Elementary zero-boundary L6 and L4 interpolation

If $\mathrm{\Omega }=\mathrm{\varnothing }$, [F9] and the local closure formulas in the Statement give $H_0^1=H=V=\{0\}$. The only linear functional on $V$ is zero, so $V^{\ast }=\{0\}$ as well; each zero space is a separable real Hilbert space (the singleton $\{0\}$ is dense). Every spatial $L^p$ class and every time-dependent map with one of these zero targets is zero, with the constant zero simple map proving strong measurability. Thus both spatial inequalities, time integrability, representative independence, the defining formula for $b_u$, and the $V^{\ast }$ norm bound all hold with zero on both sides, for the same positive universal constant as in the nonempty case. No basis, approximation sequence or zero-extension theorem for a nonempty domain is applied here. Henceforth assume $\mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }$; all uses of [F1], [F6] and [F8] below are in that case.

For real $g\in C^1([a,b])$, put $G(t)={\int }_a^t{g}^{\mathrm{\prime }}(s)ds$. The difference between its difference quotient at an interior $t$ and $g^{\mathrm{\prime }}(t)$ is bounded by ${\sup }_{\mathrm{\mid }s-t\mathrm{\mid }\le \mathrm{\mid }h\mathrm{\mid }}\mathrm{\mid }{g}^{\mathrm{\prime }}(s)-g^{\mathrm{\prime }}(t)\mathrm{\mid }$, which tends to zero. Hence $(G-g{)}^{\mathrm{\prime }}=0$ everywhere inside the interval; the mean value theorem and endpoint continuity give $g(b)-g(a)={\int }_a^b{g}^{\mathrm{\prime }}$. For $h\in C_c^{\mathrm{\infty }}({\mathbb{R}}^3)$ define $A_1(x_2,x_3)={\int }_{\mathbb{R}}\mathrm{\mid }{\mathrm{\partial }}_{1}h(x_1,x_2,x_3)\mathrm{\mid }d{x}_1$, and define $A_2(x_1,x_3)$ and $A_3(x_1,x_2)$ analogously. Applying the endpoint identity along each coordinate line from a point outside the support gives $\mathrm{\mid }h(x)\mathrm{\mid }\le A_i(x_{\widehat{i}})$ for each $i$, hence $\mathrm{\mid }h{\mathrm{\mid }}^{3\mathrm{/}2}\le (A_1{A}_2{A}_3{)}^{1\mathrm{/}2}$. Put $B_2(x_3)=\int A_2(x_1,x_3)d{x}_1$ and $B_3(x_2)=\int A_3(x_1,x_2)d{x}_1$. Cauchy–Schwarz in $x_1$ gives $\int (A_2{A}_3{)}^{1\mathrm{/}2}d{x}_1\le (B_2{B}_3{)}^{1\mathrm{/}2}$; applying Cauchy–Schwarz in $(x_2,x_3)$ to the remaining $A_1^{1\mathrm{/}2}(B_2{B}_3{)}^{1\mathrm{/}2}$ gives ${\int }_{{\mathbb{R}}^3}\mathrm{\mid }h{\mathrm{\mid }}^{3\mathrm{/}2}\le ({\int }_{{\mathbb{R}}^2}{A}_1{)}^{1\mathrm{/}2}({\int }_{\mathbb{R}}{B}_2{)}^{1\mathrm{/}2}({\int }_{\mathbb{R}}{B}_3{)}^{1\mathrm{/}2}={\prod }_{i=1}^3({\int }_{{\mathbb{R}}^3}\mathrm{\mid }{\mathrm{\partial }}_{i}h\mathrm{\mid }{)}^{1\mathrm{/}2}\mathrm{.}$ The changes of integration order are justified by [F3]. Raising to the power $2\mathrm{/}3$ yields $\mathrm{\parallel }h{\mathrm{\parallel }}_{3\mathrm{/}2}\le {\prod }_i\mathrm{\parallel }{\mathrm{\partial }}_{i}h{\mathrm{\parallel }}_1^{1\mathrm{/}3}$.

