Weak convergence does not pass quadratic products

For integers $n\ge 1$, fix the second coordinate vector $e_2\in {\mathbb{R}}^d$ and set $u_n(x)=\sin (2\pi n{x}_1\mathrm{/}L)e_2,u(x)=0.$ Each $u_n$ has only the Fourier modes $k=\pm n{e}_1$ and its vector coefficient is perpendicular to $k$, so $u_n\in H$. For a finite trigonometric polynomial $g$, $(u_n,g{)}_H=0$ for all sufficiently large $n$ because its Fourier support is finite. Approximate an arbitrary $g\in H$ by such polynomials using [F2]; the error is at most ${\sup }_n\mathrm{\parallel }{u}_n{\mathrm{\parallel }}_2$ times the approximation error. Hence $u_n\rightharpoonup 0$ in $H$.

The only nonzero entry of $u_n\otimes u_n$ is its $(2,2)$ entry, equal to ${\sin }^2(2\pi n{x}_1\mathrm{/}L)=\frac{1}{2}-\frac{1}{2}\cos (4\pi n{x}_1\mathrm{/}L)\mathrm{.}$ The oscillatory cosine has zero pairing eventually with every finite trigonometric polynomial and its $L^2$ norm is uniformly bounded; density and Cauchy–Schwarz therefore show it converges weakly to zero in $L^2$. Thus $u_n\otimes u_n\rightharpoonup \frac{1}{2}{e}_2\otimes e_2$ in $L^2$. Since the torus measure is one, every essentially bounded matrix test satisfies $\mathrm{\parallel }\mathrm{\Phi }{\mathrm{\parallel }}_2\le \mathrm{\parallel }\mathrm{\Phi }{\mathrm{\parallel }}_{\mathrm{\infty }}$, so this convergence also holds against every weak $L^1$ test. In particular the constant test $\mathrm{\Phi }=e_2\otimes e_2$ gives $\int (u_n\otimes u_n):\mathrm{\Phi }d\mu =1\mathrm{/}2$ for every $n$, whereas $\int (u\otimes u):\mathrm{\Phi }d\mu =0$. Hence the tensors do not converge weakly in $L^1$ to $u\otimes u=0$; the same smooth constant test also rules out convergence to zero as periodic distributions. This refutes the false statement. These fields are only a functional-analytic witness and are not claimed to solve Navier–Stokes.