Proof. Write \(b=b_S b_0\), where \(b_S>0\) is the product of the primes of \(S\) dividing \(b\). Then \(b_0\) is odd and prime to \(Q_S\). Choose \(\varepsilon\in\{1,-1\}\) so that \(d=\varepsilon b_0\equiv1\pmod4\), and put \(F'=F^{(\varepsilon b_S)}\). Quadratic twisting depends only on the square class and composes by multiplication of square classes. Thus \(F'\) is in \(\mathcal F\) and \((F')^{(d)}\simeq F^{(b)}\). Moreover \(|d|=|b|/b_S\le |b|\). There are at most \(2^{|S|+1}\) choices of the signed \(S\)-part. Possible coincidences among the finitely many curves only change this fixed multiplicity. ◻

Proof. We compute the inverse local Euler factors using geometric Frobenius on the inertia invariants of the dual of the rational Tate module, and the conductor as the Artin conductor (Dokchitser and Anni 2014, Definitions 17–18 and Theorem 19). The rational Tate-module representation of a quadratic twist is the original representation tensored with the corresponding quadratic character. This follows from the isomorphism over the twisting extension, whose conjugate differs by the scalar automorphism \([-1]\).

An unramified scalar twist preserves the conductor and multiplies Frobenius eigenvalues on inertia invariants by the character value. Since an admissible \(d\) is an odd fundamental discriminant, its character is unramified at every prime of \(S\). At a prime \(p\mid d\), the representation of \(F\) is unramified and the quadratic character is tamely ramified. The tensor product has no inertia invariants, Swan conductor zero, and Artin conductor exponent two. At all other primes the scalar twist is unramified. Computing the inertia-invariant Euler factors and the conductor gives (5). For an arbitrary squarefree \(b\), these same computations apply at odd primes outside \(S\), regardless of the character’s behavior at two. ◻

Proof. For coprime odd \(l_1,l_2\), the Chinese remainder theorem, followed by a change of variable in each Gauss sum, gives \[\tau_h(l_1l_2)= \left(\frac{l_1}{l_2}\right)\left(\frac{l_2}{l_1}\right) \tau_h(l_1)\tau_h(l_2).\] Quadratic reciprocity says that the displayed product of symbols is \(\epsilon_{l_1l_2}/(\epsilon_{l_1}\epsilon_{l_2})\). This proves multiplicativity. Substituting \(a\mapsto-a\) proves the first identity in (10); the second follows by considering the two residue classes of \(l\) modulo four.

For even \(v\), the Jacobi symbol modulo \(p^v\) is the indicator of the units. Thus \[\tau_h(p^v)=\sum_{a\bmod p^v}e(ah/p^v) -\sum_{b\bmod p^{v-1}}e(bh/p^{v-1}),\] which equals \(p^v-p^{v-1}\) when \(p^v\mid h\), equals \(-p^{v-1}\) when \(p^{v-1}\Vert h\), and otherwise vanishes. Here \(\epsilon_{p^v}=1\). For odd \(v\), write \(a=y+pz\) with \(y\) modulo \(p\) and \(z\) modulo \(p^{v-1}\). The \(z\)-sum vanishes unless \(p^{v-1}\mid h\); in that case \[\tau_h(p^v)=p^{v-1}\sum_{y\bmod p} \left(\frac yp\right)e\big(y(h/p^{v-1})/p\big).\] This is zero if \(p^v\mid h\). Otherwise the quadratic Gauss-sum formula makes it \(p^{v-1}\epsilon_p\sqrt p\,\bigl(\frac{h/p^{v-1}}p\bigr)\). Since \(\epsilon_{p^v}=\epsilon_p\) for odd \(v\), division by \(\epsilon_{p^v}p^{v/2}\) gives (11). ◻

Proof. For an odd prime put \(e_p=v_p(r)\) and \(t_p=v_p(h)\). Multiplicativity expresses the series as the product of the local sums \[T_p(s_1,s_2)=\sum_{a,b'\geq0} \lambda_F(p^a)\lambda_F(p^{b'}) B_{\nu h}(p^{e_p+a+b'})p^{-as_1-b's_2}.\] The symbol \(b'\) here is an integer exponent, distinct from the real shift \(b\). No division by \(B_{\nu h}(p^{e_p})\) is made; that quantity may vanish. Write \(D_p^*(s)\) for the inverse local factor of the elliptic curve \(F^{(\nu h_1)}\). The correction factor at \(p\) is \(T_p(s_1,s_2)D_p^*(s_1)D_p^*(s_2)\). At two it is just \(D_2^*(s_1)D_2^*(s_2)\), because the summation indices are odd.

There are three types of odd prime. If \(p\notin S\) and \(p\nmid rh\), put \(\chi=(\nu h/p)=(\nu h_1/p)\) and \(x_i=p^{-s_i}\). The local formulas give exactly \[T_p=1+\lambda_F(p)\chi(x_1+x_2),\qquad D_p^*(s_i)=1-\lambda_F(p)\chi x_i+x_i^2.\] Multiplication cancels both terms of total degree one. The sum of the absolute values of all polynomial coefficients before cancellation is at most \(5\cdot4\cdot4=80\). Therefore the correction is \(1+O(80p^{-1-2b})\), uniformly in the imaginary parts.

If \(p\notin S\), \(p\nmid r\), and \(t_p=1\), the twisting curve has a trivial local factor at \(p\). By (11), \[T_p=1-\lambda_F(p^2)(x_1^2+x_2^2)-\lambda_F(p)^2x_1x_2 =1+O(10p^{-1-2b}).\] In particular these primes do not produce a cost for every prime factor of the squarefree part \(h_1\).

The remaining odd primes lie in \(S\) or divide \(rh_2\). The support in (11) gives \(e_p+a+b'\leq t_p+1\) for every nonzero summand, except that \(B_{\nu h}(1)=1\) is already covered when \(e_p=a=b'=0\). Hence \[|T_p|\leq p^{e_p/2} \sum_{a+b'\leq t_p+1}(a+1)(b'+1) \leq p^{e_p/2}(t_p+3)^4.\] Each inverse local factor has modulus at most \((1+p^{-1/2})^2\), so the product of the two costs at most \(9\). Writing \(u_p=v_p(h_2)\), we have \(t_p\leq2u_p+1\). If \(u_p\geq1\), the bound \(9(2u_p+4)^4\leq(u_p+1)^{17}\) absorbs this cost into a fixed divisor power, with \(p^{e_p/2}\) absorbed by \(r\). If \(u_p=0\) and \(e_p\geq1\), the cost is at most \(9\cdot4^4p^{e_p/2}\leq p^{8e_p}\) for \(p\geq3\). If \(u_p=e_p=0\), this is one of the fixed primes of \(S\), whose cost is a fixed constant. The factor at two is bounded by \((1+2^{-1/2})^4\).

The nonexceptional product is bounded by \[\exp\left(80\sum_p p^{-1-2b}\right) \leq\zeta(1+2b)^{80}\ll b^{-80}.\] Combining the estimates proves (14) with the stated \(C_{\rm ar}\). The nonexceptional correction product converges normally on compact subsets of \(\Re s_i>1/2\); all exceptional factors are polynomials. Multiplication by the entire twist \(L\)-functions therefore gives the claimed continuation. This argument never divides by a global \(L\)-value. ◻

Proof. Our Fourier convention is \(\widehat W(y_1,y_2,\xi)=\int_{\mathbb R}W(y_1,y_2,x)e(-x\xi)\,dx\). Applying Poisson summation in each residue class modulo \(l=rn_1n_2\) gives exactly \[\frac1U\sum_m\left(\frac ml\right)W(y_1,y_2,m/U) =\sum_{h\in\mathbb Z}\frac{\tau_h(l)}l \widehat W(y_1,y_2,hU/l).\] Thus the normalized frequency factor is \[\frac{1}{\sqrt{n_1n_2}}\frac{\tau_h(rn_1n_2)}{rn_1n_2} =\frac{\epsilon_{rn_1n_2}B_h(rn_1n_2)} {\sqrt r\,n_1n_2}.\] Use (10) to replace the numerator by a fixed linear combination of \(B_h\) and \(B_{-h}\).

Put \(\rho=hU/(rN_1N_2)\) and \[V_h(y_1,y_2)=\widehat W(y_1,y_2,\rho/(y_1y_2)),\qquad \widetilde V_h(z_1,z_2)= \int_0^\infty\!\int_0^\infty V_h(y_1,y_2)y_1^{z_1}y_2^{z_2} \frac{dy_1dy_2}{y_1y_2}.\] On any fixed bounded range of real parts, Fourier integration by parts in \(x\), followed by Mellin integration by parts in \(y_1,y_2\), yields \[ |\widetilde V_h(z_1,z_2)| \ll_{J_1}\|W\|_{D,*}(1+|\rho|)^{-J_1} (1+|\Im z_1|+|\Im z_2|)^{-J_1}. \tag{17}\] To see why differentiation causes no arithmetic loss, each derivative of \(\rho/(y_1y_2)\) contributes a bounded multiple of \(\rho\) on the fixed compact support. For any fixed number of such derivatives, take that many additional integrations by parts in \(x\). The powers of \(\rho\) are then absorbed by its Fourier decay. All other factors are bounded functions of \(y_1,y_2\).

