In this section we describe the $L$-functions for which we will prove a multiplicity one result. As in other approaches to $L$-functions viewed from a classical perspective, such as that initiated by Selberg [Sel], we consider Dirichlet series with a functional equation and an Euler product. However, in contrast to Selberg, we strive to make all our axioms as specific as possible. Presumably (as conjectured by Selberg) these different axiomatic approaches all describe the same objects: $L(s,\pi)$ where $\pi$ is a cuspidal automorphic representation of $\GL(n)$. An interesing alternative is the recent approach of Booker [Boo], which describes $L$-functions in abstract terms, modeled on the explicit formula.
Before getting to $L$-functions, we recall two bits of terminology that will be used in the following discussion. An entire function $f:\C\to\C$ is said to have order at most $\alpha$ if for all $\epsilon > 0$:
\begin{equation} f(s)=\mathcal{O}(\exp(|s|^{\alpha + \epsilon})). \end{equation}Moreover, we say $f$ has order equal to $\alpha$ if $f$ has order at most $\alpha$, and $f$ does not have order at most $\gamma$ for any $\gamma < \alpha$. The notion of order is relevant because functions of finite order admit a factorization as described by the Hadamard Factorization Theorem, and the $\Gamma$-function and $L$-functions are all of order 1.
In order to ease notation, we use the normalized $\Gamma$-functions defined by:
\begin{equation} \Gamma_\R(s):=\pi^{-s/2}\,\Gamma(s/2)\ \ \ \ \text{ and }\ \ \ \ \Gamma_\C(s):=2(2\pi)^{-s}\,\Gamma(s). \end{equation}An $L$-function is a Dirichlet series
\begin{equation}L(s)=\sum_{n=1}^\infty \frac{a(n)}{n^s},\tag{2.1.1} \end{equation}where $s=\sigma+i t$ is a complex variable. We assume that $L(s)$ converges absolutely in the half-plane $\sigma>1$ and has a meromorphic continuation to all of $\C$. The resulting function is of order $1$, admitting at most finitely many poles, all of which are located on the line $\sigma = 1$. Finally, $L(s)$ must have an Euler product and satisfy a functional equation as described below.
The functional equation involves the following parameters: a positive integer $N$, complex numbers $\mu_1, \ldots, \mu_J$ and $\nu_1, \ldots, \nu_K$, and a complex number $\varepsilon$. The completed $L$-function
\begin{align} \Lambda(s) :=\mathstrut & N^{s/2} \prod_{j=1}^{J} \Gamma_\R(s+ \mu_j) \prod_{k=1}^{K} \Gamma_\C(s+ \nu_k) \cdot L(s)\tag{2.1.2} \end{align}is a meromorphic function of finite order, having the same poles as $L(s)$ in $\sigma>0$, and satisfying the functional equation
\begin{align} \Lambda(s)=\mathstrut& \varepsilon \overline{\Lambda}(1-s).\tag{2.1.3} \end{align}The number $d=J+2K$ is called the degree of the $L$-function.
We require some conditions on the parameters $\mu_j$ and $\nu_j$. The temperedness condition is the assertion that $\Re(\mu_j)\in\{0,1\}$ and $\Re(\nu_j)$ a positive integer or half-integer. With those restrictions, there is only one way to write the parameters in the functional equation, as proved in Proposition 2.1.1. This restriction is not known to be a theorem for most automorphic $L$-functions. In order to state theorems which apply in those cases, we will make use of a "partial Selberg bound," which is the assertion that $\Re(\mu_j), \Re(\nu_j) > -\frac12$.
The Euler product is a factorization of the $L$-function into a product over the primes:
\begin{equation} L(s)= \prod_p F_p(p^{-s})^{-1},\tag{2.1.4} \end{equation}where $F_p$ is a polynomial of degree at most $d$:
\begin{equation} F_p(z) = (1-\alpha_{1,p} z)\cdots (1-\alpha_{d,p} z).\tag{2.1.5} \end{equation}If $p|N$ then $p$ is a bad prime and the degree of $F_p$ is strictly less than $d$, in other words, $\alpha_{j,p}=0$ for at least one $j$. Otherwise, $p$ is a good prime, in which case the $\alpha_{j,p}$ are called the Satake parameters at $p$. The Ramanujan bound is the assertion that at a good prime $|\alpha_{j,p}|=1$, and at a bad prime $|\alpha_{j,p}| \le 1$.
The Ramanujan bound has been proven in very few cases, the most prominent of which are holomorphic forms on $\GL(2)$ and $\GSp(4)$. See [Sar] for a survey of what progress is known towards proving the Ramanujan bound. Also see [BB].
We write $|\alpha_{j,p}|\le p^\theta$, for some $\theta < \frac12$, to indicate progress toward the Ramanujan bound, referring to this as a "partial Ramanujan bound."
We will need to use symmetric and exterior power $L$-functions associated to a $L$-function $L(s)$. Let $S$ be the finite set of bad primes $p$ of $L(s)$. The partial symmetric and exterior square $L$-functions are defined as follows.
\begin{equation} L^S(s,\sym^n) = \prod_{p \not\in S}\: \prod_{i_1+\ldots+i_d=n} (1-\alpha_{1,p}^{i_1} \ldots \alpha_{d,p}^{i_d} p^{-s})^{-1}\tag{2.1.6} \end{equation} \begin{equation} L^S(s,\ext^n) = \prod_{p \not\in S}\; \prod_{1\leq i_1 < \ldots < i_n\leq d} (1-\alpha_{i_1,p} \ldots \alpha_{i_n,p} p^{-s})^{-1}.\tag{2.1.7} \end{equation}We do not define the local Euler factors at the bad primes since there is no universal recipe for these. It is conjectured that the symmetric and exterior power $L$-functions are in fact $L$-functions in the sense described above. In that case, Proposition 2.1.1 tells us that the bad Euler factors are uniquely determined. For applications that we present in this paper, the partial $L$-functions suffice.
