The mean analytic rank of quadratic twists of elliptic curves

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 [2, 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 (2.4). 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

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 (3.4); 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

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

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 (3.5).□

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

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

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 (3.5),

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 (3.5) 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

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

Combining the estimates proves (3.8) 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

Thus the normalized frequency factor is

Use (3.4) to replace the numerator by a fixed linear combination of B_h and B_{-h}.

Put \rho=hU/(rN_1N_2) and

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

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

Move both lines to \Re z_i=-1/2+b. Lemma 3.4 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 (3.11), makes the horizontal segments tend to zero. The scale factor is (N_1N_2)^{-1/2+b}. Applying (3.8) and (3.11), and discarding r^{-1/2}\leq1, proves (3.10). The final assertion follows directly from the support of \widehat W before Mellin inversion.□

Proof of Propositions 3.1 and 3.2. We give a simultaneous induction for all F\in\mathcal F. The induction hypothesis at smaller dyadic heights is

The constant K, common to the finite family, will be selected last. Since the reduction in Lemma 2.1 does not increase height and has bounded multiplicity, this hypothesis implies

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

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 [3, §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,

where we may fix C_0=512, independently of A,B and of the required smoothness orders.

To prove the claim, apply Lemma 3.5 with r=1 to the expansion of the square, first separating powers of two by (3.12). 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

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 (3.12). 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

Increasing the constant in (3.15) ensures H_0\leq M/2 for every such pair of lengths. For fixed h_2, Cauchy–Schwarz and (3.14) bound the sum over h_1 occurring in (3.10) by

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

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

Here (N_1N_2)^b\ll_G1 and \sqrt{N_1N_2}\ll_GN. The sums of the factors from two converge by (3.12). This proves (3.16). 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

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

where \mathcal B denotes the right side of (3.16). 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

show that these terms together have norm O(P^{1/2}\mathcal B^{1/2}). Summing the two choices of c proves

Here is an explicit noncircular choice of exponents. Set

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

Once A,B and the fixed weight family have been selected, a sufficiently large P_0 ensures (3.15) and gives

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

The recursive coefficient in (3.19) is therefore at most a fixed multiple of

Our value of C_1 leaves at least the losses 10,4,4 displayed in (3.22). 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

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

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

Indeed the quotient of the exponential factors is at most e^{\pi|v|/2}, and

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 (3.24) by |e^{(c+iv)^2}| absorbs its v-factors into O_R(e^{-v^2/2}) and yields

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

Apply (3.22) uniformly to the fixed smooth family G_0(y)y^{-c}, 0\leq c\leq R, and use Minkowski in (3.25). We spell out the two contributions. For the nonrecursive square-root term and N\leq MT^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

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 (3.20).

For the recursive square-root term, the short dyadic sum satisfies

The Gaussian absorbs (1+|v|)^{(A-4)/2}, so the resulting norm is at most

For the long pieces we have the stronger estimate

which gives the same bound. Squaring the total norm now proves

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

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, (3.27) gives the stated vector norm bound.

For each constant unit vector u apply the scalar strip principle to

Here A is even by (3.20). On both boundaries |1+s|\asymp1+|\Im s|, with constants independent of the number of twists. The scalar boundary bound is therefore at most

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

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 (3.22) 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 (3.28) proves (3.13) 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 (3.22), 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 (3.3), with the asserted uniformity for fixed uniformly smooth weight families. Finally use Lemma 2.1, bounded multiplicity, and a sum over admissible dyadic blocks of total length O(H) to obtain (3.2). 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

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 (3.31). 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 (3.9) 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 (3.12). 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,

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 (3.31).

It remains to estimate the nonzero frequencies. Fix an odd multiplier r and localized odd lengths N_1,N_2, and set

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 3.1 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 (3.2) imply

This inequality applies separately to each \nu; it includes both signs of h. Choose J_1>A+4 in (3.10), and then fix the corresponding smoothness order. Its vertical integral is convergent. The remaining sum is bounded, by Lemma 3.6, by

When Q\geq1 we have Q\ll X using r\leq R_0^2, (3.30), 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

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 (3.12) costs only a constant. The two multiplier coefficient sums and the bound r\leq R_0^2 cost at most

Since \sqrt{N_1N_2}/X\ll X^{-\Delta}, including the terms with an index at one, the full error is

Our choices in (3.32) 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 (3.31) with c_2=1/2.□

Proof. The completed function is

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

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 (4.3), because

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

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

Multiplication by k!/\ell^k proves (4.4). 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 (4.4) equal to two.□

Proof. It is convenient first to differentiate using D_x=x\partial_x. For

and a+b\le r, differentiation under the integral gives

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

If \delta'=\delta_j, then 0<a_t\le\ell^{-3/10} by (4.7), so

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

This proves (4.11).

