1 Introduction
The study of the extreme values of the Riemann zeta function has a long history. Extreme values on the critical line
$\sigma =1/2$
were first considered by Titchmarsh [Reference Titchmarsh7], who showed that there exist arbitrarily large t such that for any
$\alpha <1/2$
, we have
$|\zeta (1/2+{i} t)|\geqslant \exp ((\log t)^\alpha )$
. For the critical strip
$1/2<\sigma <1$
, it was also Titchmarsh [Reference Titchmarsh6] who first showed that for any
$\varepsilon>0$
and fixed
$\sigma \in (1/2,1)$
, there exist arbitrarily large t such that
$|\zeta (\sigma +{i}t)|\geqslant \exp \{(\log t)^{1-\sigma -\varepsilon }\}.$
The study of the values on the 1-line dates back to 1925 when Littlewood [Reference Littlewood4] showed that there exist arbitrarily large t for which
$$ \begin{align} |\zeta(1+{i}t)|\geqslant\{1+o(1)\}{e}^{\gamma}\log_2t. \end{align} $$
Here and throughout, we denote by
$\log _j$
the jth iterated logarithm and by
$\gamma $
the Euler constant. We refer to [Reference Dong and Wei2] for a detailed account of the historical developments.
We may also consider the extreme values of the derivatives of the Riemann zeta function. For any fixed
$\ell \in {\mathbb N}$
, denote
$$ \begin{align*}Z^{(\ell)}(T):=\max_{t\in[T,2T]}|\zeta^{(\ell)}(1+{i} t)|.\end{align*} $$
In addition to other results, Yang [Reference Yang8] recently proved that if T is sufficiently large, then uniformly for
$\ell \leqslant (\log T)/(\log _2 T)$
,
$$ \begin{align} Z^{(\ell)}(T)\geqslant \frac{{e}^{\gamma} \ell^{\ell}}{(\ell+1)^{\ell+1}}\{\log_2T-\log_3T+O(1)\}^{\ell+1}. \end{align} $$
In this note, we aim to improve the constant
$\ell ^{\ell }/(\ell +1)^{\ell +1}$
in (1.2). We prove the following theorem.
Theorem 1.1. For
$T\to \infty $
and
$\ell \leqslant (\log T)/(\log _2 T)$
,
$$ \begin{align*}Z^{(\ell)}(T)\geqslant \frac{{e}^{\gamma}}{\ell+1}(\log_2T)^{\ell+1}\{1+o(1)\}.\end{align*} $$
Remark 1.2. Compared with (1.2), we improve the lower bound by a factor
${(1+1/\ell )^\ell} $
, which tends to
${e}$
as
$\ell \rightarrow \infty $
. Further, in [Reference Dong and Wei2], we recently used the ‘long resonance’ method to show that
$$ \begin{align*}\max_{t\in[\sqrt{T},T]}|\zeta(1+{i} t)|\geqslant{e}^\gamma(\log_2T+\log_3T+c),\end{align*} $$
where c is a computable constant. Granville and Soundararajan [Reference Granville, Soundararajan, Balsubramanian and Srinivas3] predicted that this is still true for
$\max _{t\in [T,2T]}|\zeta (1+{i} t)|$
. These results seems stronger than Theorem 1.1 with
$\ell =0$
. The reason is that after taking derivatives of the Riemann zeta function, we are no longer able to make use of the multiplicativity of its Dirichlet coefficients. Nevertheless, Theorem 1.1 remains a generalisation of Littlewood’s initial bound (1.1).
