Bounds of Hankel/Bessel function Is it possible to prove that when $p > z$, the Bessel functions satisfy
$$
\vert H_p(z) \vert \leq 2|Y_p(z)|.
$$
or even stronger
$$
\vert J_p(z) \vert \leq |Y_p(z)|.
$$
I saw some author used this in paper without any references or proof.
 A: Assume $0<z<p$ and consider
$$
z \mapsto \frac{{J_p (z)}}{{Y_p (z)}}.
$$
Note that the first zeros of $J_p(z)$ and $Y_p(z)$ are larger than $p$ (cf. http://dlmf.nist.gov/10.21.E3), so this mapping is well defined and it takes negative values (because $J_p(z)>0$ and $Y_p(z)<0$ near $z=0$ when $p>0$). By the Wronskian for the Bessel functions (http://dlmf.nist.gov/10.5.E2),
$$
\frac{d}{{dz}}\frac{{J_p (z)}}{{Y_p (z)}} =  - \frac{2}{{\pi zY_p^2 (z)}} < 0.
$$
Thus, our function is strictly decreasing on $0<z<p$. Therefore, it remains to prove that $J_p(p)/Y_p(p)>-1$ for all $p>0$, since this will imply
$$
-1<\frac{{J_p (z)}}{{Y_p (z)}} =  - \frac{{\left| {J_p (z)} \right|}}{{\left| {Y_p (z)} \right|}}  \Leftrightarrow \left| {J_p (z)} \right| < \left| {Y_p (z)} \right|
$$
for $0<z<p$. By the results of the this paper, $iH_{it}^{(1)} (e^{\pi i/2} t) > 0$ for $t>0$ and
$$
J_p (p) + \frac{1}{{\sqrt 3 }}Y_p (p) =  - \frac{1}{{\sqrt 3 \pi p^{5/3} }}\int_0^{ + \infty } {\frac{{t^{2/3} e^{ - 2\pi t} }}{{1 + (t/p)^2 }}iH_{it}^{(1)} (e^{\pi i/2} t)dt}  < 0.
$$
Consequently,
$$
\frac{{J_p (p)}}{{Y_p (p)}} >  - \frac{1}{{\sqrt 3 }} >  - 1
$$
which completes the proof.
Note that in fact we obtained the stronger inequality
$$
\left| {J_p (z)} \right| < \frac{1}{{\sqrt 3 }}\left| {Y_p (z)} \right|
$$
for $0<z<p$. The constant $\frac{1}{{\sqrt 3 }}$ is the best possible, since $\left| {J_p (p)} \right| \sim \frac{1}{{\sqrt 3 }}\left| {Y_p (p)} \right|$ as $p\to +\infty$.
From the above analysis it can be inferred that the phase function $\theta_p(z)$ satisfies
$$
 - \frac{\pi }{2} < \theta _p (z) <  - \frac{\pi }{3},
$$
when $0<z<p$. Accordingly
$$
\left| {H_p^{(1,2)} (z)} \right| = \left| {Y_p (z)} \right|\left| {\csc \theta _p (z)} \right| \le \frac{2}{{\sqrt 3 }}\left| {Y_p (z)} \right|
$$
provided $0<z<p$. The constant $\frac{2}{{\sqrt 3 }}$ is again sharp.