Mellin inversion first on right lines expresses the \(\nu\) part of the mode as \[\frac1{\sqrt r(2\pi i)^2} \int\!\int \widetilde V_h(z_1,z_2)N_1^{z_1}N_2^{z_2} \mathcal G_{h,r}^{\nu}(1+it_1+z_1,1+it_2+z_2) \,dz_1\,dz_2.\] Move both lines to \(\Re z_i=-1/2+b\). Lemma 9 and the entire continuation of the twist functions show that no pole is crossed. Polynomial growth in fixed strips, together with a larger fixed order in (17), makes the horizontal segments tend to zero. The scale factor is \((N_1N_2)^{-1/2+b}\). Applying (14) and (17), and discarding \(r^{-1/2}\leq1\), proves (16). The final assertion follows directly from the support of \(\widehat W\) before Mellin inversion. ◻

Proof of Propositions 6 and 7. We give a simultaneous induction for all \(F\in\mathcal F\). The induction hypothesis at smaller dyadic heights is \[ \frac1{M'}\sum_{\substack{d\ \mathrm{admissible}\\M'\leq|d|<2M'}} |L_{F^{(d)}}(\sigma+it)|^2 \leq K\log^B(3M')(1+|t|)^A \qquad(M'<M). \tag{19}\] The constant \(K\), common to the finite family, will be selected last. Since the reduction in Lemma 4 does not increase height and has bounded multiplicity, this hypothesis implies \[ \sum_{\substack{0<|h_1|\leq H\\h_1\ \mathrm{squarefree}}} |L_{F^{(\nu h_1)}}(1/2+b+iv)|^2 \leq C_{\mathcal F}KH\log^B(3M)(1+|v|)^A \quad(1\leq H\leq M/2). \tag{20}\] Indeed all admissible images lie in earlier dyadic blocks, whose lengths have sum at most \(2H\). If \(H<1\), the sum is empty. In particular the arguments below at \(M=1\) require no induction input.

The polynomial average at inflated height. First we prove a polynomial estimate with a small coefficient in front of \(K\). Put \(L=\log(3M)\), \(T=1+|t|\), and let \(N>0\). Choose \(P\) sufficiently large, with \[ P\geq C_{\mathcal F,G}(1+N/M). \tag{21}\] Let \(\mathcal P\) consist of the primes \(p\in[P,2P]\) outside \(S\). For a sufficiently large fixed lower threshold on \(P\), the elementary dyadic prime estimate of Erdős (Erdős 1932, sec. 6, equation (10)) gives \(|\mathcal P|\gg P/\log(2P)\).

Choose a fixed nonnegative Schwartz function \(\Phi\) that is at least one on \([-8,8]\) and has compactly supported Fourier transform. For completeness, take a nonzero even real smooth bump \(g\) in frequency space with \(\int g>0\). Its inverse Fourier transform \(f\) is real, Schwartz, and nonzero near zero. A sufficiently small dilation makes \(f(\alpha x)\) nonzero throughout \([-8,8]\); a fixed multiple of \(f(\alpha x)^2\) then has the required properties. Its Fourier transform is a compactly supported convolution.

Put \(U=MP^2\). For \(c=\pm1\) use the formal character \(z_m\) with \(z_m(2)=c\), and let \(P_m^c(N',t;G)\) be the corresponding polynomial. We claim, uniformly for \(0<N'\leq N\), \[\begin{align*} \frac1U\sum_{m\in\mathbb Z}\Phi(m/U)|P_m^c(N',t;G)|^2 &\leq C_G\log^{C_0}(3M+3N) \left(1+C_{A,B,\mathcal F,G}K\frac{N}{MP^2}L^BT^A\right), \tag{22}\end{align*}\] where we may fix \(C_0=512\), independently of \(A,B\) and of the required smoothness orders.

To prove the claim, apply Lemma 10 with \(r=1\) to the expansion of the square, first separating powers of two by (18). Take \(b=1/(10\log(3M+3N))\). For the zero mode the complete-residue mean vanishes unless the odd part of \(n_1n_2\) is a square, and its absolute value is at most one. Inserting \((n_1n_2)^{-b}\) at the bounded cost \((C_GN)^{2b}\ll_G1\) majorizes the resulting sum by an Euler product. At an odd prime this product has factor \[\sum_{\substack{a,b'\geq0\\a+b'\ \mathrm{even}}} (a+1)(b'+1)p^{-(a+b')(1/2+b)} =1+O(100p^{-1-2b}).\] For example this follows by putting \(z=p^{-1/2-b}\leq3^{-1/2}\) in \(\tfrac12((1-z)^{-4}+(1+z)^{-4})\) and bounding its terms of degree at least two. The prime two costs a bounded factor by (18). The zero mode is therefore \(O_G(b^{-100})\).

For the nonzero modes, denote the two odd lengths by \(N_1,N_2\). The Fourier support restricts the frequencies to \[0<|h|\leq H_0=C_\Phi N_1N_2/(MP^2).\] Increasing the constant in (21) ensures \(H_0\leq M/2\) for every such pair of lengths. For fixed \(h_2\), Cauchy–Schwarz and (20) bound the sum over \(h_1\) occurring in (16) by \[\ll_{\mathcal F}K\frac{H_0}{h_2^2}L^B (1+|t+v_1|)^{A/2}(1+|-t+v_2|)^{A/2}.\] This formula is only used when \(H_0/h_2^2\geq1\); otherwise that sum is empty. Choose \(J_1>A+4\). Since \(1+|\pm t+v|\leq T(1+|v|)\), the integral in \(v_1,v_2\) is \(O_A(T^A)\). Also \[ \sum_{h_2\geq1}\frac{\tau(h_2)^{C_{\rm ar}}}{h_2^2}<\infty. \tag{23}\] One proof of this convergence is its Euler product: its local factor is \(1+O_{C_{\rm ar}}(p^{-2})\), since \(\sum_{a\geq1}(a+1)^{C_{\rm ar}}p^{-2a} \ll_{C_{\rm ar}}p^{-2}\) uniformly for \(p\geq2\). Consequently the nonzero contribution is bounded by \[\ll_{A,B,\mathcal F,G}b^{-C_{\rm ar}}KL^BT^A (N_1N_2)^{-1/2+b}H_0 \ll_{A,B,\mathcal F,G} \log^{C_0}(3M+3N)K\frac{N}{MP^2}L^BT^A.\] Here \((N_1N_2)^b\ll_G1\) and \(\sqrt{N_1N_2}\ll_GN\). The sums of the factors from two converge by (18). This proves (22). If the Fourier support contains no nonzero frequency, the same proof uses only its zero-mode part.

Recovering the squarefree average. Now inflate an admissible parameter by setting \(m=dp^2\). If \(c=\chi_d(2)\), multiplicativity gives the exact identity \[ P_d(N,t;G)=\sum_{a\geq0} \lambda_F(p^a)\chi_d(p)^a p^{-a(1/2+it)} P_{dp^2}^c(N/p^a,t;G). \tag{24}\] Indeed \(z_{dp^2}(n)\) deletes all positive powers of \(p\) and agrees with \(\chi_d(n)\) on integers prime to \(p\); reinstating their Euler coefficients gives the original sum. If \(p\mid d\), the terms with \(a\geq1\) vanish, so the identity also covers that case.

Equip pairs \((d,p)\), with \(M\leq|d|<2M\), \(\chi_d(2)=c\), and \(p\in\mathcal P\), with squared norm \((M|\mathcal P|)^{-1}\sum_{d,p}|\cdot|^2\). The map \((d,p)\mapsto dp^2\) is injective: in the prime factorization of its absolute value the only exponent exceeding one occurs at \(p\), and is either two or three. Moreover \(|dp^2|<8U\). Thus the \(a=0\) term has norm at most \[\left(\frac{U}{M|\mathcal P|}\right)^{1/2} \sup_{0<N'\leq N} \left(\frac1U\sum_m\Phi(m/U)|P_m^c(N',t;G)|^2\right)^{1/2} \ll(P\log(2P))^{1/2}\mathcal B^{1/2},\] where \(\mathcal B\) denotes the right side of (22). For \(a\geq1\) fix \(p\) before summing in \(d\). The weaker bound \(\sum_d|P_{dp^2}^c(N/p^a,t;G)|^2\leq U\mathcal B\) gives pair norm at most \((a+1)P^{1-a/2}\mathcal B^{1/2}\). Minkowski’s inequality and \[\sum_{a\geq1}(a+1)P^{-a/2}\ll P^{-1/2}\] show that these terms together have norm \(O(P^{1/2}\mathcal B^{1/2})\). Summing the two choices of \(c\) proves \[\begin{align*} \frac1M\sum_d|P_d(N,t;G)|^2 &\ll_G\log^{C_0}(3M+3N)P\log(2P) \left(1+C_{A,B,\mathcal F,G}K\frac{N}{MP^2}L^BT^A\right). \tag{25}\end{align*}\]

Here is an explicit noncircular choice of exponents. Set \[ \begin{split} C_0&=512,\qquad C_1=C_0+32,\qquad C_2=C_0+C_1+4,\\ R&=C_2+4,\qquad C_3=3C_2+10,\qquad A=B=2(C_3+20). \end{split} \tag{26}\] The letter \(R\) in this proof denotes a fixed contour abscissa; it is unrelated to a rank truncation threshold used later in the paper. For any prescribed \(\varepsilon>0\), choose \[ P=P_0[L(1+N/M)T]^{C_1}. \tag{27}\] Once \(A,B\) and the fixed weight family have been selected, a sufficiently large \(P_0\) ensures (21) and gives \[\begin{align*} \frac1M\sum_d|P_d(N,t;G)|^2 &\leq C_*[L(1+N/M)T]^{C_2} \\ &\quad+\varepsilon K L^{B-10}T^{A-4}(1+N/M)^{-4}. \tag{28}\end{align*}\] The constant \(C_*\) can depend on \(P_0\) and the chosen fixed data, but not on \(K,M,N,t\). For explicit verification, put \(q=1+N/M\) and \(D_0=LqT\). Then \(\log(3M+3N)\ll Lq\) and \[\frac{\log(2P)}P \leq\frac{\log(2P_0)+C_1}{P_0}D_0^{-C_1+1}.\] The recursive coefficient in (25) is therefore at most a fixed multiple of \[\frac{\log(2P_0)+C_1}{P_0} K L^{B+C_0-C_1+1}T^{A-C_1+1}q^{C_0-C_1+2}.\] Our value of \(C_1\) leaves at least the losses \(10,4,4\) displayed in (28). Its prefactor can be made smaller than \(\varepsilon\). The nonrecursive term is bounded by \(C_*D_0^{C_0+C_1+1}\) and hence by the stated \(C_2\) power. In particular \(C_2\) was fixed before \(A,B,\varepsilon,P_0,K\).