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 (4.12).

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

Retain half the exponential in the integral. Uniformly on our compact set,

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 (4.13). All integrands used for differentiation are dominated on compact sets by the displayed integrable majorants, which also justifies (4.14) under the integral.□

Proof. The bounds k\le\ell^{3/5} and \delta_j^{-1}\le\ell^{1/10} make (4.11) 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

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 (4.12). 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

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

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 4.2, differentiation on the y-scale through order r gives

uniformly in x\in\mathcal K. There is no boundary term when integrating by parts in \log y. Consequently,

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

Indeed G(n/N)T_N(n/N,x)=T_N(n/N,x). Define the normalized family norm by

We bound the kernel uniformly before applying the family moment. For each fixed t,

Minkowski’s inequality, followed by (4.18) and (4.20), therefore yields

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 (4.19), 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 (4.19) to 2^{-3Jk/2}, apart from its fixed implied constant, is

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

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

Taking logarithms proves (4.22).□

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

we obtain (5.8). 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,

Finally,

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 3.7.□

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

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

after requiring \sqrt{k_0}\ge256. Since the number of integers in these intervals is O(X), this proves (5.9) 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 (5.6), gives

The finitely many smaller primes are included using (5.5) 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},

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 (5.10) on the specified event. The number excluded, divided by X, is at most

by Lemma 5.2 and (5.6). 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

For Y_j<p\le Y_0, (5.1) 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 (5.14), these requirements imply (5.11) and (5.12). Their total exceptional proportion is at most

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

By (5.5), uniformly for every prime,

Subtract the two inverse polynomials and use this last estimate to obtain

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

The constants in the error in (5.20) 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 (5.5).

For odd primes outside the mollifier, the expansion

and centering give

Independence multiplies the local second moments exactly. Sum (5.21) by (5.3), using both a=s and a=\alpha there, and sum (5.22) by (5.2). Taking a square root proves (5.17).

For the distance estimate there is exact cancellation below Y_0. Set

This is a positive real random variable. The same expansion gives

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

Indeed (5.2) and s\log Y_0=\sqrt k give

so \Xi is uniformly small after choosing k_0. Finally,

This uses both moment estimates in (5.23) and proves (5.18).□

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

For the first assertion, at a large odd prime the exact local ratio is

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 (5.6). At each of the finitely many smaller primes, including 2, (5.5) gives the pointwise lower bound

Their finite product is also bounded below. This proves the first assertion of (5.25).

For the second, on |z|=2 the local quotient is

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 (5.6). At small primes use the direct numerator upper bound and (5.5). 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 (5.25) give

Dividing by the positive lower bound in (5.25) 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

This proves (5.24), without a factor equal to the number of blocks. Finally \sum_i2^{-m_{ji}}\le2^{1-k^{1/4}-j}.□

Proof. In (5.17), 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 (5.26) with the exact multiplier; use Lemma 5.5 to replace it by H_0. For (5.28), take \alpha=1/\ell and t=\theta/\ell. Since k\ge1,

The bound follows from (5.17) after increasing the fixed exponent C, and then from relative truncation. Finally, (5.18) bounds the distance from one for the exact multiplier and also bounds its norm. The triangle inequality and (5.24) give

□

Proof. For z<k the derivative of the power is

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

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 (5.29). 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

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

whereas

For every fixed M, the logarithmic moments

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 5.7 and differentiation under this dominated integral now imply, for 0\le h\le r and y\ge-2,

In particular the proof uses (5.29) 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 hth derivative in y is O_h(k^{-h}). Write

also for y\ge-2. By (5.32), 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

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 (5.30). 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 5.6 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 4.2 and integration by parts give, for each required fixed r,

Fourier inversion and (5.16) give the exact identity

Indeed every nonzero weight coefficient has index below the chosen cutoff and hence occurs in (5.16); 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, (5.28), and (5.35), taking r>C+2. Their integral is O((k/\delta_j)^{C_r}(k2^j)^C\delta_j^k), which proves (5.34) after fixing C_5.

