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Let $(x)_p=x(x+1)\dots(x+p-1)$ be the rising factorial function.

My question is: Is $(x)_p$ a convex function or not? And how to proof?

And what is about the falling factorial function $(x)^p=x(x-1)\dots(x-p+1)$

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up vote 5 down vote accepted

Well, not on the whole real line. $x\mapsto (x)_p$ is a polynomial of degree $p$ with the roots $0,-1,\dots,-p+1$ and has positive and negative values in between this roots. Hence, it is not convex there.

However, as already stated by Robert Israel, it is convex for $x\geq 0$. To see this you could also argue that it is the product of convex and non-decreasing functions. 

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$(x)_p$ is convex for (at least) $x > 0$, and $(x)^p$ is convex for (at least) $x > p-1$. To prove it, just take the second derivative using the product rule, and note that all terms are positive.

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