As the title suggests I need to show that the log of the modified bessel function is concave. When I graph it, certainly seems to be the case.
So far I have that: $$ y=\log I_{\nu}(x)\\ y'=\frac{I_{\nu+1}(x)}{I_{\nu}(x)}+\frac{\nu}{x}\\ y''=-\frac{I_{\nu+1}^2(x)-I_{\nu}^2(x)}{I_{\nu}^2(x)}-\frac{2\nu+1}{x}\frac{I_{\nu+1}(x)}{I_{\nu}(x)}-\frac{\nu}{x^2} $$
I have used the identites that $I_{\nu}'(x)=I_{\nu+1}(x)+\frac{\nu}{x}I_{\nu}(x)=I_{\nu-1}(x)-\frac{\nu}{x}I_{\nu}(x)$.
Basically how can I show that $y''<0$?
It's not true. For example, $\log I_1(x)$ is convex for $ x > 3.056...$.
Here's a plot of $\dfrac{d^2}{dx^2} \log I_1(x)$ for $1 < x < 5$:
In fact, asymptotically as $x \to \infty$ we have $$ \eqalign{I_\nu(x) &\sim \dfrac{e^x}{\sqrt{2\pi x}} (1 + O(1/x))\cr \log I_\nu(x) &\sim \log(1/\sqrt{2\pi}) + x - \dfrac{\log x}{2} + O(1/x)\cr \dfrac{d^2}{dx^2} \log I_\nu(x) &\sim \dfrac{1}{2x^2} + O(1/x^3)} $$ so $\log I_\nu(x)$ is convex for large $x$.