Gamma Function and Bohr-Mollerup Theorem

This post is a bit of a digression from recent area of focus–although (log-)convexity does come into play–but I wanted to give another proof of the famous result

\displaystyle\Gamma\left(\dfrac{1}{2}\right)=\int_{0}^{\infty}t^{-\frac{1}{2}}e^{-t}dt=2\int_{0}^{\infty}e^{-u^{2}}du=\int_{-\infty}^{\infty}e^{-u^{2}}=\sqrt{\pi}

Recall that we can define the gamma function on the positive real axis (0,\infty) by the integral expression

\displaystyle\Gamma(x):=\int_{0}^{\infty}t^{x-1}e^{-t}dt

This integral is indeed convergent since e^{-t}\lesssim t^{-n} for all n\in \mathbb{Z}^{\geq 0}. The gamma function satisfies has a few nice properties, which turn out to uniquely characterize it.

  1. \Gamma(x+1)=x\Gamma(x),\indent\forall x>0;
  2. \Gamma(1)=1;
  3. \Gamma is log-convex.

Proof. We prove (1) by integration by parts. For x>0, we have

\displaystyle\int_{0}^{\infty}t^{x}e^{-t}dt=-\left[t^{x}e^{-t}\right]_{t=0}^{t=\infty}+x\int_{0}^{\infty}t^{x-1}e^{-t}dt=x\int_{0}^{\infty}t^{x-1}e^{-t}dt=x\Gamma(x)

(2) is immediate from evaluating the integral expression. For (3), observe that for x,y\in(0,\infty), we have by the concavity of \log x that

\begin{array}{lcl}\displaystyle\log\left(\Gamma(\lambda x+(1-\lambda)y)\right)=\log\left(\int_{0}^{\infty}t^{\lambda x+(1-\lambda)y-1}e^{-t}dt\right)&=&\displaystyle\log\left(\int_{0}^{\infty}\left(t^{x-1}e^{-t}\right)^{\lambda}\left(t^{y-1}e^{-t}\right)^{1-\lambda}dt\right)\\&\leq&\displaystyle\log\left[\left(\int_{0}^{\infty}t^{x-1}e^{-t}dt\right)^{\lambda}\left(\int_{0}^{\infty}t^{y-1}e^{-t}dt\right)^{1-\lambda}\right]\\&=&\displaystyle\lambda\log\left(\Gamma(x)\right)+(1-\lambda)\log\left(\Gamma(y)\right)\end{array}

where the inequality follows from applying Hölder’s inequality. \Box

We now prove the Bohr-Mollerup theorem, which uniquely characterizes the gamma function in terms of the three properties listed above. Note that the Bohr here is not the 20th century physicist Niels Bohr but rather his mathematician brother, Harald Bohr.

Theorem. (Bohr-Mollerup theorem)

Suppose f: (0,\infty)\rightarrow\mathbb{R} satisfies

  1. f(x+1)=xf(x) for all $x>0$;
  2. f(1)=1;
  3. f is log-convex.

Then f=\Gamma.

Proof. I claim that f(n+1)=n! for all n\in\mathbb{Z}^{\geq 0}. Suppose equality holds for some n\in\mathbb{Z}^{\geq 1}. Then

\displaystyle f(n+1)=nf(n)=n(n-1)!=n!

Let x\in(0,1] and n\in\mathbb{Z}^{\geq 1}. Then by log-convexity,

\begin{array}{lcl}\displaystyle f(n+1+x)=f\left((1-x)(n+1)+x(n+2)\right)&\leq&\displaystyle\left[f(n+1)\right]^{1-x}\left[f(n+2)\right]^{x}\\&=&\displaystyle\left[n!\right]^{1-x}\left[(n+1)!\right]^{x}\\&=&\displaystyle (n!)(n+1)^{x}\end{array}

We can get an estimate for n! by another application of the log-convexity of f:

\begin{array}{lcl}\displaystyle n!=f(n+1)=f\left(x(n+x)+(1-x)(n+1+x)\right)&\leq&\displaystyle\left[f(n+x)\right]^{x}\left[f(n+1+x)\right]^{1-x}\\&=&\displaystyle\left[(n+x)^{-1}f(n+1+x)\right]^{x}\left[f(n+1+x)\right]^{1-x}\\&=&\displaystyle(n+x)^{-x}f(n+1+x)\end{array}

Since f(n+1+x)=(n+x)(n-1+x)\cdots xf(x), we obtain the inequalities

\displaystyle\left(1+\dfrac{x}{n}\right)^{x}=\dfrac{(n+x)^{x}}{n^{x}}\leq\dfrac{(n+x)(n-1+x)\cdots xf(x)}{(n!)n^{x}}\leq\dfrac{(n+1)^{x}}{n^{x}}=\left(1+\dfrac{1}{n}\right)^{x}

Letting n\rightarrow\infty, we obtain from the squeeze theorem that

\displaystyle\lim_{n\rightarrow\infty}\dfrac{(n+x)(n-1+x)\cdots xf(x)}{(n!)n^{x}}=1\Longleftrightarrow f(x)=\lim_{n\rightarrow\infty}\dfrac{(n!)n^{x}}{(n+x)(n-1+x)\cdots x},\indent\forall x\in (0,1]

We now show that this identity holds for all x\in (0,\infty) and therefore f is uniquely determined. Let m\in\mathbb{Z}^{\geq 0} such that x-m=x\pmod{1}\in (0,1]. Then

\begin{array}{lcl}\displaystyle f(x)&=&\displaystyle (x-1)\cdots (x-m)f(x-m)\\&=&\displaystyle(x-1)\cdots(x-m)\lim_{n\rightarrow\infty}\dfrac{(n!)n^{x-m}}{(n+x-m)(n-1+x-m)\cdots(x-m)}\\&=&\displaystyle\lim_{n\rightarrow\infty}\dfrac{(n!)n^{x-m}}{(n+x-m)(n-1+x-m)\cdots x}\\&=&\displaystyle\lim_{n\rightarrow\infty}\dfrac{(n!)n^{x}}{(n+x)(n-1+x)\cdots x}\cdot\dfrac{(n+x)(n-1+x)\cdots(n-(m-1)+x)}{n^{m}}\\&=&\displaystyle\lim_{n\rightarrow\infty}\dfrac{(n!)n^{x}}{(n+x)(n-1+x)\cdots x}\cdot\lim_{n\rightarrow\infty}\left(1+\dfrac{x}{n}\right)\left(1+\dfrac{x-1}{n}\right)\cdots\left(1+\dfrac{x-(m-1)}{n}\right)\\&=&\displaystyle\lim_{n\rightarrow\infty}\dfrac{(n!)n^{x}}{(n+x)(n-1+x)\cdots x}\end{array}

Since the gamma function satisfies all three properties in the statement of the theorem, we obtain the identity

\displaystyle\Gamma(x)=\lim_{n\rightarrow\infty}\dfrac{(n!)n^{x}}{(n+x)(n-1+x)\cdots x},\indent\forall x\in(0,\infty)

\Box

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