Both Yang’s proof and ours employ the resonance method used by Bondarenko and Seip [Reference Bondarenko, Seip, Baranov, Kisliakov and Nikolski1]. For a large x and a positive integer b, we take
$$ \begin{align} {\mathcal P}:=\prod_{p\leqslant x}p^{b-1}\quad\text{and} \quad{\mathcal M}:=\{n\in{\mathbb N}:n\mid {\mathcal P}\}. \end{align} $$
The key ingredient of the proof is a weighted reciprocal sum of the form
$$ \begin{align*} {\mathcal S}(x; \ell):=\sum_{m\in{\mathcal M}}\sum_{k\mid m}\frac{(\log k)^{\ell}}{k}, \end{align*} $$
where
$\ell \geqslant 0$
is an integer. In [Reference Yang8], Yang divided
${\mathcal M}$
as well as
${\mathcal P}$
into two subsets, according to whether
$p\leqslant x^{\ell /(\ell +1)}$
or not. Our choice is to give a finer division. Specifically, let
$J\geqslant 1$
be a positive integer. For
$0\leqslant j\leqslant J$
, denote
$$ \begin{align} {\mathcal M}_{\,j}:=\bigg\{m\in{\mathbb N}:m\mid \prod_{p\leqslant x^{j/J}}p^{b-1}\bigg\}. \end{align} $$
Thus, we divide the set
${\mathcal M}$
into J subsets:
$$ \begin{align*} \mathcal M = \bigsqcup_{j=1}^J ({\mathcal M}_{\,j}\setminus{\mathcal M}_{\,j-1}). \end{align*} $$
By this trick, we are able to enlarge the estimate of
${\mathcal S}(x; \ell )$
with a factor
$(1+1/\ell )^{\ell }$
. We summarise it in the following proposition.
Proposition 1.3. With the previous notation,
$$ \begin{align*} \frac{1}{|{\mathcal M}|}{\mathcal S}(x; \ell) \geqslant \frac{{e}^{\gamma}}{\ell+1} \bigg\{1+O\bigg(\frac{1}{J} + \frac {J\log_2 x}b + \frac{J^2}{\log x}\bigg)\bigg\}(\log x)^{\ell+1} \end{align*} $$
uniformly for
$x\geqslant 3$
,
$b\geqslant 1$
,
$J\geqslant 1$
,
$\ell \geqslant 0$
, where the implied constant is absolute.
It is worth noting that one cannot choose an extremely large J, since there are terms of the form
$(J\log _2 x)/b$
and
$J^2/{\log x}$
. In fact, we will take J to be the order of approximately
$\log _2x$
, which seems to be the limit of this approach.
2 Proof of Proposition 1.3
The following asymptotic formula plays a key role in the proof of Proposition 1.3.
Lemma 2.1. We have
$$ \begin{align*} \prod_{p\leqslant x}\sum_{\nu=0}^{b-1}\bigg(1-\frac{\nu}{b}\bigg)\frac{1}{p^\nu} = \bigg\{1+O\bigg(\frac{\log_2x}{b}+\frac{1}{\log x}\bigg)\bigg\}{e}^{\gamma}\log x \end{align*} $$
uniformly for
$x\geqslant 3$
and
$b\geqslant 1$
, where the implied constants are absolute.
Proof. See also [Reference Yang8, (15)] and [Reference Bondarenko, Seip, Baranov, Kisliakov and Nikolski1, page 129]. For a fixed prime p,
$$ \begin{align*} \sum_{\nu=0}^{b-1}\bigg(1-\frac{\nu}{b}\bigg)\frac{1}{p^\nu} & = \bigg(\sum_{\nu\geqslant0}-\sum_{\nu\geqslant b}\bigg)\bigg(1-\frac{\nu}{b}\bigg)\frac{1}{p^\nu} \\ & =\bigg(1-\frac{1}{b(p-1)}\bigg)\bigg(1-\frac1p\bigg)^{-1}+O\bigg(\frac{1}{p^b}\bigg). \end{align*} $$
Therefore,
$$ \begin{align*} \prod_{p\leqslant x}\sum_{\nu=0}^{b-1}\bigg(1-\frac{\nu}{b}\bigg)\frac{1}{p^\nu} =\bigg\{1+O\bigg(\frac1b\sum_{p\leqslant x}\frac{1}{p-1}\bigg)\bigg\}\prod_{p\leqslant x}\bigg(1-\frac1p\bigg)^{-1}. \end{align*} $$
The lemma follows from Mertens’ formula
$$ \begin{align*} \prod_{p\leqslant x}\bigg(1-\frac1p\bigg)^{-1} = \bigg\{1+O\bigg(\frac{1}{\log x}\bigg)\bigg\}{e}^{\gamma}\log x, \end{align*} $$
and the fact that
$\sum _{p\leqslant x}{1}/{(p-1)}\ll \log _2x$
.
