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A set of lecture notes I'm reading on Halasz's theorem makes the following statement in a proof, which I can't quite follow - I was hoping someone might be able to clear up what I'm missing:

Therefore, $Re(\sum_p p^{-s} = \log \log x + O_M(1)$, where $O_M$ denotes that the constant involved in the big O notation depends on $M$, and $s:= 1 + \frac{1}{\log x} +it$. Recall that $Re(\sum_p p^{-s}) = \log |\zeta(s)| +O(1)$; hence $|\zeta(s)| \geq c_M\log x$, some $c_M$.

I hopefully won't need to give any more context about the proof itself as I don't think it's relevant, but I can if needs be. The zeta referred to is the usual zeta function, and $x$ is usually some real $\geq 1$.

What I can't understand is the $\geq$ in the final line; it appears both in my handwritten notes and the lecturer's typed notes, so I'm fairly sure it's meant to be there. However, the deduction which seems to be being made is to equate $ \log \log x + O_M(1)= \log |\zeta(s)| +O(1)$ and then deduce the inequality; but wouldnt this give us (taking exponents) $|\zeta(s)|=\log x \cdot \exp(O_M(1))$? So then if anything surely that would give us an upper bound for $|\zeta|$, not a lower bound?

If it helps, I can try to provide more information to clarify. The lecture notes are online at http://www.dpmms.cam.ac.uk/~bjg23/primenumbers-2012.html courtesy of Ben Green (who I hope doesn't mind me linking to his page), namely on the penultimate page of chapter 2, though I'm not sure whether the notes are accessible outside the Cambridge domain.

Many thanks in advance for your help.

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Taking exponents you can just as easily arrive at $\log x = |\zeta(s)| \cdot \exp(O_M(1))$

which should provide the lower bound on $|\zeta(s)|$, I think

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Ah ok, yes that was silly of me. I wasn't really thinking of the big-O being a "2-sided" bound but I guess in this sense it is - thanks very much. –  Paul Jun 5 '12 at 16:32
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