Littlewood established that $2e^{\gamma} \geq \limsup_{t \to \infty} |\zeta(1+it)| / \log{\log{t}} \geq e^{\gamma}$, the lower bound unconditionally and the upper bound on RH. It now seems to be generally believed that the lower bound represents the truth, and even, in the most optimistic form, that quite possibly there is an absolute constant $C$ such that $|\zeta(1+it)| \leq e^{\gamma}(\log{\log{t}} + \log{\log{\log{t}}}+C)$ for all $t > 10$.

Suppose in these questions that we replace $|\zeta(1+it)|$ by $$ B(t) := \sup_k \Big| \sum_{n=1}^k n^{-1-it} \Big|, \quad \textrm{ resp. } \widetilde{B}(t):= \sup_{m,k} \Big| \sum_{n=m}^k n^{-1-it} \Big|, $$ the largest partial sums, respectively the largest subsum over all intervals.

Does Littlewood's RH result extend to $B$ or even $\widetilde{B}$? That is: should the exponential sums bound $\widetilde{B}(t) = O(\log{\log{t}})$ be possible to prove on RH, or what kind of bound is available in this uniformity? Should it be reasonable to expect the strongest possible bound $\widetilde{B}(t) \leq e^{\gamma}(\log{\log{t}} + \log{\log{\log{t}}}+C)$, for all $t$?

There are evident versions of this question for $L(1,\chi)$, $\frac{\zeta'}{\zeta}(1+it)$ and $\frac{L'}{L}(1,\chi)$. In the last of these, the strongest bound seems to tie well with the belief that the smallest quadratic non-residue mod $q$ is $\ll \log{q}\log{\log{q}}$ -- which it certainly implies, at least when $q$ is prime.


1 Answer 1


The short answer is yes, and this is treated explicitly for characters in the work of Granville and Soundararajan (the paper appeared in J. Amer. Math. Soc.). Their Theorem 2 gives that on GRH for $x\le q$ and a primitive character $\chi \pmod q$ one has $$ \Big| \sum_{n\le x} \chi(n) \Big| \ll \Psi(x, (\log q)^2 (\log \log q)^{20}), $$ where $\Psi(x,y)$ denotes the number of integers up to $x$ composed only of prime factors below $y$. From this and partial summation, it follows at once that for $x\le q$ $$ \Big| \sum_{n\le x} \frac{\chi(n)}{n} \Big| \ll \prod_{p\le (\log q)^2(\log \log q)^{20}} \Big(1- \frac{1}{p}\Big)^{-1} \ll \log \log q. $$ In fact their Theorem 2 is a bit more precise, and one should be able to get the Littlewood type constant also. This result will also extend to the $t$-aspect with minor changes, and establish the bound that you want.

Alternatively you could look at another paper of Granville and Soundararajan (also in JAMS, and also in the case of characters); the argument given there in Section 5 would give a slightly simpler approach to such results.

Here's a quick sketch proof based on the second approach. Let $|t|$ be large, and assume that $x\le |t|$ (for larger $x$ the partial sum simply approximates $\zeta(1+it)$). Let $y$ be a parameter to be chosen, and note that $n\le x$ is either $y$-smooth, or may be written as $n=mp$ where $p\ge y$ is the largest prime factor of $n$ (so that $m$ is $p$-smooth). Thus, with $P(m)$ denoting the largest prime factor of $m$,
$$ \sum_{n\le x} \frac{1}{n^{1+it}} = \sum_{\substack{n\le x\\ p|n \implies p\le y}} \frac{1}{n^{1+it}} + \sum_{m\le x/y} \frac{1}{m^{1+it}} \sum_{\max(y, P(m)) \le p \le x/m} \frac{1}{p^{1+it}}. $$ The first term in the RHS is simply bounded by $\prod_{p\le y} (1-1/p)^{-1}$. As for the second term, RH can be used (usual contour shift argument) to show that the sum over $p$ there is $$ \ll \frac{(\log |t|)}{\sqrt{y}}, $$ and bounding the sum over $m$ trivially, the second term is $$ \ll \frac{(\log |t|) \log x}{\sqrt{y}} \ll \frac{(\log |t|)^2}{\sqrt{y}}. $$ Now choosing $y=(\log |t|)^4$ gives a bound of $\ll \log \log |t|$ for your partial sums. With more care the product may be truncated at $y=(\log |t|)^{2+\epsilon}$. One would expect that truncation at $y=(\log |t|)^{1+\epsilon}$ (or even $y=C \log |t| \log \log |t|$) is the truth.

  • $\begingroup$ Dear Lucia: thanks a lot! Regarding the logarithmic derivative case I was again thinking primarily of the Dedekind zeta of a general number field $K$ with discriminant $D$. For $L := \zeta_K / \zeta$, while $(L'/L)(1)$ can be very negative in the case of growing degree and small discriminant, Ihara proves in ["On the Euler-Kronecker constant..."] that the same type of GRH upper bound holds, uniformly in the degree: $(L'/L)(1) \leq 2\log{\log{|D|}} + O(1)$. Here too the expected bound will presumably be $\log{\log{|D|}}+\log{\log{\log{|D|}}} + O(1)$... (continued), $\endgroup$ Jun 24, 2016 at 16:23
  • $\begingroup$ ...though it is less clear to me if the $O(1)$ should still be an absolute constant, as in the GRH bound, or must have a dependence on the degree. If it is absolute, and if this upper bound persists for all the partial sums, then the sum of $\Lambda(n)/n$ over the ideals $I \subset O_K$ of norm $n = N(I) \ll \log{|D|}\log{\log{|D|}}$ would be $\gg \log{\log{|D|}}$ with absolute implied constants, which is just what I'd need to improve Dobrowolski's bound on Mahler measures from $(\log{\log{d}}/\log{d})^3$ to $\gg 1/\log{d}$. $\endgroup$ Jun 24, 2016 at 16:27
  • $\begingroup$ Lucia (and @VesselinDimitrov) - How far down can we hope to go assuming RH but not GRH? Thm 14.9 of Titchmarsh (due to Littlewood) gives us an upper bound of $2 e^\gamma \cdot 6/pi^2$ instead of $2 e^\gamma$, assuming RH but not GRH, unless I am very mistaken. $\endgroup$ Oct 8, 2021 at 7:57
  • $\begingroup$ @HAHelfgott: Hi Harald, I don't understand your question. In my answer for the questions on $\zeta$, only RH is used. What exactly are you asking about? $\endgroup$
    – Lucia
    Oct 9, 2021 at 3:31
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    $\begingroup$ @HAHelfgott: No the constant is just that (basically the Euler product up to $(\log t)^2$). Take a look at the paper of Lamzouri, Li and Soundararajan. arxiv.org/pdf/1309.3595.pdf On page 4 you'll find an explicit bound stated (proved for characters in the paper because of the class number connectioon). $\endgroup$
    – Lucia
    Oct 9, 2021 at 16:24

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