Proof of Proposition 1.3.
By the construction, the set
${\mathcal M}$
is divisor-closed which means
$k\mid m, \, m\in {\mathcal M}$
implies
$k\in {\mathcal M}$
. By the definition of
${\mathcal M}_{\,j}$
in (1.4),
$$ \begin{align*} \sum_{m\in{\mathcal M}}\sum_{k\mid m}\frac{(\log k)^{\ell}}{k} = \sum_{j=1}^J\sum_{k\in{\mathcal M}_{\,j}\setminus{\mathcal M}_{\,j-1}}\frac{(\log k)^{\ell}}{k}\sum_{\substack{m\in{\mathcal M}\\ k\mid m}} 1. \end{align*} $$
Note that
$k\in {\mathcal M}_{\,j}\setminus {\mathcal M}_{\,j-1}$
implies
$k\geqslant x^{(\,j-1)/J}$
. Therefore,
$$ \begin{align} \sum_{m\in{\mathcal M}}\sum_{k\mid m}\frac{(\log k)^{\ell}}{k} \geqslant(\log x)^{\ell} \sum_{j=1}^J\bigg(\frac {j-1}{J}\bigg)^{\ell} \sum_{k\in{\mathcal M}_{\,j}\setminus{\mathcal M}_{\,j-1}}\frac{1}{k}\sum_{\substack{m\in{\mathcal M}\\ k\mid m}} 1. \end{align} $$
For each i with
$0\leqslant i\leqslant J$
, we rewrite uniquely
$m=m_1m_2$
where
$P_+(m_1)\leqslant x^{i/J}$
and
$P_-(m_2)>x^{i/J}$
. Here,
$P_+(\cdot )$
and
$P_-(\cdot )$
denote the largest and smallest prime factors separately, with
$P_+(1)=P_-(1)=\infty $
for convenience. Then,
$$ \begin{align*} \sum_{k\in{\mathcal M}_{i}}\frac{1}{k}\sum_{\substack{m\in{\mathcal M}\\ k\mid m}} 1 =\sum_{m_1\in{\mathcal M}_{i}}\sum_{k|{m_1}}\frac{1}{k} \sum_{\substack{m_2\in{\mathcal M}\\ P_-(m_2)>x^{i/J}}}1. \end{align*} $$
For the sum over
$m_1$
,
$$ \begin{align*} \sum_{m_1\in{\mathcal M}_{i}}\sum_{k\mid m_1}\frac{1}{k} =\prod_{p\in{\mathcal M}_i}\sum_{\substack{m_1|p^{b-1}\\ k\mid m_1}} \frac{1}{k} =\prod_{p\leqslant x^{i/J}}\sum_{\nu=0}^{b-1} \frac{b-\nu}{p^\nu}. \end{align*} $$
For the sum over
$m_2$
,
$$ \begin{align*} \sum_{m_2\in{\mathcal M}\setminus{\mathcal M}_i}1=\prod_{x^{i/J}<p\leqslant x}b. \end{align*} $$
Therefore, we deduce that
$$ \begin{align*} \sum_{k\in{\mathcal M}_{i}}\frac{1}{k}\sum_{\substack{m\in{\mathcal M}\\ k\mid m}} 1 ={b^{\pi(x)}}\prod_{p\leqslant x^{i/J}}\sum_{\nu=0}^{b-1}\frac{1}{p^\nu}\bigg(1-\frac{\nu}{b}\bigg). \end{align*} $$
Note that
${b^{\pi (x)}}=|{\mathcal M}|$
. By Lemma 2.1,
$$ \begin{align} \sum_{k\in{\mathcal M}_{i}} \frac{1}{k} \sum_{\substack{m\in{\mathcal M}\\ k\mid m}} 1 = \frac iJ |{\mathcal M}| \bigg\{1+O\bigg(\frac{\log_2 x}{b} + \frac{J}{\log x}\bigg)\bigg\} {e}^{\gamma}\log x. \end{align} $$
In view of (2.2), by taking the difference of
${\mathcal M}_{\,j-1}$
and
${\mathcal M}_{\,j}$
, we obtain
$$ \begin{align*} \sum_{k\in{\mathcal M}_{\,j}\setminus{\mathcal M}_{\,j-1}}\frac{1}{k}\sum_{\substack{m\in{\mathcal M}\\ k\mid m}} 1 = \frac {|{\mathcal M}|}J \bigg\{1+O\bigg(\frac{J\log_2 x}{b} + \frac{J^2}{\log x}\bigg)\bigg\}{e}^{\gamma}\log x. \end{align*} $$
Inserting this into (2.1),
$$ \begin{align*} \sum_{m\in{\mathcal M}}\sum_{k\mid m}\frac{(\log k)^{\ell}}{k} \geqslant \frac{|{\mathcal M}|}J\sum_{j=1}^J \bigg(\frac {j-1}{J}\bigg)^{\ell} \bigg\{1+O\bigg(\frac{J\log_2 x}{b}+\frac{J^2}{\log x}\bigg)\bigg\}{e}^{\gamma}(\log x)^{\ell+1}, \end{align*} $$
where the implied constant is absolute.
