I need to prove that the modified bessel function of the second kind is log convex in the square of the argument. Specifically I'm interested in showing, $\log \mathcal{K}_0(\sqrt{x})$ (zero order) is convex.
Any ideas of proving this? Visually it seems to be the case:
x vs $\mathcal{K}_0(\sqrt{x})$
x vs $\log\mathcal{K}_0(\sqrt{x})$
Define $f(x):= \ln \left(K_0(\sqrt{x})\right)$ and compute (e.g. with the help of a CAS) $$f'(x)=-\frac{1}{2}\frac{K_1(\sqrt{x})}{\sqrt{x}K_0(\sqrt{x})}$$ $$f''(x) = \frac{1}{4} \frac{\sqrt{x}K_0(\sqrt{x})^2+2K_0(\sqrt{x})K_1(\sqrt{x})-\sqrt{x}K_1(\sqrt{x})^2} {x^{3/2}K_0(\sqrt{x})^2}$$ From http://dlmf.nist.gov/10.37 we have $K_1(x)>K_0(x)>0\;$ for $x>0$. Therefore both numerator and denominator of $f''(x)$ are positive and $\ln \left(K_0(\sqrt{x})\right)$ is strictly convex.