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 $t$.
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The function is smooth according to Gamma Function is Smooth on Positive Reals, and
- $\ds \forall k \in \N: \map {\Gamma^{\paren k} } z = \int_0^{\infty} \map \ln t^k t^{z - 1} e^{-t} \rd t$
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Let $\map f z := \map \ln {\map \Gamma z}$.
- $f$ is smooth because $\Gamma$ is smooth and positive.
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Then:
- $\map {f'} z = \dfrac {\map {\Gamma'} z} {\map \Gamma z}$
- $\map {f^{\paren 2} } z = \dfrac {\map {\Gamma^{\paren 2} } z \map \Gamma z - \map {\Gamma'} z^2} {\map \Gamma z^2} > 0$
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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 Real Function with Strictly Positive Second Derivative is Strictly Convex.
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $11.14 \ \text{(iv)}$