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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 $x > 0$ by

\Gamma (x)=\int _{0}^{\infty }t^{x-1}e^{-t}\,\mathrm {d} t

as the only positive function $f$ , with domain on the interval $x > 0$ , that simultaneously has the following three properties:

$f (1) = 1$ , and
$f (x + 1) = x f (x)$ for $x > 0$ and
$f$ 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.

Γ(x)

is the only function that satisfies

f (x + 1) = x f (x)

with

log(f (x))

convex and also with

f (1) = 1

.

Proof

Let $Γ(x)$ be a function with the assumed properties established above: $Γ(x + 1) = x Γ(x)$ and $log(Γ(x))$ is convex, and $Γ(1) = 1$ . From $Γ(x + 1) = x Γ(x)$ we can establish

<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/b6c700ef4f146772ae8c5e7525fb6a7e9344c3a1" class="mwe-math-fallback-image-inline mw-invert" aria-hidden="true" style="vertical-align: -0.838ex; width:62.617ex; height:2.843ex;" alt="{\displaystyle \Gamma (x+n)=(x+n-1)(x+n-2)(x+n-3)\cdots (x+1)x\Gamma (x)}">

The purpose of the stipulation that $Γ(1) = 1$ forces the $Γ(x + 1) = x Γ(x)$ property to duplicate the factorials of the integers so we can conclude now that $Γ(n) = (n - 1)!$ if $n \in N$ and if $Γ(x)$ exists at all. Because of our relation for $Γ(x + n)$ , if we can fully understand $Γ(x)$ for $0 < x \leq 1$ then we understand $Γ(x)$ for all values of $x$ .

For $x 1$ , $x 2$ , the slope $S (x 1, x 2)$ of the line segment connecting the points $(x 1, log(Γ (x 1)))$ and $(x 2, log(Γ (x 2)))$ is monotonically increasing in each argument with $x 1 < x 2$ since we have stipulated that $log(Γ(x))$ is convex. Thus, we know that

<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/2b8a1b0d0b6d90e540efbb5aaab05415df56f382" class="mwe-math-fallback-image-inline mw-invert" aria-hidden="true" style="vertical-align: -0.838ex; width:58.222ex; height:2.843ex;" alt="{\displaystyle S(n-1,n)\leq S(n,n+x)\leq S(n,n+1)\quad {\text{for all }}x\in (0,1].}">

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

<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/6f7671bf17a5ec57e8ae1fba788b94acfb8a89fb" class="mwe-math-fallback-image-inline mw-invert" aria-hidden="true" style="vertical-align: -0.838ex; width:42.325ex; height:2.843ex;" alt="{\displaystyle (n-1)^{x}(n-1)!\leq \Gamma (n+x)\leq n^{x}(n-1)!.}">

From previous work this expands to

<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/dbef7caf9b53973e2298464463b43d3bb9945e5a" class="mwe-math-fallback-image-inline mw-invert" aria-hidden="true" style="vertical-align: -0.838ex; width:72.812ex; height:2.843ex;" alt="{\displaystyle (n-1)^{x}(n-1)!\leq (x+n-1)(x+n-2)\cdots (x+1)x\Gamma (x)\leq n^{x}(n-1)!,}">

and so

<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/3f6a6c6045ac9628734a688efe3dfc4f65e07d38" class="mwe-math-fallback-image-inline mw-invert" aria-hidden="true" style="vertical-align: -2.671ex; width:89.147ex; height:6.509ex;" alt="{\displaystyle {\frac {(n-1)^{x}(n-1)!}{(x+n-1)(x+n-2)\cdots (x+1)x}}\leq \Gamma (x)\leq {\frac {n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}}\left({\frac {n+x}{n}}\right).}">

The last line is a strong statement. In particular, it is true for all values of $n$ . That is $Γ(x)$ is not greater than the right hand side for any choice of $n$ and likewise, $Γ(x)$ is not less than the left hand side for any other choice of $n$ . 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 $n$ for the RHS and the LHS. In particular, if we keep $n$ for the RHS and choose $n + 1$ for the LHS we get:

<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/50208b58a7c2bd54efc1c11702954768f216e75f" class="mwe-math-fallback-image-inline mw-invert" aria-hidden="true" style="vertical-align: -5.838ex; width:100.489ex; height:12.843ex;" alt="{\displaystyle {\begin{aligned}{\frac {((n+1)-1)^{x}((n+1)-1)!}{(x+(n+1)-1)(x+(n+1)-2)\cdots (x+1)x}}&amp;\leq \Gamma (x)\leq {\frac {n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}}\left({\frac {n+x}{n}}\right)\\{\frac {n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}}&amp;\leq \Gamma (x)\leq {\frac {n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}}\left({\frac {n+x}{n}}\right)\end{aligned}}}">

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 $n \to \infty$ :

