Log of Gamma Function is Convex on Positive Reals/Proof 1

Proof
By definition, the Gamma function $\Gamma: \R_{> 0} \to \R$ is defined as:
 * $\displaystyle \Gamma \left({z}\right) = \int_0^{\infty} t^{z - 1} e^{-t} \rd t$


 * $\forall z > 0: \Gamma \left({z}\right) > 0$, as an integral of a strictly positive function in $t$.

The function is smooth according to Gamma Function is Smooth on Positive Reals, and
 * $\displaystyle \forall k \in \N: \Gamma^\left({k}\right) \left({z}\right) = \int_0^{\infty} \ln \left({t}\right)^k t^{z - 1} e^{-t} \, \mathrm d t$

Let $f \left({z}\right) := \ln \left({\Gamma \left({z}\right) }\right)$.


 * $f$ is smooth because $\Gamma$ is smooth and positive.

Then:
 * $f' \left({z}\right) = \dfrac {\Gamma' \left({z}\right)} {\Gamma \left({z}\right)}$


 * $f^{\left({2}\right)} \left({z}\right) = \dfrac {\Gamma^{\left(2\right)} \left({z}\right) \Gamma \left({z}\right) - \Gamma' \left({z}\right)^2} {\Gamma \left({z}\right)^2} > 0$

The numerator is positive due to the Cauchy-Bunyakovsky-Schwarz Inequality applied to the scalar products:
 * $\displaystyle \left \langle {g, h} \right \rangle = \int_0^\infty g \left({t}\right) h \left({t}\right) t^{z - 1} e^{-t} \rd t \quad \forall z \gt 0$

applied to $g = \ln$ and $h = 1$.


 * $\forall z \in \R_{>0}: f^{\left({2}\right)} \left({z}\right) \gt 0 \implies$ $f$ is convex.