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It is a theorem that any function $f$ defined for positive real numbers satisfying

  1. $f(1)=1$
  2. $f(x+1)=x\cdot f(x)$
  3. $f$ is log convex

is identically equal to the gamma function. (Condition 2 means that this function interpolates a shifted factorial function.)

Now, a beginner (such as myself) might ask: What if we weaken condition 2 by instead requiring $f$ to be merely convex, not log convex?

I would imagine that such functions would look not too different, since intuitively, I can't wildly deviate the graph of the gamma function if I want to maintain condition 2 and stay convex.

Just a follow-up musing---What if instead of condition 3, we require convexity and infinite differentiability? Do we still uniquely determine the gamma function?

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  • $\begingroup$ See this. $\endgroup$
    – Neves
    Mar 13, 2011 at 9:46
  • $\begingroup$ For completeness, this is the Bohr–Mollerup theorem. $\endgroup$
    – lhf
    Apr 12, 2011 at 10:48
  • $\begingroup$ Artin's book The Gamma Function contains some other results on uniqueness that depend on continuity or continuous differentiability only but require Legendre functional equation. $\endgroup$
    – lhf
    Apr 12, 2011 at 10:59
  • $\begingroup$ See also mathoverflow.net/questions/23229/… . $\endgroup$
    – lhf
    Apr 12, 2011 at 11:01

1 Answer 1

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(This should be either a comment or CW.)

Peter Luschny studied a number of gamma-like functions that do not have the log-convex imposition; you might want to look into them for inspiration.

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    $\begingroup$ I do not think Luschny's examples satisfy condition 2. Any functions satisfying 1 and 2 must have a singularity at the origin. $\endgroup$
    – Johan
    Mar 13, 2011 at 10:20

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