Chapter 7

We have seen in this section, that any DIRICHLET series $$ D(s)=\sum_{n=1}^{\infty} a_{n} n^{-s} $$ admits a (maximal) convergence half-plane of equation \(\sigma>\sigma_{0} .\) More exactly, this is a half-plane where the required convergence is an absolute convergence. Show that there also exists a maximal right half-plane for the ordinary convergence, $$ \left\\{s \in \mathbb{C} ; \quad \text { Re } s>\sigma_{1}\right\\} \quad\left(\sigma_{1} \geq-\infty\right) $$ if the series \(D(s)\) converges (possibly not absolutely) in at least one point. (From the simple convergence in a point \(\sigma+i t, \sigma, t \in \mathbb{R}\), one can deduce the simple convergence in all points \(\sigma^{\prime}+i t, \sigma^{\prime}, t \in \mathbb{R}\) with \(\sigma^{\prime}>\sigma .\) ) Then \(D(s)\) converges normally in this half-plane, and represents there an analytic function. The series does not converge for any \(s\) with \(\sigma<\sigma_{1}\). Hint. Use ABEL's partial summation technique. Supplement. If the DIFICHLET series \(D(s)\) converges in at least one point, i.e. if \(\sigma_{1}<\infty\) exists, then it converges abeolutely in at least one point, i.e. \(\sigma_{0}<\infty\) exists, and we have the double inequality $$ \sigma_{0} \geq \sigma_{1} \geq \sigma_{0}-1 $$

Let \(D\) be a meromorphic function in the whole plane, which has finite order in any vertical strip, and which can be represented as a DIRICHLET series in a suitable right half-plane. We suppose that there exists a natural number \(k\), such that the following functional equation is satisfied. $$ R(s)=(-1)^{k} R(2 k-s) \text { with } R(s)=(2 \pi)^{-s} \Gamma(s) D(s) $$ We assume that \(D(s)\) is analytic excepting the point \(s=2 k\), which is either regular, or a simple pole. Show: In the case \(k=1\) the function \(D\) identically vanishes. In the cases \(k=2,3,4\) we have $$ D(s)=C C(s) \zeta(s+1-2 k), \quad C \in \mathbb{C} $$

The FOURIER coefficients of the normed EISENSTEIN series $$ \frac{(k-1) !}{2(2 \pi \mathrm{i})^{k}} G_{k}(z)=\sum_{n=0}^{\infty} a(n) e^{2 \pi \ln x}, \quad k \geq 4 $$ satisfy the equations (a) $$ \begin{aligned} a(n) a(m) &=a(n m), \quad \text { if }(n, m)=1 \\ a\left(p^{\nu+1}\right) &=a(p) a\left(p^{\nu}\right)-p^{k-1} a\left(p^{\nu-1}\right) \end{aligned} $$ Deduce from this $$ \begin{aligned} \sum_{n=1}^{\infty} a(n) n^{-s} &=\prod_{p} \sum_{\nu=0}^{\infty} a\left(p^{\nu}\right) p^{-\nu s} \\ &=\prod_{P} \frac{1}{\left(1-p^{-s}\right)\left(1-p^{k-1-s}\right)} \\ &=\zeta(s) \zeta(s+1-k) \text { for } \sigma>k \end{aligned} $$

Let \(D\) be a meromorphic function in the whole plane, which has finite order in any vertical strip, and which can be represented as a DIRICHLET series in a suitable right half-plane. We suppose that there exists a natural number \(r\), such that the following functional equation is satisfied, $$ R(s)=R(r / 2-s) \text { with } R(s)=\pi^{-s} \Gamma(s) D(s) $$ We assume that \(D(s)\) is analytic excepting the point \(s=r / 2\), which is either regular, or a simple pole. Show in the cases \(r<8\), that this DiriCHLET series is up to a constant factor of the shape $$ D_{r}(s)=\sum_{n=1}^{\infty} A_{r}(n) n^{-s} $$ where \(A_{r}(n)\) is the number of representations of \(n\) as a sum of \(r\) squares. In the case \(r=1\) we have \(D_{1}(s)=2 \zeta(2 s) .\) The DIRICHLET series \(D_{2}(s)\) can also be written in the form $$ \zeta_{K}(s)=D_{2}(s)=\sum_{a \in Z+12}|a|^{-2 s} $$