<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/a7ed338ecd737035c5c5abf8d6f0dd0de93ead6e" class="mwe-math-fallback-image-inline mw-invert" aria-hidden="true" style="vertical-align: -1.838ex; width:15.321ex; height:5.009ex;" alt="{\displaystyle \lim _{n\to \infty }{\frac {n+x}{n}}=1}">

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

<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/9f47cb9bb010429e0488ab336e49428c02fed2d3" class="mwe-math-fallback-image-inline mw-invert" aria-hidden="true" style="vertical-align: -2.671ex; width:31.556ex; height:6.176ex;" alt="{\displaystyle {\frac {n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}}}">

is sandwiched in between. This can only mean that

<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c5d45ef4d5a5614201e975e506a8219350e33e44" class="mwe-math-fallback-image-inline mw-invert" aria-hidden="true" style="vertical-align: -2.671ex; width:44.553ex; height:6.176ex;" alt="{\displaystyle \lim _{n\to \infty }{\frac {n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}}=\Gamma (x).}">

In the context of this proof this means that

<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/708d9c67b2d581aac1632f157e2f8f2d0d79e9ce" class="mwe-math-fallback-image-inline mw-invert" aria-hidden="true" style="vertical-align: -2.671ex; width:36.216ex; height:6.176ex;" alt="{\displaystyle \lim _{n\to \infty }{\frac {n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}}}">

has the three specified properties belonging to $Γ(x)$ . Also, the proof provides a specific expression for $Γ(x)$ . 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 $0 < x \leq 1$ only one possible number $Γ(x)$ can exist. Therefore, there is no other function with all the properties assigned to $Γ(x)$ .

The remaining loose end is the question of proving that $Γ(x)$ makes sense for all $x$ where

<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/708d9c67b2d581aac1632f157e2f8f2d0d79e9ce" class="mwe-math-fallback-image-inline mw-invert" aria-hidden="true" style="vertical-align: -2.671ex; width:36.216ex; height:6.176ex;" alt="{\displaystyle \lim _{n\to \infty }{\frac {n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}}}">

exists. The problem is that our first double inequality

<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/a7720889e9e5b8827b5c59f67cc58d9d64476268" class="mwe-math-fallback-image-inline mw-invert" aria-hidden="true" style="vertical-align: -0.838ex; width:39.768ex; height:2.843ex;" alt="{\displaystyle S(n-1,n)\leq S(n+x,n)\leq S(n+1,n)}">

was constructed with the constraint $0 < x \leq 1$ . If, say, $x > 1$ then the fact that $S$ is monotonically increasing would make $S (n + 1, n) < S (n + x, n)$ , contradicting the inequality upon which the entire proof is constructed. However,

<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/097077f0ff5940f1cbea1eee8e705ddb29be8433" class="mwe-math-fallback-image-inline mw-invert" aria-hidden="true" style="vertical-align: -5.671ex; width:65.881ex; height:12.509ex;" alt="{\displaystyle {\begin{aligned}\Gamma (x+1)&amp;=\lim _{n\to \infty }x\cdot \left({\frac {n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}}\right){\frac {n}{n+x+1}}\\\Gamma (x)&amp;=\left({\frac {1}{x}}\right)\Gamma (x+1)\end{aligned}}}">

which demonstrates how to bootstrap $Γ(x)$ to all values of $x$ where the limit is defined.

References

^ "Bohr–Mollerup theorem", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
^ Weisstein, Eric W. "Bohr–Mollerup Theorem". MathWorld.
^ ^a ^b Mollerup, J., Bohr, H. (1922). Lærebog i Kompleks Analyse vol. III, Copenhagen.{{cite book}}: CS1 maint: multiple names: authors list (link)
^ Artin, Emil (1964). The Gamma Function. Holt, Rinehart, Winston.
^ Rosen, Michael (2006). Exposition by Emil Artin: A Selection. American Mathematical Society.
^ J.-L. Marichal; N. Zenaïdi (2022). A Generalization of Bohr-Mollerup's Theorem for Higher Order Convex Functions. Developments in Mathematics, Vol. 70. Springer, Cham, Switzerland.

[1] "Bohr–Mollerup theorem", Encyclopedia of Mathematics, EMS Press, 2001 [1994]

[2] Weisstein, Eric W. "Bohr–Mollerup Theorem". MathWorld.

[BM-3] Mollerup, J., Bohr, H. (1922). Lærebog i Kompleks Analyse vol. III, Copenhagen.{{cite book}}: CS1 maint: multiple names: authors list (link)

[4] Artin, Emil (1964). The Gamma Function. Holt, Rinehart, Winston.

[5] Rosen, Michael (2006). Exposition by Emil Artin: A Selection. American Mathematical Society.

[6] J.-L. Marichal; N. Zenaïdi (2022). A Generalization of Bohr-Mollerup's Theorem for Higher Order Convex Functions. Developments in Mathematics, Vol. 70. Springer, Cham, Switzerland.

[1]

[2]

[3]

[4]

[5]

[6]

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Contents

Statement

Proof

See also

References