Английская Википедия:Bohr–Mollerup theorem
Шаблон:Short description In mathematical analysis, the Bohr–Mollerup theorem[1][2] is a theorem proved by the Danish mathematicians Harald Bohr and Johannes Mollerup.[3] The theorem characterizes the gamma function, defined for Шаблон:Math by
- <math>\Gamma(x)=\int_0^\infty t^{x-1} e^{-t}\,\mathrm{d}t</math>
as the only positive function Шаблон:Mvar, with domain on the interval Шаблон:Math, that simultaneously has the following three properties:
- Шаблон:Math, and
- Шаблон:Math for Шаблон:Math and
- Шаблон:Mvar is logarithmically convex.
A treatment of this theorem is in Artin's book The Gamma Function,[4] which has been reprinted by the AMS in a collection of Artin's writings.[5]
The theorem was first published in a textbook on complex analysis, as Bohr and Mollerup thought it had already been proved.[3]
The theorem admits a far-reaching generalization to a wide variety of functions (that have convexity or concavity properties of any order).[6]
Statement
- Bohr–Mollerup Theorem. Шаблон:Math is the only function that satisfies Шаблон:Math with Шаблон:Math convex and also with Шаблон:Math.
Proof
Let Шаблон:Math be a function with the assumed properties established above: Шаблон:Math and Шаблон:Math is convex, and Шаблон:Math. From Шаблон: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 Шаблон:Math forces the Шаблон:Math property to duplicate the factorials of the integers so we can conclude now that Шаблон:Math if Шаблон:Math and if Шаблон:Math exists at all. Because of our relation for Шаблон:Math, if we can fully understand Шаблон:Math for Шаблон:Math then we understand Шаблон:Math for all values of Шаблон:Mvar.
For Шаблон:Math, Шаблон:Math, the slope Шаблон:Math of the line segment connecting the points Шаблон:Math and Шаблон:Math is monotonically increasing in each argument with Шаблон:Math since we have stipulated that Шаблон: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 Шаблон:Mvar. That is Шаблон:Math is not greater than the right hand side for any choice of Шаблон:Mvar and likewise, Шаблон:Math is not less than the left hand side for any other choice of Шаблон: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 Шаблон:Mvar for the RHS and the LHS. In particular, if we keep Шаблон:Mvar for the RHS and choose Шаблон: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 Шаблон: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 Шаблон:Math. Also, the proof provides a specific expression for Шаблон: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 Шаблон:Math only one possible number Шаблон:Math can exist. Therefore, there is no other function with all the properties assigned to Шаблон:Math.
The remaining loose end is the question of proving that Шаблон:Math makes sense for all Шаблон: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 Шаблон:Math. If, say, Шаблон:Math then the fact that Шаблон:Mvar is monotonically increasing would make Шаблон: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 Шаблон:Math to all values of Шаблон:Mvar where the limit is defined.
See also
References