For the first piece, Lemma 5.8 and the same finite-product expansion give

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

which is exactly the coefficient identity needed in (5.37). Use (5.26) and (5.30) with r>C+2 to bound the norm of (5.37) after multiplication by H_0 by O(k^{-1}). By (5.27),

This proves (5.33).

All smoothness orders used here are fixed after the absolute exponent in Proposition 5.6, 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 5.3 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 4.3 and Lemma 5.1 permit Proposition 3.7 to be applied to H_jS_j, for 0\le j\le J, with \Delta=\delta_j/4. By the first assertion of Proposition 5.9, outside O(X/k^2) integers in X\le|m|<2X,

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

The last inequality follows first by increasing fixed k_0 to absorb the polynomial factor. For the comparison error, use

and then increase one X_0 for all the permitted k,j. The terminal estimate of Lemma 4.5 gives

Chebyshev and summation of the resulting geometric probabilities imply, outside O(X/k^2) further admissible parameters,

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 5.3. Its additional exceptional set is also O(X/k^2). On the intersection,

Choose a fixed a_*>0 with A_x\ge a_* for all the parameters. Equation (6.1) yields

Equation (6.2) yields

For a sufficiently large fixed k_0, the last parenthesis is less than a_*/4. The entire weighted sum is therefore nonzero. Lemma 4.1 excludes the simultaneous conditions a(F^{(d)})>k and \epsilon_{F^{(d)}}=(-1)^k. Including both values of c, we have proved

For the other functional-equation sign, apply (6.3) 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 (2.2) give a uniform bound. On \Re s=-1, the functional equation is

At s=-1+it, the modulus of the gamma quotient equals

Since L(2-it) is uniformly bounded,

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

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

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 1.2. We first work in \mathcal A(X) for a fixed F\in\mathcal F. Let

For an integer R\ge k_0+1 and X large enough that K>R, the elementary integer-tail bound gives

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 6.2. Proposition 6.1 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 (6.4) 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 (6.4) 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

After division by Y and passage to the limsup, the desired C_E/R bound holds for each F\in\mathcal F. Lemma 2.1, 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 1.1. Write N(Y)=\#\mathcal D(Y) and let N_j(Y) count the parameters of analytic rank j. The elementary squarefree sieve gives

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 (1.1), 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,

Theorem 1.2 now implies

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.□

Proof of Corollary 1.3. Put N(Y)=\#\mathcal D(Y), r_d=r(E^{(d)}), and let N_j^{\mathrm{alg}}(Y) count the parameters with r_d=j. The rank-equality conclusion of the companion density theorem [12, Theorem 1.2], together with its analytic rank-zero and rank-one densities, gives

Indeed, analytic and algebraic ranks agree outside a density-zero set, so their rank-j counts differ by o(N(Y)). Consequently

For the next bound, E^{(v)} for any nonzero integer v denotes the twist determined by its squareclass, so the integer sum may repeat squareclasses. Theorem 1.4 of Koymans and Smith [9], specialized to F=\Q, A=E, one variable, and P(u)=u, gives a constant C_E>0 such that for every fixed \kappa>0,

for all sufficiently large Y, with the threshold allowed to depend on \kappa. Their integer box already includes both signs, and P(v)\ne0 excludes exactly v=0. Restricting this nonnegative sum to \mathcal D(Y), and using N(Y)\sim2Y/\zeta(2), gives for each fixed \kappa>0

For fixed t>0, Cauchy–Schwarz and (7.1) at \kappa=2t give

For fixed t<0, the same exceptional average is at most \#\mathcal B(Y)/N(Y), since r_d\ge0; for t=0 the full average is identically one. The contributions from ranks zero and one are N_0^{\mathrm{alg}}(Y)/N(Y) and e^tN_1^{\mathrm{alg}}(Y)/N(Y), proving the exponential limit.

Finally, for each fixed positive integer m, one has x^{2m}\ll_m e^x for x\ge0. A second Cauchy–Schwarz application and (7.1) at \kappa=1 therefore give

The rank-zero contribution is zero and the rank-one contribution is N_1^{\mathrm{alg}}(Y)/N(Y), proving the power-moment limit without differentiating a limiting exponential-moment formula.□