In most cases it is not necessary to specify the local factors at the bad primes because, by almost any version of the strong multiplicity one theorem, an $L$-function is determined by its Euler factors at the good primes. For completeness we state a simple version of the result.
In the following proposition we use the term "$L$-function" in a precise sense, referring to a Dirichlet series which satisfies a functional equation of the form (2.1.2)-(2.1.3) with the restrictions $\Re(\mu_j)\in\{0,1\}$ and $\Re(\nu_j)$ a positive integer or half-integer, and having an Euler product satisfying (2.1.4)-(2.1.5). We refer to the quadruple $(d,N,(\mu_1,\ldots,\mu_J:\nu_1,\ldots,\nu_K),\varepsilon)$ as the functional equation data of the $L$-function.
Suppose that $L_j(s)=\prod_p F_{p,j}(p^{-s})^{-1}$, for $j=1,2$, are $L$-functions which satisfy a partial Ramanujan bound for some $\theta < \frac12$. If $F_{p,1}=F_{p,2}$ for all but finitely many $p$, then $F_{p,1}=F_{p,2}$ for all $p$, and $L_1$ and $L_2$ have the same functional equation data.
In particular, the proposition shows that the functional equation data of an $L$-function is well defined. There are no ambiguities arising, say, from the duplication formula of the $\Gamma$-function. Also, we remark that the partial Ramanujan bound is essential. One can easily construct counterexamples to the above proposition using Saito-Kurokawa lifts, which do not satisfy the partial Ramanujan bound.
Let $\Lambda_j(s)$ be the completed $L$-function of $L_j(s)$ and consider
\begin{align} \lambda(s)=\mathstrut&\frac{\Lambda_1(s)}{\Lambda_2(s)}\cr =\mathstrut& \Bigl(\frac{N_1}{N_2}\Bigr)^{s/2} \frac{\prod_{j} \Gamma_\R(s+ \mu_{j,1}) \prod_{k} \Gamma_\C(s+ \nu_{k,1})} {\prod_{j} \Gamma_\R(s+ \mu_{j,2}) \prod_{k} \Gamma_\C(s+ \nu_{k,2})} \prod_p \frac{F_{p,1}(p^{-s})^{-1}}{F_{p,2}(p^{-s})^{-1}}.\tag{2.1.8} \end{align}By the assumption on $F_{p,j}$, the product over $p$ is really a finite product. Thus, (2.1.8) is a valid expression for $\lambda(s)$ for all $s$.
By the partial Ramanujan bound and the assumptions on $\mu_j$ and $\nu_j$, we see that $\lambda(s)$ has no zeros or poles in the half-plane $\Re(s)>\theta$. But by the functional equations for $L_1$ and $L_2$ we have $\lambda(s) = (\varepsilon_1/\varepsilon_2)\overline{\lambda}(1-s)$. Thus, $\lambda(s)$ also has no zeros or poles in the half-plane $\Re(s) < 1-\theta$. Since $\theta < \frac12$, we conclude that $\lambda(s)$ has no zeros or poles in the entire complex plane.
If the product over $p$ in (2.1.8) were not empty, then the fact that $\{\log(p)\}$ is linearly independent over the rationals implies that $\lambda(s)$ has infinitely many zeros or poles on some vertical line. Thus, $F_{p,1}=F_{p,2}$ for all $p$.
The $\Gamma$-factors must also cancel identically, because the right-most pole of $\Gamma_\R(s+\mu)$ is at $-\mu$, and the right-most pole of $\Gamma_\C(s+\nu)$ is at $-\nu$. This leaves possible remaining factors of the form $\Gamma_\C(s+1)/\Gamma_\R(s+1)$, but that also has poles because the $\Gamma_\R$ factor cancels the first pole of the $\Gamma_\C$ factor, but not the second pole. Note that the restriction $\Re(\mu)\in\{0,1\}$ is a critical ingredient in this argument.
This leaves the possibility that $\lambda(s)=(N_1/N_2)^{s/2}$, but such a function cannot satisfy the functional equation $\lambda(s) = (\varepsilon_1/\varepsilon_2)\overline{\lambda}(1-s)$ unless $N_1=N_2$ and $\varepsilon_1=\varepsilon_2$.
In this section we state a version of strong multiplicity one for $L$-functions which is stronger than Proposition 2.1.1 because it only requires the Dirichlet coefficients $a(p)$ and $a(p^2)$ to be reasonably close. This is a significantly weaker condition than equality of the local factor.
Although the main ideas behind the proof appear in Kaczorowski-Perelli [KP] and Soundararajan [S], we give a slightly stronger version which assumes a partial Ramanujan bound $\theta < \frac16$, plus an additional condition, instead of the full Ramanujan conjecture. We provide a self-contained account because we also wish to bring awareness of these techniques to people with a more representation-theoretic approach to $L$-functions.
