Lemma: Window, zero moment, and finite occupations

For sufficiently large $L$:

1. $f_L\in {C_c}^{\infty }((-2L,2L))$, ${\Vert f_L\Vert }_2=1$, and $\int f_L=0$.

2. $f_L\to p$ in the Schwartz topology and ${\hat{f}}_L\to h$ in that topology. In particular, ${\theta }_L\to 0$.

3. The occupations

   $$E_f={{\int }_0}^{\infty }n(\omega )|{\hat{f}}_L(\omega ){|}^2\mathrm{d}\omega ,E_g={{\int }_0}^{\infty }n(\omega )|g_L(\omega ){|}^2\mathrm{d}\omega$$

   are finite for every nondegenerate participating plane.

4. With

   $$w_L={\hat{f}}_L-{\alpha }_{L}h,\mathcal{N}(w)={{\int }_0}^{\infty }n(\omega )|w(\omega ){|}^2\mathrm{d}\omega ,$$

   one has $\mathcal{N}(w_L)\to 0$, $E_f\to {\nu }_h$, and

   $${{\theta }_L}^2{E}_g⟶0.$$

Proof. The support and zero integral follow from (eq:windowconstruction). The inverse Fourier transform $p$ is Schwartz. Also $\int p=0$, because $h$ vanishes near zero. Hence

which decays faster than every inverse power of $L$. Differentiating the window shows that $(1-{\chi }_L)p\to 0$ in every Schwartz seminorm. The scaled bump $m_L{\psi }_L$ has the same property. Normalization and the phase alignment therefore preserve Schwartz convergence.

For positive frequencies,

The zero integral gives ${\hat{f}}_L(0)=0$. Thus ${\hat{f}}_L(\omega )=O(\omega )$ at zero; the same holds for $w_L$ and, for fixed $L$, $g_L$. Their weighted integrals are finite near zero. At high frequencies, the occupation is bounded and the tests are square integrable.

More quantitatively, for any differentiable $w$ with $w(0)=0$,

On $[1,\infty )$, $n$ is bounded. Schwartz convergence proves $\mathcal{N}(w_L)\to 0$ and $E_f\to {\nu }_h$. Finally,