Yes.

]]>In order to prove RH, it is sufficient (by Hurwitz theorem) to find a sequence of non-vanishing analytic functions in the critical strip which converge locally uniformly to the xi function in the critical strip.

]]>I have just submitted the Polymath15 paper to the arXiv and to Research in the Mathematical Sciences.

]]>For complicated calculations it may be better to use another web site, such as the polymath wiki, and then add a link here.

One issue though is that it appears that you are trying to use the Riemann-von Mangoldt formula. Currently this is only established (with effective constants) at t=0, however we will need also some effective version of this formula for positive t, and this has not yet been done.

]]>If , it is known that

Where denotes the set of indices for which

If for all , , then

Hence

where , and

Now observe that because the contribution of each with real to this sum is zero (since ), while the contribution of any with non-real is canceled(!) by the contribution of its conjugate.) This gives the basic inequality

In order to lower bound , define for all

where denotes the set of indices in for which . Hence for all

where

Hence

Now we extend the definition of to be an odd function for all real x by . It easy to verify that the number of indices in satisfies

with equality if are not for some . This gives the more explicit bound

For simplicity, we lower bound by

where

and

Since . Therefore

For sufficiently large , we may approximate this maximization problem by maximizing the function

The maximum is attained (approximately) for , giving the approximate lower bound

.

Remark: This elementary bounding method is crude, improving very slightly de Bruijn bound. Fortunately, there is a much better (nearly optimal) analytical bounding method, giving a much better differential inequality of the form

where the "main term" of the expression for can be calculated precisely and is very close to . I'll give the improved analytical bounding method and its improved de Bruijn bound for in my next comment.

]]>

where is a simple non-real zero of such that for all zeros of .

It follows that is a growing positive function of (which grows like ) and is lower bounded by a positive absolute constant. The improved differential inequality implies the new bound

Which is slightly better than de Bruijn bound. (the best constant is dependent on as well on any given lower bound on but, as remarked, it is still lower bounded by an absolute positive constant)

I hope to give the details of this (simple) derivation in a day or two.

]]>Note also that the coefficients (defined in (15)) grow faster than any power of . Does it means that the summands in the first sum of (14) are growing in absolute value?

*[Asymptotics corrected. is indeed increasing, but the expression in practice will be decreasing up to , though it will eventually diverge if we instead took . -T]*

Note however that this discussion partially reappear(!) (with more details) in the proof of theorem 1.5(iv) (page 61).

*[Explanation added, thanks – T.]*