Tauberian theorems and the PNT

 A common theme in number theory is to study the properties of a sequence $(a_i)_{i=0}^{\infty}$ via the properties of its associated Dirichlet series $\sum_{i=0}^{\infty}\frac{a_i}{i^s}$. A famous example of such a series is the Riemann zeta function $\zeta(s)$ associated to the constant sequence $a_i=1$ for all $i$. Recall that a function $f: \mathbb{N} \to \mathbb{C}$ is said to be multiplicative if $f(mn)=f(m)f(n)$ whenever $m,n$ are coprime. Multiplicativity of a sequence is then reflected in the existence of an Euler product: $\sum_{i=0}^{\infty}\frac{a_i}{i^s}= \prod_p \sum_{i=0}^{\infty}\frac{a_{p^i}}{p^{is}}$. 

Some undergraduate analysis proves that if such series diverges somewhere and converges somewhere, then there is some real number $\sigma$ such that the series converges if and only if $Re(s) > \sigma$. This region of the complex plane is called the abscissa of convergence. One can then hope to gain information about the growth of the sequence based on the value of $\sigma$, e.g. for a constant sequence $\sigma=1$. Such theorems are usually called Tauberian theorems, and have been a workhorse of analytic number theory. See here for a recent survey which, in addition to being much more expository, attempts to highlight the connections to number theory and algebraic geometry, since older texts usually emphasise the analytic aspects. Below we discuss a relatively simple example.

Theorem Let $f(t)$, defined for $t \geq 0$, be a bounded and locally integrable function, and suppose that the function

\[g(z) = \int_0^\infty f(t) e^{-zt} \, dt \quad (\text{Re}(z) > 0)\]

extends holomorphically to $\text{Re}(z) \geq 0$. Then

\[\int_0^\infty f(t) \, dt\]

exists (and equals $g(0)$).

Proof: For $T > 0$, set

\[g_T(z) = \int_0^T f(t) e^{-zt} \, dt.\]

This is holomorphic for all $z$. We must show that

\[\lim_{T \to \infty} g_T(0) = g(0).\]

Let $R$ be large and let $C$ be the boundary of the region

\[\left\{ z \in \mathbb{C} : |z| < R, \ \text{Re}(z) \geq -\delta \right\},\]

where $\delta > 0$ is small enough (depending on $R$) so that $g(z)$ is holomorphic in and on $C$. Then, by Cauchy's theorem:

\[g(0) - g_T(0) = \frac{1}{2\pi i} \int_C (g(z) - g_T(z)) e^{zT}\frac{1 + z^2/R^2}{z} \, dz.\]

Let $B = \sup_{t \geq 0} |f(t)|$. On the semicircle $C_+ = C \cap \{ \text{Re}(z) \geq 0 \}$, the integrand is bounded by $2B/R^2$, because

\[|g(z) - g_T(z)| = \left| \int_T^\infty f(t) e^{-zt} dt \right| \leq B \int_T^\infty e^{-\text{Re}(z)t} dt = \frac{Be^{-\text{Re}(z)t}}{\text{Re}(z)},\]

and one can compute the modulus of the other terms in the integral exactly. Therefore, the contribution to $g(0) - g_T(0)$ from the integral over $C_+$ is bounded in absolute value by $B/R$.

For the integral over $C_- = C \cap \{ \text{Re}(z) < 0 \}$, we analyze $g(z)$ and $g_T(z)$ separately. Since $g_T$ is entire, we can replace the path of integration for $g_T$ by the semicircle

\[C'_- = \{ z \in \mathbb{C} : |z| = R, \ \text{Re}(z) < 0 \},\]

and estimate the integral over $C'$ as bounded by $2\pi B/R$ by the same method as above.

Finally, the remaining integral over $C_-$ tends to $0$ as $T \to \infty$ because the integrand is the product of the function

\[\frac{g(z)(1 + z^2/R^2)}{z},\]

which is independent of $T$, and the factor $e^{zT}$, which goes to $0$ rapidly and uniformly on compact sets as $T \to \infty$ in the half-plane $\text{Re}(z) < 0$. Hence,

\[\limsup_{T \to \infty} |g(0) - g_T(0)| \leq \frac{2B}{R},\]

and since $R$ is arbitrary, this proves the theorem. $\blacksquare$

Often mathematicians are interested in proving an analytic continuation a little bit beyond the abscissa of convergence for such series because this would give a lot of arithmetic data. For example, the crackpot statement, popular on the internet, that $1+ 1/2 +1/3 \cdots = -1/12$ is really the statement that the analytic continuation of $\zeta(s)$ at 1 satisfying nice properties like a functional equation of some sort takes the value $-1/12$ at $s=1$. In fact $\zeta(s)$ has a simple pole at 1, and $\zeta(s)-\frac{1}{s-1}$ admits a holomorphic continuation in $\text{Re}(s)>0$. Check that the following expression is well-defined: $\sum_{n=1}^{\infty} \int_n^{n+1} (\frac{1}{n^s}-\frac{1}{x^s} dx$.

To illustrate an example of how Tauberian theorems can be used to prove important results in number theory, we discuss how to use this one to prove the Prime Number Theorem (PNT), a crown jewel of 19th century mathematics. In what follows $f \ sim g$ means that the two functions are asymptotically equal, i.e. their ratio tends to 1 in the limit, and $f=O(g)$ means $f$ is asymptotically bounded above by a multiple of $g$. Denote by $\pi(x)$ the number of prime numbers at most $x$.