Now Proposition 1.3 follows given that
$$ \begin{align*} \frac{1}{J} \sum_{j=1}^J \bigg(\frac {j-1}{J}\bigg)^{\ell} = \frac{1}{\ell+1}+O\bigg(\frac{1}{J}\bigg), \end{align*} $$
which is an easy consequence of the inequalities
$$ \begin{align*} \frac{1}{J}\sum_{j=1}^J\bigg(\frac {j-1}{J}\bigg)^{\ell} \leqslant \int_0^1u^{\ell} \,d u \leqslant \frac{1}{J}\sum_{j=1}^J\bigg(\frac {j-1}{J}\bigg)^{\ell}+\frac{1}{J}.\\[-48pt] \end{align*} $$
3 Proof of Theorem 1.1
We start with the following lemma, which helps approximate the derivatives of the Riemann zeta function by Dirichlet polynomials.
Lemma 3.1. For
$T\to \infty $
,
$T\leqslant t\leqslant 2T$
and
$\ell \leqslant (\log T)/(\log _2 T)$
,
$$ \begin{align*} (-1)^{\ell}\zeta^{(\ell)}(1+{i} t) = \sum_{n\leqslant T}\frac{(\log n)^{\ell}}{n^{1+{i} t}} + O((\log_2 T)^{\ell}), \end{align*} $$
where the implied constant is absolute.
Proof. This is [Reference Yang8, Lemma 1], where we have taken
$\sigma =1$
and
$\varepsilon =(\log _2 T)^{-1}$
as Yang did. See also [Reference Titchmarsh7, Theorem 4.11].
Proof of Theorem 1.1.
To employ the resonance method, we choose the same weight function
$\phi (\cdot )$
as that used by Soundararajan [Reference Soundararajan5, page 471]. Thus, let
$\phi (t)$
be a smooth function compactly supported in
$[1,2]$
, such that
$0\leqslant \phi (t)\leqslant 1$
always and
$\phi (t)=1$
for
$t\in (5/4,7/4)$
. Then the Fourier transform of
$\phi $
satisfies
$\widehat {\phi }(u)\ll _\alpha {|u|^{-\alpha }}$
for any integer
$\alpha \geqslant 1$
.
