
MHB Journeyman
#1
October 30th, 2013,
15:59
Introduction :
This thread is dedicated to discuss about HardyLittlewood'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 HardyLittlewood 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! 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 :
Balarka
.
Last edited by mathbalarka; October 30th, 2013 at 16:16.

October 30th, 2013 15:59
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MHB Journeyman
#2
October 31st, 2013,
08:53
Thread Author
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 HardyLittlewood eta has a very neat functional equation $ \displaystyle \pi^{s/2} \Gamma(s/2) \eta(s) = \pi^{(1s)/2} \Gamma((1s)/2) \eta(1s)$ 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 nontrivial lemma regarding this etaintegral :
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[s] = 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 wellknown theorem on integral meanvalue 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}$.
Last edited by mathbalarka; October 31st, 2013 at 09:47.