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I think it is probably well known that, for every $1/2<\sigma\leq 1$, the function $1/\zeta(\sigma+it)$ is unbounded.

Yet, I cannot decide how deep this is. I imagine it could be proved using a universality argument, or by extending the $\sigma=1$ case by a convexity argument on the $\mu$-function, or perhaps directly by Kronecker's theorem, or maybe even Euler-MacLaurin, but I don't yet know.

Therefore I would like to know about proofs and, importantly, anything you would like to say about how "deep" this is. For example, can you explain that it follows from something quite trivial, or perhaps that it is relatively as weak/strong as something else?

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    $\begingroup$ I looked this up in Titchmarsh: The theory of the Riemann zeta-function, 2nd edition. A good reference seems to be Theorem 11.9, whose proof is less than 3 pages (an elementary but tricky argument using Kronecker's theorem). An explicit $\Omega$-result can be found in the notes for Chapter 8, with a reference to Montgomery: Extreme values of the Riemann zeta-function, Comment. Math. Helv. 52 (1977), 511-518. $\endgroup$
    – GH from MO
    Dec 2, 2013 at 13:36
  • $\begingroup$ Thanks GH, I'll look that up. I only remember seeing the special case $\sigma=1$ in Titchmarsh. $\endgroup$ Dec 2, 2013 at 14:46
  • $\begingroup$ Ah yes- A long time ago I had thought that it would follow from a Picard like argument, but I forgot about it! Indeed it is a tricky argument. I have a much simpler proof, and I am wondering if it would be appropriate to post it here? I would still like to hear people's thoughts on whether this implies / is implied by other facts. $\endgroup$ Dec 3, 2013 at 12:49
  • $\begingroup$ You should also check out the Balsubramanian-Ramachandra method which gives nice proofs of these results. Ramachandra's book "Mean value estimates and omega theorems for the Riemann zeta function" is available online: math.tifr.res.in/~publ/ln/tifr85.pdf $\endgroup$ Dec 3, 2013 at 15:23
  • $\begingroup$ @Kevin: It is not appropriate to post new work here. Use the arXiv, your blog, or some journal for that purpose (e.g. the Amer. Math. Monthly likes to publish new proofs of old theorems). $\endgroup$
    – GH from MO
    Dec 3, 2013 at 17:32

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If for some $\beta>1/2$, $1/\zeta(\beta+it)$ were bounded, then the function would certainly have no poles on this vertical line, and thus $\zeta(s)$ would have no zeros on the line. So perhaps the big picture answer to 'how deep is it?' is this: it's what keeps us from making any easy progress on the Riemann hypothesis.

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  • $\begingroup$ Absolutely. But I ask this in the way I do above because, on its own, it is not as strong as the Riemann hypothesis being false, and therefore I wonder if it can be seen to be equivalent to some weaker statement. It is actually true that Littlewood's bound $\zeta(s)=O(t^{\epsilon})$, $\sigma>1/2$, on RH, cannot be improved to an inequality valid for all $t>0$. This seems to be even worse than unboundedness to me. $\endgroup$ Dec 3, 2013 at 16:53
  • $\begingroup$ This should be $1/\zeta(s)=O(t^{\epsilon})$. $\endgroup$ Dec 3, 2013 at 18:06

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