Fourier compactness with coefficient time control

For each fixed $k$, the coefficients are bounded by $A$ and have common Lipschitz constant $C_k$. The following deterministic rule extracts a convergent subsequence from any sequence in the closed cube $[-A,A{]}^{2d}$ underlying ${\mathbb{C}}^d$. Order the finitely many closed half-sized subcubes lexicographically, retain the first containing infinitely many terms of the current index set, and repeat inside that cube. Select from the $j$th retained infinite index set the least index greater than the previously selected index. The selected terms converge because the cubes are nested, closed, and have diameters tending to zero. Enumerate the countable pairs $(k,q)$ with $k\in {\mathbb{Z}}^d$ and $q\in \mathbb{Q}\cap [0,T]$. Apply this fixed rule recursively to the preceding subsequence for each successive pair, and take diagonal indices, each the least available index greater than the preceding one. For any fixed pair the diagonal is eventually a subsequence of its convergent extraction. Thus it converges at every such pair without any dependent sequence of arbitrary choices. For each fixed $k$, the common Lipschitz bound and a finite rational mesh make the coefficients uniformly Cauchy on $[0,T]$, including its endpoints. Thus one subsequence converges uniformly in every Fourier coefficient.

At each time, every finite Fourier sum of squared limiting coefficients is at most $A^2$; Parseval therefore defines $z(t)\in H$ with norm at most $A$. Pairings with finite Fourier sums are continuous in time. For arbitrary $h\in H$, approximate $h$ by its finite Fourier truncations; the pairing error is at most $2A\mathrm{\parallel }h-{\mathrm{\Pi }}_{K}h{\mathrm{\parallel }}_2$, uniformly in time, so $t\mapsto (z(t),h)$ is continuous. Finite gradient sums integrated in time pass to the limit and are bounded by $B$; monotone convergence gives a finite gradient square sum for almost every $t$. On that full-measure set, the finite Fourier truncations ${\mathrm{\Pi }}_{M}z(t)$ converge to $z(t)$ in $V$. Each ${\mathrm{\Pi }}_{M}z$ is continuous with values in a finite-dimensional subspace of $V$, hence is a uniform limit of measurable simple maps by uniform continuity on $[0,T]$. Keep the actual map $z:[0,T]\to H$ defined by its limiting coefficients at every time. Define a separate $V$-valued representative $\tilde{z}(t)=z(t)$ where the gradient square sum is finite and $\tilde{z}(t)=0$ on the exceptional null set. The continuous finite-dimensional maps ${\mathrm{\Pi }}_{M}z$ converge to $\tilde{z}$ in $V$ almost everywhere, so $\tilde{z}$ is strongly measurable. Its $V$ norm is integrable in square because the integrated $H$ norm is at most $T{A}^2$ and the integrated gradient square sum is at most $B$. Thus $\tilde{z}$ defines a class in $L^2(0,T;V)$ whose $H$-valued class is represented by the unchanged weakly continuous map $z$. The asserted pointwise $H$ bound and every limiting Fourier coefficient continue to refer to that unchanged map; only the Bochner representative has been assigned zero on the exceptional set.

If $\mathrm{\mid }k{\mathrm{\mid }}_{\mathrm{\infty }}>K$, then $\mathrm{\mid }{\kappa }_k\mathrm{\mid }\ge 2\pi K\mathrm{/}L$, so Parseval and the gradient bound give ${\int }_0^T{\sum }_{\mathrm{\mid }k{\mathrm{\mid }}_{\mathrm{\infty }}>K}\mathrm{\mid }{\widehat{z}}_n(k,t){\mathrm{\mid }}^{2}dt\le B{L}^2\mathrm{/}(4{\pi }^2{K}^2)$; the same estimate holds for $z$. On the finitely many modes $\mathrm{\mid }k{\mathrm{\mid }}_{\mathrm{\infty }}\le K$, uniform coefficient convergence gives strong $L^2$ convergence. Letting $K\to \mathrm{\infty }$ proves strong convergence in $L^2(0,T;H)$.

The product estimate $\mathrm{\parallel }{z}_n\otimes z_n-z\otimes z{\mathrm{\parallel }}_{L_{t,x}^1}\le (\mathrm{\parallel }{z}_n{\mathrm{\parallel }}_{L_{t,x}^2}+\mathrm{\parallel }z{\mathrm{\parallel }}_{L_{t,x}^2})\mathrm{\parallel }{z}_n-z{\mathrm{\parallel }}_{L_{t,x}^2}$ proves strong $L^1$ convergence of the quadratic tensors. Recursively choose the least index larger than its predecessor for which the squared $L_t^{2}H$ error is at most $2^{-j}$ at stage $j$; strong convergence ensures such an index exists. The resulting squared errors have finite sum. By [F3], their pointwise sum is finite for almost every time, so the subsequence converges strongly in $H$ at each such time.