Suppose $L_1(s)$, $L_2(s)$ are Dirichlet series with Dirichlet coefficients $a_1(n)$, $a_2(n)$, respectively, which continue to meromorphic functions of order 1 satisfying functional equations of the form (2.1.2)-(2.1.3) with a partial Selberg bound $\Re(\mu_j), \Re(\nu_j)>-\frac12$ for both functions, and having Euler products satisfying (2.1.4)-(2.1.5). Assume a partial Ramanujan bound for some $\theta < \frac16$ holds for both functions, and that the Dirichlet coefficients at the primes are close to each other in the sense that
\begin{equation} \sum_{p\le X} p\,\log(p) |a_1(p)-a_2(p)|^2\ll X .\tag{2.1.9} \end{equation}We have $L_1(s)=L_2(s)$ if either of the following two conditions are satisfied
Note that condition (2.1.9) is satisfied if $|a_1(p)-a_2(p)|\ll 1/\sqrt{\mathstrut p}$, in particular, if $a_1(p)=a_2(p)$ for all but finitely many $p$, or more generally if $a_1(p)=a_2(p)$ for all but a sufficiently thin set of primes. In particular, $a_1(p)$ and $a_2(p)$ can differ at infinitely many primes. Also, by the prime number theorem [Ap, Theorem 4.4] in the form
\begin{equation} \sum\limits_{p < X} \log(p) \sim X,\tag{2.1.10} \end{equation}condition 2.1 for both $L$-functions implies condition 1.
The condition $\theta < \frac16$ arises from the $p^{-3s}$ terms in the proof of Lemma 2.2.2. Those terms do not seem to give rise to a naturally occuring $L$-function at $3s$, so it may be difficult to replace the $\theta < \frac16$ condition by a statement about the average of certain Dirichlet coefficients.
In this section we provide the lemmas required for the proof of Theorem 2.1.2. There are two types of lemmas we require. The first deals with manipulating Euler products and establishing zero-free half-planes via the convergence of those products. The second deals with possible zeros at the edge of the half-plane of convergence.
If $L(s)=\sum a(n)n^{-s}$ then for $\rho=\sym^n$ or $\ext^n$ we write
\begin{equation}L(s,\rho) = \sum_j a(j,\rho) \,j^{-s}.\tag{2.2.1} \end{equation}If $p$ is a good prime then
Let $p$ be a good prime. Expanding the Euler factor $L_p(s)$ for $L(s)$ we have
\begin{equation} L_p(s)= \prod_{j=1}^d \frac 1{(1-\alpha_{j,p} \, p^{-s})} = \prod_{j=1}^d \sum_{\ell=0}^\infty \alpha_{j,p}^\ell \, p^{-\ell s} = \sum_{\ell=0}^\infty p^{-\ell s} \!\!\! \sum_{n_1+\cdots+n_d=\ell} \alpha_{1,p}^{n_1}\cdots \alpha_{d,p}^{n_d}\, ,\tag{2.2.2} \end{equation}where the $n_j$ are restricted to non-negative integers. Expanding the Euler factor for $L^S(s,\sym^n)$ we have
\begin{align} L_p(s,\sym^n)=& \prod_{i_1+ \cdots +i_d=n} (1-\alpha_{1,p}^{i_1} \cdots \alpha_{d,p}^{i_d} p^{-s})^{-1}\cr =&\prod_{i_1+ \cdots +i_d=n} \sum_{\ell=0}^\infty \left(\alpha_{1,p}^{i_1} \cdots \alpha_{d,p}^{i_d}\right)^\ell p^{-\ell s}.\tag{2.2.3} \end{align}The coefficient of $p^{-s}$ in (2.2.3) is
\begin{equation}\sum_{i_1+ \cdots +i_d=n} \alpha_{1,p}^{i_1} \cdots \alpha_{d,p}^{i_d},\tag{2.2.4} \end{equation}which equals the coefficient of $p^{-ns}$ in (2.2.2).
The other identities in the lemma just involve expanding the definitions and checking particular coefficients.
The next lemma tells us that if there are zeros in the critical strip for $\sigma\geq \tfrac12$, those zeros come from Euler factors involving the coefficients $a(p)$ or $a(p^2)$ of the Dirichlet series or the Euler factors of the symmetric or exterior square $L$-functions.
Suppose
\begin{align}L(s) =\mathstrut & \sum_{n} a_n n^{-s} \cr =\mathstrut &\prod_{p\ \mathrm{bad}} \:\prod_{j=1}^{d_p} (1-\alpha_{j,p} p^{-s})^{-1} \prod_{p\ \mathrm{good}}\:\prod_{j=1}^d (1-\alpha_{j,p} p^{-s})^{-1},\tag{2.2.5} \end{align}where $|\alpha_{j,p}|\le p^\theta$ for some $\theta\in \R$. Then for $\sigma > 1+\theta$,
\begin{align} L(s)=\mathstrut & \prod_p (1+a(p) p^{-s}) \cdot \prod_p (1+a(p^2) p^{-2s}) \cdot h_0(s),\cr =\mathstrut & \prod_p (1+a(p) p^{-s}) \cdot L^S(2s,\sym^2) \cdot h_1(s),\cr =\mathstrut & \prod_p (1+a(p) p^{-s} + a(p)^2 p^{-2s} ) \cdot L^S(2s,\ext^2)^{-1} \cdot h_2(s),\tag{2.2.6} \end{align}where $h_j(s)$ is regular and nonvanishing for $\sigma > \frac13 + \theta$.
