$\Gamma(x) = \int_0^\infty t^{x-1}e^{-t}dt.$
Here is the Hölder's inequality:
(1/p) + (1/q) = 1.
|$\int_a^bfg$ $d\alpha$| $\leq$ {$\int_a^b$ $|f|^p$ $d\alpha$}$^{1/p}$ {$\int_a^b$ $|g|^q$ $d\alpha$}$^{1/q}$. The 8.18 theorem says that log $\Gamma$ is convex on (0,$\infty$).
Here is the proof of this:
If 1 < p < $\infty$ and (1/p) + (1/q) = 1, apply Hölder's inequality to $\Gamma(x)$ and obtain $\Gamma(\frac{x}{p} + \frac{y}{q})$ $\leq$ $\Gamma(x)^{1/p}\Gamma(y)^{1/q}.$ This is equivalent to this : "log $\Gamma$ is convex on (0,$\infty$)."
This means that $\int_0^\infty t^{\frac{x}{p} + \frac{y}{q} - 1}e^{-t}dt$ $\leq$ ${\int_0^\infty t^{x-1}e^{-t}dt}^{1/p}{\int_0^\infty t^{y-1}e^{-t}}dt^{1/q}$. I couldn't relate this inequality to the Hölder's inequality. I don't understand how did we obtain this. I also don't understand the idea of the convexity of log $\Gamma$. As I know the definition of convexity of function I couldn't relate this inequality: $\Gamma(\frac{x}{p} + \frac{y}{q})$ $\leq$ $\Gamma(x)^{1/p}\Gamma(y)^{1/q}$ to the idea of $log\Gamma(x)$ convexity on (0,$\infty$). Any help would be appreciated.