Re: The proof of the Riemann hypothesis

Why would you have to say things like "Do you have a
reason or should I start thinking that perhaps this
is why the paper is ignored?
" The last sentence is particularly hurting. But, of
course, if the idea is proved to be wrong, I will
recommend you to ignore it.

Sorry if I hurt you. As I told you, I really hope there are no flaws in the argument. But I don't think you should be so sensitive to critics other than the one from a qualified referee. What matters what I or anyone has to say if the paper was revised by a referee? That's why for me it's more important that type of recognition than collecting nice comments in a newsgroup.

Well, I have revised the paper over the past week. In
the revised version, you will see more clearly that
we

1. first calculate the fraction AT (pardon my
capitalization) x = delta. In replying to your
comment, if we consider the double limiting process
(of x and y), then the first limit is x -> delta, and
the second is y -> infinity. The purpose, however, is
to calculate the fraction at x = delta, and to do so,
the first limit MUST be x -> delta.

2. In the revised version, it is briefly shown that
the fraction is continuous on some interval (delta -
epsilon, delta + epsilon) epsilon > 0. Well, if we
know that a function is continuous on an interval (a,
b) except possibly for c in (a, b) and is equal to a
finite value at c, then the whole argument will
proceed smoothly.

Not so fast...the continuity of the function is not enough to justify the interchange of limits, because the second limit is taken when y-->oo in the integral from 0 to y. So, in order to prove that you can really interchange limits there, you need something stronger than continuity in (a,b), you will need uniform continuity in the interval 0<y<oo. Continuity in (a,b) certainly implies uniform continuity in the same interval, but the argument fails if teh interval in question is infinite, as happens to be the case here.

So I really think there's a problem here, unless it can be proven uniform continuity on (0,oo) in some other way.


Hisanobu
.

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