We can write the Euler product in the form
\begin{equation}L(s) = \prod_p \sum_{j=0}^\infty a(p^j) p^{-j s} ,\tag{2.2.7} \end{equation}where
\begin{equation} a(p^j) \ll_j p^{j\theta} \ll p^{j(\theta+\varepsilon)},\tag{2.2.8} \end{equation}with the implied constant depending only on $\varepsilon$. We manipulate the Euler product, introducing coefficients $A_j$, and $B_j$ where $A_j(p), B_j(p) \ll_j p^{j\theta} \ll p^{j(\theta+\varepsilon)}$. We have
\begin{align} L(s) =\mathstrut & \prod_p \sum_{j=0}^\infty a(p^j) p^{-j s} \cr =\mathstrut & \prod_p \Bigl(1+a(p)p^{-s}+ A_2(p) p^{-2s} + \sum_{j=3}^\infty A_j(p)p^{-js} \Bigr)\cr &\phantom{xxx}\times \Bigl(1+ (a(p^2) - A_2(p)) p^{-2s} + \sum_{j=2}^\infty B_{2j}(p)p^{-2js} \Bigr)\cr &\phantom{xxx}\times \bigl(1+ O(p^{3\theta}p^{-3s})\bigr) \\ =\mathstrut & F_1(s) F_2(s) F_3(s),\tag{2.2.9} \end{align}say. Note that by the assumptions on $A_j$ and $B_j$ we have
\begin{equation}\sum_{j=3}^\infty A_j(p)p^{-js} = O(p^{3\theta} p^{-3 s}) \ \ \ \ \ \text{ and } \ \ \ \ \ \sum_{j=2}^\infty B_{2j}(p)p^{-2js} = O(p^{4\theta} p^{-4 s}).\tag{2.2.10} \end{equation}Combining this with (2.2.8) justifies (2.2.9).
For the first assertion, set $A_j(p)=0$ and $B_j(p)=0$ and note that $F_1(s)$ converges absolutely for $\sigma>1+\theta$ and $F_2(s)$ converges absolutely for $\sigma>\frac12+\theta$, and $F_3(s)$ converges absolutely for $\sigma>\frac13+\theta$.
For the second assertion, set $A_j(p)=0$. For good primes $p$, choose $B_j(p)$ so that $F_2(s)= L^S(2s,\sym^2)$. For bad primes $p$, choose $B_j(p) = 0$. Note that (by the construction of the symmetric square) this choice of $B_j$ satisfies the required bounds. The finitely many factors at the bad primes together with $F_3(s)$ give $h_1(s)$.
For the third assertion, the only modification is to set $A_2(p)=a(p)^2$, $A_j(p)=0$ for $j\ge 3$, and use the second identity in Lemma 2.2.1 and appropriate choices for $B_j(p)$ so that $F_2(s)=L^S(2s,\ext^2)^{-1}$.
The absolute convergence of an Euler product in a half-plane $\sigma>\sigma_0$ implies that the function has no zeros or poles in that region. If the Euler product has a meromorphic continuation to a larger region, it could possibly have zeros or poles on the $\sigma_0$-line. The lemma in this section, which is standard and basically follows the proof of Lemma 1 of [KP], says that if the Dirichlet coefficients $a(p)$ are small on average then there are finitely many zeros or poles on the $\sigma_0$-line. Our modification is that we only require the $L$-function to satisfy a partial Ramanujan bound.
Note that the lemma is stated with $\sigma_0=1$ as the boundary of convergence. Applying the lemma in contexts with a different line of convergence, as in the proof of Theorem 2.1.2, just involves a simple change of variables $s \to s+A$.
Let
\begin{equation}L(s)=\prod_p \sum_{j=0}^\infty a({p^j}) p^{-j s}\tag{2.4.1} \end{equation}and suppose there exists $M_1, M_2 \ge 0$ and $\theta < \frac23$ so that $|a({p^j})|\ll p^{\theta j}$ and
\begin{align} \sum_{p\le X} |a(p)|^2 \log p \le\mathstrut & M_1^2 X + o(X), \\ \sum_{p\le X} p^{-2} |a(p^2)|^2 \log p \le\mathstrut & M_2^2 X + o(X).\tag{2.2.12} \end{align}Then $L(s)$ is a nonvanishing analytic function in the half-plane $\sigma>1$. Furthermore, if $L(s)$ has a meromorphic continuation to a neighborhood of $\sigma\ge 1$, then $L(s)$ has at most $(M_1 + 2 M_2)^2$ zeros or poles on the $\sigma=1$ line.
Note that, by the prime number theorem (2.1.10), the condition (2.2.12) on $a(p)$ is satisfied if $|a(p)|\le M_1$. Also, if $\theta < \frac12$ then condition (2.2.12) on $a(p^2)$ holds with $M_2=1$.
The proof of Lemma 2.2.3 is in Section 2.4.
Now we have the ingredients to prove Theorem 2.1.2. The proof begins the same as that of Proposition 2.1.1, by considering the ratio of completed $L$-functions:
\begin{equation} \lambda(s) := \frac{\Lambda_1(s)}{\Lambda_2(s)},\tag{2.3.1} \end{equation}which is a meromorphic function of order 1 and satisfies the functional equation $\lambda(s)=\varepsilon \overline{\lambda}(1-s)$, where $\varepsilon = \varepsilon_1/\varepsilon_2$.
$\lambda(s)$ has only finitely many zeros or poles in the half-plane $\sigma\ge\frac12$.
Assuming the lemma, we complete the proof of Theorem 2.1.2 as follows. By the functional equation, $\lambda(s)$ has only finitely many zeros or poles, so by the Hadamard factorization theorem
\begin{equation} \lambda(s) = e^{A s} r(s)\tag{2.3.2} \end{equation}where $r(s)$ is a rational function.
