm{/}4$; its strict supercritical hypotheses hold since $0<1\mathrm{/}4<2-n\mathrm{/}2$. This produces a representative of $u{\mathrm{\mid }}_G$ continuous on $\bar{G}$, to which the bounded $H^1(G)$ trace theorem applies and identifies the boundary values with the trace. Applying this to $u=F(w(t),z(t))\in V_2$ from step 3.2, for each $t$, proves the claimed continuous-boundary-value interpretation of its zero obstacle trace, without requiring a zero trace of $w$ or a global exterior trace theorem.

Finally, in the energy case absorb $C\mathrm{\mid }\lambda \mathrm{\mid }{M}_1{T}^{3\mathrm{/}4}$ and $C\mathrm{\mid }\lambda \mathrm{\mid }{T}^{1\mathrm{/}2}$ into $C_w$; in the order-two case absorb $C_{\mathrm{\Omega }}\mathrm{\mid }\lambda \mathrm{\mid }T(M_2+1)$. The sharper estimates then imply exactly the two approved quadratic-cubic tame inequalities for $s=1,2$, with the stated dependence and all membership and measurability assertions proved above. This completes the proof. ∎