Reconstructing the central-line values. We next reconstruct \(L\) on the central line. Choose a fixed smooth dyadic partition \(G_0\) with common compact support in \((0,\infty)\), so that \(\sum_{N=1,2,4,\ldots}G_0(n/N)=1\) for every integer \(n\geq1\). Let \(C_d=\sqrt{q_F}|d|/(2\pi)\) and define \[I_d(t)=\frac1{2\pi i}\int_{(R)} L_{F^{(d)}}(1/2+it+w)C_d^w \frac{\Gamma(1+it+w)}{\Gamma(1+it)}\frac{e^{w^2}}w\,dw.\] Moving the contour to \(\Re w=-R\) crosses only the residue \(L_{F^{(d)}}(1/2+it)\), because the completed function is entire. Apply its functional equation on the new line and replace \(w\) by \(-w\). One obtains the exact approximate functional equation \[ L_{F^{(d)}}(1/2+it)=I_d(t)+\omega_d(t)I_d(-t),\qquad \omega_d(t)=\epsilon_d C_d^{-2it} \frac{\Gamma(1-it)}{\Gamma(1+it)}, \quad |\omega_d(t)|=1. \tag{29}\] The Gaussian makes the contour displacement legitimate. Initially, on \(\Re w=R\), the Dirichlet series is absolutely convergent, so we may insert the dyadic partition.

For \(0\leq c\leq R\), Stirling’s formula uniformly gives \[ \left|\frac{\Gamma(1+c+i(t+v))}{\Gamma(1+it)}\right| \ll_R T^c(1+|v|)^{R+1/2}e^{\pi|v|/2}. \tag{30}\] Indeed the quotient of the exponential factors is at most \(e^{\pi|v|/2}\), and \[\frac{(1+|t+v|)^{c+1/2}}{(1+|t|)^{1/2}} \leq T^c(1+|v|)^{c+1/2}.\] The same estimates with \(1+|t+v|\) also cover bounded ordinates, including \(v\) near \(-t\). Thus the bound is uniform as \(c\) tends to zero. For a fixed dyadic piece the polynomial is finite, so its contour can be moved to any \(0<c\leq R\). Multiplying (30) by \(|e^{(c+iv)^2}|\) absorbs its \(v\)-factors into \(O_R(e^{-v^2/2})\) and yields \[ |L_{F^{(d)}}(1/2+it)| \ll_{R,\mathcal F}\sum_{\nu=\pm1}\sum_N \int_{\mathbb R}\frac{e^{-v^2/2}}{|c_N+iv|} \left(\frac{MT}N\right)^{c_N} |P_d(N,\nu t+v;G_0(y)y^{-c_N})|\,dv. \tag{31}\] There is no use of the desired moment to justify this representation: on the initial fixed line \(R\), the elementary bound \(|P_d(N,t;G)|\ll_G\sqrt N\log(2+N)\) makes the sum of the long pieces converge, since \(R>1/2\). Finitely many shorter pieces can then be moved separately. The moment estimates below also show convergence after taking the family norm.

Use \[c_N=\frac1{\log(3MT)}\quad(N\leq MT^2),\qquad c_N=R\quad(N>MT^2).\] Apply (28) uniformly to the fixed smooth family \(G_0(y)y^{-c}\), \(0\leq c\leq R\), and use Minkowski in (31). We spell out the two contributions. For the nonrecursive square-root term and \(N\leq MT^2\), \[[L(1+N/M)(1+|\nu t+v|)]^{C_2/2} \ll L^{C_2/2}T^{3C_2/2}(1+|v|)^{C_2/2}.\] There are \(O(\log(3MT^2))\) such dyadic pieces, \((MT/N)^{c_N}\leq e\) for \(N\geq1\), and \(1/|c_N+iv|\leq\log(3MT)\). Since both logarithms are \(O(LT)\), their squared total cost gives at most \(C_{**}(LT)^{3C_2+4}\) in the mean square. For the long pieces put \(q'=N/M>T^2\). Then \[\sum_{q'>T^2}\left(\frac T{q'}\right)^R(q')^{C_2/2} \ll_R T^{-R+C_2},\] where \(q'\) runs over a dyadic progression. Including the outside factor \(L^{C_2/2}T^{C_2/2}\) again fits within \(C_{**}(LT)^{C_3}\) by (26).

For the recursive square-root term, the short dyadic sum satisfies \[\sum_{N\leq MT^2}(1+N/M)^{-2}\ll\log(3M)=L.\] The Gaussian absorbs \((1+|v|)^{(A-4)/2}\), so the resulting norm is at most \[C_{A,R,\mathcal F}\sqrt{\varepsilon K}\, L^{B/2-5}T^{A/2-2}\,L\log(3MT) \ll C_{A,R,\mathcal F}\sqrt{\varepsilon K}\, L^{B/2}T^{A/2}.\] For the long pieces we have the stronger estimate \[ \sum_{q'>T^2}\left(\frac T{q'}\right)^R(1+q')^{-2} \ll_R T^{-R-4}, \tag{32}\] which gives the same bound. Squaring the total norm now proves \[ \frac1M\sum_d|L_{F^{(d)}}(1/2+it)|^2 \leq C_{**}(LT)^{C_3} +C_{\dagger}\varepsilon K L^BT^A. \tag{33}\] Here \(C_{\dagger}\) depends on \(A,R,\mathcal F\) but not on \(P_0,K,M,t\); \(C_{**}\) may depend on \(P_0\). This independence is essential when choosing the constants.

Closing the induction in the strip. It remains to extend the estimate to the strip without a factor for the number of twists. Form the vector \[\mathbf V_F(s)=\big(M^{-1/2}L_{F^{(d)}}(s)\big)_d\] in the finite-dimensional Euclidean space indexed by the current admissible block. On \(\Re s=2\) its squared norm is bounded by an absolute constant: each \(L\)-value has modulus at most \(\sum_n\tau(n)n^{-2}=\zeta(2)^2\), and the number of parameters is \(O(M)\). On \(\Re s=1/2\), (33) gives the stated vector norm bound.

For each constant unit vector \(u\) apply the scalar strip principle to \[f_u(s)=\frac{\langle\mathbf V_F(s),u\rangle}{(1+s)^{A/2}}.\] Here \(A\) is even by (26). On both boundaries \(|1+s|\asymp1+|\Im s|\), with constants independent of the number of twists. The scalar boundary bound is therefore at most \[C_A\big(C_{**}L^{C_3}+C_{\dagger}\varepsilon K L^B+1\big)^{1/2}.\] For clarity, finite-order growth suffices for the strip principle: multiply \(f_u(s)\) by \(\exp\{-\rho\cos(s-5/4)\}\), with \(\rho>0\). Its modulus is at most one on the strip boundaries and it decays faster than every finite-order growth bound on horizontal sides, since \(\cos(\Re s-5/4)>0\) for \(1/2\leq\Re s\leq2\). Apply the maximum principle on rectangles and then let their heights tend to infinity and \(\rho\) decrease to zero. Taking the supremum over unit vectors at the desired point gives \[ \|\mathbf V_F(\sigma+it)\|_2^2 \leq C_{\rm st} \big(C_{**}L^{C_3}+C_{\dagger}\varepsilon K L^B+1\big)T^A, \qquad 1/2\leq\sigma\leq2. \tag{34}\] This argument has no dimension-dependent loss.

The choice order is now explicit. Fix the family, the elementary arithmetic exponent, \(C_0,C_1,C_2,R,C_3,A,B\) as above, and then all needed fixed kernel orders. Next choose \(0<\varepsilon<(4C_{\rm st}C_{\dagger})^{-1}\), choose \(P_0\) to obtain (28) for the entire weight family, and finally choose \(K\) so large that \(C_{\rm st}(C_{**}L^{C_3}+1)\leq K L^B/2\) for every \(L\geq\log3\). This is possible because \(B>C_3\) and \(C_{**}\) is independent of \(K\). Equation (34) proves (19) at height \(M\). At \(M=1\), all dual sums used above are empty, so this also establishes the initial step. The simultaneous induction is complete.