PNT: $\pi(x) \sim \frac{x}{\log x}$.

The original proof of this was long and intricate, but the argument we present below is due to Donald Newman. We introduce the following important functions in number theory:

\[\Phi(s)= \sum_{p} \frac{\log p}{p^s}, \vartheta(x) =\sum_{p \leq x} \log p  (s \in \mathbb{C}, x \in \mathbb{R}) \]

The strategy is to apply the Tauberian theorem above to the functions \[f(t) = \vartheta(e^t) e^{-t }-1, \quad g(z) = \frac{\Phi(z + 1)}{(z + 1)} -\frac{1}{z} \] 

to show that \(\int_0^\infty \frac{\vartheta(x)-x}{x^{2}} dx\) is a convergent integral. This implies, after a little bit of calculation that $\vartheta(x) \sim x$ and we deduce PNT from this as follows:

\[\vartheta(x) =  \sum_{p \leq x} \log p \leq \sum_{p < x} \log x = \pi(x) \log x,\]

\[\vartheta(x) \geq \sum_{x^{1-\epsilon} \leq p \leq x} \log p \geq \sum_{x^{1-\epsilon} \leq p \leq x} (1- \epsilon) \log x = (1-\epsilon)\log x (\pi(x) +O(x^{1-\epsilon})\]

Here's how to verify that $f,g$ satisfy the hypotheses:

 Lemma 1: \(\vartheta(x) = O(x)\)

Proof: For \(n \in \mathbb{N}\) we have

\[2^{2n} \geq \binom{2n}{n}\geq \prod_{n <p \leq 2n} p =e^{\vartheta(2n)-\vartheta(n)}\]

and hence, since \(\vartheta(x)\) changes by \(O(\log x)\) if \(x\) changes by \(O(1)\), it follows that

\[\vartheta(x) - \vartheta(x/2) < Cx\]

for any \(C > \log 2\) and all \(x \geq x_0 = x_0(C)\). Summing this over \(x, x/2, \dots, x/2^r\), where \(x/2^r \geq x_0 > x/2^{r+1}\), we obtain $\vartheta(x) < 2Cx + O(1).$ $\blacksquare$

Lemma 2: $(\zeta(s) \neq 0$, and \(\Phi(s) - \frac{1}{s-1}\) is holomorphic for \(\text{Re}(s) \geq 1\).

Proof: For \(\text{Re}(s) > 1\), the convergent Euler product for the Riemann zeta function implies that \(\zeta(s) \neq 0\) and that

\[-\frac{\zeta'}{\zeta}(s) = \sum_p \frac{\log p}{p^s - 1}.\]

The final sum converges for \(\text{Re}(s) > \frac{1}{2}\). Therefore this and the holomorphic extension of $\zeta$ imply that \(\Phi(s)\) extends meromorphically to \(\text{Re}(s) > \frac{1}{2}\), with poles only at \(s = 1\) and at the zeros of \(\zeta(s)\).

If \(\zeta(s)\) has a zero of order \(\mu\) at \(s = 1 + i\omega\) (for \(\omega \in \mathbb{R}\), \(\omega \neq 0\)), and a zero of order \(\nu\) at \(s = 1 + 2i\omega\) (\(\mu, \nu \geq 0\) by holomorphic extension), then

\[\lim_{\varepsilon \to 0} \varepsilon \Phi(1 + \varepsilon) = 1, \quad\lim_{\varepsilon \to 0} \varepsilon \Phi(1 + \varepsilon \pm i\omega) = -\mu, \quad\lim_{\varepsilon \to 0} \varepsilon \Phi(1 + \varepsilon \pm 2i\omega) = -\nu.\]

Let 

\[\sum_{r=-2}^2 \binom{4}{2+r} \Phi(1+ \varepsilon +ir \omega)= \sum_p \frac{\log p}{p^{1+\epsilon}} (p^{i\omega/2}+p^{-i\alpha/2})^4 \geq 0 \]

which then implies that $6 - 8\mu - 2\nu \geq 0 \Rightarrow \mu = 0,$ i.e., \(\zeta(1 + i\omega) \neq 0\).

Lemma 3 \(\int_0^\infty \frac{\vartheta(x)-x}{x^{2}} dx\) is a convergent integral.

Proof: For \(\text{Re}(s) > 1\), we have

\[ \Phi(s)= \sum_p \frac{\log p}{p^s} = \int_0^\infty \frac{d\vartheta(x)}{x^{s}}  = s \int_0^\infty \frac{\vartheta(x)}{x^{s+1}} dx=s\int_0^\infty e^{-st} t \vartheta(e^t) dt.\]

We conclude by applying the Tauberian theorem to \[f(t) = \vartheta(e^t) e^{-t }-1, \quad g(z) = \frac{\Phi(z + 1)}{(z + 1)} -\frac{1}{z} \]

which satisfy the theorem’s hypotheses by Lemmas 1 and 2. $\blacksquare$

These methods can be pushed a little further to show a strengthening of Dirichlet's theorem on arithmetic progressions. Denote by $\pi(x;q,a)$ the number of primes less than $x$ which are congruent to $a \mod q$. One can show that, if $(a,q)=1$, that $\pi(x;q,a) \sim \frac{Li(x)}{\varphi(q)}$. Philosophically, this is a statement that the primes behave randomly: they are uniformly distributed in every congruence class for every modulus that they could be. However, the various strengthenings of this statement, e.g. the second order term can be computed using the Siegel Walfsiz theorem, require serious analytic machinery, much more than what we have discussed here.