By (2.3.2), as $\sigma\to\infty$,
\begin{equation}\lambda(\sigma) = C_0 \sigma^{m_0} e^{A \sigma} \bigl(1 + C_1 \sigma^{-1} + O(\sigma^{-2})\bigr),\tag{2.3.3} \end{equation}for some $C_0\not=0$ and $m_0\in \Z$. On the other hand, if $b(n_0)$ is the first non-zero Dirichlet coefficient (with $n_0>1$) of $L_1(s)/L_2(s)$, then by (2.3.1) and Stirling's formula, as $\sigma\to\infty$,
\begin{equation}\lambda(\sigma) = \bigl(B_0 \sigma^{B_1} e^{B_2 \sigma\log \sigma + B_3 \sigma }(1 +o(1))\bigr)\bigl(1 + b(n_0) n_0^{-\sigma} + O((n_0+1)^{-\sigma}).\tag{2.3.4} \end{equation}Comparing those two asymptotic formulas, the leading terms must be equal, so $B_0=C_0$, $B_1=m_0$, $B_2=0$, and $B_3=A$. Comparing second terms, we have polynomial decay equal to exponential decay, which is impossible unless $b(n_0)=0$ and $C_1=0$. But $b(n_0)$ was the first nonzero coefficient of $L_1(s)/L_2(s)$, so we conclude that $L_1(s)=L_2(s)$, as claimed.\qed
\vspace{3ex} The rest of this section is devoted to the proof of Lemma 2.3.1. By (2.1.8) and the partial Selberg bound assumed on $\mu$ and $\nu$, only the product
\begin{equation} P(s)=\prod_p\frac{ F_{p,1}(p^{-s})^{-1} }{ F_{p,2}(p^{-s})^{-1} } = \prod_p\frac{ 1+a_1(p)p^{-s}+a_1(p^{2})p^{-2s}+\cdots} {1+a_2(p)p^{-s}+a_2(p^{2})p^{-2s}+\cdots} \end{equation}could contribute any zeros or poles to $\lambda(s)$ in the half-plane $\sigma\ge\frac12$.
By the first line in equation (2.2.6) of Lemma 2.2.2 we have
\begin{align} P(s) =\mathstrut & \prod_p \frac{1+a_1(p)p^{-s}}{1+a_2(p)p^{-s}} \cdot \prod_p \frac{1+a_1(p^2)p^{-2s}}{1+a_2(p^2)p^{-2s}} \cdot H_1(s)\cr =\mathstrut & A_1(s) H_1(s),\tag{2.3.5} \end{align}say, where $H_1(s)$ is regular and nonvanishing for $\sigma>\frac13+\theta$.
Assuming $\theta < \frac16$, bound (2.1.9), and condition 1) in Theorem 2.1.2, with $A_1(s)$ as defined in (2.3.5) we have
\begin{align} A_1(s)=\mathstrut & \prod_p (1+(a_1(p)-a_2(p))p^{-s}) \cdot \prod_p (1+(a_1(p^2)-a_2(p^2))p^{-2s}) \cdot H_2(s),\tag{2.3.6} \end{align}where $H_2(s)$ is regular and nonvanishing for $\sigma > \frac{5}{12}$.
We finish the proof of Lemma 2.3.1 and then conclude with the proof of Lemma 2.3.2.
Using the notation of Lemma 2.3.2, write (2.3.6) as $A_1(s)=A_2(s)H_2(s)$. Since $A_1(s)$ and $H_2(s)$ are meromorphic in a neighborhood of $\sigma\ge\frac12$, so is $A_2(s)$. Changing variables $s\mapsto s+\frac12$, which divides the $n$th Dirichlet coefficient by $1/\sqrt{n}$, we can apply Lemma 2.2.3, using the estimate (2.1.9) and condition 1) to conclude that $A_2(s)$ has only finitely many zeros or poles in $\sigma\ge\frac12$. Since the same is true of $H_1(s)$ and $H_2(s)$, we have shown that $P(s)$ has only finitely many zeros or poles in $\sigma\ge\frac12$. This completes the proof for conditions 2.1) and 1).
In the other cases, the proof is almost the same, using Lemma 2.2.2 to rewrite equation (2.3.5) in terms of $L_j^S(s,\sym^2)$ or $L_j^S(s,\ext^2)$, and using Lemma 2.2.3 for the factors that remain. This concludes the proof of Lemma 2.3.1.\qed
\vspace{3ex} Proof of Lemma 2.3.2. Using the identities
\begin{equation}\frac{1+a x}{1+b x} = 1+(a-b)x - \frac{b(a-b)x^2}{1+b x}\tag{2.3.7} \end{equation}and
\begin{equation}1+ax+bx^2 = (1+ax)\left(1+ \frac{b x^2}{1+ax} \right)\tag{2.3.8} \end{equation}we have
\begin{equation}\frac{1+a x}{1+b x} = (1+(a-b)x)\left(1-\frac{b(a-b)x^2}{(1+(a-b)x)(1+bx)}\right).\tag{2.3.9} \end{equation}Thus
\begin{align}\prod_p \frac{1+a_1(p)p^{-s}}{1+a_2(p)p^{-s}} =\mathstrut & \prod_p \bigl(1+(a_1(p)-a_2(p))p^{-s} \bigr) \cr &\phantom{xx}\times \prod_p \biggl( 1- \frac{a_2(p)(a_1(p)-a_2(p)) p^{-2s}}{ (1+(a_1(p)-a_2(p))p^{-s})(1+a_2(p)p^{-s})}\biggr) \cr =\mathstrut & \prod_p \bigl(1+(a_1(p)-a_2(p))p^{-s} \bigr) \cdot h(s)\tag{2.3.10} \end{align}say. We wish to apply Lemma 2.2.3 to show that $h(s)$ is regular and nonvanishing for $\sigma>\sigma_0$ for some $\sigma_0 < \frac12$. Since $\theta < \frac16$, if $\sigma\ge \frac16$ and $p > P_0$ where $P_0$ depends only on $\theta$, then $|1+a_2(p)p^{-\sigma}| \geq \frac12$ and $|1+(a_1(p)-a_2(p))p^{-\sigma}| \geq \frac12$. Using those inequalities and $|a_2(p)|\ll p^\theta$ we have
\begin{align}\sum_{P_0\le p\le X}& \left|\frac{a_2(p)(a_1(p)-a_2(p)) }{ (1+(a_1(p)-a_2(p))p^{-\sigma})(1+a_2(p)p^{-\sigma})}\right|^2 \log p\cr &\phantom{xxxxxxxxxxxxxxxxxxx}\le 16 \mathstrut \sum_{P_0\le p\le X} \left|{a_2(p)(a_1(p)-a_2(p)) }\right|^2 \log p\cr &\phantom{xxxxxxxxxxxxxxxxxxx}\ll\mathstrut X^{2\theta} \sum_{P_0\le p\le X} \left|(a_1(p)-a_2(p)) \right|^2 \log p \cr &\phantom{xxxxxxxxxxxxxxxxxxx}\ll\mathstrut X^{\frac12+2\theta}.\tag{2.3.11} \end{align}Changing variables $s\to \frac{s}{2}-\frac{1}{12}$ and applying Lemma 2.2.3, we see that $h(s)$ is regular and nonvanishing for $\sigma>\frac{5}{12}$.
