Bagchi’s Theorem is a functional version of earlier results of Bohr and Jessen related to the statistical properties of the Riemann zeta function on a vertical line between the critical line and the region of absolute convergence. It seems that it is not as well-known as it could, partly because Bagchi proved it in his thesis, and did not publish a paper with this result (his only related paper explicitly states that he removed the probabilistic language that a referee did not like). It seems therefore useful to describe the result. I will then sketch the proof I gave last semester…
Consider an open disc $latex D$ contained in the region $latex 1/2<\mathrm{Re}(s)< 1$ (other compact regions may be considered, for instance an open rectangle). For any real number $latex t$, we can look at the function $latex \zeta_t\colon s\mapsto \zeta(s+it)$ on $latex D$. This is a holomorphic function on $latex D$, continuous on the closed disc $latex \bar{D}$. What kind of functions arise this way? Bagchi proved the following (this is essentially Theorem 3.4.11 in his thesis):
Theorem. Let $latex H$ denote the Banach space of holomorphic functions on $latex D$ which are continuous on the closed disc. For $latex T>0$, define a probability measure $latex \mu_T$ on $latex H$ to be the law of the random variable $latex t\mapsto \zeta_t$, where $latex t$ is uniformly distributed on $latex [-T,T]$. Then $latex \mu_T$ converges in law, as $latex T\to +\infty$, to the random holomorphic function
$latex Z(s)=\prod_{p}(1-X_pp^{-s})^{-1}$,
where $latex (X_p)$ is a sequence of independent random variables indexed by primes, all uniformly distributed on the unit circle.
This is relatively easy to motivate: if we could use the Euler product
$latex \zeta(s+it)=\prod_p (1-p^{-s-it})^{-1}$
in $latex D$, then we would be led to an attempt to understand the probabilistic behavior of the sequence $latex (p^{-it})_p$, viewed as a random variable on $latex [-T,T]$ with values in the infinite product $latex \widehat{U}$ of copies of the unit circle indexed by primes. This is a compact topological group, and the easy answer (using the Weyl criterion) is simply that this sequence converges to the Haar measure on $latex \widehat{U}$. In other words, the random sequence $latex (p^{-it})$ converges in law to a sequence $latex (X_p)$ of independent, uniform, random variables on the unit circle. Then it is natural to expect that $latex Z_t$ should converge to the random function $latex Z(s)$, which is obtained formally by replacing $latex (p^{-it})$ by its limit $latex (X_p)$.
Bagchi’s proof is somewhat intricate, in comparison with this heuristic justification, especially if one notices that if $latex D$ is replaced by a compact region in the domain of absolute convergence, then the same idea applies, and is a completely rigorous proof (one need only observe that the assignment of an Euler product
$latex \prod_p (1-x_pp^{-s})^{-it}$
to a sequence $latex (x_p)$ of complex numbers of modulus one is a continuous operation in the region of absolute convergence.)
The proof I give in my script tries to remain closer to the basic intuition, and is indeed less involved (it avoids both a use of the pointwise ergodic theorem that Bagchi required and any use of tightness or weak-compactness). It makes it easy to see exactly what arithmetic ingredients are needed, beyond the convergence in law of $latex (p^{-it})_p$ to the Haar measure on $latex \widehat{U}$. Roughly speaking, it goes as follows:
- One checks that the random Euler product $latex Z(s)$ does exist (as an $latex H$-valued random variable), and that it has the Dirichlet series expansion
$latex Z(s)=\sum_{n\geq 1} X_nn^{-s}$
converging for $latex \mathrm{Re}(s)> 1/2$ almost surely, where $latex (X_n)_{n\geq 1}$ is defined as the totally multiplicative extension of $latex (X_p).$ This is done as Bagchi did using fairly standard probability theory and elementary facts about Dirichlet series.
- One shows that $latex Z(s)$ has polynomial growth on vertical lines for $latex \mathrm{Re}(s)> 1/2$. This is again mostly elementary probability with a bit of Dirichlet series theory.
