Two-scale compactness for bounded L2 sequences

Let $M={\sup }_n\mathrm{\parallel }{u}_n{\mathrm{\parallel }}_2$. By [F7], extend functions by zero to see that $L^2(Q\times Y)$ is separable; [F3] says the allowed test subspace is dense there. Intersect a countable base of its metric topology with that subspace and, using [F4], choose one point from every nonempty intersection. The rational span of these points is a countable dense family $D$. For each fixed $\mathrm{\Phi }\in D$, Cauchy--Schwarz gives $\mid {\int }_Q{u}_n(x)\mathrm{\Phi }(x,x\mathrm{/}{ϵ}_n)dx\mid \le M\mathrm{\parallel }\mathrm{\Phi }(\cdot ,\cdot \mathrm{/}{ϵ}_n){\mathrm{\parallel }}_2,$ which is bounded by [F2]. For a bounded real sequence $(b_n)$ fix this extraction rule: let $m$ be the least positive integer bounding every $\mathrm{\mid }{b}_n\mathrm{\mid }$, start with $[-m,m]$, bisect repeatedly, and retain the left closed half when it contains infinitely many terms, otherwise the right closed half. From the $r$th retained interval take the term with least index larger than the previously selected index. Such indices exist; nestedness and vanishing interval lengths make the selected sequence Cauchy, and [F5] makes it convergent. This uniquely specified extraction map requires no choice. Enumerate $D$, recursively apply this map to each scalar pairing on the current subsequence, and take as $n_r$ the $r$th index of the $r$th nested subsequence. The indices increase strictly and their tail lies in every fixed earlier subsequence, so all pairings with $D$ converge. Recursion of the specified map does not use Dependent Choice.

Denote the limit on $D$ by $L_0$. For each $\mathrm{\Phi }\in D$, step 1.1 and [F2] give $\mathrm{\mid }{L}_0(\mathrm{\Phi })\mathrm{\mid }\le M\mathrm{\parallel }\mathrm{\Phi }{\mathrm{\parallel }}_{L^2(Q\times Y)}$. Limits preserve rational linearity; the bound makes $L_0$ uniformly continuous, so it has a unique continuous extension $L$ to all of the real Hilbert space $L^2(Q\times Y)$, with $\mathrm{\parallel }L\mathrm{\parallel }\le M$. To verify real homogeneity, first fix $\mathrm{\Phi }\in D$ and $t\in \mathbb{R}$, and choose rationals $q_m\to t$. Since $q_m\mathrm{\Phi }\in D$, $L(q_m\mathrm{\Phi })=q_m{L}_0(\mathrm{\Phi })$; continuity and $q_m\mathrm{\Phi }\to t\mathrm{\Phi }$ give $L(t\mathrm{\Phi })=tL(\mathrm{\Phi })$. For arbitrary $h\in L^2(Q\times Y)$, choose ${\mathrm{\Phi }}_j\in D$ with ${\mathrm{\Phi }}_j\to h$; then $t{\mathrm{\Phi }}_j\to th$, and continuity gives $L(th)=tL(h)$. Additivity also passes from $D$ by continuity, so $L$ is bounded and real-linear.

By [F6], choose $u\in L^2(Q\times Y)$ such that $L(\mathrm{\Phi })={\int }_{Q\times Y}u\mathrm{\Phi }$. Fix any allowed smooth product test $\mathrm{\Psi }$ and choose ${\mathrm{\Phi }}_j\in D$ with ${\mathrm{\Phi }}_j\to \mathrm{\Psi }$ in product $L^2$. Along the extracted subsequence, [F2] applied to $\mathrm{\Psi }-{\mathrm{\Phi }}_j$ gives ${\mathrm{lim\hspace{0.17em}sup}}_n\mid {\int }_Q{u}_n(\mathrm{\Psi }-{\mathrm{\Phi }}_j)(x,x\mathrm{/}{ϵ}_n)dx\mid \le M\mathrm{\parallel }\mathrm{\Psi }-{\mathrm{\Phi }}_j{\mathrm{\parallel }}_2\mathrm{.}$ Also $\mathrm{\mid }L(\mathrm{\Psi }-{\mathrm{\Phi }}_j)\mathrm{\mid }\le M\mathrm{\parallel }\mathrm{\Psi }-{\mathrm{\Phi }}_j{\mathrm{\parallel }}_2$. Since the pairings for ${\mathrm{\Phi }}_j$ converge to $L({\mathrm{\Phi }}_j)$, letting $j\to \mathrm{\infty }$ proves convergence for $\mathrm{\Psi }$ to $L(\mathrm{\Psi })=\int u\mathrm{\Psi }$. This is precisely the two-scale convergence in [F1].