# One line Proof of the Prime Number Theorem

Whenever I am not doing anything, I generally happen to see pages of some good Mathematical Institutes in India, so as to know more about the faculty members and see what they are working on.

While doing so, in one of the faculty webpage he has written this statement.

• The prime number theorem has an one-line proof : The Riemann zeta function does not vanish on the one line "real part of $s$ equals 1".

Can anyone explain more about this proof. Is he assuming the Riemann Hypothesis? I am curious to know.

• Right, this is one line, but it's not a proof. It's just one important ingredient in a multipage proof. (And it requires a proof!) – Hendrik Vogt Jan 10 '11 at 14:36
• Check out Shakarchi and Stein volume 2, chapter 7. – Matt Calhoun Jan 10 '11 at 16:27

The assertion you made about $\zeta(s)$ does not assume the Riemann hypothesis and can, in fact, be proved reasonably quickly using reasonably elementary manipulations of trigonometric functions. The actual deduction of the prime number theorem from this is not trivial, and can be done (edit: as Matt E observes below, this is not the only -- and, indeed, not the original -- way) by some sort of "Tauberian" theorem: in these notes by Ash (which are based off a short paper by Donald Newman), the argument is to express $\zeta'(s)/\zeta(s)$ as the "Mellin transform" (basically, a variant of the Laplace transform) of partial sums of the von Mangoldt function, denoted $\psi$. The point of the "Tauberian" theorem is that you can deduce properties about the original function from the Mellin transform: since we have some information about the Mellin transform (namely, because $\zeta$ is zero-free on the line $Re(s)=1$), we can deduce some information about $\psi$. This is enough to imply the prime number theorem.
• Dear Akhil, One doesn't need Tauberian theorems. E.g., if one reads Riemann's original paper, this is essentially the ony missing ingredient to deduce the PNT from what he writes, and it is the key to the arguments of Hadamard and de la Vallee Poussin. Basically, one starts with the Euler product, does some Fourier analysis to get explicit formulas for $\pi(x)$, or $\psi(x)$, or some other related counting function, in terms of the zeroes of $\zeta(s)$, and then uses the fact that the real parts of the zeroes are $< 1$ to conclude the correct asymptotic from the explicit formula. – Matt E Jan 10 '11 at 16:31
• @Matt E: Are you sure that Riemman's formulas are good enough to get the PNT from non-vanishing on $Re(s)=1$? My understanding is that non-vanishing shows that no single term of Riemman's formula is large enough to break PNT, but you either need a Tauberian theorem or a bound on the growth of zeta in the critical strip to show that all the terms together aren't large enough. That said, you are certainly morally right -- Riemann's paper makes it clear that the main obstacle in proving PNT would be showing there are no zeroes on $Re(s)=1$. – David E Speyer Sep 8 '11 at 16:39
• @David: Dear David, You are right; my statement is a (deliberate, for ease of exposition) oversimplification. My memory (from reading Edwards's book) is that Hadamard's argument involved proving no zeroes on the line $Re(s) = 1$, combined with an explicit formula with better convergence properties than Riemann's. Perhaps for the integral of $\psi(x)$ from $2$ ot $x$ ... ? Something similar, anyway. Going back from the resulting asymptotic to the desired asymptotic for $\pi(x)$ itself is a mild Tauberian result. And of course growth estimates on the $\zeta$-function itself are used ... – Matt E Sep 8 '11 at 21:13