# 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:

$\displaystyle \Gamma \left({z}\right) = \int_0^{\infty} t^{z - 1} e^{-t} \rd t$
$\forall z > 0: \Gamma \left({z}\right) > 0$, 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.

$\blacksquare$

## 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:

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

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

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

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 $\ln \Gamma \left({z}\right)$.

 $\displaystyle \Gamma \left({z}\right)$ $=$ $\displaystyle \lim_{m \mathop \to \infty} \frac {m^z m!} {z \left({z + 1}\right) \left({z + 2}\right) \dotsb \left({z + m}\right)}$ $\displaystyle \implies \ \$ $\displaystyle \dfrac {\d^2 \ln \Gamma \left({z}\right)} {\d z^2}$ $=$ $\displaystyle \dfrac {\d^2} {\d z^2} \ln \left({\lim_{m \mathop \to \infty} \frac {m^z m!} {z \left({z + 1}\right) \left({z + 2}\right) \dotsb \left({z + m}\right)} }\right)$ $\displaystyle$ $=$ $\displaystyle \lim_{m \mathop \to \infty} \dfrac {\d^2} {\d z^2} \ln \left({\frac {m^z m!} {z \left({z + 1}\right) \left({z + 2}\right) \dotsb \left({z + m}\right)} }\right)$

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

 $\displaystyle \lim_{m \mathop \to \infty} \dfrac {\d^2} {\d z^2} \left({z \ln \left({m}\right) + \ln \left({m!}\right) - \sum_{n \mathop = 0}^m \ln \left({z + n}\right)}\right)$ $=$ $\displaystyle \lim_{m \mathop \to \infty} \sum_{n \mathop = 0}^m \dfrac 1 {\left({z + n}\right)^2}$ $\displaystyle \implies \ \$ $\displaystyle \dfrac {\d^2 \ln \Gamma \left(z\right)} {\d z^2}$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \dfrac 1 {\left({z + n}\right)^2} > 0$

Logarithmic convexity then follows from Second Derivative of Strictly Convex Real Function is Strictly Positive.

$\blacksquare$