Thursday morning update: After many hours, I decided that there is a critical error in the otherwise cleverly constructed proof. On page 138 (discussing Lemma 3), second part, he says "whence the function converges absolutely" essentially for any \(z\) with a real positive part. But it seems he hasn't really established that (except for circular reasoning) because if RH is false, and it may be false, the numerator \(|\psi(e^t)-e^t|\) goes like \(e^{at}\) for some positive \(a\) and the region of convergence is shifted by \(a\). So the "absolute" part of the convergence isn't correctly proven, it seems to me. Maybe it's enough to prove the "ordinary" convergence but I suspect that there could be a similar error in the \(g_1\) part of Lemma 3, too. Apologies if I am making a mistake.Some people talk about the proof of "almost twin" prime integers separated by at most 70 million or something like that. I am not terribly excited by this result even if it is true. It's always more interesting to talk about somewhat promising proofs to the Riemann Hypothesis, not only because of the $1 million that will be given to the first person who solves the old puzzle.
Many people have thought that they had a proof but the candidate proofs have always failed so far. So you must understand it is extremely likely that we have another example of a failure here. But I am going to tell you, anyway. It would be great if some readers spend a sufficient time and energy by reading the paper. Please don't be repelled by the idiosyncratic Chinese English. Even I can recognize that it's not how a native speaker would formulate the ideas. ;-)吴豪聪
That's his real name. Today, Hao-cong [first name] Wu [surname] of China sent me his new paper with a somewhat strange title (linguistically)
Showing How to Imply Proving The Riemann Hypothesis (PDF full)published in the European Journal of Mathematical Sciences. How does the proof work?
It's likely that I won't quite reproduce everything that is needed for the proof in this blog entry even though I may try. Teaching things is the best way to learn them. ;-)
Wu elaborates upon some ideas initiated by Serge Lang, a famous mathematician. But that's the last comment about the sociological context. Now, let us look at the ideas which don't seem to require any esoteric new branches of mathematics.
The proof reduces the Riemann Hypothesis to a claim about the absolute convergence of an integral that is related to the Riemann \(\zeta\)-function in a simple way. Let's roll.
The function that Wu finds more convenient is called \(\psi(x)\), pronounce "psi of ex". It is related to the Riemann \(\zeta\)-function by the following identities\[
\eq{
\phi(s) &= -\frac{\zeta'(s)}{\zeta(s)} = \sum_{n=1}^\infty \frac{\Lambda(n)}{n^s} =\sum_p \frac{\log p}{p^s-1}=\\
&= s \int_1^\infty \frac{\psi(x)}{x^{s+1}}\dd x = \frac{s}{s-1}+s\int_1^\infty \frac{\psi(x)-x}{x^{s+1}}\dd x
}
\] where the sum over \(p\) goes over the primes \(2,3,5,\dots\). The first step you should be able to verify if you want to validate Wu's proof is that the identities above are satisfied if \(\psi(x)\) is defined as the manifestly convergent sum\[
\psi(x) = \sum_{p^m\leq x} \log p = \sum_{n\leq x} \Lambda(n)
\] where \(\Lambda(n)=\log p\) if \(\exists m\geq 1: \,n=p^m\) for a prime \(p\) and otherwise it is set to zero. Note that this \(\psi(x)\) is defined in such a way that for a large \(x\), it's expected to be very close to \(x\) because the "probability to be prime" \(1/\log x\) is cancelled by the factor \(\log p\) from the definition of \(\psi(x)\) – it's close enough already when we allow \(m=1\) only.
The second step is to realize that the presence of a zero or zeroes of \(\zeta(s)\) also implies (or would imply) a pole of \(\phi(s)\), the [minus] "logarithmic derivative of the \(\zeta\)-function", at the same location of the complex plane. To prove the Riemann hypothesis, it is sufficient to prove that \(\phi(s)\) has no poles for \[
\frac 12 \lt {\rm Re}(s) \lt 1
\] (in the "right half-strip", as I will call it) because the hypothetical "RH-violating" zeroes (and singularities) come in pairs symmetrically distributed relatively to the critical axis \(s=1/2+it\) for \(t\in\RR\). Note that \(\phi(s)\) has a pole (or would have a pole) even for a higher-order zero of \(\zeta(s)\).
