Signs of Fourier coefficients of half-integral weight modular forms

Let $g$ be a Hecke cusp form of half-integral weight, level $4$ and belonging to Kohnen's plus subspace. Let $c(n)$ denote the $n$th Fourier coefficient of $g$, normalized so that $c(n)$ is real for all $n \geq 1$. A theorem of Waldspurger determines the magnitude of $c(n)$ at fundamental discriminants $n$ by establishing that the square of $c(n)$ is proportional to the central value of a certain $L$-function. The signs of the sequence $c(n)$ however remain mysterious. Conditionally on the Generalized Riemann Hypothesis, we show that $c(n)<0$ and respectively $c(n)>0$ holds for a positive proportion of fundamental discriminants $n$. Moreover we show that the sequence $\{c(n)\}$ where $n$ ranges over fundamental discriminants changes sign a positive proportion of the time. Unconditionally, it is not known that a positive proportion of these coefficients are non-zero and we prove results about the sign of $c(n)$ which are of the same quality as the best known non-vanishing results. Finally we discuss extensions of our result to general half-integral weight forms $g$ of level $4 N$ with $N$ odd, square-free.

The Fourier coefficients c(n) encode arithmetic information. For instance under certain hypotheses, Waldspurger's Theorem shows that for fundamental discriminants d, |c(|d|)| 2 is proportional to the central value of an L-function, so that the magnitude of the L-function essentially determines the size of the coefficient c(n). However for g with real Fourier coefficients, their signs remain mysterious. In this article we contribute towards understanding the sign of such coefficients at fundamental discriminants through examining the number of coefficients which are positive (respectively negative) as well as the number of sign changes in-between.
Date: March 15, 2019. 1 Signs of Fourier coefficients of half-integral weight forms have been studied by many authors following the works of Knopp-Kohnen-Pribitkin [12] and Bruinier-Kohnen [3], the former of which showed that such forms have infinitely many sign changes. Subsequent works [8,16,19] showed that the sequence {c(n)} exhibits many sign changes under suitable conditions (such as the form having real Fourier coefficients). Notably, Jiang-Lau-Lü-Royer-Wu [10] showed for suitable g that for every ε > 0 there are more than ≫ X 2/9−ε sign changes along square-free integers n ∈ [1, X]. They also showed this result can be improved assuming the Generalized Lindelöf Hypothesis 1 with an exponent of 1/4 in place of 2/9.
For an integral weight Hecke cusp form f the second named author and Matomäki [21] proved a stronger result, establishing a positive proportion of sign changes along the positive integers. This uses the multiplicativity of the Fourier coefficients of f in a fundamental way. Fourier coefficients of half-integral weight Hecke cusp forms lack this property, except at squares. So one may wonder whether the Fourier coefficients of half-integral weight Hecke cusp forms also exhibit a positive proportion of sign changes, along the sequence of fundamental discriminants.
In this article we answer this question in the affirmative. We show that under the assumption of the Generalized Riemann Hypothesis (GRH) there exists a positive proportion of fundamental discriminants at which the Fourier coefficients of a suitable half-integral weight form are positive as well as a positive proportion at which the coefficients are negative. Moreover, we show under GRH that the coefficients exhibit a positive proportion of sign changes along the sequence of fundamental discriminants.