The induction just completed uses the fixed weight family \(G_0(y)y^{-c}\), \(0\leq c\leq R\). For any other fixed smooth compactly supported \(G\), repeat the polynomial argument through (28), using the established \(L\)-function bound at smaller heights. The constants \(P_0\) and \(C_*\) may now depend on \(G\), while \(K,A,B\) remain fixed. Since \(A,B>C_2\), this proves (9), with the asserted uniformity for fixed uniformly smooth weight families. Finally use Lemma 4, bounded multiplicity, and a sum over admissible dyadic blocks of total length \(O(H)\) to obtain (8). No analytic-density input has entered the argument. ◻

Proof. If \(Q<1\), bound \((1+T/Q)^{-J_1}\) by \((Q/T)^{J_1}\). The remaining series is \(\sum_{j\geq0}2^{-(J_1-1)j}\log^B(3\cdot2^j)<\infty\). For \(Q\geq1\), the part \(T\leq Q\) is bounded by \(\log^B(3Q)\sum_{T\leq Q}T\ll Q\log^B(3Q)\). For the remaining part choose the first dyadic \(T_0>Q\), so \(Q<T_0\leq2Q\), and write \(T=2^jT_0\). The summand is at most \[C_B Q\,2^{-(J_1-1)j} \big(\log^B(3Q)+(j+1)^B\big).\] Both resulting geometric series converge. This estimates the infinite frequency tail before any logarithm is replaced by \(\log X\). ◻

Proof. Put \(L=X^{1-\Delta}\). If the length hypothesis specifies only that the integer coefficients vanish for \(n>L\), multiply the interpolating weight by \(\rho_+(n/L)\), where \(\rho_+\) is a fixed smooth function equal to one on \((-\infty,1]\) and zero on \([2,\infty)\). This preserves every integer coefficient and gives real support in \(n\leq2L\). On a dyadic scale meeting the transition, \(N/L\) is bounded, so derivatives of \(\rho_+(Ny/L)\) add only fixed constants to the assumed scaled seminorm bounds. Thus all nonempty localized lengths satisfy \(N_i\ll L\).

Expand the square on the left of (37). A term involving two multiplier indices \(r_1',r_2'\) has character factor \(z_m(r_1'r_2'n_1n_2)\). Its odd multiplier part is an integer \(r\leq R_0^2\). Powers of two contribute only fixed signs. Split powers of two out of \(n_1,n_2\), and insert smooth dyadic partitions in their odd parts. Each resulting term has exactly the form (15) with \(U=X\), \(t_1=t_2=0\), and a weight \(W\) whose required seminorm is \(O(\ell^{D_1})\) for a fixed \(D_1\). The independent factors coming from the powers of two satisfy (18). Cross terms with \(A(x)\) are covered by using an index supported just at one: choose a fixed smooth function supported in \((1/2,3/2)\) and equal to one at one. The identity \(\lambda_F(1)=1\) then puts them in the same form. The term \(|A|^2\) is covered by two such indices. Its derivatives obey the same fixed polynomial bounds.

The zero frequency is exactly the model expression. In fact, for an odd prime and \(v\geq1\), \[\frac1{p^v}\sum_{a\bmod p^v}\left(\frac a{p^v}\right) =\begin{cases}0&v\text{ odd},\\1-p^{-1}&v\text{ even},\end{cases} =\mathbb E Z(p)^v.\] For \(v=0\) both sides are one. Chinese remaindering gives independence at distinct odd primes, and the value at two is already fixed. The integral of the zero-frequency smooth amplitude is thus the corresponding term on the right of (37).

It remains to estimate the nonzero frequencies. Fix an odd multiplier \(r\) and localized odd lengths \(N_1,N_2\), and set \[b=\frac1{10\ell},\qquad Q=\frac{rN_1N_2}{X}.\] Only nonempty lengths need be considered, so \(N_i\) is bounded below by a fixed positive constant. Also \(N_i\ll X^{1-\Delta}\), whence \((N_1N_2)^b\ll1\). Proposition 6 is available at every height. For a dyadic frequency range \(T\leq|h|<2T\), write \(h=h_1h_2^2\). For fixed \(h_2\), the parameter \(h_1\) lies in \(0<|h_1|<2T/h_2^2\). Cauchy–Schwarz and (8) imply \[\begin{align*} &\sum_{\substack{T\leq|h|<2T\\h=h_1h_2^2}} \tau(h_2)^{C_{\rm ar}} \prod_{i=1}^2 |L_{F^{(\nu h_1)}}(1/2+b+iv_i)|\\ &\quad\ll T\log^B(3T)(1+|v_1|)^{A/2}(1+|v_2|)^{A/2} \sum_{h_2\leq\sqrt{2T}}\frac{\tau(h_2)^{C_{\rm ar}}}{h_2^2}\\ &\quad\ll T\log^B(3T) (1+|v_1|)^{A/2}(1+|v_2|)^{A/2}. \end{align*}\] This inequality applies separately to each \(\nu\); it includes both signs of \(h\). Choose \(J_1>A+4\) in (16), and then fix the corresponding smoothness order. Its vertical integral is convergent. The remaining sum is bounded, by Lemma 11, by \[\ll \ell^{D_2}r^{C_{\rm ar}}(N_1N_2)^{-1/2} \begin{cases}Q^{J_1}&Q<1,\\Q\log^B(3Q)&Q\geq1. \end{cases}\] When \(Q\geq1\) we have \(Q\ll X\) using \(r\leq R_0^2\), (36), and \(\eta<1\); hence \(\log(3Q)\ll\ell\). When \(Q<1\) use \(Q^{J_1}\leq Q\). In both regimes the error for this localized pair is therefore \[ \ll\ell^{D_3}r^{C'}\frac{\sqrt{N_1N_2}}X, \qquad C'=C_{\rm ar}+1. \tag{39}\] The increase from \(C_{\rm ar}\) to \(C'\) is only the factor \(r\) in \(Q\), not a smoothness loss. Notice that this argument controls arbitrarily large dual frequencies as well as the case \(Q<1\).

There are \(O(\ell)\) possible nonempty dyadic pieces for each main index. Their cost is a fixed logarithmic power. Summing the coefficients from powers of two by (18) costs only a constant. The two multiplier coefficient sums and the bound \(r\leq R_0^2\) cost at most \[(1+\textstyle\sum|h(r')|)^2R_0^{2C'}\ll_{C_H}R_0^{4+2C'}.\] Since \(\sqrt{N_1N_2}/X\ll X^{-\Delta}\), including the terms with an index at one, the full error is \[ \ll\ell^{D_4}X^{-\Delta}R_0^{4+2C'} \leq\ell^{D_4}\exp(-3\Delta\ell/4). \tag{40}\] Our choices in (38) give the last inequality. The power \(D_4\) is fixed before choosing \(X\). Because \(\Delta\ell\geq c_1\ell^{9/10}\), for one sufficiently large \(X\) the factor \(\ell^{D_4}\) is at most \(\exp(\Delta\ell/4)\). This proves (37) with \(c_2=1/2\). ◻

Proof. The completed function is \[\Lambda_d(s)=(C_F|d|)^s\Gamma(s+1/2)L_{F^{(d)}}(s).\] Thus \(g_d(z)=(C_F|d|)^{-1/2}\Lambda_d(1/2+z)\). Its entireness and reflection law follow from those of \(\Lambda_d\).

For \(T>0\), Mellin inversion gives \[\frac{1}{2\pi i}\int_{(2)}\frac{T^z}{z^{k+1}}\,dz =\frac{(\log T)_+^k}{k!}.\] Insert the absolutely convergent Dirichlet series for \(L_{F^{(d)}}(1/2+z)\), and use \(\Gamma(1+z)=\int_0^\infty e^{-v}v^z\,dv\). On this line all exchanges are absolutely convergent: the series is dominated by \(\sum_n\tau(n)n^{-5/2}\), and the other two factors to be integrated are dominated by \(e^{-v}v^2\) and \(|2+it|^{-k-1}\). The result is (43), because \[\ell^{-1}\log(C_F|d|v/n) =1-\ell^{-1}\log n+\ell^{-1}\log(C_F|x|v).\] This argument, or the same computation with absolute coefficient values, also proves absolute convergence of the series.

Let \(I_+\) and \(I_-\) denote the integrals of \(g_d(z)z^{-k-1}/(2\pi i)\) on the upward oriented lines \(\operatorname{Re}z=2\) and \(\operatorname{Re}z=-2\). Shifting between these lines crosses only the pole at zero, and gives \[I_+-I_-=\frac{g_d^{(k)}(0)}{k!}.\] The horizontal integrals tend to zero: in this fixed strip the uncompleted \(L\)-function has polynomial growth, while the gamma factor has exponential decay on horizontal segments tending to infinity. Apparent gamma poles on the real axis are removable in \(g_d\). Changing \(z\) to \(-z\), with orientations included, gives \[I_-=(-1)^{k+1}\epsilon_d I_+.\] Multiplication by \(k!/\ell^k\) proves (44). Finally, the completing factor is holomorphic and nonzero at zero. Hence \(a(F^{(d)})>k\) implies \(g_d^{(k)}(0)=0\), and the stipulated sign makes the coefficient on the left of (44) equal to two. ◻

Proof. It is convenient first to differentiate using \(D_x=x\partial_x\). For \[h=1-u+\ell^{-1}\log(C_F|x|v)\] and \(a+b\le r\), differentiation under the integral gives \[ \partial_u^aD_x^b h_+^k =(-1)^a\ell^{-b}(k)_{a+b}h_+^{k-a-b}, \tag{54}\] where \((k)_t=k(k-1)\cdots(k-t+1)\). The choice \(k\ge r+2\) ensures enough continuous derivatives at \(h=0\); domination for differentiating follows from the estimates below. Ordinary \(x\)-derivatives are fixed linear combinations of \(x^{-b}D_x^i\), \(i\le b\), and therefore satisfy the same bounds on our compact set.

