Suppose $H^{(1)}_p(z)$ is the Hankel function of the first kind with argument $z$ and order $p$. I found the inequality below in the paper "Multi-level fast multipole solution of the scattering problem" by S. Amini and A.T.J. Profit. $$ \text{For } 1 \leq p \leq z, \,\, \vert H^{(1)}_p(z) \vert \leq 1. $$
The paper did not point any reference to it, and I could not find any textbooks or literatures stating this. This inequality seems to be true in numerical experiments. Does anyone have any idea on how to prove this?
Thanks!!
Assume that $1\leq p \leq z$. By Nicholson’s Integral , $$ \left| {H_p^{(1)} (z)} \right|^2 = J_p^2 (z) + Y_p^2 (z) = \frac{8}{{\pi ^2 }}\int_0^{ + \infty } {\cosh (2pt)K_0 (2z\sinh t)\mathrm{d}t} . $$ Differentiation with respect to $p$ shows that the modulus square is monotonically increasing with respect to $p$. Consequently, $| H_p^{(1)} (z)| \leq | H_z^{(1)} (z)| $ whenever $1\leq p \leq z$. By a results of this paper, $$ \left| {H_z^{(1)} (z)} \right| \!\le\! \frac{2}{{3\pi }}\frac{{\sqrt 3 }}{2}\left( {6^{1/3} \frac{{\Gamma (1/3)}}{{z^{1/3} }} + \frac{3}{{10}}\frac{1}{z}} \right) \!\le\! \frac{2}{{3\pi }}\frac{{\sqrt 3 }}{2}\left( {6^{1/3} \Gamma \!\left( {\frac{1}{3}} \right) + \frac{3}{{10}}} \right) \!=0.9497\ldots\!<1, $$ provided $z\geq 1$.