Applying the same reasoning to the second factor in (2.3.6) completes the proof.\qed
Two basic results which are used in this section are:
If $\sum_{n \le X} |a(n)| \ll X^{1+\epsilon}$ for every $\epsilon>0$, then $\displaystyle \sum_{n=1}^\infty \frac{a(n)}{n^s} $ converges absolutely for all $\sigma>1$.
If $\sum_{n \le X} |a(n)| \le C X $ as $X\to\infty$, then
\begin{equation} \sum_{n=1}^\infty \frac{a(n)}{n^\sigma} \le \frac{C}{\sigma-1} + O(1) \end{equation}as $\sigma \to 1^+$.
Both of those results follow by partial summation.
We first state and prove a simplified version of Lemma 2.2.3.
Let
\begin{equation}L(s)=\prod_p \sum_{j=0}^\infty a({p^j}) p^{-j s}\tag{2.4.1} \end{equation}and suppose there exists $M\ge 0$ and $\theta < \frac 12$ so that $|a({p^j})|\ll p^{j \theta}$ and
\begin{equation} \sum_{p\le X} |a(p)|^2 \log p \le (1+o(1)) M^2 X.\tag{2.4.2} \end{equation}Then $L(s)$ is a nonvanishing analytic function in the half-plane $\sigma>1$. Furthermore, if $L(s)$ has a meromorphic continuation to a neighborhood of $\sigma\ge1$, then $L(s)$ has at most $M^2$ zeros or poles on the $\sigma=1$ line.
Note that, by the prime number theorem (2.1.10), the condition on $a(p)$ is satisfied if $|a(p)|\le M$.
We have
\begin{align} L(s)=\mathstrut &\prod_p \sum_{j=0}^\infty a({p^j}) p^{-j s} \cr =\mathstrut & \prod_p \biggl(1+a(p) p^{-s} + \sum_{j\ge 2} a({p^j}) p^{-j s}\biggr) \cr =\mathstrut & \prod_p \left(1+a(p) p^{-s} \right) \cr &\times \prod_p \bigl(1+a(p^2)p^{-2s} +(a(p^3)-a(p) a(p^2))p^{-3s}\cr &\phantom{xxxxxx} + (a(p^4)-a(p)a(p^3)+a(p)^2a(p^2))p^{-4s}+\cdots\bigr)\cr =\mathstrut & \prod_p \left(1+a(p) p^{-s} \right) \prod_p \biggl(1+\sum_{j=2}^\infty b({p^j}) p^{-j s}\biggr),\tag{2.4.3} \end{align}say, where $b(p^j)\ll j M^j p^{j \theta} \ll p^{j(\theta+\epsilon)}$ for any $\epsilon>0$.
Writing (2.4.3) as $L(s)=f(s)g(s)$ we have
\begin{equation} \log g(s) = \sum_p \log(1+Y) = \sum_p \left( Y + O(Y^2) \right),\tag{2.4.4} \end{equation}where $\displaystyle Y=\sum_{j=2}^\infty b({p^j}) p^{-j s}$. Now,
\begin{align}|Y| \le \mathstrut& \sum_{j=2}^\infty |b({p^j})| p^{-j \sigma} \cr \ll \mathstrut& \sum_{j=2}^\infty p^{j(\theta-\sigma+\epsilon)} \cr = \mathstrut& \frac{p^{2(\theta-\sigma+\epsilon)}}{1-p^{\theta-\sigma+\epsilon}}.\tag{2.4.5} \end{align}If $\sigma > \frac12 + \theta$ we have $|Y|\ll 1/p^{1+\delta}$ for some $\delta>0$. Therefore the series (2.4.4) for $\log(g(s))$ converges absolutely for $\sigma > \frac12 + \theta$, so $g(s)$ is a nonvanishing analytic function in that region. By (2.4.2), Cauchy's inequality, and Lemma 2.4.1, $f(s)$ is a nonvanishing analytic function for $\sigma>1$, so the same is true for $L(s)$. This establishes the first assertion in the lemma.