Suppose \(1-u\le\delta'\), where \(\delta'>0\). For \(0\le t\le r\), write \(m=k-t>0\) and \(a_t=m/(\ell\delta')\). The elementary inequality \((1+q)_+^m\le e^{mq}\), valid for every real \(q\), gives \[ h_+^m \le(\delta')^m(C_F|x|v)^{a_t}. \tag{55}\] If \(\delta'=\delta_j\), then \(0<a_t\le\ell^{-3/10}\) by (47), so \[\int_0^\infty e^{-v}(C_F|x|v)^{a_t}\,dv =(C_F|x|)^{a_t}\Gamma(1+a_t)\le C.\] The bound is uniform once \(X\) is large enough that \(a_t\le1\). Thus, on \(1-u\le\delta_j\), a derivative of \(W_{k,x}\) involving at most \(r\) differentiations is bounded by a constant times \(k^r\delta_j^{k-r}\). The cutoff derivatives of order \(q\) cost \(O_r(\delta_j^{-q})\). More precisely, each Leibniz term with \(t\) derivatives hitting \(W\) and \(q\) hitting a cutoff has \(t+q\le r\), and is at most \[C_r k^t\delta_j^{k-t-q} \le C_r(k/\delta_j)^r\delta_j^k.\] This proves (51).

For \(u\ge-\log2/\ell\), use instead \(\delta'=1+\log2/\ell\). Now \((\delta')^k\le\exp(k\log2/\ell)\le2\) for large \(X\), and \(a_t\le k/\ell\le\ell^{-2/5}\). The same argument and the fixed derivative bounds for \(F_0\) prove (52).

On the support of the terminal cutoff, or any nonzero derivative of it, \(1-u\le\delta_J/2\le\delta_J\). Moreover \(h>0\) implies \[v>\frac{n}{C_F|x|X}.\] Retain half the exponential in the integral. Uniformly on our compact set, \[ \begin{split} \int_{\{h>0\}}e^{-v}(C_F|x|v)^{a_t}\,dv &\le e^{-n/(2C_F|x|X)} (C_F|x|)^{a_t} \int_0^\infty e^{-v/2}v^{a_t}\,dv\\ &=e^{-n/(2C_F|x|X)} (C_F|x|)^{a_t}2^{1+a_t}\Gamma(1+a_t) \le C e^{-cn/X}. \end{split} \tag{56}\] Here \(a_t=(k-t)/(\ell\delta_J)\le\ell^{-3/10}\), so \(2^{1+a_t}\Gamma(1+a_t)\) is bounded independently of \(k\). The same Leibniz calculation now proves (53). All integrands used for differentiation are dominated on compact sets by the displayed integrable majorants, which also justifies (54) under the integral. ◻

Proof. The bounds \(k\le\ell^{3/5}\) and \(\delta_j^{-1}\le\ell^{1/10}\) make (51) polynomial in \(\ell\). Choose a fixed smooth function \(\chi\) equal to zero on \((-\infty,-\log2]\) and equal to one on \([0,\infty)\). For the comparison alone, replace \(w_{0,x}(u)\) on the real line by \[\widetilde w_{0,x}(u)=\chi(\ell u)w_{0,x}(u).\] It agrees with the original weight at every positive integer \(n=e^{\ell u}\), and its extra derivatives cost only fixed powers of \(\ell\), by (52). Its left support is \(u\ge-\log2/\ell\), exactly the enlargement \(n\ge1/2\). For \(j\ge1\), the weights already vanish on a neighborhood of \(u\le0\). After a fixed smooth dyadic localization \(n=Ny\), the operator \(y\partial_y\) acts as \(\ell^{-1}\partial_u\), so the same bounds give all the required seminorms in the scaled variables. The endpoint extension has therefore taken place in \(\log n\); no extension to a fixed negative value of \(u\) is used.

Finally \(0<s\le\ell^{-2/5}\), while \(C_F|x|\) lies in a fixed positive compact interval. The formula for \(A_x\), its derivatives in \(x\), and the continuity and positivity of \(\Gamma(1+s)\) give the assertions concerning \(A_x\). ◻

Proof. Choose a smooth dyadic partition with a fixed \(\rho\in C_c^\infty((1/2,2))\) satisfying \[\sum_{N\in\{1,2,4,\ldots\}}\rho(n/N)=1\qquad(n\ge1).\] For example one may start with a smooth nonincreasing function equal to one on \((0,1]\) and zero on \([2,\infty)\), and take its difference at arguments \(y\) and \(2y\); choosing its transition strictly inside \((1,2)\) gives the asserted compact support. Fix also \(G\in C_c^\infty((0,\infty))\) equal to one on \(\operatorname{supp}\rho\). Put \[T_N(y,x)=\rho(y)w_{*,x}(\ell^{-1}\log(Ny)),\qquad \widehat T_N(t,x)=\int_0^\infty T_N(y,x)y^{it}\frac{dy}{y}.\] Select once and for all integers \(r>A/2+2\) and \(D>A/2+2\), and then choose \(k_0\ge r+2\). If derivatives required elsewhere are of higher order, increase this fixed \(r\) first. By Lemma 14, differentiation on the \(y\)-scale through order \(r\) gives \[\left|(y\partial_y)^a T_N(y,x)\right| \ll_r \ell^{C_r}\delta_J^k e^{-c'N/X} \qquad(0\le a\le r),\] uniformly in \(x\in\mathcal K\). There is no boundary term when integrating by parts in \(\log y\). Consequently, \[ \sup_{x\in\mathcal K}|\widehat T_N(t,x)| \ll_{r,D}\ell^{C_r}\delta_J^k (1+N/X)^{-D}(1+|t|)^{-r}. \tag{60}\] In passing from the exponential to \((1+N/X)^{-D}\), the constant depends on the fixed \(D\), not on \(k\).

Mellin inversion now expresses the \(N\)-piece of the terminal sum as \[\frac1{2\pi}\int_{\mathbb R} \widehat T_N(t,d/X)N^{it} P_d(N,t)\,dt,\qquad P_d(N,t)=\sum_{n\ge1}\frac{\lambda(n)\chi_d(n)}{\sqrt n}\, n^{-it}G(n/N).\] Indeed \(G(n/N)T_N(n/N,x)=T_N(n/N,x)\). Define the normalized family norm by \[\|B_d\|_{2,X} =\left(\frac1X\sum_{\substack{d\ {\rm admissible}\\X\le|d|<2X}} |B_d|^2\right)^{1/2}.\] We bound the kernel uniformly before applying the family moment. For each fixed \(t\), \[\|\widehat T_N(t,d/X)P_d(N,t)\|_{2,X} \le \sup_{x\in\mathcal K}|\widehat T_N(t,x)| \|P_d(N,t)\|_{2,X}.\] Minkowski’s inequality, followed by (58) and (60), therefore yields \[\begin{split} \|S_{*,\chi_d}(d/X)\|_{2,X} &\ll \ell^{C_r+B/2}\delta_J^k \sum_{N=1,2,4,\ldots}(1+N/X)^{-D+A/2} \int_{\mathbb R}(1+|t|)^{-r+A/2}\,dt\\ &\ll \ell^{C_r+B/2+1}\delta_J^k. \end{split}\] The integral converges by the fixed choice of \(r\). There are \(O(\ell)\) dyadic terms with \(N\le X\), and those with \(N>X\) form a convergent geometric tail by the choice of \(D\). All sums and integrals can first be truncated; the displayed integrable majorants justify passage to the limit. Squaring proves (59), for example with \(C_T=2C_r+B+2\). This exponent depends only on choices made before \(k_0\). Restricting the family can only decrease the nonnegative sum. ◻

Proof. The ratio of the right side of (59) to \(2^{-3Jk/2}\), apart from its fixed implied constant, is \[\ell^{C_T}2^{-Jk/2}.\] For \(J\ge1\) this decreases with \(k\), so throughout the permitted range it is at most its value at the fixed integer \(k_0\). The floor in the definition of \(J\) gives \[\ell^{C_T}2^{-Jk/2} \le \ell^{C_T}2^{-Jk_0/2} \le 2^{k_0/2}\ell^{C_T-k_0/20}.\] Choose \(k_0>20C_T\) after \(C_T\) is fixed. The last expression then tends to zero with \(X\). In particular the choice of a single \(X\)-threshold works for the entire \(k\)-range.

For the second assertion, \(\Delta_j\ell\ge\ell^{9/10}/4\), whereas uniformly in the same range \[L\log k+Ljk\log2 =O_L(\ell^{3/5}\log\ell)=o(\ell^{9/10}).\] Taking logarithms proves (62). ◻

Proof. In a block, a chosen monomial has prime exponents \(e_p\in\{0,1,2\}\) whose sum is at most \(m_{ji}\). Its integer index is therefore at most \(\exp(m_{ji}2^{-i}\log Y_j)\). Since \[\sum_{i\ge0}m_{ji}2^{-i} \le\sum_{i\ge0}(k^{1/4}+j+i+1)2^{-i} =2k^{1/4}+2j+4,\] we obtain (70). Unique factorization shows that each index occurs in exactly one prime-exponent pattern. The truncations only delete patterns. A surviving coefficient is a product of local coefficients \(1\), \(-\lambda(p)p^{-1/2-s}\), or \(\xi_pp^{-1-2s}\), and hence is bounded by \(\tau(r)r^{-1/2-s}\le\tau(r)r^{-1/2}\). For example, \[\sum_{r\le R}\tau(r)r^{-1/2} =\sum_{ab\le R}(ab)^{-1/2} \le2\sqrt R\sum_{a\le R}\frac1a \ll\sqrt R\log(2R)\ll R^2\qquad(R\ge1).\] Finally, \[\frac{\log R_j}{\ell\delta_j/4} \le4C_{\rm len}\bigl(k^{-1/4}+(j+1)2^{-j}k^{-1/2}\bigr) \le8C_{\rm len}k^{-1/4}.\] Here \((j+1)2^{-j}\le1\). Thus the fixed choice \(k_0\ge(8C_{\rm len}/\eta)^4\) suffices simultaneously for all \(j\). The coefficient estimate and this length bound verify both multiplier conditions of Proposition 12. ◻

Proof. First omit the deterministic prime \(2\) and write \(a_p=\lambda(p)p^{-1/2-s}\). On expanding the model moment, any odd prime that occurs just once has zero expectation. Group the remaining terms by their partition of the \(64\) positions into \(q\le32\) parts of sizes at least two. Since \(|a_p|\le2\), for every part of size \(r\ge2\) we have \(|a_p|^r\le2^{r-2}|a_p|^2\). Summing the distinct-prime choices and then discarding distinctness bounds the moment by \[C\sum_{q=1}^{32}\left(\sum_{p\in\mathcal I}|a_p|^2\right)^q \ll\left(1+\sum_{p\in\mathcal I}\frac1p\right)^{32}.\] Restoring the prime \(2\) changes only the constant, by \(|u+v|^{64}\le2^{63}(|u|^{64}+|v|^{64})\) and \(|a_2c|\le\sqrt2\).