For sufficiently large T, we set
$$ \begin{align*}x=\frac{\log T}{3\log_2T}\quad\text{and}\quad b=\lfloor\log_2T\rfloor.\end{align*} $$
Furthermore, we take
$\mathcal P$
and
$\mathcal M$
as in (1.3). Note that
${\mathcal P}\leqslant \sqrt T$
by the prime number theorem. Then we define the resonator
$$ \begin{align*}R(t):=\sum_{m\in{\mathcal M}}m^{{i} t}.\end{align*} $$
Denote
$$ \begin{align*}M_1(R,T):=\int_{\mathbb R}|R(t)|^2\phi\bigg(\frac tT\bigg)\,d t,\end{align*} $$
$$ \begin{align*}M_2(R,T):=\int_{\mathbb R}(-1)^{\ell}\zeta^{(\ell)}(1+{i} t)|R(t)|^2\phi\bigg(\frac tT\bigg)\,d t.\end{align*} $$
Since
$ \mathrm {supp}(\phi )\subset [1,2]$
,
$$ \begin{align} Z^{(\ell)}(T)\geqslant\frac{|M_2(R,T)|}{M_1(R,T)}. \end{align} $$
For
$M_1(R,T)$
,
$$ \begin{align*} M_1(R,T)=\sum_{m,n\in{\mathcal M}}\int_{\mathbb R}\bigg(\frac mn\bigg)^{{i} t}\phi\bigg(\frac tT\bigg)\,d t =T\sum_{m,n\in{\mathcal M}}\widehat\phi(T\log(n/m)). \end{align*} $$
When
$m\neq n$
, the choice of
${\mathcal P}$
guarantees that
$|\kern-0.8pt\log (n/m)|\gg 1/\sqrt T$
, and consequently
$$ \begin{align} \widehat\phi(T\log(n/m))\ll\frac{1}{T^2}. \end{align} $$
Thus, the off-diagonal terms contribute
$$ \begin{align*} T\sum_{\substack{m,n\in{\mathcal M}\\ m\neq n}} \widehat\phi(T\log(n/m))\ll\frac{1}{T}|{\mathcal M}|^2. \end{align*} $$
Therefore,
$$ \begin{align} M_1(R,T)=T\widehat\phi(0)|{\mathcal M}|+O\bigg(\frac{1}{T}|{\mathcal M}|^2\bigg). \end{align} $$
For
$M_2(R,T)$
, by Lemma 3.1,
$$ \begin{align} M_2(R,T) & = \int_{\mathbb R}\bigg(\sum_{k\leqslant T}\frac{(\log k)^{\ell}}{k^{1+{i} t}}\bigg)|R(t)|^2\phi\bigg(\frac tT\bigg)\,d t+O((\log_2 T)^{\ell}M_1(R,T))\nonumber\\ & = T\sum_{k\leqslant T}\frac{(\log k)^{\ell}}{k}\sum_{m,n\in{\mathcal M}}\widehat\phi(T\log(kn/m))+O((\log_2 T)^{\ell}M_1(R,T)). \end{align} $$
Similar to (3.2), for
$kn\neq m$
,
$$ \begin{align*} \widehat\phi(T\log(kn/m))\ll\frac{1}{T^2}. \end{align*} $$
Consequently,
$$ \begin{align*} \sum_{k\leqslant T}\frac{(\log k)^{\ell}}{k}\sum_{\substack{m,n\in{\mathcal M}\\ kn\neq m}} \widehat\phi(T\log(kn/m)) \ll\frac{(\log T)^{\ell+1}}{T^2}|{\mathcal M}|^2. \end{align*} $$
Inserting this into (3.4),
$$ \begin{align*} M_2(R,T)=T\widehat\phi(0)\sum_{m\in{\mathcal M}}\sum_{k\mid m}\frac{(\log k)^{\ell}}{k} +O\bigg(\frac{(\log T)^{\ell+1}}{T}|{\mathcal M}|^2\bigg)+O((\log_2 T)^{\ell}M_1(R,T)). \end{align*} $$
Combining this with (3.1) and (3.3),
$$ \begin{align*} Z^{(\ell)}(T)\geqslant\frac{1}{|{\mathcal M}|}\sum_{m\in{\mathcal M}}\sum_{k\mid m}\frac{(\log k)^{\ell}}{k}+O((\log_2 T)^{\ell}). \end{align*} $$
By taking
$J=\lfloor \tfrac 12\log _3T\rfloor $
in Proposition 1.3, we deduce that
$$ \begin{align*} \frac{1}{|{\mathcal M}|}\sum_{m\in{\mathcal M}}\sum_{k\mid m}\frac{(\log k)^{\ell}}{k}\geqslant \frac{{e}^{\gamma}}{\ell+1} \{1+o(1)\}(\log x)^{\ell+1}. \end{align*} $$
The theorem follows by recalling that
$x=({\log T})/({3\log _2T})$
.
Acknowledgements
The authors appreciate the comments and suggestions provided by the reviewer. The authors would like to thank Professor Jie Wu for his suggestion to explore this subject and Daodao Yang for some discussion.
