Can one show $h(x)=|2(\zeta'(x))^2-\zeta''(x)\zeta(x)|$ is a decreasing function for $x\in\mathbb{R}\cap [1,\infty)$?

Question

This question is related to This question. When I tried to approach it I couldn't even proof that the LHS is a decreasing function on the given domain using regular methods. I have tried to write the dirichlet series and got some messy coefficients. I calculated Their values upto $1000$ and it seems the are not all positive.

Answer 1

$\newcommand\z\zeta\newcommand{\ga}{\gamma}\newcommand{\de}{\delta}$Yes, $h$ is a decreasing function (on $(1,\infty)$).

Indeed, let \begin{equation} g:=2{\z'}^2-\z''\z, \end{equation} so that $h=|g|$. It is easy to see that $g(\infty-)=0$. So, it is enough to show that $g$ is increasing.

Note that \begin{equation} g'=3\z'\z''-\z\z'''. \end{equation} We have the Laurent series expansion \begin{equation} \z(x)=\frac1{x-1}+\sum_{n\ge0}\frac{(-1)^n}{n!}\,\ga_n(x-1)^n \end{equation} for $x$ in a neighborhood of $1$, where the $\ga_n$'s are the Stieltjes constants, with the bounds \begin{equation} |\ga_n|\le\frac{4(n-1)!}{\pi^n} \end{equation} for $n\ge1$ and $\ga_0=\ga$, the Euler constant. So, for $k\ge1$ and $x\in(1,1+\pi)$, \begin{equation} \Big|\z^{(k)}(x)-\frac{(-1)^k k!}{(x-1)^{k+1}}\Big| \le r_k(x):=\sum_{n\ge k}\frac{4(n-1)!}{\pi^n}\frac{(x-1)^{n-k}}{(n-k)!} =\frac{4\pi(k-1)!}{(\pi+1-x)^k}. \end{equation} Also, for $x\in(1,1+\pi)$, \begin{equation} \z(x)\ge\frac1{x-1}+\ga-\tilde r_0(x), \end{equation} where $\tilde r_0(x):=4\ln\frac\pi{\pi+1-x}$. So, \begin{equation} g'(x)\ge\Big(\frac1{x-1}+\ga-\tilde r_0(x)\Big) \Big(\frac6{(x-1)^4}-r_3(x)\Big) \\ -\Big(\frac1{(x-1)^2}+r_1(x)\Big)\Big(\frac2{(x-1)^3}+r_2(x)\Big)>0 \tag{20}\label{20} \end{equation} for \begin{equation} x\in\Big(1,\frac{119}{100}\Big]. \end{equation}

Next, for $k\ge1$, \begin{equation} \z^{(k)}(x)=\sum_{n\ge2}\frac{\ln^k n}{n^x}. \end{equation} Note that $\frac{\ln n}{n^x}$ and $\frac{\ln^2 n}{n^x}$ are convex in $n\ge5/2$ for each real $x\ge7/2$. So, \begin{equation} |\z'(x)|\le\frac{\ln2}{2^x}+I_1(x),\quad |\z''(x)|\le\frac{\ln^2 2}{2^x}+I_2(x), \end{equation} where \begin{equation} I_1(x):=\int_{5/2}^\infty dn\,\frac{\ln n}{n^x} <\frac{132}{100}\,\frac1{(5/2)^x}, \end{equation} \begin{equation} I_2(x):=\int_{5/2}^\infty dn\,\frac{\ln^2 n}{n^x} <\frac{19}{10}\,\frac1{(5/2)^x}. \end{equation} Also, $\z\ge1$ and $-\z'''\ge\frac{\ln^3 2}{2^x}$ for $x>1$. So, \begin{equation} g'(x)\ge\frac{\ln^3 2}{2^x} -\Big(\frac{\ln2}{2^x}+\frac{132}{100}\,\frac1{(5/2)^x}\Big) \Big(\frac{\ln^2 2}{2^x}+\frac{19}{10}\,\frac1{(5/2)^x}\Big) >0 \end{equation} for $x\ge7/2$.

It remains to show that $g'>0$ on the interval $(\frac{119}{100},\frac72]=(\frac{119}{100},\frac{350}{100}]$. This can be done by the interval method. Indeed, note that \begin{equation} g''=3{\z''}^2+2\z' \z'''-\z \z''''>b:=-\z \z''''. \end{equation} Therefore and because $b$ is decreasing,
\begin{equation} g''(x)\ge b(x_*) \text{ if } x\ge x_*>1. \end{equation} So, on any nonempty interval $(a,a+\de]\in(1,\infty)$ we have $g'\ge g'(a)+b(a)\de$. Using the latter fact for intervals

• $(\frac{119}{100}+\frac{j-1}{250},\frac{119}{100}+\frac{j}{250}]$ with $j=1,\dots,(\frac{127}{100}-\frac{119}{100})/\frac1{250}$

• $(\frac{127}{100}+\frac{j-1}{100},\frac{127}{100}+\frac{j}{100}]$ with $j=1,\dots,(\frac{190}{100}-\frac{127}{100})/\frac1{100}$

• $(\frac{190}{100}+\frac{j-1}{10},\frac{190}{100}+\frac{j}{10}]$ with $j=1,\dots,(\frac{350}{100}-\frac{190}{100})/\frac1{10}$

we complete the proof. $\quad\Box$

(To verify that $g'>0$ on each of these many intervals of the form $(a,a+\de]\in(1,\infty)$, we just have to check that for each such interval we have $g'(a)+b(a)\de>0$ and recall that $g'\ge g'(a)+b(a)\de$ on $(a,a+\de]$. )