I wish to show that


$|j\_n(x)| < \frac{1}{\sqrt{x}}$


for $n=0,1,2,\ldots$ and $x>0$, where $j\_n$ is the spherical Bessel function of the first kind.


Experimenting with Matlab I am sure that this is the case by I am unable to prove it analytically.


Since


$j\_n(x) = \frac{1}{\sqrt{x}}\sqrt{\frac{\pi}{2}}J\_{n+\frac{1}{2}}(x)$,


where $J\_n$ is the standard Bessel function of the first kind, it is sufficient to show that


$|J\_{n+\frac{1}{2}}(x)| < \sqrt{\frac{2}{\pi}}$


for $n=0,1,2,\ldots$ and $x>0$.



Spherical Bessel functions
In Abramowitz and Stegun, Handbook of Mathematical Functions, you will find the formula 10.1.52
$$\sum\_0^\infty j\_n^2(x)={Si(2x)\over 2x}.$$
Consequently
$$|j\_n(x)|\le {1\over\sqrt{x}}\sqrt{{\max\_{[0,\infty)}|Si(x)|\over 2}}\sim {0.962\over\sqrt{x}}.$$


