\vhalf_0 T (i\sigma)\vhalf_0 P_l\|.$

Recall $iT(i\sigma)= R_0(e^{i\pi }i\sigma)-R_0(i\sigma)$. Set $$\psi_l(r\theta)= \chi_a(r\theta)Y^\mu_l(\theta)r^{-(d-2)/2} J_{\nu_l}(i\sigma r)$$ for any $\mu\in \{ 1,...,\mu(l)\},$ and note that $$\| \vhalf_0 T \vhalf_0 P_l\| 
\geq  \frac{\left| \langle \vhalf_0 T \vhalf_0 \psi_l, \psi_l \rangle
\right| }{\|\psi_l\|^2} .$$ By (\[eq:gnudiff\]) and (\[eq:shexp12\]), $$\begin{aligned}
\label{eq:lb1}
\frac{|\langle \vhalf_0 T \vhalf_0 \psi_l, \psi_l \rangle |}
{\|\psi_l\|^2} & 
= \frac{ \pi \left( \int_0^a \epsilon^{1/2} |J_{\nu_l}(i\sigma r)|^2 r^{-(d-2)}r^{d-1}dr\right)^{2}}
{ \int_0^a  |J_{\nu_l}(i\sigma r)|^2 r^{-(d-2)}r^{d-1}dr}  \nonumber \\
& = \pi \epsilon \int_0^a  |J_{\nu_l}(i\sigma r)|^2 r dr \nonumber \\
& \geq \pi \epsilon \int_{a/2}^a |J_{\nu_l}(i\sigma r)|^2 r dr.\end{aligned}$$

As in [@ab-st 9.6.3], setting $$I_\nu(z){\stackrel{\rm{def}}{=}}e^{-\nu \pi i/2}J_{\nu}(z e^{i\pi/2}),\; -\pi <\arg z \leq \pi/2,$$ from [@ab-st 9.7.7] there is a constant $c>0$ so that for $\nu$ sufficiently large $$|I_{\nu}(\nu s)| \geq c \frac{e^{c\nu}}{\sqrt{\nu}},\; 3\leq s \leq M.$$ Here and below we denote by $c$ a positive constant, independent of $\nu$, $l$, and $\sigma$, which may change from line to line. Now we use that $| J_{\nu_l}(i\sigma z)|= | I_{\nu_l}(\sigma z)|$ and apply these to (\[eq:lb1\]). We find that if $3\leq \sigma r/\nu_l\leq M$ for all $r$ with $a/2\leq r \leq a$, then $$\left| \langle \vhalf_0 T \vhalf_0 \psi_l, \psi_l \rangle \right|
\geq 
 c \int_{a/2}^a \frac{e^{2\nu_l c}}{\nu_l} dr \geq c \frac{e^{2\nu_l 