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This is a follow-up to the MathStackexchange question 3305 and MathOverflow question 23229.

The Bohr-Mollerup theorem states that the Gamma function is the unique function that satisfies:

$(1)\ f(x+1) = xf(x), \ (2)\ f(1) = 1, \ (3)\ \ln \circ f \ $ is convex.

Assume a function $f\colon \mathbb{R^{+}} \to \mathbb{R}$ that satisfies:

$(1)\ f(x+1) = xf(x), \ (2)\ f(1) = 1, \ (3)\ $ f is superadditive.

Is there a (natural) condition which makes this function unique?

Edit 1: The answer is 'no' as Moron explains. And if we assume $f$ superadditive only for $x,y \ge a$ for some real $a$?

Edit 2: I accept Moron's answer because it the correct answer to my question. Intended was the question in the sense of my first edit (and my first question). I am also curious to see an answer if condition (3) reads: (3'') $\ \ln \circ f \ $ is superadditive. Thanks to whuber for suggesting this.

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This (and the related questions) would be far more interesting if you replaced "f is superadditive" by "ln(f) is superadditive". – whuber Aug 26 '10 at 16:04

1 Answer 1

up vote 4 down vote accepted

There is no such function.

$f(2) = f(1+1) = 1×f(1) = 1$.

By superadditivity $f(1+1) \ge f(1) + f(1)$ i.e. $ 1 = f(2) \ge 2f(1) = 2$.

Not possible.

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