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Assume that $f(z)$ is holomorphic in $\{z=x+iy:\ 0\leq x\leq 1,\ t>2\}$ and it satisfies $|f(x +it)| = O(t^{(1-x)/2}\log t)$ for $0 \le x \le1$. In particular, $|f(it)| = O(t^{1/2}\log t)$. Is it true that $|f(x +it)| \le |f(it)|$ for $t$ sufficiently large and all $0\leq x\leq 1$?

Your answer is much appreciated.

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  • $\begingroup$ The MO website is for questions of mathematical research. Is there a research angle to your question? $\endgroup$
    – Gerry Myerson
    Commented Mar 1, 2016 at 21:53
  • $\begingroup$ Yes: to show that for the Riemann Zeta function in the critical strip: modulus[zeta(x+it) <= modulus[Zeta(it)]. It seems that there is no proof of this relation in the literature, although the relation can be proved for x = 1. Maybe this relation can be directly proved. I believe it will complement the behavior of the Zeta function in the critical strip, thus the research angle to the question. I researched, asked a lot of people and tried to prove it, but I found no proof so far. This inequality is supported by numerical results. $\endgroup$
    – Hass Saidane
    Commented Mar 1, 2016 at 22:04
  • $\begingroup$ Numerical/graphical support can be found in: dml.cz/bitstream/handle/10338.dmlcz/136881/… page 150 $\endgroup$
    – Hass Saidane
    Commented Mar 1, 2016 at 22:13
  • $\begingroup$ OK. I have formatted the question in TeX. I hope I haven't introduced any mistakes. Please edit into the body of your question the explanation of the Zeta motivation – people shouldn't have to go to the comments to see where a question is coming from. And please do it using TeX. $\endgroup$
    – Gerry Myerson
    Commented Mar 1, 2016 at 22:15
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    $\begingroup$ I edited your question to avoid easy conterexamples. $\endgroup$
    – GH from MO
    Commented Mar 2, 2016 at 23:22

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Based on your comment below the original post, you are mainly interested in the case of $f(s)=\zeta(s)$. Let me restrict to that case.

Then, for a given $t$, your inequality is implied by Conjecture 1 in Filip Saidak and Peter Zvengrowski, On the modulus of the Riemann zeta function in the critical strip, Math. Slovaca 53 (2003), no. 2, 145-172 (MR1986257), using the main result of this paper. Note that this conjecture is stated for $t\geq 2\pi +1$, and in fact it fails for $t$ small. For example, $|\zeta(0)|<1<|\zeta(0.5)|$ and $|\zeta(6i)|<0.9<|\zeta(0.5+6i)|$.

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    $\begingroup$ @HassSaidane: Well, then you should state your question that way. Note also that without any analycity assumption, your function can be pretty arbitrary. $\endgroup$
    – GH from MO
    Commented Mar 1, 2016 at 22:43
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    $\begingroup$ @HassSaidane: The inequality fails for $f(s)=\zeta(s)$ and $t=6$. See my updated response. Also, it is not clear what you mean by "Big O equality". Your question is about monotonicity, it has nothing to do with "Big O". $\endgroup$
    – GH from MO
    Commented Mar 1, 2016 at 22:45
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    $\begingroup$ @HassSaidane: Yes, but your question makes no assumption on $t$. Note that for $f(s)=\zeta(s/100)$, say, you need to make $t>600$. So please try to ask the question as you mean it, because as things stand I answered it in the negative. $\endgroup$
    – GH from MO
    Commented Mar 1, 2016 at 22:49
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    $\begingroup$ @HassSaidane: The authors mention that their condition $t\geq 2\pi +1$ can be improved, e.g. to $t>7$. So it is not analytically based, it was just convenient for them in certain steps of their proof. $\endgroup$
    – GH from MO
    Commented Mar 2, 2016 at 13:44
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    $\begingroup$ @HassSaidane: Their Theorem 1 implies that $\max_{0\leq x\leq 1}|\zeta(x+it)|=\max_{0\leq x\leq 1/2}|\zeta(x+it)|$, and their Conjecture 1 implies that $\max_{0\leq x\leq 1/2}|\zeta(x+it)|=|\zeta(it)|$. Putting these two statements together, $\max_{0\leq x\leq 1}|\zeta(x+it)|=|\zeta(it)|$, and this is what you asked. I assume $t\geq 2\pi+1$ of course. $\endgroup$
    – GH from MO
    Commented Mar 2, 2016 at 21:55

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