I think that the following inequality holds for all $x > 0$ and all $\nu \ge \frac{1}{2}$:

$$ K_\nu(2 x) \le \frac{2^{2 - 2 \nu}}{\Gamma(\nu)} x^\nu K_\nu^2(x) ,$$ where $K$ is a modified Bessel function of the second kind.

Here's a plot of $\frac{2^{2 - 2 \nu}}{\Gamma(\nu)} x^\nu K_\nu^2(x) - K_\nu(2 x)$ for different $\nu$s:

x for each nu

And the minimum of these curves over 10,000 points $0 < x \le 10$ while varying $\nu$ (which is of course just the curve's value at 10):

minima over x for each nu

I've proved the relationship when $\nu$ is a half-integer on math.stackexchange; I'd like to show it for all other $\nu \ge \tfrac12$. So showing that the derivative in $\nu$ is always positive would be sufficient, but that derivative is gross: Letting $L_\nu(x) := \frac{\partial}{\partial \nu} K_\nu(x)$ and $\psi$ be the digamma function, the derivative is

$$ 4 x^{\nu } K_{\nu }(x) \left(K_{\nu }(x) \left(\psi(\nu )+\log \left(\frac{4}{x}\right)\right)-2 L_\nu(x)\right)+4^{\nu } \Gamma (\nu ) L_\nu(2 x) .$$

Neither I nor Mathematica can prove that it's positive – though it seems numerically like it is. Is there some duplication relation for $L_\nu$ or other property that will help prove this? Or a completely different approach?


1 Answer 1


The proposer’s inequality can be improved by a factor of two, that is,

$ (I) \quad K_\nu(2x) \le \, \dfrac{2^{1-2\nu}}{\Gamma(\nu)} \,x^\nu K_\nu^{\,2}(x) \,\, ,x>0, \nu\ge 1/2 .$

Two integral identities and one inequality is needed: $ (a) \quad K_\nu^{\,2}(x) =\frac{1}{2}\int_{0}^\infty\dfrac{dt}{t} e^{\large-t/2-x^2/t} K_\nu(x^2/t)$

$ (b) \quad K_\nu(2x) =\frac{1}{2}x^\nu\int_{0}^\infty\dfrac{dt}{t^{\nu+1}} e^{\large-t-x^2/t}$

$ (c) \quad x^\nu K_\nu(x)>2^{\nu-1} \Gamma(\nu)\, e^{\large-x}$

The references are (a) DLMF 10.32.18 with x=y; (b) DLMF 10.32.10 with 2x as the argument; (c) J. Math. Anal. Appl. 420 (2014) 373-386. It should be noted that the author of (c) claims that the region of validity is for x>0, $\nu > 0,$ but it appears that the proper region is for $\nu\ge 1/2 .$ (The derivation uses a formula that is good only for $\nu\ge 1/2,$ and plots indicate that the inequality breaks down for $\nu\lt 1/2$. Besides, the problem requires only $\nu\ge 1/2$.) All formulas are valid for $x>0.$

In $(I)$, use $(a)$ on the RHS (right-hand side). Use $(c)$ within the resulting integral to get

$ \dfrac{2^{1-2\nu}}{\Gamma(\nu)} \,x^\nu K_\nu^{\,2}(x) \ge \dfrac{(2x)^\nu}{2}\int_{0}^\infty\dfrac{dt}{t^{\nu+1}} e^{\large-t/2-2x^2/t} \Bigl(\dfrac{t}{2x}\Bigr)^{2\nu} \,.$

Now the RHS of the previous equation is, by (b), $K_{-\nu}(2x)=K_{\nu}(2x),$ where the well-known index reflection formula has been used. $\square$

  • $\begingroup$ Thank you! It took me a while to reproduce your steps because of a dumb mistake, but this seems to be correct, and I'll be citing you in my thesis soon. :) [Let me know if you'd prefer to be cited under some other name...] $\endgroup$
    – Danica
    Commented Aug 29, 2016 at 3:21
  • $\begingroup$ In terms of the flawed claim of (c): it seems that (c) was shown for $\nu > \tfrac12$ in (1.4) of Ismail, Complete Monotonicity of Modified Bessel Functions (doi). It is also trivially true for $K_{1/2}(x) = \sqrt{\frac{\pi}{2 x}} e^{-x}$, since $\sqrt{x} \sqrt{\frac{\pi}{2 x}} e^{-x} = \sqrt{\pi / 2} e^{-x}$ is identically more than $1/\sqrt{2} \Gamma(2) e^{-x} = \frac1{\sqrt 2} e^{-x}$. So I'll just cite it from there. $\endgroup$
    – Danica
    Commented Aug 29, 2016 at 3:33

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