Bohr–Mollerup theorem

Template:Short description In mathematical analysis, the Bohr–Mollerup theorem<ref>Template:Springer</ref><ref>{{#invoke:Template wrapper|{{#if:|list|wrap}}|_template=cite web |_exclude=urlname, _debug, id |url = https://mathworld.wolfram.com/{{#if:Bohr-MollerupTheorem%7CBohr-MollerupTheorem.html}} |title = Bohr–Mollerup Theorem |author = Weisstein, Eric W. |website = MathWorld |access-date = |ref = Template:SfnRef }}</ref> is a theorem proved by the Danish mathematicians Harald Bohr and Johannes Mollerup.<ref name="BM">Template:Cite book</ref> The theorem characterizes the gamma function, defined for Template:Math by

<math>\Gamma(x)=\int_0^\infty t^{x-1} e^{-t}\,\mathrm{d}t</math>

as the only positive function Template:Mvar, with domain on the interval Template:Math, that simultaneously has the following three properties:

• Template:Math, and
• Template:Math for Template:Math and
• Template:Mvar is logarithmically convex.

A treatment of this theorem is in Artin's book The Gamma Function,<ref>Template:Cite book</ref> which has been reprinted by the AMS in a collection of Artin's writings.<ref>Template:Cite book</ref>

The theorem was first published in a textbook on complex analysis, as Bohr and Mollerup thought it had already been proved.<ref name="BM"/>

The theorem admits a far-reaching generalization to a wide variety of functions (that have convexity or concavity properties of any order).<ref>Template:Cite book</ref>

Statement

Bohr–Mollerup Theorem.     Template:Math is the only function that satisfies Template:Math with Template:Math convex and also with Template:Math.

Proof

Let Template:Math be a function with the assumed properties established above: Template:Math and Template:Math is convex, and Template:Math. From Template:Math we can establish

<math>\Gamma(x+n)=(x+n-1)(x+n-2)(x+n-3)\cdots(x+1)x\Gamma(x)</math>

The purpose of the stipulation that Template:Math forces the Template:Math property to duplicate the factorials of the integers so we can conclude now that Template:Math if Template:Math and if Template:Math exists at all. Because of our relation for Template:Math, if we can fully understand Template:Math for Template:Math then we understand Template:Math for all values of Template:Mvar.

For Template:Math, Template:Math, the slope Template:Math of the line segment connecting the points Template:Math and Template:Math is monotonically increasing in each argument with Template:Math since we have stipulated that Template:Math is convex. Thus, we know that

<math>S(n-1,n) \leq S(n,n+x)\leq S(n,n+1)\quad\text{for all }x\in(0,1].</math>

After simplifying using the various properties of the logarithm, and then exponentiating (which preserves the inequalities since the exponential function is monotonically increasing) we obtain

<math>(n-1)^x(n-1)! \leq \Gamma(n+x)\leq n^x(n-1)!.</math>

From previous work this expands to

<math>(n-1)^x(n-1)! \leq (x+n-1)(x+n-2)\cdots(x+1)x\Gamma(x)\leq n^x(n-1)!,</math>

and so

<math>\frac{(n-1)^x(n-1)!}{(x+n-1)(x+n-2)\cdots(x+1)x} \leq \Gamma(x) \leq \frac{n^xn!}{(x+n)(x+n-1)\cdots(x+1)x}\left(\frac{n+x}{n}\right).</math>

The last line is a strong statement. In particular, it is true for all values of Template:Mvar. That is Template:Math is not greater than the right hand side for any choice of Template:Mvar and likewise, Template:Math is not less than the left hand side for any other choice of Template:Mvar. Each single inequality stands alone and may be interpreted as an independent statement. Because of this fact, we are free to choose different values of Template:Mvar for the RHS and the LHS. In particular, if we keep Template:Mvar for the RHS and choose Template:Math for the LHS we get:

<math>\begin{align}

\frac{((n+1)-1)^x((n+1)-1)!}{(x+(n+1)-1)(x+(n+1)-2)\cdots(x+1)x}&\leq \Gamma(x)\leq\frac{n^xn!}{(x+n)(x+n-1)\cdots(x+1)x}\left(\frac{n+x}{n}\right)\\ \frac{n^xn!}{(x+n)(x+n-1)\cdots(x+1)x}&\leq \Gamma(x)\leq\frac{n^xn!}{(x+n)(x+n-1)\cdots(x+1)x}\left(\frac{n+x}{n}\right) \end{align}</math>

It is evident from this last line that a function is being sandwiched between two expressions, a common analysis technique to prove various things such as the existence of a limit, or convergence. Let Template:Math:

<math>\lim_{n\to\infty} \frac{n+x}{n} = 1</math>

so the left side of the last inequality is driven to equal the right side in the limit and

<math>\frac{n^xn!}{(x+n)(x+n-1)\cdots(x+1)x}</math>

is sandwiched in between. This can only mean that

<math>\lim_{n\to\infty}\frac{n^xn!}{(x+n)(x+n-1)\cdots(x+1)x} = \Gamma (x).</math>

In the context of this proof this means that

<math>\lim_{n\to\infty}\frac{n^xn!}{(x+n)(x+n-1)\cdots(x+1)x}</math>

has the three specified properties belonging to Template:Math. Also, the proof provides a specific expression for Template:Math. And the final critical part of the proof is to remember that the limit of a sequence is unique. This means that for any choice of Template:Math only one possible number Template:Math can exist. Therefore, there is no other function with all the properties assigned to Template:Math.

The remaining loose end is the question of proving that Template:Math makes sense for all Template:Mvar where

<math>\lim_{n\to\infty}\frac{n^xn!}{(x+n)(x+n-1)\cdots(x+1)x}</math>

exists. The problem is that our first double inequality

<math>S(n-1,n)\leq S(n+x,n)\leq S(n+1,n)</math>

was constructed with the constraint Template:Math. If, say, Template:Math then the fact that Template:Mvar is monotonically increasing would make Template:Math, contradicting the inequality upon which the entire proof is constructed. However,

<math>\begin{align}

\Gamma(x+1)&= \lim_{n\to\infty}x\cdot\left(\frac{n^xn!}{(x+n)(x+n-1)\cdots(x+1)x}\right)\frac{n}{n+x+1}\\ \Gamma(x)&=\left(\frac{1}{x}\right)\Gamma(x+1) \end{align}</math>

which demonstrates how to bootstrap Template:Math to all values of Template:Mvar where the limit is defined.

See also

• Wielandt theorem