# how to prove the two functions are convex or not?

The modified Bessel function of the second kind, denoted by $K_{\nu}(x)$. Is it convex in $x\in\mathbb{R}_{++}$? Further more, is $logK_{\nu}(x)$ convex in $x\in\mathbb{R}_{++}$?

I have been struggling on these questions for a long time. I really need your help. Thank you very much in advance!

• Moments are log-convex by the Cauchy-Schwarz inequality, and $K_\nu(x)$ is a moment. – Jack D'Aurizio Oct 25 '17 at 16:31

If we use the integral representation $$K_{\nu}(x) = \int_0^{\infty} e^{-x\cosh{t}}\cosh{\nu t} \, dt,$$ these are both quite simple to show directly. We have $$K_{\nu}(\lambda x+(1-\lambda)y) = \int_0^{\infty} (e^{-x\cosh{t}})^{\lambda}(e^{-y\cosh{t}})^{1-\lambda}\cosh{\nu t} \, dt.$$ AM–GM gives $$(e^{-x\cosh{t}})^{\lambda}(e^{-y\cosh{t}})^{1-\lambda} \leq \lambda e^{-x\cosh{t}} + (1-\lambda)e^{-y\cosh{t}},$$ from which $$K_{\nu}(\lambda x+(1-\lambda)y) \leq \lambda K_{\nu}(x) + (1-\lambda) K_{\nu}(y)$$ is immediate. Log-convexity can be proven by using Hölder's inequality in the form $$\int \lvert f g h \rvert \leq \left( \int f^{1/\lambda} h \right)^{\lambda} \left( \int g^{1/(1-\lambda)} h \right)^{1-\lambda}$$ for $0 \leq \lambda \leq 1$ (put $p=1/\lambda$ to recover the usual form).