Now we consider the zeros of $L(s)$ on $\sigma=1$. Since $\theta < \frac12$, the zeros or poles of $L(s)$ on the $\sigma=1$ line are the zeros or poles of $f(s)$. Furthermore, by (2.4.3) and the properties of $g(s)$, for $\sigma>1$ we have
\begin{equation} \frac{L'}{L}(s) = \sum_p \frac{-a(p)\log(p)}{p^s} + h(s),\tag{2.4.6} \end{equation}where $h(s)$ is bounded in $\sigma > \frac12+\theta+\epsilon$ for any $\epsilon>0$. Suppose $s_1,\ldots,s_J$ are zeros or poles of $L(s)$, with $s_j = 1+i t_j$ having multiplicity $m_j$. We have
\begin{equation}\frac{L'}{L}(\sigma+i t_j) \sim \frac{m_j}{\sigma-1}, \ \ \ \ \ \ \ \ \mathrm{as} \ \ \sigma \to 1^+,\tag{2.4.7} \end{equation}therefore
\begin{equation} \sum_p \frac{-a(p)\log(p)}{p^{\sigma + it_j}} \sim \frac{m_j}{\sigma-1}, \ \ \ \ \ \ \ \ \mathrm{as} \ \ \sigma \to 1^+.\tag{2.4.8} \end{equation}Now write
\begin{equation}k(s) = \sum_{j=1}^J m_j \sum_p \frac{-a(p)\log(p)}{p^{s+it_j}}.\tag{2.4.9} \end{equation}By (2.4.8) we have
\begin{equation} k(\sigma) \sim \frac{\sum_{j=1}^J m_j^2}{\sigma-1}, \ \ \ \ \ \ \ \ \mathrm{as} \ \ \sigma \to 1^+.\tag{2.4.10} \end{equation}On the other hand, for $\sigma>1$ we have
\begin{align} |k(\sigma)| = \mathstrut & \left| \sum_p \frac{a(p)\log(p)}{p^{\sigma}} \sum_{j=1}^J m_j p^{-it_j} \right| \cr \le \mathstrut & \left(\sum_p \frac{|a(p)|^2 \log(p)}{p^{\sigma}} \right)^{\frac12} \left( \sum_p \frac{\log p}{p^\sigma} \left| \sum_{j=1}^J m_j p^{-it_j} \right|^2 \right)^{\frac12} \cr \le \mathstrut & (1+o(1)) \left(\frac{M^2}{\sigma-1} \right)^{\frac12} \left( \sum_{j=1}^J \sum_{\ell=1}^J m_j m_\ell \sum_p \frac{\log p}{p^{\sigma + i(t_j - t_\ell)} } \right)^{\frac12} \cr \sim \mathstrut & \left(\frac{M^2}{\sigma-1} \right)^{\frac12} \left( \sum_{j=1}^J \frac{m_j^2}{\sigma-1} \right)^{\frac12} \ \ \ \ \ \ \ \ \mathrm{as} \ \ \sigma \to 1^+.\tag{2.4.11} \end{align}On the first line we used the Cauchy-Schwartz inequality, on the next-to-last line we wrote the sum over $a(p)$ as a Stieltjes integral and used (2.4.2) and Lemma 2.4.2, and on the last line we used the fact that the Riemann zeta function has a simple pole at $s=1$ and no other zeros or poles on the $\sigma=1$ line.
Combining (2.4.8) and (2.4.11) we have $\sum_{j=1}^J m_j^2 \le M^2$. Since $m_j^2\ge 1$, we see that $J\le M^2$, as claimed.
The proof of Lemma 2.2.3 is similar to Lemma 2.4.3.
We have
\begin{align} L(s)=\mathstrut &\prod_p \sum_{j=0}^\infty a({p^j}) p^{-j s} \cr =\mathstrut & \prod_p \biggl(1+a(p) p^{-s} + a(p^2)p^{-2s} + \sum_{j\ge 3} a({p^j}) p^{-j s}\biggr) \cr =\mathstrut & \prod_p \left(1+a(p) p^{-s} \right) \left(1+a(p^2) p^{-2s} \right) \cr &\phantom{xxx}\times \bigl(1+(a(p^3)-a(p) a(p^2))p^{-3s}\cr &\phantom{xxxxxx} + (a(p^4)-a(p)a(p^3)+a(p)^2a(p^2))p^{-4s}+\cdots\bigr)\cr = \mathstrut & \prod_p \left(1+a(p) p^{-s} \right)\left(1+a(p^2) p^{-2s} \right) \prod_p \biggl(1+\sum_{j=3}^\infty c({p^j}) p^{-j s}\biggr)\cr =\mathstrut & f(s)g(s),\tag{2.4.12} \end{align}say.
We have $c({p^j}) \ll j M^j p^{j\theta} \ll p^{j(\theta+\epsilon)}$ for any $\epsilon>0$. We use this to show that $g(s)$ is a nonvanishing analytic function in $\sigma>\frac13+\theta$. Writing $g(s) = \prod_p(1+Y)$ we have
\begin{equation} \log g(s) = \sum_p \log(1+Y) = \sum_p \left( Y + O(Y^2) \right),\tag{2.4.13} \end{equation}where $\displaystyle Y=\sum_{j=3}^\infty b({p^j}) p^{-j s}$. Now,
\begin{align}|Y| \le \mathstrut& \sum_{j=3}^\infty |b({p^j})| p^{-j \sigma} \cr \ll \mathstrut& \sum_{j=3}^\infty p^{j(\theta-\sigma+\epsilon)} \cr = \mathstrut& \frac{p^{3(\theta-\sigma+\epsilon)}}{1-p^{\theta-\sigma+\epsilon}}.\tag{2.4.14} \end{align}If $\sigma > \frac13 + \theta$ we have $|Y|\ll 1/p^{1+\delta}$ for some $\delta>0$. Therefore by Lemma 2.4.1 the series (2.4.13) for $\log(g(s))$ converges absolutely for $\sigma > \frac13 + \theta$, so $g(s)$ is a nonvanishing analytic function in that region. By the same argument, using (2.2.12) and (2.2.12), $f(s)$ is a nonvanishing analytic function for $\sigma>1$, so the same is true for $L(s)$. This establishes the first assertion in the lemma.
