# Upper bounds for Bessel functions

Cosider the K-Bessel function $$K_\nu(x):= \frac\pi 2 \frac{I_{-\nu}(x)-I_\nu(x)}{\sin(\nu\pi)}.$$

See also Watson, G. N., A treatise on the theory of Bessel functions., Cambridge: University Press, Chapter III, p.78.

This book contains many results on the asymptotic behaviour of functions of this kind. However, I am interested in upper bound estimates against simpler terms. Something similar is used in a another paper to show resolvent estimates by estimating K-Bessel functions against exponential and polynomial terms. At the moment I try to reproduce this estimates as a part of my research. However, no explanation on these kinds of estimated is provided and it falls short in providing the literature so I am looking for a reference on this topic.

• What properties do you want the upper bounds to have? For what values of $\nu$ and $x$? What is the "another paper"? Dec 9, 2020 at 18:56
• I am sorry for not being precise enough; originally, I have not wanted to go into details, but now I will be more "concrete". Giving a look to the estimate (14) in the proof of Theorem 3 of the 2016 paper "Criterion for the Solvability of the Cauchy Problem for an Abstract Euler–Poisson–Darboux Equation" written by A.V. Glushak and O.A. Pokruchin, I have the impression that for real $\nu$ and positive $x$ the estimate $|K_{\nu}(x)|\leq \frac{1}{\sqrt{x}}e^{-x}$ is valid. I do not understand this result and I wonder where estimates of this kind can be found.
– mape
Dec 9, 2020 at 22:36

\begin{align*} &\Big\|\int_0^\infty t^{\nu+n}K_{\nu-n+1}(t\sqrt\mu)Y_k(t)x\,dt\Big\| \\ &\le M\|x\|\int_0^\infty t^{\nu+n-1/2}\exp((\om-\sqrt\mu)t)\,dt \tag{14} \end{align*} given that, by formula (3) in the paper, $$$$\|Y_k(t)\|\le M\exp(\om t) \tag{0}$$$$ for some real $$\om\ge0$$ and all real $$t>0$$. Here, $$\nu=(k-1)/2$$ (by p. 40 of the paper), $$k\ge0$$, and $$n$$ is a natural number. One can easily see that inequality (14) is false in general (or, maybe, always).

However, one can save the conclusion (at the bottom of p. 42 of the paper) that $$\begin{equation*} \|R^k(\mu)\|\le M(k)/(\mu-\om_1)^n \tag{1} \end{equation*}$$ for some real $$M(k)>0$$ not depending on $$\mu$$ or $$n$$, some real $$\om_1$$, and all real $$\mu>\om_1$$, which makes the Hille--Yosida theorem applicable.

Indeed, for real $$u>0$$ and $$\al$$ $$\begin{equation*} K_\al(u)=\int_0^\infty e^{-u\cosh z}\cosh\al z\,dz \tag{2} \end{equation*}$$ by the second display under the picture in Section Modified Bessel functions: $$I_\al$$, $$K_\al$$. By rescaling, without loss of generality, $$\|x\|=1$$ and (0) holds with $$M=1$$. So, by (2), with $$$$m:=n-\nu+1,\quad l:=2\nu+2=k+1\ge1,\quad\mu>4\om^2,\quad r:=\om/\sqrt\mu\in(0,1/2),$$$$ the left-hand side of (14) is no greater than \begin{align*} &\int_0^\infty dt\, t^{\nu+n}e^{\om t} \int_0^\infty dz\,e^{-t\sqrt\mu\,\cosh z}\cosh mz \\ &=\int_0^\infty dz\,\cosh mz \int_0^\infty dt\, t^{\nu+n}e^{-t(\sqrt\mu\,\cosh z-\om)} \\ &=\Ga(\nu+n+1)\int_0^\infty \frac{dz\,\cosh mz}{(\sqrt\mu\,\cosh z-\om)^{m+l}} \\ &=\frac{\Ga(\nu+n+1)}{(\sqrt\mu)^{m+l}}\int_0^\infty \frac{dz\,\cosh mz}{(\cosh z-r)^{m+l}} \\ &\ll\frac{\Ga(\nu+n+1)}{(\sqrt\mu)^{m+l}}\int_0^\infty dz\,f(z)^m e^{-lz}, \tag{3} \end{align*} where $$a\ll b$$ means that $$|a|\le Cb$$ for some real $$C>0$$ not depending on $$n$$ or $$\mu$$ or $$\om$$ (as long as $$r$$ is small enough) and $$\begin{equation*} f(z):=\frac{e^z}{\cosh z-r}. \end{equation*}$$

Note that $$\max_{z>0}f(z)=(1-r^2)^{-1}$$. So, the left-hand side of (14) is \begin{align*} &\ll\frac{\Ga(\nu+n+1)}{(\sqrt\mu)^{m+l}}\, (1-r^2)^{\nu-1-n} \ll\frac{\Ga(\nu+n+1)}{(\sqrt\mu)^{n+\nu+3}}\, (1-r^2)^{-n}=:C_n(\mu). \end{align*}

So, by the equality in formula (14) in the paper, for $$\mu>\om_1:=1+4\om^2$$, \begin{align*} \|R^k(\mu)\|&\ll\frac{C_n(\mu)}{2^{n+\nu-1}(n-1)!\Gamma(\nu+1)\mu^{n/2-k/4}} \\ &\ll\frac{\Ga(\nu+n+1)(1-r^2)^{-n}}{2^{n+\nu-1}(n-1)!\Gamma(\nu+1)\mu^n} \\ &\ll\frac{\Ga(\nu+n+1)(4/3)^n}{2^{n+\nu-1}(n-1)!\Gamma(\nu+1)\mu^n} \\ &\ll\frac{M(k)}{(\mu-\om_1)^n}, \end{align*} which proves (1).

Edit: In fact, $$\max_{z>0}f(z)=2(1-r^2)^{-1}=:c>2$$ -- I previously missed the factor $$2$$ here. So, the last integral in (3) is $$\asymp c^m/\sqrt m$$. Since the bounding was lossless (up to inessential constant factors) everywhere in the above reasoning, this appears to invalidate the entire result in the paper by Glushak and Pokruchin.

• This is a very clarifying answer! I only have few supplementary considerations: 1) As $K_\alpha=K_{-\alpha}$, I suppose that you actually want to put $m:=n-\nu-1$ above, such that $m+l=n+\nu+1$. 2) I agree with you on the fact that the argument of Glushak and Prokruchin ist quite unclear. Anyway, it is true that the generator of a Bessel operator family generates also a strongly continuous operator semigroup. This is not trivial and it is a consequence of a more general result. What I was interested in, was to obtain a resolvent estimate for such operators.
– mape
Dec 21, 2020 at 16:32

There are some estimates and corresponding resolvent estimates for the Laplacian in exterior cylindric domains in

J. Appell, C.-J. Chen, S. Tseng, M. Väth, Iterative Approximation for a Boundary Value Problem Arising for the Electrical Potential on a Cylindrical Double Layer, Journal of Analysis and its Applications 27, 2008 (3), 283-300.