For simplicity, our results are stated for the Kohnen space S + k+1/2 , which consists of all weight k +1/2 modular forms on Γ 0 (4) whose nth Fourier coefficient equals zero whenever (−1) k n ≡ 2, 3 (mod 4). In this space Shimura's correspondence between half-integral weight forms and integral weight forms is well understood. Kohnen proved [13] there exists a Hecke algebra isomorphism between S + k+1/2 and the space of level 1 cusp forms of weight 2k. Also, every Hecke 2 cusp form g ∈ S + k+1/2 can be normalized so that it has real coefficients 3 and from here on we assume that g has been normalized in this way.
Let N ♭ denote the set of fundamental discriminants of the form 8n with n > 0 odd, square-free. Also, let N ♭ g (X) = {n ∈ N ♭ ∩ [1, X] : c(n) = 0}. Theorem 1. Assume the Generalized Riemann Hypothesis. Let k ≥ 2 be an integer and g ∈ S + k+1/2 be a Hecke cusp form. Then for all X sufficiently large the number of sign changes of the sequence {c(n)} n∈N ♭ g (X) is ≫ X.
In particular, for all X sufficiently large we can find ≫ X integers d ∈ N ♭ ∩ [1, X] such that c(d) < 0, and we can find ≫ X integers d ∈ N ♭ ∩ [1, X] such that c(d) < 0.
The proof of Theorem 1 uses the explicit form of Waldspurger's Theorem due to Kohnen and Zagier [17]. Given a Hecke cusp form g ∈ S + k+1/2 they show that for each fundamental discriminant d with (−1) k d > 0 that Here f is a weight 2k Hecke cusp form of level 1 which corresponds to g as described above, L(s, f ⊗ χ d ) is the L-function where λ f (n) are the Hecke eigenvalues of f , χ d (·) is the Kronecker symbol. Also, f, f = SL 2 (Z)\H y 2k |f (z)| 2 dxdy y 2 , g, g = 1 6 Γ 0 (4)\H y k+1/2 |g(z)| 2 dxdy y 2 .
In Theorem 1 we assume GRH for L(s, f ⊗ χ d ) for every fundamental discriminant d with (−1) k d > 0. The restriction to d ∈ N ♭ is important. As the proof of Theorem 3 below will show, it is easy to produce many positive (resp. negative) coefficients along the integers, assuming a suitable non-vanishing result.
1.1. Unconditional results. We also are able to prove a quantitatively weaker yet unconditional result.
Theorem 2. Let k ≥ 2 be an integer and g ∈ S + k+1/2 be a Hecke cusp form. Then for any ε > 0 and all X sufficiently large the sequence {c(n)} n∈N ♭ g (X) has ≫ X 1−ε sign changes.
Theorems 1 and 2 quantitatively match the best known non-vanishing results for Fourier coefficients of half-integral weight forms that are proved using analytic techniques, conditionally under GRH and unconditionally (resp.). In particular, Theorem 1 gives a different proof that a positive proportion of these coefficients are non-zero, for Hecke forms in the Kohnen space. It should be noted that Ono-Skinner [23] have shown that there exist ≫ X/ log X fundamental discriminants at which these coefficients are non-zero for such forms. 4 However, their result does not give a quantitative lower bound on the size of the coefficients, whereas the analytic estimates do provide such information, which is crucial for our argument.
On the other hand, using the result of Ono-Skinner it is not difficult to produce both many positive Fourier coefficients and many negative ones, at integers. Theorem 3. Let k ≥ 2 be an integer and g ∈ S + k+1/2 be a Hecke cusp form. Then, for all sufficiently large X #{n ≤ X : c(n) > 0} ≫ X log X and #{n ≤ X : c(n) < 0} ≫ X log X .

Extensions beyond the Kohnen plus space.
Since by now Shimura's correspondence is fairly well-understood (see [2]) we show in the Appendix that the conclusion of Theorems 1 and 2 holds for every half-integral weight Hecke cusp form on Γ 0 (4) 5 . Additionally, we can prove analogs of Theorems 1 and 2 which hold for weight k + 1 2 (k ≥ 2) cusp forms g of level 4N with N odd and square-free provided that g corresponds (through Shimura's correspondence) to an integral weight newform. The necessary modifications to our argument and precise statements of the results are in the Appendix.

