Let \begin{equation*} \Gamma(z)=\int_0^{\infty}t^{z-1}\textrm{e}^{-t}\textrm{d}t, \quad \Re(z)>0 \end{equation*} and $$ \psi(z)=[\ln\Gamma(z)]'=\frac{\Gamma'(z)}{\Gamma(z)}. $$ In the literature, these two functions are respectively called the Euler gamma function and the digamma function. The derivatives $\psi^{(k)}(z)$ for $k=1,2,3,4$ are respectively called the trigamma, tetragamma, pentagamma, and hexagamma functions. For non-specific positive integer $n\in\mathbb{N}$, we call $\psi^{(n)}(z)$ the polygamma function.

I have proved that, for $n\in\mathbb{N}$ and $\alpha>0$, the ratio $\frac{\psi^{(n)}(x+\alpha)}{\psi^{(n)}(x)}$ is increasing from $(0,\infty)$ onto $(0,1)$. This is equivalent to saying that, for $n\in\mathbb{N}$ and $\alpha>0$, the function $$ \Phi_{n,\alpha}(x)=\frac{1}{1-\frac{\psi^{(n)}(x+\alpha)}{\psi^{(n)}(x)}} $$ is increasing from $(0,\infty)$ onto $(1,\infty)$.

After considering the increasing property of $\Phi_{n,\alpha}(x)$, I would like to ask the following question.

For $n\in\mathbb{N}$ and $\alpha>0$, is the function $\Phi_{n,\alpha}(x)$ convex on $(0,\infty)$?

The answer to this question should be YES.

This problem is perhaps much easy for you, but it is very difficult for me right now. Please have a try to solve it. Thank you very mcuh.

I have asked this question at the site on the ResearchGate, but till now I haven't gotten any correct and useful answer.


1 Answer 1


I have proved that, for $n\in\mathbb{N}$, the function $\Phi_{n,1}(x)$ is convex on $(0,\infty)$.

In the paper "H. Alzer, Sharp inequalities for the digamma and polygamma functions, Forum Math. 16 (2004), no. 2, 181--221; available online at http://dx.doi.org/10.1515/form.2004.009 ", the function $x^c\lvert\psi^{(n)}(x)\lvert$ for $n\in\mathbb{N}$ and $c\in\mathbb{R}$ was proved to be convex on $(0,\infty)$ if and only if either $c\le n$, or $c=n+1$, or $c\ge n+2$. This is the main base of my proof.

The next special case I want to prove is the possibly correct conclusion that the function $\Phi_{1,1/2}(x)$ is convex on $(0,\infty)$.


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