Bohr–Mollerup theorem

:Bohr–Mollerup Theorem.     {{math|Γ(x)}} is the only function that satisfies {{math| f (x + 1) {{=}} x f (x)}} with {{math|log( f (x))}} convex and also with {{math| f (1) {{=}} 1}}.

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

:$\backslash Gamma(x+n)=(x+n-1)(x+n-2)(x+n-3)\backslash cdots(x+1)x\backslash Gamma(x)$

The purpose of the stipulation that {{math|Γ(1) {{=}} 1}} forces the {{math|Γ(x + 1) {{=}} xΓ(x)}} property to duplicate the factorials of the integers so we can conclude now that {{math|Γ(n) {{=}} (n − 1)!}} if {{math|n ∈ N}} and if {{math|Γ(x)}} exists at all. Because of our relation for {{math|Γ(x + n)}}, if we can fully understand {{math|Γ(x)}} for {{math|0 < x ≤ 1}} then we understand {{math|Γ(x)}} for all values of {{mvar|x}}.

For {{math|x₁}}, {{math|x₂}}, the slope {{math|S(x₁, x₂)}} of the line segment connecting the points {{math|(x₁, log(Γ (x₁)))}} and {{math|(x₂, log(Γ (x₂)))}} is monotonically increasing in each argument with {{math|x₁ < x₂}} since we have stipulated that {{math|log(Γ(x))}} is convex. Thus, we know that

:$S(n-1,n)\; \backslash leq\; S(n,n+x)\backslash leq\; S(n,n+1)\backslash quad\backslash text\{for\; all\; \}x\backslash in(0,1].$

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

:$(n-1)^x(n-1)!\; \backslash leq\; \backslash Gamma(n+x)\backslash leq\; n^x(n-1)!.$

From previous work this expands to

:$(n-1)^x(n-1)!\; \backslash leq\; (x+n-1)(x+n-2)\backslash cdots(x+1)x\backslash Gamma(x)\backslash leq\; n^x(n-1)!,$

and so

:$\backslash frac\{(n-1)^x(n-1)!\}\{(x+n-1)(x+n-2)\backslash cdots(x+1)x\}\; \backslash leq\; \backslash Gamma(x)\; \backslash leq\; \backslash frac\{n^xn!\}\{(x+n)(x+n-1)\backslash cdots(x+1)x\}\backslash left(\backslash frac\{n+x\}\{n\}\backslash right).$

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

:$\backslash begin\{align\}$

\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}

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 {{math|n → ∞}}:

:$\backslash lim\_\{n\backslash to\backslash infty\}\; \backslash frac\{n+x\}\{n\}\; =\; 1$

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

:$\backslash frac\{n^xn!\}\{(x+n)(x+n-1)\backslash cdots(x+1)x\}$

is sandwiched in between. This can only mean that

:$\backslash lim\_\{n\backslash to\backslash infty\}\backslash frac\{n^xn!\}\{(x+n)(x+n-1)\backslash cdots(x+1)x\}\; =\; \backslash Gamma\; (x).$

In the context of this proof this means that

:$\backslash lim\_\{n\backslash to\backslash infty\}\backslash frac\{n^xn!\}\{(x+n)(x+n-1)\backslash cdots(x+1)x\}$

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

The remaining loose end is the question of proving that {{math|Γ(x)}} makes sense for all {{mvar|x}} where

:$\backslash lim\_\{n\backslash to\backslash infty\}\backslash frac\{n^xn!\}\{(x+n)(x+n-1)\backslash cdots(x+1)x\}$

exists. The problem is that our first double inequality

:$S(n-1,n)\backslash leq\; S(n+x,n)\backslash leq\; S(n+1,n)$

was constructed with the constraint {{math|0 < x ≤ 1}}. If, say, {{math|x > 1}} then the fact that {{mvar|S}} is monotonically increasing would make {{math|S(n + 1, n) < S(n + x, n)}}, contradicting the inequality upon which the entire proof is constructed. However,

:$\backslash begin\{align\}$

\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}

which demonstrates how to bootstrap {{math|Γ(x)}} to all values of {{mvar|x}} where the limit is defined.

• Wielandt theorem

{{reflist}}

{{DEFAULTSORT:Bohr-Mollerup theorem}}

Category:Gamma and related functions

Category:Theorems in complex analysis