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
contained in the region
(other compact regions may be considered, for instance an open rectangle). For any real number
, we can look at the function
on
. This is a holomorphic function on
, continuous on the closed disc
. What kind of functions arise this way? Bagchi proved the following (this is essentially Theorem 3.4.11 in his thesis):
Theorem. Let
denote the Banach space of holomorphic functions on
which are continuous on the closed disc. For
, define a probability measure
on
to be the law of the random variable
, where
is uniformly distributed on
. Then
converges in law, as
, to the random holomorphic function
,
where
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
![\zeta(s+it)=\prod_p (1-p^{-s-it})^{-1} \zeta(s+it)=\prod_p (1-p^{-s-it})^{-1}](https://s0.wp.com/latex.php?latex=%5Czeta%28s%2Bit%29%3D%5Cprod_p+%281-p%5E%7B-s-it%7D%29%5E%7B-1%7D&bg=ffffff&fg=000000&s=0)
in
, then we would be led to an attempt to understand the probabilistic behavior of the sequence
, viewed as a random variable on
with values in the infinite product
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
. In other words, the random sequence
converges in law to a sequence
of independent, uniform, random variables on the unit circle. Then it is natural to expect that
should converge to the random function
, which is obtained formally by replacing
by its limit
.
Bagchi’s proof is somewhat intricate, in comparison with this heuristic justification, especially if one notices that if
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
![\prod_p (1-x_pp^{-s})^{-it} \prod_p (1-x_pp^{-s})^{-it}](https://s0.wp.com/latex.php?latex=%5Cprod_p+%281-x_pp%5E%7B-s%7D%29%5E%7B-it%7D&bg=ffffff&fg=000000&s=0)
to a sequence
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
to the Haar measure on
. Roughly speaking, it goes as follows:
- One checks that the random Euler product
does exist (as an
-valued random variable), and that it has the Dirichlet series expansion
![Z(s)=\sum_{n\geq 1} X_nn^{-s} Z(s)=\sum_{n\geq 1} X_nn^{-s}](https://s0.wp.com/latex.php?latex=Z%28s%29%3D%5Csum_%7Bn%5Cgeq+1%7D+X_nn%5E%7B-s%7D&bg=ffffff&fg=000000&s=0)
converging for
almost surely, where
is defined as the totally multiplicative extension of
This is done as Bagchi did using fairly standard probability theory and elementary facts about Dirichlet series.
- One shows that
has polynomial growth on vertical lines for
. This is again mostly elementary probability with a bit of Dirichlet series theory.
-
Consider next smoothed partial sums of
, of the type
![Z^{(N)}(s)=\sum_{n\geq 1}X_n\varphi(n/N)n^{-s}, Z^{(N)}(s)=\sum_{n\geq 1}X_n\varphi(n/N)n^{-s},](https://s0.wp.com/latex.php?latex=Z%5E%7B%28N%29%7D%28s%29%3D%5Csum_%7Bn%5Cgeq+1%7DX_n%5Cvarphi%28n%2FN%29n%5E%7B-s%7D%2C&bg=ffffff&fg=000000&s=0)
where
is a compactly supported test function with
. Using again standard techniques (including Cauchy’s formula for holomorphic functions), one proves that
![\mathbf{E}(\sup_{s\in D}|Z(s)-Z^{(N)}(s)|)\ll N^{-\delta} \mathbf{E}(\sup_{s\in D}|Z(s)-Z^{(N)}(s)|)\ll N^{-\delta}](https://s0.wp.com/latex.php?latex=%5Cmathbf%7BE%7D%28%5Csup_%7Bs%5Cin+D%7D%7CZ%28s%29-Z%5E%7B%28N%29%7D%28s%29%7C%29%5Cll+N%5E%7B-%5Cdelta%7D&bg=ffffff&fg=000000&s=0)
for some
.