For completeness the passage to integers needs no estimate for primes in progressions. For each ordered \(64\)-tuple of primes, the product of its character values is periodic modulo the product of its odd primes, which is at most \(Y_0^{64}\). Its absolute value is at most one. The arithmetic mean over the two integer intervals \(X\le|m|<2X\) differs from its complete-residue mean by \(O(Y_0^{64}/X)\). The complete-residue mean is exactly the independent-model mean, by the Chinese remainder theorem and the counts of \(0,1,-1\) for the Legendre symbol. There are at most \(Y_0^{64}\) tuples and each coefficient has absolute value at most \(2^{64}\). Thus the discrepancy between that arithmetic mean and the model moment is \[O(Y_0^{128}/X)=O(X^{-1+128/\sqrt k})=O(X^{-1/2})\] after requiring \(\sqrt{k_0}\ge256\). Since the number of integers in these intervals is \(O(X)\), this proves (71) with its normalization by \(X\). If endpoints are not integers, the same period-counting argument changes only its absolute constant. ◻

Proof. For a block \(I=I_{ji}\) write \(B=B_I(z_m)\). The elementary expansion \(\log(1+w)=w+O(|w|^2)\) at all sufficiently large primes, together with (68), gives \[ \sup_{|z|=2}|K_{ji}(z_m,z)|\le e^{2|B|+C}, \qquad K_{ji}(z_m,1)\ge e^{-|B|-C}. \tag{75}\] The finitely many smaller primes are included using (67) at \(z=1\) and the direct upper bound \(\prod_{\nu}(1+2p^{-1/2})\) on \(|z|=2\); they change \(C\) by an absolute amount. Thus no nonvanishing assertion on the radius-two circle is required. Cauchy’s coefficient estimate now gives, with \(m=m_{ji}\), \[\frac{|T_{ji}(z_m)-K_{ji}(z_m,1)|}{K_{ji}(z_m,1)} \le C2^{-m}e^{3|B|}.\] If every block satisfies \(|B_{I_{ji}}(z_m)|\le m_{ji}/100\), the last quantity is at most \(Ce^{-c m_{ji}}\) for a fixed \(c>0\). The sum over \(i\) is \(O(e^{-c(k^{1/4}+j)})\). By enlarging \(k_0\), the product over blocks of the ratios \(T_{ji}(z_m)/K_{ji}(z_m,1)\) lies between \(1/2\) and \(2\). Since \(\prod_iK_{ji}(z_m,1)=(P_j^{\mathrm{eu}}(z_m))^{-1}\), this proves (72) on the specified event. The number excluded, divided by \(X\), is at most \[C\sum_{j=0}^J\sum_{i\ge0}(k^{1/4}+j+i)^{-64} \ll (k^{1/4})^{-62}=k^{-31/2}\ll k^{-2},\] by Lemma 19 and (68). The sum may be extended to all \(i\) because only nonempty blocks impose an event.

The real logarithm is available by positivity, and the same local expansion, including the finitely many small primes separately, gives for every prime set \(\mathcal I\) \[ \log\prod_{p\in\mathcal I}D_p(s,z_m)^{-1} =B_{\mathcal I}(z_m)+O\left(\sum_{p\in\mathcal I}p^{-1}\right). \tag{76}\] For \(Y_j<p\le Y_0\), (63) bounds this reciprocal mass by \(C(1+j)\), since \(\log Y_0/\log Y_j=4^j\). For \(p\le Y_0\) it is at most \(C(1+J)\): indeed \(\log\log Y_0\le\log\ell\), and \(\log\ell<10(J+1)\log2\). Require that the corresponding prime sums have absolute values at most \((\log2)jk/16\) and \((\log2)Jk/16\), respectively. Once \(k_0\) dominates the fixed drift constants in (76), these requirements imply (73) and (74). Their total exceptional proportion is at most \[C\sum_{j=1}^J\frac{(1+j)^{32}}{(jk)^{64}} +C\frac{(1+J)^{32}}{(Jk)^{64}} \ll k^{-64}.\] Combining the exceptional sets proves the proposition. ◻

Proof. We first give the local cancellation with its dependence on \(b\). Write \(r=p^{-1/2}\), \(u=p^{-s}\), \(v=p^{-b}\), and \(M=\min(1,h\log p)\). The mean-value formula for the exponential and the trivial bound imply \[ |u-v|\ll M(u+|v|). \tag{81}\] By (67), uniformly for every prime, \[D_p(b,Z)^{-1}-1=O(r|v|).\] Subtract the two inverse polynomials and use this last estimate to obtain \[ \frac{D_p(s,Z)}{D_p(b,Z)} =1+\lambda(p)Z(p)r(v-u) +O\big(r^2|v-u|(u+|v|)\big). \tag{82}\] For odd \(p\), the displayed linear term has expectation zero. Expanding the squared absolute value, and using \(M^2\le M\) and \((u+|v|)^2\le2(u^2+|v|^2)\), therefore gives \[ \mathbb E\left|\frac{D_p(s,Z)}{D_p(b,Z)}\right|^2 \le \exp\left\{\frac Cp\min(1,h\log p) (p^{-2s}+p^{-2\alpha})\right\}. \tag{83}\] The constants in the error in (82) are uniform even at the small primes, because \((1-2^{-1/2})^2\) is a common positive lower bound for the denominator. At the deterministic prime \(2\) the local ratio has a fixed upper bound by (67).

For odd primes outside the mollifier, the expansion \[D_p(b,Z)^{-1}=1+\lambda(p)Z(p)p^{-1/2-b} +O(p^{-1-2\alpha})\] and centering give \[ \mathbb E|D_p(b,Z)^{-1}|^2 \le \exp(Cp^{-1-2\alpha}). \tag{84}\] Independence multiplies the local second moments exactly. Sum (83) by (65), using both \(a=s\) and \(a=\alpha\) there, and sum (84) by (64). Taking a square root proves (79).

For the distance estimate there is exact cancellation below \(Y_0\). Set \[U=(P_0^{\mathrm{eu}})^{-1}\mathcal L_Q(s) =\prod_{Y_0<p\le Q}D_p(s,Z)^{-1},\qquad \Xi=\sum_{p>Y_0}p^{-1-2s}.\] This is a positive real random variable. The same expansion gives \[\mathbb E D_p(s,Z)^{-1}=1+O(p^{-1-2s}),\qquad \mathbb E D_p(s,Z)^{-2}=1+O(p^{-1-2s}).\] The first error may have either sign. Nevertheless, the elementary product bound \(|\prod(1+e_p)-1|\le\exp(\sum|e_p|)-1\) and independence show that \[ \mathbb EU=1+O(\Xi),\qquad \mathbb EU^2=1+O(\Xi). \tag{85}\] Indeed (64) and \(s\log Y_0=\sqrt k\) give \[\Xi\ll\int_{2\sqrt k}^{\infty}e^{-v}\frac{dv}v \ll e^{-2\sqrt k}/\sqrt k,\] so \(\Xi\) is uniformly small after choosing \(k_0\). Finally, \[\|U-1\|_2^2=\mathbb EU^2-2\mathbb EU+1\ll\Xi.\] This uses both moment estimates in (85) and proves (80). ◻

Proof. For a block \(I=I_{ji}\) let \(\mathcal L_I(b)=\prod_{p\in I}D_p(b,Z)^{-1}\). We claim there are fixed positive constants \(c_*,C_*\) such that \[ \|K_{ji}(Z,1)\mathcal L_I(b)\|_2\ge c_*,\qquad \sup_{|z|=2}\|K_{ji}(Z,z)\mathcal L_I(b)\|_2\le C_*. \tag{87}\] For the first assertion, at a large odd prime the exact local ratio is \[\frac{D_p(s,Z)}{D_p(b,Z)} =1+\lambda(p)Z(p)p^{-1/2}(p^{-b}-p^{-s})+O(p^{-1}).\] The centered linear term shows that its local second moment is \(1+O(p^{-1})\); for all sufficiently large primes it is at least \(1-C/p\ge1/2\). Their product is bounded below by \(\exp(-2C\sum_{p\in I}1/p)\), which is a fixed positive constant by (68). At each of the finitely many smaller primes, including \(2\), (67) gives the pointwise lower bound \[\left|\frac{D_p(s,Z)}{D_p(b,Z)}\right| \ge\frac{(1-p^{-1/2})^2}{(1+p^{-1/2})^2}>0.\] Their finite product is also bounded below. This proves the first assertion of (87).

For the second, on \(|z|=2\) the local quotient is \[\frac{1-\lambda(p)Z(p)p^{-1/2-s}z +\xi_p Z(p)^2p^{-1-2s}z^2}{D_p(b,Z)} =1+\lambda(p)Z(p)p^{-1/2}(p^{-b}-zp^{-s})+O(p^{-1})\] at large odd primes, uniformly in \(z\) and \(b\). Its second moment is \(1+O(p^{-1})\), and therefore its product is bounded above, again by (68). At small primes use the direct numerator upper bound and (67). The numerator is permitted to vanish; only an upper bound has been used on this circle.

