Log of Gamma Function is Convex on Positive Reals

Theorem
Let $\Gamma: \R_{>0} \to \R$ be the Gamma function, restricted to the strictly positive real numbers.

Let $\log$ be the natural logarithm function.

Then the composite mapping $\log \circ \operatorname \Gamma$ is a convex function.

Proof
According to Gamma Function, the Gamma function $\Gamma: \R \to \R \ $ is defined, for z > 0, as:
 * $\displaystyle \Gamma \left({z}\right) = \int_0^{\infty} t^{z-1} e^{-t} \ \mathrm d t$
 * $\displaystyle \forall z > 0: \, \Gamma \left({z}\right) > 0$, as an integral of a strictly positive function in $t$.

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

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


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

Then $\displaystyle f^{'}\left({z}\right) = \Gamma^{'} \left({z}\right) \,/\, \Gamma \left({z}\right)$


 * $\displaystyle f^{\left(2\right)}\left({z}\right) = \left\{ \Gamma^{\left(2\right)} \left({z}\right) \,*\, \Gamma \left({z}\right) \,-\, \Gamma^{'} \left({z}\right)^2 \right\} \,/\, \Gamma \left({z}\right)^2 \,\gt\,0$

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

applied to $ g = log$ and $h = 1$

$f^{\left(2\right)}\left({z}\right) \gt 0 \quad \forall z \gt 0 \implies$ f is convex.