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Consider next smoothed partial sums of $latex Z(s)$, of the type
$latex Z^{(N)}(s)=\sum_{n\geq 1}X_n\varphi(n/N)n^{-s},$
where $latex \varphi$ is a compactly supported test function with $latex \varphi(0)=1$. Using again standard techniques (including Cauchy’s formula for holomorphic functions), one proves that
$latex \mathbf{E}(\sup_{s\in D}|Z(s)-Z^{(N)}(s)|)\ll N^{-\delta}$
for some $latex \delta>0$.
- One next shows that the smoothed partial sums of the zeta function
$latex \zeta^{(N)}(s)=\sum_{n\geq 1}\varphi(n/N)n^{-s}$
satisfy
$latex \mathbf{E}_T(\sup_{s\in D}|\zeta(s+it)-\zeta^{(N)}(s+it)|)\ll N^{-\delta}+NT^{-1}$
(the second term arises because of the pole), where $latex \mathbf{E}_T(\cdot)$ denotes the expectation with respect to the uniform measure on $latex [-T,T]$. This step is also in Bagchi’s proof, and is essentially the only place where a specific property of the Riemann zeta function is needed: one requires the boundedness on average of $latex \zeta(s)$ in vertical strips to the right of the critical line. The standard proof of this uses the Cauchy inequality and the mean-value property
$latex \frac{1}{2T}\int_{-T}^T|\zeta(\sigma+it)|^2dt\to \zeta(2\sigma)$
for any fixed $latex \sigma$ with $latex \sigma> 1/2$. It is here that the bottleneck lies if one wishes to generalize Bagchi’s Theorem to any “reasonable” family of $latex L$-functions.
- Finally, we just use the definition of convergence in law: for any continuous bounded function $latex f\colon H\to\mathbf{C}$, we should prove that
$latex \mathbf{E}_T(f(\zeta_T))\to \mathbf{E}(f(Z)),$
where $latex \zeta_T$ is the $latex H$-valued random variable giving the translates of $latex \zeta(s)$, and $latex Z$ is the random Dirichlet series. The minor tweak that is useful to notice (and that I wasn’t consciously aware of before) is that one may assume that $latex f$ is Lipschitz: there exists a constant $latex C$ such that
$latex |f(g_1)-f(g_2)|\leq C\sup_{s\in D}|g_1(s)-g_2(s)|$
(this is hidden in standard references — e.g., Billingsley’s — in the proof that one may assume that $latex f$ is uniformly continuous; the functions used to prove this are in fact Lipshitz…).
Now pick some parameter $latex N>0$, and write
$latex |\mathbf{E}_T(f(\zeta_T))-\mathbf{E}(f(Z))|\leq A_1+A_2+A_3$,
where
$latex A_1=|\mathbf{E}_T(f(\zeta_T))\to \mathbf{E}_T(f(\zeta_T^{(N)}))|\leq C\ \mathbf{E}_T(\sup_{s\in D}|\zeta(s+it)-\zeta^{(N)}(s+it)|),$
$latex A_2=|\mathbf{E}_T(f(\zeta_T^{(N)}))\to \mathbf{E}(f(Z^{(N)}))|,$
$latex A_3=|\mathbf{E}(f(Z^{(N)}))\to \mathbf{E}(f(Z))|\leq C\ \mathbf{E}(\sup_{s\in D}|Z(s)-Z^{(N)}(s)|).$
Fix $latex \varepsilon>0$. For some fixed $latex N=N_0$ big enough, $latex A_3$ is less than $latex \varepsilon$ by Step 3, and $latex A_1$ is at most $latex \varepsilon+N_0T^{-1}$. For this fixed $latex N_0$, $latex A_2$ tends to $latex 0$ as $latex T$ tends to infinity because of the convergence in law of $latex (p^{-it})$ to $latex (X_p)$ — the sum defining the truncations are finite, so there is no convergence issue. So for all $latex T$ large enough, we will get
$latex |\mathbf{E}_T(f(\zeta_T))\to \mathbf{E}(f(Z))|\leq 4\varepsilon.$