05/23/2008, 12:07 PM
Ok, now the promised proof for the uniqueness of the gamma function, which was point 1. here. I nearly only translate it from the mentioned book:
We consider the function \( v=F-\Gamm \) on the right half plane, it also satisfies \( v(z+1)=zv(z) \). Hence \( v \) has a meromorphic continuation to \( \mathbb{C} \). Poles can be at most at 0, -1, -2 , ....
As \( v(1)=0 \) we have \( \lim_{z\to 0} zv(z)=0 \), hence \( v \) has a holomorphic continuation to 0 and also to to each \( -n \), \( n\in\mathbb{N} \) by \( v(z+1)=zv(z) \).
\( v(z) \) restricted to \( 1\le \Re(z)<2 \) is bounded because \( \Gamma \) is bounded there. Then \( v(z) \) is also restricted on \( S \) given by \( 0\le \Re(z)\le 1 \). \( q(z) \) defined by \( v(z)v(1-z) \) is bounded on \( S \) because \( v(z) \) and \( v(1-z) \) have the same values on \( S \). Now \( q(z+1)=-q(z) \), hence \( q \) is bounded on whole \( \mathbb{C} \) and by Liouville \( q(z)=q(1)=0 \). Hence \( v\equiv 0 \) and \( F\equiv\Gamma \).
We consider the function \( v=F-\Gamm \) on the right half plane, it also satisfies \( v(z+1)=zv(z) \). Hence \( v \) has a meromorphic continuation to \( \mathbb{C} \). Poles can be at most at 0, -1, -2 , ....
As \( v(1)=0 \) we have \( \lim_{z\to 0} zv(z)=0 \), hence \( v \) has a holomorphic continuation to 0 and also to to each \( -n \), \( n\in\mathbb{N} \) by \( v(z+1)=zv(z) \).
\( v(z) \) restricted to \( 1\le \Re(z)<2 \) is bounded because \( \Gamma \) is bounded there. Then \( v(z) \) is also restricted on \( S \) given by \( 0\le \Re(z)\le 1 \). \( q(z) \) defined by \( v(z)v(1-z) \) is bounded on \( S \) because \( v(z) \) and \( v(1-z) \) have the same values on \( S \). Now \( q(z+1)=-q(z) \), hence \( q \) is bounded on whole \( \mathbb{C} \) and by Liouville \( q(z)=q(1)=0 \). Hence \( v\equiv 0 \) and \( F\equiv\Gamma \).