Apply Cauchy’s coefficient formula to the polynomial \(z\mapsto K_{ji}(Z,z)\mathcal L_I(b)\) with values in \(L^2\). For \(m=m_{ji}\), Minkowski’s inequality and (87) give \[\|(T_{ji}(Z)-K_{ji}(Z,1))\mathcal L_I(b)\|_2 \le C_*\sum_{h>m}2^{-h}\le C_*2^{-m}.\] Dividing by the positive lower bound in (87) makes this a relative error \(\varepsilon_i=C2^{-m_{ji}}\).

To see explicitly how relative errors combine, put \(A_i=K_{ji}(Z,1)\mathcal L_I(b)\) and \(B_i=T_{ji}(Z)\mathcal L_I(b)\). Then \(\|B_i-A_i\|_2\le\varepsilon_i\|A_i\|_2\) and \(\|B_i\|_2\le(1+\varepsilon_i)\|A_i\|_2\). Variables belonging to distinct blocks and to the residual primes \(Y_j<p\le Q\) are independent. In the telescoping identity for \(\prod B_i-\prod A_i\), the norm of each product of factors on distinct blocks is the product of their norms. Thus its relative norm, also including the independent residual product, is at most \[\sum_i\varepsilon_i\prod_{h<i}(1+\varepsilon_h) =\prod_i(1+\varepsilon_i)-1 \le\exp\left(\sum_i\varepsilon_i\right)-1.\] This proves (86), without a factor equal to the number of blocks. Finally \(\sum_i2^{-m_{ji}}\le2^{1-k^{1/4}-j}\). ◻

Proof. In (79), first take \(\alpha=s/2\), \(t=\theta s\), and \(j=0\). Then \(h/\min(\alpha,s)=3+2|\theta|\) and \(\alpha\log Y_0=\sqrt k/2\). This proves (88) with the exact multiplier; use Lemma 22 to replace it by \(H_0\). For (90), take \(\alpha=1/\ell\) and \(t=\theta/\ell\). Since \(k\ge1\), \[h/\min(\alpha,s)=1+k+|\theta|, \qquad \alpha\log Y_j=\delta_j^2/\sqrt k.\] The bound follows from (79) after increasing the fixed exponent \(C\), and then from relative truncation. Finally, (80) bounds the distance from one for the exact multiplier and also bounds its norm. The triangle inequality and (86) give \[\|H_0\mathcal L_Q(s)-1\|_2 \ll 2^{-k^{1/4}}+e^{-\sqrt k}k^{-1/4} \ll e^{-c_0k^{1/4}}.\] ◻

Proof. For \(z<k\) the derivative of the power is \[(-1)^r\frac{(k)_r}{k^r}(1-z/k)^{k-r},\qquad (k)_r=k(k-1)\cdots(k-r+1),\] and it is zero for \(z\ge k\). The formula is continuous through \(z=k\) in the derivative orders under consideration. For \(z<k\), the logarithm of its ratio in absolute value to \(e^{-z}\) is \[\log((k)_r/k^r)+z+(k-r)\log(1-z/k).\] The last two terms have derivative \((r-z)/(k-z)\), so their maximum over \(z<k\) is at \(z=r\). It follows at once that the absolute ratio is at most \(e^r\). For \(|z|\ge\sqrt k\) this bound and \(k^{-1}(1+|z|)^2\ge1\) prove (91). For \(|z|<\sqrt k\) we have \(|z/k|\le1/2\). Taylor’s formula for \(\log(1-z/k)\) and \(\log((k)_r/k^r)=O_r(k^{-1})\) bound the displayed logarithm by \[O_r\big(k^{-1}(1+|z|)^2\big).\] This quantity is bounded in this range; applying \(|e^w-1|\le |w|e^{|w|}\) proves the result there too. All estimates hold for negative \(z\) as well as positive \(z\). ◻

Proof. Write \(q=C_F|x|\), which lies in a fixed compact subinterval of \((0,\infty)\). The exact weight definition gives \[W_{k,x}(y/k)=\int_0^\infty e^{-v} \left(1-\frac{y-s\log(qv)}k\right)_+^k\,dv,\] whereas \[\int_0^\infty e^{-v}e^{-y+s\log(qv)}\,dv=A_xe^{-y}.\] For every fixed \(M\), the logarithmic moments \[ \int_0^\infty e^{-v}(qv)^s (1+|s\log(qv)|)^M\,dv\le C_M \tag{93}\] are uniform for \(0<s\le1\): on \(0<v\le1\), bound \(v^s\le1\) and \(s|\log v|\le|\log v|\); on \(v\ge1\), use \(v^s\le v\) and the exponential decay. Lemma 24 and differentiation under this dominated integral now imply, for \(0\le h\le r\) and \(y\ge-2\), \[ \left|\frac{d^h}{dy^h} \left(W_{k,x}(y/k)-A_xe^{-y}\right)\right| \ll_r k^{-1}(1+y^2)e^{-y}. \tag{94}\] In particular the proof uses (91) when \(y-s\log(qv)\) is negative and arbitrarily large in absolute value.

The cutoff \(F_0(y/k)\) is one for \(y\le k/2\), zero for \(y\ge3k/4\), and its \(h\)th derivative in \(y\) is \(O_h(k^{-h})\). Write \[E_x(y)=F_0(y/k)\big(W_{k,x}(y/k)-A_xe^{-y}\big) +(F_0(y/k)-1)A_xe^{-y}\] also for \(y\ge-2\). By (94), after multiplication by \(e^{y/2}\) the first term and its first \(r\) derivatives have \(L^1([-2,\infty))\) norm \(O_r(k^{-1})\). The second term is supported in \(y\ge k/2\) and has these norms \(O_r(e^{-k/4})\). Choose a fixed smooth function \(\rho\) equal to zero for \(y\le-2\) and one for \(y\ge-1\), and extend by \[g_x(y)=\rho(y)e^{y/2}E_x(y),\] with zero value to the left of \(-2\). All derivatives through order \(r\) have \(L^1(\mathbb R)\) norm \(O_r(k^{-1})\). Integration by parts \(r\) times and the bound without integration by parts give (92). Lastly, continuity and positivity of \(q^s\Gamma(1+s)\) on the compact parameter set \(q\in[q_-,q_+]\), \(0\le s\le1\), give the two bounds for \(A_x\). ◻

Proof. Choose a finite cutoff \(Q\) exceeding \(Y_0\) and the lengths of all the short polynomials \(S_{j,Z}\), \(0\le j\le J\). This is the only requirement on \(Q\); Proposition 23 is uniform in its choice. For \(j\ge1\), set \(g_{j,x}(u)=e^u w_{j,x}(u)\). The support is contained in \([1-\delta_j,1-\delta_j/4]\subset[1/2,1]\). The fixed-order derivative bounds in Lemma 14 and integration by parts give, for each required fixed \(r\), \[ |\widehat g_{j,x}(\theta)| \ll_r (k/\delta_j)^{C_r}\delta_j^k(1+|\theta|)^{-r}. \tag{97}\] Fourier inversion and (78) give the exact identity \[ S_{j,Z}(x)=\frac1{2\pi}\int_{\mathbb R} \widehat g_{j,x}(\theta) \mathcal L_Q((1+i\theta)/\ell)\,d\theta. \tag{98}\] Indeed every nonzero weight coefficient has index below the chosen cutoff and hence occurs in (78); all other indices contribute zero after inversion. At this fixed finite \(Q\), absolute convergence of the Dirichlet expansion and the integrable kernel justify interchanging the sum and integral. Multiply by \(H_j\) and use Minkowski’s inequality, (90), and (97), taking \(r>C+2\). Their integral is \(O((k/\delta_j)^{C_r}(k2^j)^C\delta_j^k)\), which proves (96) after fixing \(C_5\).

For the first piece, Lemma 25 and the same finite-product expansion give \[ S_{0,Z}(x)-A_x\mathcal L_Q(s) =\frac1{2\pi}\int_{\mathbb R}\widehat g_x(\theta) \mathcal L_Q(s/2+i\theta s)\,d\theta. \tag{99}\] To check the scaling, at an integer index \(n\) put \(y=k\log n/\ell=s\log n\ge0\). Fourier inversion for \(g_x\) then reads \[w_{0,x}(\log n/\ell)-A_x n^{-s} =\frac1{2\pi}\int\widehat g_x(\theta) n^{-s/2-i\theta s}\,d\theta,\] which is exactly the coefficient identity needed in (99). Use (88) and (92) with \(r>C+2\) to bound the norm of (99) after multiplication by \(H_0\) by \(O(k^{-1})\). By (89), \[\|H_0S_{0,Z}-A_x\|_2 \le \|H_0(S_{0,Z}-A_x\mathcal L_Q(s))\|_2 +A_x\|H_0\mathcal L_Q(s)-1\|_2 \ll k^{-1}+e^{-c_0k^{1/4}}\ll k^{-1}.\] This proves (95).

All smoothness orders used here are fixed after the absolute exponent in Proposition 23, and before choosing \(k_0\). The derivative estimates for the weights, the inequality \(s\le\ell^{-2/5}\), and the lower bound \(\delta_J\ge\ell^{-1/10}\) then give a single \(X\) threshold for the whole parameter range. We have not used the event of Proposition 20 anywhere in these model norm arguments. Nor is a passage to an infinite Euler cutoff required: the coefficient identities hold for every sufficiently large finite \(Q\), and all norm estimates are uniform in that cutoff. ◻

Proof. Write \(\ell=\log X\), and first consider the functional-equation sign \((-1)^k\), with \(k_0\le k\le\ell^{3/5}\). Fix \(c=\pm1\), and use the weights and partition of Section 4. We write \(P_j^{\rm eu}\) for the positive Euler products of Section 5.

