Lemma: Variable regulator near a nonzero null point (audit 3)

Let $z_{\ast }\in \Omega$ with $Q(z_{\ast })=0$ and put $s_{\ast }:=\mathrm{sgn}({z_{\ast }}^0)$. Then ${z^0}_{\ast }\ne 0$, and there are an open neighbourhood $U⋐\Omega$ of $z_{\ast }$ and a chart $\Phi =(Q,y^{\prime }):U\to (-\delta ,\delta )\times V^{\prime }$ such that, with $a_0:=|{z^0}_{\ast }|$, $s_{\ast }{z}^0\ge a_0/2>0$ on $U$. For $0<\tau \le 1$ put $g_{\tau }(z):=Q(z)-2i\tau z^0-{\tau }^2$. Then $g_{\tau }$ has no zero on ${\mathbb{R}}^4$, and

For $s_{\ast }=+1$ the right-hand side is $T_1{|}_U$ of Lemma lem:bv.

Proof. If ${z^0}_{\ast }=0$ then $Q(z_{\ast })=-|{\mathbf{z}}_{\ast }{|}^2=0$ forces $z_{\ast }=0\notin \Omega$. The chart with first coordinate $Q$ exists by the proof of Lemma lem:bv; shrinking it, we may assume $\overline{U}$ compact in $\Omega$ and $s_{\ast }{z}^0\ge a_0/2$ on $\overline{U}$, with $a_0:=|{z^0}_{\ast }|$. A zero of $g_{\tau }$ needs $\mathrm{Im}{g}_{\tau }=-2\tau z^0=0$ and then $Q=-|\mathbf{z}{|}^2={\tau }^2>0$, which is impossible. In the chart, $g_{\tau }\circ {\Phi }^{-1}=w_{\tau }:=s-{\tau }^2-i\tau a$ with $a:=2z^0\circ {\Phi }^{-1}$ smooth, $s_{\ast }a\ge a_0$, $|a|\le a_1$, $|{\partial }_{s}a|\le a_2$ on the chart (after a further shrinking). Pullback by the diffeomorphism $\Phi$ is continuous on ${\mathcal{D}}^{\prime }$ and the Jacobian is absorbed in the test function, so it suffices to show, for $\psi \in {C^{\infty }}_c((-\delta ,\delta )\times V^{\prime })$, $\int \psi {w_{\tau }}^{-1}\mathrm{d}s\mathrm{d}{y}^{\prime }\to ⟨(s-i0s_{\ast }{)}^{-1}\otimes 1,\psi ⟩$.

Since $\mathrm{Im}{w}_{\tau }=-\tau a$ has the fixed sign $-s_{\ast }$, $w_{\tau }$ stays in one open half-plane, on which the principal logarithm $L_{\tau }:=\mathrm{Log}{w}_{\tau }$ is smooth, and ${\partial }_s{L}_{\tau }=(1-i\tau {\partial }_{s}a)/w_{\tau }$. Hence

(i) Remainder. $|\tau {\partial }_{s}a/w_{\tau }|\le \tau a_2/|\mathrm{Im}{w}_{\tau }|\le a_2/a_0$, and for $s\ne 0$ the term tends to $0$ (since $w_{\tau }\to s\ne 0$). By dominated convergence on the compact support of $\psi$, $\int \psi i\tau {\partial }_{s}a/w_{\tau }\to 0$.

(ii) Main term. Integrating by parts in $s$, $\int \psi {\partial }_s{L}_{\tau }=-\int ({\partial }_s\psi )L_{\tau }$. For $s\ne 0$, $L_{\tau }\to {\mathrm{Log}}_{s_{\ast }}(s):=\log |s|-i\pi s_{\ast }\theta (-s)$ (for $s<0$ the point $w_{\tau }$ approaches the negative axis from the half-plane $\mathrm{sgn}\mathrm{Im}=-s_{\ast }$). Moreover $|\mathrm{Im}{L}_{\tau }|\le \pi$ and $|s-{\tau }^2|\le |w_{\tau }|\le C:=\delta +1+a_1$, so $|\mathrm{Re}{L}_{\tau }|\le |\log C|+|\log |s-{\tau }^2||$. The family $\{\log |s-{\tau }^2|{\}}_{0<\tau \le 1}$ is bounded in $L^2(-\delta ,\delta )$ (each is a translate of $\log |u|$ restricted to $(-\delta -1,\delta +1)$), hence uniformly integrable on $\mathrm{supp}\psi$; by Vitali's theorem $\int ({\partial }_s\psi )L_{\tau }\to \int ({\partial }_s\psi ){\mathrm{Log}}_{s_{\ast }}(s)$.

(iii) Identification. The case $a\equiv 2s_{\ast }$ without the shift ${\tau }^2$ (i.e.\ $w=s-i\eta s_{\ast }$, remainder $\equiv 0$) gives by the same argument $(s-i\eta s_{\ast }{)}^{-1}\to {\partial }_s{\mathrm{Log}}_{s_{\ast }}(s)$ in ${\mathcal{D}}^{\prime }$; hence both limits equal ${\partial }_s{\mathrm{Log}}_{s_{\ast }}$. Finally ${\partial }_s\log |s|=\mathrm{PV}\frac{1}{s}$ and ${\partial }_s(-i\pi s_{\ast }\theta (-s))=i\pi s_{\ast }\delta (s)$. For $s_{\ast }=+1$, $(s-i\eta {)}^{-1}\otimes 1$ is $(Q-i\eta {)}^{-1}$ in the chart, whose limit defines $T_1{|}_U$.