Spherical Hankel modulus and uniform radial bounds

Direct differentiation of $h_0=-i{e}^{iz}\mathrm{/}z$ gives the finite expression $h_{\mathrm{\ell }}(z)=(-i{)}^{\mathrm{\ell }+1}{e}^{iz}{\sum }_{m=0}^{\mathrm{\ell }}{i}^m{A}_{\mathrm{\ell }m}{z}^{-m-1},A_{\mathrm{\ell }m}=\frac{(\mathrm{\ell }+m)!}{m!(\mathrm{\ell }-m)!2^m}\mathrm{.}$ Indeed it holds at $\mathrm{\ell }=0$; substituting this expression in $h_{\mathrm{\ell }+1}=\mathrm{\ell }{h}_{\mathrm{\ell }}\mathrm{/}z-h_{\mathrm{\ell }}^{\mathrm{\prime }}$ gives $A_{\mathrm{\ell }+1,m}=A_{\mathrm{\ell }m}+(\mathrm{\ell }+m)A_{\mathrm{\ell },m-1}$ (out-of-range coefficients are zero), and the displayed factorial formula satisfies that recurrence and the $m=0$ base.

Let $F_{\mathrm{\ell }}=\mathrm{\mid }{h}_{\mathrm{\ell }}{\mathrm{\mid }}^2$. The radial recurrence gives $h_{\mathrm{\ell }+1}=(2\mathrm{\ell }+1)h_{\mathrm{\ell }}\mathrm{/}z-h_{\mathrm{\ell }-1}$. Also $F_{\mathrm{\ell }}^{\mathrm{\prime }}=2\mathrm{Re}({\bar{h}}_{\mathrm{\ell }}{h}_{\mathrm{\ell }-1})-2(\mathrm{\ell }+1)F_{\mathrm{\ell }}\mathrm{/}z$ by [F2]. Expanding the squared modulus and eliminating the real cross term yields $F_{\mathrm{\ell }+1}=F_{\mathrm{\ell }-1}-(2\mathrm{\ell }+1)F_{\mathrm{\ell }}^{\mathrm{\prime }}\mathrm{/}z-(2\mathrm{\ell }+1)F_{\mathrm{\ell }}\mathrm{/}{z}^2\mathrm{.}$ From $F_0=z^{-2}$ and $F_1=z^{-2}+z^{-4}$, coefficient comparison yields $S_{\mathrm{\ell }+1,0}=S_{\mathrm{\ell }-1,0}=1$ and, for $1\le q\le \mathrm{\ell }$, $S_{\mathrm{\ell }+1,q}=S_{\mathrm{\ell }-1,q}+(2\mathrm{\ell }+1)(2q-1)S_{\mathrm{\ell },q-1}$. The displayed factorial coefficients satisfy this: the ratios are $S_{\mathrm{\ell }-1,q}\mathrm{/}{S}_{\mathrm{\ell }+1,q}=(\mathrm{\ell }-q)(\mathrm{\ell }-q+1)\mathrm{/}[(\mathrm{\ell }+q)(\mathrm{\ell }+q+1)]$ and $S_{\mathrm{\ell }+1,q}\mathrm{/}{S}_{\mathrm{\ell },q-1}=(2q-1)(\mathrm{\ell }+q)(\mathrm{\ell }+q+1)\mathrm{/}(2q)$, whose difference gives the recurrence. At the top index $q=\mathrm{\ell }+1$, $S_{\mathrm{\ell }+1,\mathrm{\ell }+1}=(2\mathrm{\ell }+1{)}^2{S}_{\mathrm{\ell },\mathrm{\ell }}$; thus induction proves the stated sum at every degree.

Every $S_{\mathrm{\ell }m}$ is positive, so differentiating the finite sum gives $F_{\mathrm{\ell }}^{\mathrm{\prime }}(z)<0$. For $0\le m\le \mathrm{\ell }-1$, $S_{\mathrm{\ell }m}\mathrm{/}{S}_{\mathrm{\ell }-1,m}=(\mathrm{\ell }+m)\mathrm{/}(\mathrm{\ell }-m)\ge 1$, and the degree-$\mathrm{\ell }$ sum has one additional positive term. Hence $\mathrm{\mid }{h}_{\mathrm{\ell }}\mathrm{\mid }$ decreases with $z$ and $\mathrm{\mid }{h}_{\mathrm{\ell }-1}\mathrm{\mid }\le \mathrm{\mid }{h}_{\mathrm{\ell }}\mathrm{\mid }$. For $\mathrm{\ell }\ge 1$, [F2] now implies $\mathrm{\mid }{h}_{\mathrm{\ell }}^{\mathrm{\prime }}\mathrm{\mid }\le \mathrm{\mid }{h}_{\mathrm{\ell }-1}\mathrm{\mid }+(\mathrm{\ell }+1)\mathrm{\mid }{h}_{\mathrm{\ell }}\mathrm{\mid }\mathrm{/}z\le (1+(\mathrm{\ell }+1)\mathrm{/}z)\mathrm{\mid }{h}_{\mathrm{\ell }}\mathrm{\mid }$; for $\mathrm{\ell }=0$ the same bound follows by differentiating $h_0=-i{e}^{iz}\mathrm{/}z$.

Differentiating the finite sum in step 1.1 cancels its $i{e}^{iz}$ derivative in $h_{\mathrm{\ell }}^{\mathrm{\prime }}-i{h}_{\mathrm{\ell }}$, so $\mathrm{\mid }z(h_{\mathrm{\ell }}^{\mathrm{\prime }}-i{h}_{\mathrm{\ell }})\mathrm{\mid }\le {\sum }_{m=0}^{\mathrm{\ell }}(m+1)A_{\mathrm{\ell }m}{z}^{-m-1}$. The lowest-power term of the positive modulus sum gives $\mathrm{\mid }{h}_{\mathrm{\ell }}(z_0)\mathrm{\mid }\ge A_{\mathrm{\ell }\mathrm{\ell }}{z}_0^{-\mathrm{\ell }-1}$. For $j=\mathrm{\ell }-m$, factorial cancellation gives $\frac{A_{\mathrm{\ell },\mathrm{\ell }-j}}{A_{\mathrm{\ell }\mathrm{\ell }}}=\frac{2^j}{j!}{\prod }_{s=0}^{j-1}\frac{\mathrm{\ell }-s}{2\mathrm{\ell }-s}\le \frac{2^j}{j!}\mathrm{.}$ Using $z\ge z_0$ and $m+1\le \mathrm{\ell }+1$, the quotient in the claim is at most $(\mathrm{\ell }+1){\sum }_{j=0}^{\mathrm{\ell }}(2z_0{)}^j\mathrm{/}j!\le (\mathrm{\ell }+1)e^{2z_0}$.