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Let, $a \in \Bbb R$ be any real number with $a>1$. Define, $f:\Bbb R\to \Bbb R$ by $$f(x)=-\sum_{n=1}^\infty\frac{\sin (x\log n)}{n^a},\forall x\in \Bbb R.$$ I have managed to find a way to show it has a fixed point in $\Bbb R$ by Brower's fixed point theorem [which is obviously zero].

I want to show it is unique. I am unable to proceed with the known fixed point theorems. After some regular homework on that series, I have concluded that if we are able to show that, "$f(x)$ attains its upper bound/maximum after $x=\zeta(a)$ where $\zeta$- is the Riemann Zeta function then the fixed point will be unique [Not verified yet!]."

Loosely speaking, the problem is directly related to the fact that the function $f_a(x)=\operatorname{Im}(\zeta(a+ix)),x\in\Bbb R$ has a unique fixed point [precisely zero] for every $a>1$.

How should I proceed? Any related paper/note/answer is highly appreciated.

Thanks in advance!

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  • $\begingroup$ Note that $|f'(x)| \le \sum_{n=2}^\infty\frac{\log n}{n^a}$ so already for $a$ where the series is less than $1$ you are done (for example for $a \ge 2$ as at $a=2$ the series is $.937..$ $\endgroup$
    – Conrad
    Aug 7, 2019 at 3:02
  • $\begingroup$ Thanks I agree with it. But what about $a\in (1,2)$? $\endgroup$ Aug 7, 2019 at 4:38
  • $\begingroup$ That's the interesting part of course - actually i checked with Wolfram alpha and it seems $1.97...$ is where the series is $1$ (at $1.97$ it is slightly bigger); obviously you can absolutely bound $f$ too by a series, so you get bounds on possible fixed points $x(a)\ne 0$ that way depending on the sum of $\sum_{n=2}^\infty\frac{1}{n^a}$, so that's another reduction; $\endgroup$
    – Conrad
    Aug 7, 2019 at 11:39
  • $\begingroup$ hmm, that seems good. I am trying to find that. If you get anything pls inform me! $\endgroup$ Aug 7, 2019 at 12:41

1 Answer 1

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Is $(1,\infty)$ the whole vanishing set of $\Im(\zeta(s) - s), \Re(s) > 1$ ?

  • For $\Re(s)$ large enough it follows from that $|\zeta'(s)| < 1$.

  • For $\Re(s) = 1+\epsilon$ use that $\zeta(s) = \frac{1}{s-1}+O(\log (2+|\Im(s)|))$ uniformly on $\Re(s) > 1$ which restricts the possible zeros of $\Im(\zeta(s) - s)$ to a bounded domain where you can do the numerical checks

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  • $\begingroup$ Thanks. I would like to avoid numerical computations. Is there any other approach? $\endgroup$ Aug 7, 2019 at 4:57
  • $\begingroup$ Why would you want to avoid numerical checks ? They will give a proof.. $\endgroup$
    – reuns
    Aug 7, 2019 at 18:10
  • $\begingroup$ There is no particular reason for that. If you suggest I will do numerical checks. I just want to ask one last question. Can I do the following, $$\Im(\zeta(\sigma +it)-\sigma -it)=\Im(\frac {1}{\sigma -1+it}-\sigma -it+O(...))$$. Using that estimate of zeta function. $\sigma >1, t\in \Bbb R$ $\endgroup$ Aug 8, 2019 at 2:40
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    $\begingroup$ If you mean around the pole at $s=1$ then it is a simple pole so $\zeta(s)$ is locally bijective and its inverse maps the large reals to a single curve which in our case in $(1-\epsilon,1+\epsilon)$, thus the bounded domain to check is of the form $\Re(s) \in (1,a), \Im(s) \in (0,T), |s-1| > r$, using that $\zeta'(\sigma)> \zeta'(a), |\zeta''(s)| \le \zeta(1+r)$ you can rule out the neighborhood of $s\in (1,a)$ so the domain is now $\Re(s) \in (1,a), \Im(s) \in (b,T)$. If $|\Im(\zeta(s)-s)|$ doesn't get too small there then it is non-zero for all $\Re(s) > 1$ $\endgroup$
    – reuns
    Aug 8, 2019 at 4:47

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