- One next shows that the smoothed partial sums of the zeta function
![\zeta^{(N)}(s)=\sum_{n\geq 1}\varphi(n/N)n^{-s} \zeta^{(N)}(s)=\sum_{n\geq 1}\varphi(n/N)n^{-s}](https://s0.wp.com/latex.php?latex=%5Czeta%5E%7B%28N%29%7D%28s%29%3D%5Csum_%7Bn%5Cgeq+1%7D%5Cvarphi%28n%2FN%29n%5E%7B-s%7D&bg=ffffff&fg=000000&s=0)
satisfy
![\mathbf{E}_T(\sup_{s\in D}|\zeta(s+it)-\zeta^{(N)}(s+it)|)\ll N^{-\delta}+NT^{-1} \mathbf{E}_T(\sup_{s\in D}|\zeta(s+it)-\zeta^{(N)}(s+it)|)\ll N^{-\delta}+NT^{-1}](https://s0.wp.com/latex.php?latex=%5Cmathbf%7BE%7D_T%28%5Csup_%7Bs%5Cin+D%7D%7C%5Czeta%28s%2Bit%29-%5Czeta%5E%7B%28N%29%7D%28s%2Bit%29%7C%29%5Cll+N%5E%7B-%5Cdelta%7D%2BNT%5E%7B-1%7D&bg=ffffff&fg=000000&s=0)
(the second term arises because of the pole), where
denotes the expectation with respect to the uniform measure on
. 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
in vertical strips to the right of the critical line. The standard proof of this uses the Cauchy inequality and the mean-value property
![\frac{1}{2T}\int_{-T}^T|\zeta(\sigma+it)|^2dt\to \zeta(2\sigma) \frac{1}{2T}\int_{-T}^T|\zeta(\sigma+it)|^2dt\to \zeta(2\sigma)](https://s0.wp.com/latex.php?latex=%5Cfrac%7B1%7D%7B2T%7D%5Cint_%7B-T%7D%5ET%7C%5Czeta%28%5Csigma%2Bit%29%7C%5E2dt%5Cto+%5Czeta%282%5Csigma%29&bg=ffffff&fg=000000&s=0)
for any fixed
with
. It is here that the bottleneck lies if one wishes to generalize Bagchi’s Theorem to any “reasonable” family of
-functions.
- Finally, we just use the definition of convergence in law: for any continuous bounded function
, we should prove that
![\mathbf{E}_T(f(\zeta_T))\to \mathbf{E}(f(Z)), \mathbf{E}_T(f(\zeta_T))\to \mathbf{E}(f(Z)),](https://s0.wp.com/latex.php?latex=%5Cmathbf%7BE%7D_T%28f%28%5Czeta_T%29%29%5Cto+%5Cmathbf%7BE%7D%28f%28Z%29%29%2C&bg=ffffff&fg=000000&s=0)
where
is the
-valued random variable giving the translates of
, and
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
is Lipschitz: there exists a constant
such that
![|f(g_1)-f(g_2)|\leq C\sup_{s\in D}|g_1(s)-g_2(s)| |f(g_1)-f(g_2)|\leq C\sup_{s\in D}|g_1(s)-g_2(s)|](https://s0.wp.com/latex.php?latex=%7Cf%28g_1%29-f%28g_2%29%7C%5Cleq+C%5Csup_%7Bs%5Cin+D%7D%7Cg_1%28s%29-g_2%28s%29%7C&bg=ffffff&fg=000000&s=0)
(this is hidden in standard references — e.g., Billingsley’s — in the proof that one may assume that
is uniformly continuous; the functions used to prove this are in fact Lipshitz…).
Now pick some parameter
, and write
,
where
![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)|), 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)|),](https://s0.wp.com/latex.php?latex=A_1%3D%7C%5Cmathbf%7BE%7D_T%28f%28%5Czeta_T%29%29%5Cto+%5Cmathbf%7BE%7D_T%28f%28%5Czeta_T%5E%7B%28N%29%7D%29%29%7C%5Cleq+C%5C+%5Cmathbf%7BE%7D_T%28%5Csup_%7Bs%5Cin+D%7D%7C%5Czeta%28s%2Bit%29-%5Czeta%5E%7B%28N%29%7D%28s%2Bit%29%7C%29%2C&bg=ffffff&fg=000000&s=0)
![A_2=|\mathbf{E}_T(f(\zeta_T^{(N)}))\to \mathbf{E}(f(Z^{(N)}))|, A_2=|\mathbf{E}_T(f(\zeta_T^{(N)}))\to \mathbf{E}(f(Z^{(N)}))|,](https://s0.wp.com/latex.php?latex=A_2%3D%7C%5Cmathbf%7BE%7D_T%28f%28%5Czeta_T%5E%7B%28N%29%7D%29%29%5Cto+%5Cmathbf%7BE%7D%28f%28Z%5E%7B%28N%29%7D%29%29%7C%2C&bg=ffffff&fg=000000&s=0)
![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)|). 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)|).](https://s0.wp.com/latex.php?latex=A_3%3D%7C%5Cmathbf%7BE%7D%28f%28Z%5E%7B%28N%29%7D%29%29%5Cto+%5Cmathbf%7BE%7D%28f%28Z%29%29%7C%5Cleq+C%5C+%5Cmathbf%7BE%7D%28%5Csup_%7Bs%5Cin+D%7D%7CZ%28s%29-Z%5E%7B%28N%29%7D%28s%29%7C%29.&bg=ffffff&fg=000000&s=0)
Fix
. For some fixed
big enough,
is less than
by Step 3, and
is at most
. For this fixed
,
tends to
as
tends to infinity because of the convergence in law of
to
— the sum defining the truncations are finite, so there is no convergence issue. So for all
large enough, we will get