In mathematical analysis, the Bohr–Mollerup theorem is a theorem proved by the Danish mathematicians Harald Bohr and Johannes Mollerup.[1] The theorem characterizes the gamma function, defined for by
| infty | |
| \Gamma(x)=\int | |
| 0 |
tx-1e-tdt
as the only positive function, with domain on the interval, that simultaneously has the following three properties:
A treatment of this theorem is in Artin's book The Gamma Function,[2] which has been reprinted by the AMS in a collection of Artin's writings.[3]
The theorem was first published in a textbook on complex analysis, as Bohr and Mollerup thought it had already been proved.[1]
The theorem admits a far-reaching generalization to a wide variety of functions (that have convexity or concavity properties of any order).[4]
Bohr–Mollerup Theorem. is the only function that satisfies with convex and also with .
Let be a function with the assumed properties established above: and is convex, and . From we can establish
\Gamma(x+n)=(x+n-1)(x+n-2)(x+n-3) … (x+1)x\Gamma(x)
The purpose of the stipulation that forces the property to duplicate the factorials of the integers so we can conclude now that if and if exists at all. Because of our relation for, if we can fully understand for then we understand for all values of .
For,, the slope of the line segment connecting the points and is monotonically increasing in each argument with since we have stipulated that is convex. Thus, we know that
S(n-1,n)\leqS(n,n+x)\leqS(n,n+1) forallx\in(0,1].
(n-1)x(n-1)!\leq\Gamma(n+x)\leqnx(n-1)!.
(n-1)x(n-1)!\leq(x+n-1)(x+n-2) … (x+1)x\Gamma(x)\leqnx(n-1)!,
| (n-1)x(n-1)! | |
| (x+n-1)(x+n-2) … (x+1)x |
\leq\Gamma(x)\leq
| nxn! | \left( | |
| (x+n)(x+n-1) … (x+1)x |
| n+x | |
| n |
\right).
| \begin{align} | ((n+1)-1)x((n+1)-1)! |
| (x+(n+1)-1)(x+(n+1)-2) … (x+1)x |
&\leq\Gamma(x)\leq
| nxn! | \left( | |
| (x+n)(x+n-1) … (x+1)x |
| n+x | \right)\\ | |
| n |
| nxn! | |
| (x+n)(x+n-1) … (x+1)x |
&\leq\Gamma(x)\leq
| nxn! | \left( | |
| (x+n)(x+n-1) … (x+1)x |
| n+x | |
| n |
\right) \end{align}
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 :
\limn\toinfty
| n+x | |
| n |
=1
so the left side of the last inequality is driven to equal the right side in the limit and
| nxn! | |
| (x+n)(x+n-1) … (x+1)x |
is sandwiched in between. This can only mean that
\limn\toinfty
| nxn! | |
| (x+n)(x+n-1) … (x+1)x |
=\Gamma(x).
In the context of this proof this means that
\limn\toinfty
| nxn! | |
| (x+n)(x+n-1) … (x+1)x |
has the three specified properties belonging to . Also, the proof provides a specific expression for . 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 only one possible number can exist. Therefore, there is no other function with all the properties assigned to .
The remaining loose end is the question of proving that makes sense for all where
\limn\toinfty
| nxn! | |
| (x+n)(x+n-1) … (x+1)x |
exists. The problem is that our first double inequality
S(n-1,n)\leqS(n+x,n)\leqS(n+1,n)
was constructed with the constraint . If, say, then the fact that is monotonically increasing would make, contradicting the inequality upon which the entire proof is constructed. However,
\begin{align} \Gamma(x+1)&=\limn\toinftyx ⋅ \left(
| nxn! | \right) | |
| (x+n)(x+n-1) … (x+1)x |
| n | \\ \Gamma(x)&=\left( | |
| n+x+1 |
| 1 | |
| x |
\right)\Gamma(x+1) \end{align}
which demonstrates how to bootstrap to all values of where the limit is defined.