For a smooth compactly supported vector field $w$, set $h=\mathrm{\mid }w{\mathrm{\mid }}^4$, which is smooth, and note $\mathrm{\mid }{\mathrm{\partial }}_{i}h\mathrm{\mid }\le 4\mathrm{\mid }w{\mathrm{\mid }}^3\mathrm{\mid }{\mathrm{\partial }}_{i}w\mathrm{\mid }$. Hölder gives $\mathrm{\parallel }{\mathrm{\partial }}_{i}h{\mathrm{\parallel }}_1\le 4\mathrm{\parallel }w{\mathrm{\parallel }}_6^3\mathrm{\parallel }{\mathrm{\partial }}_{i}w{\mathrm{\parallel }}_2$. Substitution into step 1.2 yields $\mathrm{\parallel }w{\mathrm{\parallel }}_6^4\le 4\mathrm{\parallel }w{\mathrm{\parallel }}_6^3{\prod }_i\mathrm{\parallel }{\mathrm{\partial }}_{i}w{\mathrm{\parallel }}_2^{1\mathrm{/}3}\le 4\mathrm{\parallel }w{\mathrm{\parallel }}_6^3\mathrm{\parallel }\mathrm{\nabla }w{\mathrm{\parallel }}_2$. Divide when $\mathrm{\parallel }w{\mathrm{\parallel }}_6>0$; if it is zero the inequality is immediate.

By [F1] and [F5], choose $w_n\in C_c^{\mathrm{\infty }}(\mathrm{\Omega };{\mathbb{R}}^3)$ converging to $a$ in $H^1$. The inequality of step 2.1 applied to differences makes $(w_n)$ Cauchy in $L^6$; on bounded $\mathrm{\Omega }$, convergence in $L^6$ also implies convergence in $L^2$, so its $L^6$ limit is the given $L^2$ class $a$. Passing to the limit proves the first inequality. Finally, $\int \mathrm{\mid }a{\mathrm{\mid }}^4\le (\int \mathrm{\mid }a{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}(\int \mathrm{\mid }a{\mathrm{\mid }}^6{)}^{1\mathrm{/}2}$ by Cauchy–Schwarz, which proves the interpolation inequality. For almost every $t$, raising it to power $8\mathrm{/}3$ gives $\mathrm{\parallel }u(t){\mathrm{\parallel }}_4^{8\mathrm{/}3}\le \mathrm{\parallel }u(t){\mathrm{\parallel }}_2^{2\mathrm{/}3}\mathrm{\parallel }u(t){\mathrm{\parallel }}_6^2\le C\mathrm{\parallel }u(t){\mathrm{\parallel }}_H^{2\mathrm{/}3}\mathrm{\parallel }u(t){\mathrm{\parallel }}_V^2$. Integrating and using $u\in L_t^{\mathrm{\infty }}H\cap L_t^{2}V$ proves the stated $L_t^{8\mathrm{/}3}{L}_x^4$ bound.

Fix the strongly measurable representative $\tilde{u}$ chosen in the Statement. For almost every time, Hölder gives $\mathrm{\mid }{b}_u(t)(\varphi )\mathrm{\mid }\le \mathrm{\parallel }\tilde{u}(t){\mathrm{\parallel }}_4^2\mathrm{\parallel }\mathrm{\nabla }\varphi {\mathrm{\parallel }}_2$, so $b_u(t)\in V^{\ast }$ and its norm is at most $\mathrm{\parallel }\tilde{u}(t){\mathrm{\parallel }}_4^2$. The map $a\mapsto b_a$ is continuous from $L^4$ to $V^{\ast }$ because $\mathrm{\parallel }{b}_a-b_c{\mathrm{\parallel }}_{V^{\ast }}\le (\mathrm{\parallel }a{\mathrm{\parallel }}_4+\mathrm{\parallel }c{\mathrm{\parallel }}_4)\mathrm{\parallel }a-c{\mathrm{\parallel }}_4$. Composing simple-map approximations to $\tilde{u}$ with the continuous embedding $V\to L^4$ and this map therefore gives strongly measurable $V^{\ast }$-valued $b_u$, up to the permitted null-set assignment; [F6,F8] provide the separable Hilbert target. Two representatives of the same $L_t^{2}V$ class agree in $V$ almost everywhere, hence agree in $L^4$ there and define the same $b_u$ almost everywhere. Thus its Bochner class is independent of the representative. Finally ${\int }_0^T\mathrm{\parallel }{b}_u(t){\mathrm{\parallel }}_{V^{\ast }}^{4\mathrm{/}3}dt\le {\int }_0^T\mathrm{\parallel }\tilde{u}(t){\mathrm{\parallel }}_4^{8\mathrm{/}3}dt<\mathrm{\infty }$ by step 3.1; [F7] gives $b_u\in L^{4\mathrm{/}3}(0,T;V^{\ast })$.