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:
 * $\ds \map \Gamma z = \int_0^{\infty} t^{z - 1} e^{-t} \rd t$


 * $\forall z > 0: \map \Gamma z > 0$, as an integral of a strictly positive function in $t$.

The function is smooth according to Gamma Function is Smooth on Positive Reals, and
 * $\ds \forall k \in \N: \map {\Gamma^{\paren k} } z = \int_0^{\infty} \map \ln t^k t^{z - 1} e^{-t} \rd t$

Let $\map f z := \map \ln {\map \Gamma z}$.


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

Then:
 * $\map {f'} z = \dfrac {\map {\Gamma'} z} {\map \Gamma z}$


 * $\map {f^{\paren 2} } z = \dfrac {\map {\Gamma^{\paren 2} } z \map \Gamma z - \map {\Gamma'} z^2} {\map \Gamma z^2} > 0$

The numerator is positive due to the Cauchy-Bunyakovsky-Schwarz Inequality for Inner Product Spaces applied to the scalar products:
 * $\ds \innerprod g h = \int_0^\infty \map g t \map h t t^{z - 1} e^{-t} \rd t \quad \forall z \gt 0$

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


 * $\forall z \in \R_{>0}: \map {f^{\paren 2} } z > 0 \implies f$ is convex.