Log of Gamma Function is Convex on Positive Reals/Proof 2
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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.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 17.4 \ (6)$