Let \(a: \mathbb{N} \rightarrow \mathbb{C}\) be an arbitrary sequence of complex numbers, and let $$ A(x):=\sum_{n \leq x} a(n) \quad(A(0)=0) $$ be the associated sum function. Then for any continuously differentiable function \(f:[x, y] \rightarrow \mathbb{C}, 0

Let \(p\) be a prime number. For any integer number \(\nu, 1 \leq \nu

The functional equation of the \(\zeta\)-function can also be written in the following form: $$ \zeta(1-s)=2(2 \pi)^{-s} \Gamma(s) \cos \left(\frac{\pi s}{2}\right) \zeta(s) $$ Deduce from this: In the half-plane \(\sigma \leq 0\), the function has \(\zeta(s)\) exactly the zeros \(s=-2 k, k \in \mathbb{N}\). All other zeros of the \(\zeta\)-function are located in the vertical strip \(0<\) Re \(s<1\).

Let $$ f(z)=\sum_{n=0}^{\infty} a(n) e^{2 \pi \ln z} $$ be a modular form of weight \(k\), and let $$ T(p) f(z)=\sum_{n=0}^{\infty} b(n) e^{2 \pi i n z} $$ be its image under \(T(p)\). We formally define $$ a(n):=0 \text { for non-integer numbers } n $$ Show: $$ b(n)=a(p n)+p^{k-1} a(n / p) $$

For \(\sigma>1\) it holds the following integral representation (B. RIEMANN, 1859) $$ \Gamma(s) \cdot \zeta(s)=\int_{0}^{\infty} \frac{t^{s-1} e^{-t}}{1-e^{-t}} d t $$

Let \(p_{n}\) be the \(n\).th prime number with respect to the natural order. Then the Prime Number Theorem VII. \(4.5\) is equivalent to $$ \lim _{n \rightarrow \infty} \frac{p_{n}}{n \log n}=1 $$

For \(|q|<1\) prove the identity $$ \sum_{n=-\infty}^{\infty}(-1)^{n} q^{\frac{n(3 n+1)}{2}}=\prod_{n=1}^{\infty}\left(1-q^{n}\right) $$

The operators \(T(p)\) map cusp forms to cusp forms. Because of this, the discriminant \(\Delta(z)\) is an eigenform of all \(T(p) .\) As a special case of Exercise 8 we obtain (for \(\sigma>7\) ) $$ \sum_{n=1}^{\infty} \frac{\tau(n)}{n^{s}}=\prod_{p} \frac{1}{1-\tau(p) p^{-s}+p^{11-2 s}} $$ Here, \(\tau(n)\) is the RAMANUJAN \(\tau\)-function, i.e. \(\tau(n)\) is the n.th FOURIER coefficient of \(\Delta /(2 \pi)^{12}\). The relations for \(\tau(n)\) reflected by the above product representation were conjectured by S. RAMANUJAN (1916), and proven by L.J. MORDELL (1917). The RAMANUJAN Conjecture, which was formulated in Exercise 1 of Sect. VII.1, is equivalent to the fact that both zeros of the polynomial $$ 1-\tau(p) X+p^{11} X^{2} $$ are complex conjugated.

(1) Let \(f \in[\Gamma, k]_{0}\) be a cusp form, let \(p\) be a prime number, and set \(\tilde{f}=T(p) f\). The functions $$ g(z)=|f(z)| y^{k / 2} \quad \text { and } \quad \tilde{g}(z)=|\tilde{f}(z)| y^{k / 2} $$ have maximal values \(m, \tilde{m}\) in \(\mathbb{H}\) (see Exercise 2 in VI.4). Show: $$ \tilde{m} \leq p^{\frac{k}{2}-1}(1+p) m $$ VII Analytic Number Theory We further suppose that \(f \not \neq 0\) is an eigenform of the HECKE operator \(T(p)\) with eigenvalue \(\lambda(p) .\) Show: $$ |\lambda(p)| \leq p^{\frac{k}{2}-1}(1+p) $$ (2) If, contrarily, \(f \in[\Gamma, k]\) is (modular but) not a cusp form, with the property \(T(p) f=\lambda(p) f\), then by Exercise 5 $$ \lambda(p)=1+p^{k-1} $$ Deduce from this (J. ELSTRODT, \(1984,[\mathrm{El}]\) ): The EISENSTEIN series \(G_{k}, k \geq 4, k \equiv 0 \quad\) mod 2, is \(u p\) to a constant factor the only non-cusp form, which is the eigenform of at least one HECKE operator.