The third step, and it's the only hard one, is to actually prove that one of the integrals involving \(\psi(x)\) used to calculate \(\psi(s)\) above\[
\int_1^\infty \frac{\psi(x)-x}{x^{s+1}}\dd x
\] is analytic in the right half-strip so it has no poles over there. Consequently, the \(\zeta\)-function has no zeroes in the right half-strip and, by the left-right symmetry, no zeroes in the left half-strip, either.
Wu reduces the claimed analyticity of the integral above to the absolute convergence (convergence even if the integrand is replaced by its absolute value) and uniform convergence (the speed of convergence may be taken to be \(\varepsilon\)-independent), \(\forall\varepsilon\gt 0\), of the integral\[
\int_1^\infty \frac{\psi(x)-x}{x^{3/2+\varepsilon}}\dd x.
\] It shouldn't be hard to see that the absolute and uniform convergence of the integral above (here) is enough for the analyticity of the previous integral, and therefore for the absence of the non-trivial zeroes. Note that the exponents \(s+1\) for \(s\) in the right half-strip and \(3/2+\varepsilon\) for a positive \(\varepsilon\) are the same objects.
So aside from the claims that should be straightforward, the beef of the proof should be the demonstration of the absolute and uniform convergence of the integral in the last displayed equation.
Note that Wu's approach is linked both to the "complex analytic" interpretation of the Riemann Hypothesis as well as the prime-integer-counting, "number-theoretical" interpretation. It's because sufficient experts know that the Riemann Hypothesis is equivalent to the statement\[
\forall \varepsilon\gt 0: \, \psi(x) = x+ O(x^{1/2+\varepsilon})
\] which says that if we accept that the probability for a "rough number \(x\)" to be a prime is \(1/\log(x)\), then the estimated number of primes up to \(n\) deviates from the actual one at most by a power law (that is producing the \(O(\dots)\) term above.
Proving the convergence
OK, so how does Wu want to prove the uniform and absolute convergence? He offers some introduction to the theory of functions of real and complex variables together with some lemmas that are not quite well-known and that may even be new. Finally, the proof boils down to the existence (for any \(s\) with a real positive part) of the Laplace transforms \(g_{1,2}(s)\) of a function called \(f_{1,2}(t)\) related to \(\psi(e^t)-e^t\) for the subscript \(1\) or its absolute value for the subscript \(2\).
If you quickly want to focus on claims related to the \(\zeta\)-function and ignore various theorems and lemmas about completely general functions and their convergence etc. (assuming that these things are harmless and perhaps known to you, explicitly or intuitively), you may find it helpful for me to say that only Theorem 5 (among 7 theorems) and Lemma 3 (among 3 lemmas) is what you want to read. If there is some circularity in Wu's argument (secretly assuming RH), it's probably somewhere in Theorem 5 or Lemma 3.
In particular, I believe that Theorem 5 contains the main trick that allows us to show the convergence in the right half-strip. This theorem claims the absence of poles (except for the \(s=1\) pole) of the function\[
\eq{
\Phi(s) &= \sum_p \frac{\log p}{p^s} = \phi(s)-\sum_p h_p(s),\\
|h_p(s)|&\leq B\frac{\log p}{|p^{2s}|}
}
\] On one hand, this capital \(\Phi(s)\) is shown to be rather close to the lowercase \(\phi(s)\), using an argument based on geometric series. On the other hand, the \(2s\)-th power of something appears in the difference between \(\Phi\) and \(\phi\) which makes \(\sum\log n/n^{2s}\) converge for \({\rm Re}(s)\geq 1/2+\delta\). So the coefficient \(2\) in \(2s\) here is the ultimate reason why the meromorphic character of \(\Phi(s)\) starts at \({\rm Re}(s)\gt 1/2\), how we get the one-half somewhere, and why the critical axis becomes a decisive boundary for the well-definedness of \(\phi(s)\), too.
I don't see any mistake so far but I haven't really devoured all the beef of the proof yet, either, so no complete confirmation from your humble correspondent yet. But it is apparently making more sense every minute!
See the previous TRF blog entries mentioning the Riemann Hypothesis.
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