Corollary 15 and Lemma 18 permit Proposition 12 to be applied to \(H_jS_j\), for \(0\le j\le J\), with \(\Delta=\delta_j/4\). By the first assertion of Proposition 26, outside \(O(X/k^2)\) integers in \(X\le|m|<2X\), \[ |H_0(z_m)S_{0,z_m}(m/X)-A_{m/X}|<\frac12 A_{m/X}. \tag{100}\] Indeed \(A_x=(C_F|x|)^s\Gamma(1+s)\) is bounded above and below by positive constants on the fixed \(x\)-range, uniformly for \(s=k/\ell\le\ell^{-2/5}\). Chebyshev’s inequality applies to the transferred \(O(k^{-2})\) mean-square error. The exponentially small comparison error is absorbed uniformly.

For \(1\le j\le J\), the second model estimate gives \[\frac1X\sum_m\Psi(m/X)|H_j(z_m)S_{j,z_m}(m/X)|^2 \ll (k2^j)^{2C_5}2^{-2jk}+e^{-c\delta_j\ell} \ll 2^{-3jk/2}.\] The last inequality follows first by increasing fixed \(k_0\) to absorb the polynomial factor. For the comparison error, use \[\delta_j\ell\ge\ell^{9/10},\qquad jk\ll\ell^{3/5}\log\ell,\] and then increase one \(X_0\) for all the permitted \(k,j\). The terminal estimate of Lemma 17 gives \[\frac1X\sum_{d\in\mathcal A(X)} |S_{*,\chi_d}(d/X)|^2\ll 2^{-3Jk/2}.\] Chebyshev and summation of the resulting geometric probabilities imply, outside \(O(X/k^2)\) further admissible parameters, \[ |H_j(\chi_d)S_{j,\chi_d}(d/X)|\le2^{-jk/2}\quad(1\le j\le J), \qquad |S_{*,\chi_d}(d/X)|\le2^{-Jk/2}. \tag{101}\] Here \(\sum_{j\ge1}2^{-jk/2}\ll k^{-2}\) after fixing \(k_0\), and \(J\ge1\) for sufficiently large \(X\).

Intersect these events with the event in Proposition 20. Its additional exceptional set is also \(O(X/k^2)\). On the intersection, \[\frac12\le H_jP_j^{\rm eu}\le2,\qquad P_j^{\rm eu}\le 2^{jk/8}P_0^{\rm eu},\qquad 1\le 2^{Jk/8}P_0^{\rm eu}.\] Choose a fixed \(a_*>0\) with \(A_x\ge a_*\) for all the parameters. Equation (100) yields \[|S_{0,\chi_d}(d/X)|\ge\frac{a_*}{4}P_0^{\rm eu}.\] Equation (101) yields \[\sum_{j=1}^J|S_{j,\chi_d}(d/X)|+|S_{*,\chi_d}(d/X)| \le P_0^{\rm eu} \left(2\sum_{j=1}^J2^{-3jk/8}+2^{-3Jk/8}\right).\] For a sufficiently large fixed \(k_0\), the last parenthesis is less than \(a_*/4\). The entire weighted sum is therefore nonzero. Lemma 13 excludes the simultaneous conditions \(a(F^{(d)})>k\) and \(\epsilon_{F^{(d)}}=(-1)^k\). Including both values of \(c\), we have proved \[ \#\{d\in\mathcal A(X): a(F^{(d)})>k,\ \epsilon_{F^{(d)}}=(-1)^k\} \ll Xk^{-2}. \tag{102}\] For the other functional-equation sign, apply (102) at \(k-1\). A rank exceeding \(k\) also exceeds \(k-1\), and \((k-1)^{-2}\ll k^{-2}\). This proves the proposition. ◻

Proof. Put \(L=L_{F^{(d)}}\) and \(C_d=C_F|d|\). On \(\Re s=5\), absolute convergence and (3) give a uniform bound. On \(\Re s=-1\), the functional equation is \[L(s)=\epsilon_{F^{(d)}} C_d^{\,1-2s} \frac{\Gamma(3/2-s)}{\Gamma(1/2+s)}L(1-s).\] At \(s=-1+it\), the modulus of the gamma quotient equals \[|(-1/2+it)(1/2+it)(3/2+it)|.\] Since \(L(2-it)\) is uniformly bounded, \[|L(-1+it)|\ll_E X^3(1+|t|)^3.\] Apply the strip principle to \(L(s)/(s+2)^4\) on \(-1\le\Re s\le5\). To see that the possibly conductor-dependent finite-order growth in the strip introduces no new constant, first multiply by \[\exp\!\left(-\varepsilon \cos\frac{\pi(s-2)}{12}\right),\qquad \varepsilon>0.\] Its modulus is at most one on this strip and decays faster than any finite-order growth as \(|\Im s|\to\infty\). The maximum principle on expanding rectangles gives the bound from the two vertical sides. Letting \(\varepsilon\) decrease to zero shows that \(L(s)/(s+2)^4\ll_E X^3\) throughout the strip. In particular, the maximum of \(|L|\) on the closed disk \(|s-2|\le5/2\) is \(O_E(X^3)\).

At the center, the Euler factors and their parameter bounds give \[|L(2)|\ge \prod_p(1+p^{-2})^{-2} =\left(\frac{\zeta(4)}{\zeta(2)}\right)^2>0.\] Jensen’s formula, with inner radius \(2\) and outer radius \(5/2\), therefore bounds the number of zeros in the inner disk, counted with multiplicity, by \(O_E(\log(3X))\). The point \(1/2\) lies in that inner disk. This proves the result. ◻

Proof of Theorem 2. We first work in \(\mathcal A(X)\) for a fixed \(F\in\mathcal F\). Let \[T_{F,X}(k)=\#\{d\in\mathcal A(X):a(F^{(d)})>k\}, \qquad K=\lfloor(\log X)^{3/5}\rfloor.\] For an integer \(R\ge k_0+1\) and \(X\) large enough that \(K>R\), the elementary integer-tail bound gives \[\begin{align*} \sum_{\substack{d\in\mathcal A(X)\\a(F^{(d)})>R}}a(F^{(d)}) &\le R T_{F,X}(R) +\sum_{k=R}^{K-1}T_{F,X}(k) +C_E\log X\,T_{F,X}(K) \\ &\ll_E \frac X R+X(\log X)^{-1/5}. \tag{103}\end{align*}\] For ranks between \(R+1\) and \(K\), the first two terms count the rank exactly. For ranks exceeding \(K\), the last term bounds the remaining mass by Lemma 28. Proposition 27 supplies every counting estimate used here. In particular the last term is \(O_E(X\log X/K^2)=O_E(X(\log X)^{-1/5})\). The constant in (103) is independent of \(R\).

For completeness, fix \(R\) and cover \(0<|d|\le Y\) by dyadic blocks. Enlarge the last partial block to a full block, using nonnegativity. The sum of their left endpoints is less than \(2Y\). The finitely many blocks below the threshold for (103) contribute a finite number \(B_{E,R}\). For the error term in the remaining blocks, split at \(X=\sqrt Y\). The smaller blocks contribute \(O_E(\sqrt Y)\), and in the larger ones \(\log X\ge\frac12\log Y\). Consequently \[\sum_{\substack{d\ {\rm admissible}\\0<|d|\le Y\\a(F^{(d)})>R}} a(F^{(d)}) \le B_{E,R}+\frac{C_EY}{R} +O_E\bigl(\sqrt Y+Y(\log Y)^{-1/5}\bigr).\] After division by \(Y\) and passage to the limsup, the desired \(C_E/R\) bound holds for each \(F\in\mathcal F\). Lemma 4, its bounded multiplicity, and the finite maximum of these constants prove the same bound for every original signed squarefree parameter of \(E\). ◻

Proof of Theorem 1. Write \(N(Y)=\#\mathcal D(Y)\) and let \(N_j(Y)\) count the parameters of analytic rank \(j\). The elementary squarefree sieve gives \[N(Y)=2\sum_{n\le Y}\mu(n)^2 =\frac{2Y}{\zeta(2)}+O(\sqrt Y).\] Indeed, expand \(\mu(n)^2=\sum_{r^2\mid n}\mu(r)\), interchange the finite sums, and bound the omitted tail of \(\sum_r\mu(r)/r^2=1/\zeta(2)\).

By the analytic density theorem (1), \(N_0(Y)/N(Y)\to1/2\) and \(N_1(Y)/N(Y)\to1/2\). Thus \(N(Y)-N_0(Y)-N_1(Y)=o(Y)\). For every fixed integer \(R\ge R_E\), \[\sum_{\substack{d\in\mathcal D(Y)\\2\le a(E^{(d)})\le R}} a(E^{(d)}) \le R\bigl(N(Y)-N_0(Y)-N_1(Y)\bigr)=o_{E,R}(Y).\] Theorem 2 now implies \[\limsup_{Y\to\infty}\frac1Y \sum_{\substack{d\in\mathcal D(Y)\\a(E^{(d)})\ge2}}a(E^{(d)}) \le\frac{C_E}{R}.\] Let \(R\to\infty\). The left side is zero by nonnegativity. The total rank sum is consequently \(N_1(Y)+o_E(Y)\). Dividing by \(N(Y)\) proves the limit \(1/2\). ◻