Commentary for "Hardy & Littlewood's Result"

[mathbalarka]

The topic is for discussion of the tutorial :

http://mathhelpboards.com/math-notes-49/hardy-littlewoods-result-7374.html

 

[mathbalarka]

I had been thinking about posting this tutorial for a while. I often hear from peoples that they don't even get what RH is basically saying, let alone the fact that they understand this major result in the theory of L-functions (an extension of this to L-forms exists, I think).

So I decided I would master myself in the theory of zeta functions first and then come back here to post the tutorial. Although there are a whole lot of things I don't understand even after a year of study, I think I am now quite capable of posting a good and explicit tutorial here.

Anyways, I don't think I would be able to get started too soon, for obvious reasons (many peoples at this site know my so called "obvious reasons" so I wont repeat and get frowned at )

Well, the upper bound on the date of the next post is 20th Nov, however, I believe I would get plenty of times before that, I am quite sure.

PS I know many peoples would be wondering then why I didn't start in in 20th Nov if I don't have time right now. The answer from me would be to keep our readers in a little suspense

Balarka
.

 

[ZaidAlyafey]

I will be interesting to see your tutorial ( disappointed it will take that long ) .Anyways , I am sure it will be so interesting for me as long as you don't include so much number theory stuff . It would be interesting if you post an outline of the topics you are going to discuss and the background needed for such tutorial .

 

[mathbalarka]

It would be interesting if you post an outline of the topics you are going to discuss and the background needed for such tutorial .

Actually, if you peer at Titchmarsh & Health-Brown, you'll see several complicated proof by ACing etas and applying Real and Complex analysis to it. The estimates are also gigantic. But the proof I am posting must be a refinement by Iwaniec & Kowalski; it's pretty neat, you know.

The basic things you will need is a very good understanding of the analytic continuation of zeta on the critical strip. Furthermore, you need to understand the functional equation of zeta, some basic estimates of zeta on the critical line. Along with these, you need to understand the basics of asymptotic analysis and integral analysis (the whole proof relies mostly on integral analysis and estimates).

PS After this tutorial is complete, I'd think about posting another tutorial which will consist of the Selberg's result (a positive proportion of the zeros of zeta lies in the 1/2-line) and merge those tutorials. Would it be a good idea?

 

[ZaidAlyafey]

PS After this tutorial is complete, I'd think about posting another tutorial which will consist of the Selberg's result (a positive proportion of the zeros of zeta lies in the 1/2-line) and merge those tutorials. Would it be a good idea?

Of course ! all these seem interesting topics !

 

[mathbalarka]

Now, about the approach, we would be considering the real-valued eta function instead of the zeta, i.e.

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

Next, we consider the integrals \(\displaystyle I_0(z) = \int_{t}^{t + \Delta} \eta(s) \, \mathrm{d}s\) and \(\displaystyle I_1(z) = \int_{t}^{t + \Delta} \left | \eta(s) \right | \, \mathrm{d}s\) for some positive \(\displaystyle \Delta\)

The basic idea of the proof is to show that \(\displaystyle \left | I_0(t) \right | < I_1(t)\) infinitely often regardless of \(\displaystyle \Delta\); in fact, this is equivalent to show that there are infinitely many zeros of zeta which lies in in the 1/2-line.

 

[mathbalarka]

Hopefully, I had time before Nov 20 and posted a lemma already.

I have omitted a proof of a certain theorem related to the result. I was wondering whether to give it or not. What is the reader's opinion on it?

 

[ZaidAlyafey]

It would be so nice if you can sketch the proof of the eta representation using the zeta function also I think you have a typo in the functional equation it should be

\(\displaystyle \displaystyle \pi^{-s/2} \Gamma(s/2) \eta(s) = \pi^{-(1-s)/2} \Gamma((1-s)/2) \eta(1-s)\)

Right ?

 

[mathbalarka]

It would be so nice if you can sketch the proof of the eta representation using the zeta

I am not sure exactly what proof you are asking for. Could you elaborate?

 

[ZaidAlyafey]

I meant the relation

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

I think you are some how using the reflection formula of the Gamma function and \(\displaystyle f\cdot \overline{f}=|f|^2\)

 

[mathbalarka]

I meant the relation ...

It's not a relation, it's a definition. Hardy & Littlewood defined a function (Hardy Littlewood eta) which is related to zeta like that.

I think you are some how using the reflection formula of the Gamma function

Yes, I used the reflection formula of gamma and zeta to get a functional equation of H-L eta.

 

[mathbalarka]

The second result that'd be shown in the tutorial thread would require the use of convexity bounds for zeta. I'll post the proof of that particular non-trivial bound later as a note and also give a proof of subconvexity bound as it could be useful for Selberg's result, as well as Conrey's approach (*)

(*) : The theorem shows what Selberg's doesn't -- the precise proportion of zeros in the critical line. It can be shown that there are 40% of the zeros of zeta which lies on the 1/2-line. Note that to show Conrey's approach, one must go through Levinson's (33.3% of the zeros of zeta lies on the 1/2-line).

 

[mathbalarka]

I was thinking about posting it since 19-th Nov, but because of unfortunate events, the second one post is postponed till 24-th. I have an appointment with a professor at 24-th, so you know, preparing a bit by revising my skills (not confident).

I am leaving this notification here for an announcement.

Balarka
.