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Hardy & Littlewood's Result

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Introduction :

This thread is dedicated to discuss about Hardy-Littlewood's estimate of the \(\displaystyle N_0(T)\), i.e., the number of critical zeros of the Riemann zeta function with imaginary part smaller than \(\displaystyle T + 1\). The final result of Hardy-Littlewood estimate shows that there are infinitely many zeros of zeta that lies in the critical line.

This thread would be continued in more than one post, each of them showing different lemmas. The estimated number of post required would be 8 or less. There could be prolonged gaps in between the posts which are to be continued, so it would certainly be disappointing not to be able to look at it all at once; but I think a serious subject like this could take days to digest, so go slow and you won't throw up! :D Nevertheless, I think this note would be of particular interest once completed.

And as a final note, I would be putting it up altogether in a very short, less elegant and more understandable format. I have put quite a few things from here and there, so there could possibly be many typos. I hope members of MHB would certainly help me to point those out.

The commentary thread for this note is here : http://mathhelpboards.com/commentary-threads-53/commentary-hardy-littlewoods-result-7375.html

Balarka
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mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Introducing The Lemmas (1) :

First, we introduce a transform of zeta at the half line, i.e., the eta function -

\(\displaystyle \eta(s)= \frac{\pi^{-1/4 - is/2} \, \Gamma(1/4 + is/2)}{\left | \pi^{-1/4 - is/2} \, \Gamma(1/4 + is/2) \right |} \zeta\left (\frac{1}{2} + i u \right )\)

Note that the Hardy-Littlewood eta has a very neat functional equation \(\displaystyle \pi^{-s/2} \Gamma(s/2) \eta(s) = \pi^{-(1-s)/2} \Gamma((1-s)/2) \eta(1-s)\) which implies that eta is also even.

The first integral we are interested in is \(\displaystyle I_1(t) = \int_{t}^{t + \Delta} \left | \eta(s) \right | \, \mathrm{d}s\). We preasent a non-trivial lemma regarding this eta-integral :

Proposition : There exists a function \(\displaystyle K(t, \Delta)\) such that \(\displaystyle I_1(t) \ge \Delta - K(t, \Delta)\) and \(\displaystyle \int_{T}^{2T} \left | K(t, \Delta) \right |^2 \,\, \mathrm{d}t \ll T\) where \(\displaystyle T \ge \Delta^2 \ge 1\)

The first portion can be proved very easily, by elementary integral analysis :

\(\displaystyle \begin{aligned}I_1(t) = \int_{t}^{t + \Delta} \left | \eta(s) \right | \, \mathrm{d}s &= \int_{0}^{\Delta} \left | \zeta(1/2 + it + iu) \right | \, \mathrm{d}u\\ &\ge \left | \int_{0}^{\Delta} \zeta(1/2 + it + iu) \, \mathrm{d}u \right |\\ &\ge \Delta - \left | \int_{0}^{\Delta} (\zeta(1/2 + it + iu) - 1) \, \mathrm{d}u \right |\\ &= \Delta - K(t, \Delta) \end{aligned}\)

Thus, proving the first part. We now proceed to prove the second part, which is considerably harder than the former -

We choose the formula \(\displaystyle \zeta(s) = \sum_{n \leq T} \frac{1}{n^s} + \mathcal{O} \left (T^{-1/2} \right )\) for \(\displaystyle \Re = 1/2\) to estimate \(\displaystyle K(t, \Delta)\) as

\(\displaystyle K(t, \Delta) = \left | \sum_{n \leq T} n^{\frac{-1}{2} - it} \frac{1 - n^{-i \Delta}}{\log n} \right | + O(\Delta T^{-1/2})\)

Now we use a well-known theorem on integral mean-value estimates over Dirichlet polynomials. The proof can be easily worked out by using tools of integral analysis and asymptotic analysis, thus we do not omit it here :

Theorem : For any complex sequence \(\displaystyle a_n\), \(\displaystyle \int_{0}^{T} \left | \sum_{n \leq N} a_n n^{i t} \right |^2 \,\, \mathrm{d}t = (T + \mathcal{O}(N)) \sum_{n \leq N} | a_n |^2\)

By a little tweaking of the above formula, we get the desired result. In fact, we even have the asymptotic constant \(\displaystyle \sum_{n \leq T} \frac{1}{n (\log n)^2}\).
 
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