Log of Gamma Function is Convex on Positive Reals/Proof 2

Theorem

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

Let $\ln$ denote the natural logarithm function.

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

Proof

The strategy is to show that:

$\map \ln {\map \Gamma {\dfrac x 2 + \dfrac y 2} } \le \dfrac 1 2 \map \ln {\map \Gamma x} + \dfrac 1 2 \map \ln {\map \Gamma y}$

Let $0 < \delta < \Delta$.

Then:

 $\ds \paren {\int_\delta^\Delta t^{\paren {x + y - 2} / 2} e^{-t} \rd t}^2$ $=$ $\ds \paren {\int_\delta^\Delta \paren {t^{\paren {x - 1} / 2} e^{-t/2} } \paren {t^{\paren {y - 1} / 2} e^{-t/2} } \rd t}^2$ $\text {(1)}: \quad$ $\ds$ $\le$ $\ds \paren {\int_\delta^\Delta t^{x - 1} e^{-t} \rd t} \paren {\int_\delta^\Delta t^{y - 1} e^{-t} \rd t}$ Cauchy-Bunyakovsky-Schwarz Inequality for Definite Integrals

Letting $\delta \to 0$ and $\Delta \to \infty$, $(1)$ becomes equivalent to:

$\paren {\map \Gamma {\dfrac {x + y} 2} }^2 \le \paren {\map \Gamma x} \paren {\map \Gamma y}$
 $\ds \paren {\map \Gamma {\dfrac {x + y} 2} }^2$ $\le$ $\ds \paren {\map \Gamma x} \paren {\map \Gamma y}$ $\ds \leadsto \ \$ $\ds \map \ln {\paren {\map \Gamma {\dfrac {x + y} 2} }^2}$ $\le$ $\ds \map \ln {\paren {\map \Gamma x} \paren {\map \Gamma y} }$ $\ds \leadsto \ \$ $\ds 2 \map \ln {\map \Gamma {\dfrac {x + y} 2} }$ $\le$ $\ds \map \ln {\map \Gamma x} + \map \ln {\map \Gamma y}$ $\ds \leadsto \ \$ $\ds \map \ln {\map \Gamma {\dfrac x 2 + \dfrac y 2} }$ $\le$ $\ds \frac 1 2 \map \ln {\map \Gamma x} + \frac 1 2 \map \ln {\map \Gamma y}$

The result follows by definition of convex function.

$\blacksquare$