# 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 $\ln$ denote the natural logarithm function.

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

## Proof 1

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.

$\blacksquare$

## Proof 2

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.

$\blacksquare$

## Proof 3

The strategy is to use the Euler Form of the Gamma function and directly calculate the second derivative of $\ln \map \Gamma z$.

 $\ds \map \Gamma z$ $=$ $\ds \lim_{m \mathop \to \infty} \frac {m^z m!} {z \paren {z + 1} \paren {z + 2} \dotsm \paren {z + m} }$ $\ds \leadsto \ \$ $\ds \dfrac {\d^2 \ln \map \Gamma z} {\d z^2}$ $=$ $\ds \dfrac {\d^2} {\d z^2} \map \ln {\lim_{m \mathop \to \infty} \frac {m^z m!} {z \paren {z + 1} \paren {z + 2} \dotsm \paren {z + m} } }$ $\ds$ $=$ $\ds \lim_{m \mathop \to \infty} \dfrac {\d^2} {\d z^2} \map \ln {\frac {m^z m!} {z \paren {z + 1} \paren {z + 2} \dotsm \paren {z + m} } }$

The limit interchange is justified because Gamma Function is Smooth on Positive Reals.

 $\ds \lim_{m \mathop \to \infty} \dfrac {\d^2} {\d z^2} \paren {z \map \ln m + \map \ln {m!} - \sum_{n \mathop = 0}^m \map \ln {z + n} }$ $=$ $\ds \lim_{m \mathop \to \infty} \sum_{n \mathop = 0}^m \dfrac 1 {\paren {z + n}^2}$ $\ds \leadsto \ \$ $\ds \dfrac {\d^2 \ln \map \Gamma z} {\d z^2}$ $=$ $\ds \sum_{n \mathop = 0}^\infty \dfrac 1 {\paren {z + n}^2} > 0$

Logarithmic convexity then follows from Twice Differentiable Real Function with Positive Second Derivative is Strictly Convex.

$\blacksquare$