Let Γ(x) be a function with the assumed properties established above: Γ(x + 1) = xΓ(x) and log(Γ(x)) is convex, and Γ(1) = 1. From Γ(x + 1) = xΓ(x) we can establish
The purpose of the stipulation that Γ(1) = 1 forces the Γ(x + 1) = xΓ(x) property to duplicate the factorials of the integers so we can conclude now that Γ(n) = (n − 1)! if n ∈ 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 ≤ 1 then we understand Γ(x) for all values of x.
The slope of a line connecting two points (x1, log(Γ (x1))) and (x2, log(Γ (x2))), call it S(x1, x2), is monotonically increasing in each argument with x1 < x2 since we have stipulated log(Γ(x)) is convex. Thus, we know that
The last line is a strong statement. In particular, it is true for all values ofn. 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:
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 → ∞:
so the left side of the last inequality is driven to equal the right side in the limit and
is sandwiched in between. This can only mean that
In the context of this proof this means that
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 ≤ 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
exists. The problem is that our first double inequality
was constructed with the constraint 0 < x ≤ 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. But notice
which demonstrates how to bootstrap Γ(x) to all values of x where the limit is defined.