Now we consider the zeros of $L(s)$ on $\sigma=1$. Since $\theta < \frac23$, the zeros or poles of $L(s)$ on the $\sigma = 1$ line are the zeros or poles of $f(s)$. Taking the logarithmic derivative of (2.4.12) and using the same argument as above for the lower order terms, we have
\begin{align} \frac{L'}{L}(s) =\mathstrut & \sum_p \frac{-a(p)\log(p)}{p^s} + 2\,\frac{a(p)^2\log(p)}{p^{2s}} -2\, \frac{a(p^2)\log(p)}{p^{2s}} + h_1(s)\cr =\mathstrut & \sum_p \frac{-a(p)\log(p)}{p^s} -2 \,\frac{a(p^2)\log(p)}{p^{2s}} + h_2(s),\tag{2.4.15} \end{align}where $h_j(s)$ is bounded in $\sigma > \frac13+\theta+\epsilon$ for any $\epsilon>0$. By (2.2.12) and Lemma 2.4.1, the middle term in the sum over primes in (2.4.15) converges absolutely for $\sigma>\frac12$, so it was incorporated into $h_1(s)$.
Suppose $s_1,\ldots,s_J$ are zeros or poles of $L(s)$, with $s_j = 1+i t_j$ having multiplicity $m_j$. We have
\begin{equation}\frac{L'}{L}(\sigma+i t_j) \sim \frac{m_j}{\sigma-1}, \ \ \ \ \ \ \ \ \mathrm{as} \ \ \sigma \to 1^+,\tag{2.4.16} \end{equation}therefore
\begin{equation} \sum_p \left( \frac{-a(p)\log(p)}{p^{\sigma + it_j}} - 2\, \frac{a(p^2)\log(p)}{p^{2(\sigma + it_j)}} \right) \sim \frac{m_j}{\sigma-1}, \ \ \ \ \ \ \ \ \mathrm{as} \ \ \sigma \to 1^+.\tag{2.4.17} \end{equation}Now write
\begin{equation} k(s) = \sum_{j=1}^J m_j \sum_p \left( \frac{-a(p)\log(p)}{p^{s+i t_j}} -2\, \frac{a(p^2)\log(p)}{p^{2(s+i t_j)}} \right)\tag{2.4.18} \end{equation}By (2.4.17) we have
\begin{equation} k(\sigma) \sim \frac{\sum_{j=1}^J m_j^2}{\sigma-1}, \ \ \ \ \ \ \ \ \mathrm{as} \ \ \sigma \to 1^+.\tag{2.4.19} \end{equation}We will manipulate (2.4.18) so that so that we can use (2.2.12) and (2.2.12) to give a bound on $\sum m_j^2$ in terms of $M_1$ and $M_2$.
By Cauchy's inequality and Lemma 2.4.2 we have
\begin{align} |k(\sigma)| \le \mathstrut & \left| \sum_p \frac{a(p)\log(p)}{p^{\sigma}} \sum_{j=1}^J \frac{m_j}{p^{it_j}}\right| + 2 \left|\sum_p \frac{p^{-\sigma} a(p^2)\log(p)}{p^{\sigma}} \sum_{j=1}^J \frac{m_j}{p^{2it_j}} \right| \cr \le \mathstrut &\left(\sum_p \frac{|a(p)|^2 \log(p)}{p^{\sigma}}\right)^{\frac12} \left( \sum_p \frac{\log p}{p^\sigma} \biggl|\sum_{j=1}^J m_j p^{-it_j} \biggr|^2\right)^{\frac12}\nonumber\\ &\ \ \ \ \ \ +2 \left(\sum_p \frac{p^{-2\sigma}|a(p^2)|^2 \log(p)}{p^{\sigma}} \right)^{\frac12} \left( \sum_p \frac{\log p}{p^\sigma}\biggl| \sum_{j=1}^J m_j p^{-2it_j} \biggr|^2\right)^{\frac12}\nonumber\\ \le & (1+o(1))\Biggl( \left(\frac{M_1^2}{\sigma-1} \right)^{\frac12} \biggl( \sum_{j=1}^J \sum_{\ell=1}^J m_j m_\ell \sum_p \frac{\log p}{p^{\sigma + i(t_j - t_\ell)} } \biggr)^{\frac12} \nonumber\\ &\ \ \ \ \ +2 \left(\frac{M_2^2}{\sigma-1} \right)^{\frac12} \biggl( \sum_{j=1}^J \sum_{\ell=1}^J m_j m_\ell \sum_p \frac{\log p}{p^{\sigma + 2i(t_j - t_\ell)} } \biggr)^{\frac12} \Biggr)\nonumber\\ \sim & \frac{M_1 + 2 M_2}{(\sigma-1)^\frac12 } \left( \sum_{j=1}^J \frac{m_j^2}{\sigma-1} \right)^{\frac12} \ \ \ \ \ \ \ \ \mathrm{as} \ \ \sigma \to 1^+.\tag{2.4.20} \end{align}In the last step we used the fact that the Riemann zeta function has a simple pole at 1 and no other zeros or poles on the $1$-line.
Combining (2.4.19) and (2.4.20) we have $\displaystyle \sum_{j=1}^J m_j^2 \le (M_1 +2 M_2)^2. $ Since $m_j\ge 1$, the proof is complete.