Numerical examples.
To illustrate our results above with a concrete example consider the following weight 13 2 Hecke cusp form σ k−1 (n)e(nz) and θ(z) = n∈Z e(n 2 z).
The modular form δ(z) corresponds to the modular discriminant ∆(z) under the Shimura lift. Assuming GRH for L( 1 2 , ∆ ⊗ χ d ) for every fundamental discriminant d, Theorem 1 implies that there is a positive proportion of sign changes of α δ (n) along the sequence of fundamental discriminants of the form 8d. In fact numerical evidence below suggests that the Fourier coefficients of δ(z) change sign approximately one 4 As discussed above for forms in the Kohnen space the Fourier coefficients are algebraic integers which lie in a number field [18,Proposition 4.2] so that the Fundamental Lemma of [23] applies. 5 For g / ∈ S + k+1/2 it is possible that c(8n)µ 2 (2n) is zero for each n ∈ N, in this case we detect sign changes of {c(n)} where n ranges over the set {n ≤ X : n is even and µ 2 (n)c(n) = 0}. half of the time. Given a subset L, let S δ,L (X) denote the number of sign changes of α δ (n) along n ∈ L ∩ [1, X], and denote by N δ,L (X) the cardinality of L ∩ [1, X]. We then have the following numerical data. If we restrict to N ♭ = {8d : d odd, square-free} then we find the following data.

Main Estimates.
The main results follow from the following three propositions. The first two of the propositions allow us to control the size of c(·) by introducing a mollifier M(·; ·) which is defined in (4.18). In our application the specific shape of the mollifier is crucial to the success of the method. We have constructed this mollifier to counteract the large values of that contribute to the bulk of the moments of L( 1 2 , f ⊗ χ d ). Essentially, we are mollifying the L-function through an Euler product, as opposed to traditional methods which use a Dirichlet series. This approach was sparked by innovations in understanding of the moments of L-functions, such as the works of Soundararajan [28], Harper [6], and Radziwi l l -Soundararajan [24]. Proposition 1.1. Assume GRH. Let g ∈ S + k+1/2 be a Hecke cusp form. Also, let M(·; ·) be as defined in (4.18). Then Proposition 1.2. Let g ∈ S + k+1/2 be a Hecke cusp form. Also, let M(·; ·) be as defined in (4.18). Then The other key ingredient of our results is an estimate for sums of Fourier coefficients summed against a short Dirichlet polynomial over short intervals. This is proved through estimates for shifted convolution sums of half-integral weight forms. Proposition 1.3. Let g be a cusp form of weight k + 1 2 on Γ 0 (4). Let (β(n)) n≥1 be complex coefficients such that |β(n)| ≪ ε n 1 2 +ε for all ε > 0. Let In particular, Proposition 1.3 holds for M(·) = M(·; 1 2 ) as defined in (4.18).

The Proof of Theorem 1
The basic method of proof follows a straightforward approach. Observe that since M((−1) k n; 1 2 ) > 0 (see the discussion before and after (4.18)), if then the sequence {µ 2 (2n)c(8n)} must have at least one sign change in the interval [x, x + y]. To analyze the sums above we use a direct approach, which was developed in [21], where sign changes of integral weight forms was studied. The key input is Proposition 1.3.
Proof. This follows from Markov's inequality combined with Proposition 1.
Proof. For sake of brevity write C(n) = |c(8n)|M((−1) k 8n; 1 2 ). By Hölder's inequality . Applying Proposition 1.2 it follows that the LHS is ≫ X. Applying Proposition 1.1 the second sum on the RHS is ≪ X. Hence, we conclude that X≤8n≤2X 2n is -free Also, by Proposition 1.2 we have X≤8n≤2X 2n is -free So that for ε sufficiently small Since the first term above is of size εX we must have that the second term is ≫ X since ε is sufficiently small. So #{X ≤ x ≤ 2X : x / ∈ U} ≫ εX.

The Proofs of Theorem 2 and Theorem 3
3.1. The proof of Theorem 2. Throughout we will need the following estimate.
Proof. This follows immediately by applying (1.1) along with Proposition 5.1 below with u = 1.
We are now ready to start the preparations for the proof of Theorem 2, which as it turns out is considerably easier to prove than Theorem 1.
Proof. Note that using (1.1) and Heath-Brown's result [7, Theorem 2] we get that X≤8n≤2X 2n is -free Hence, using the above estimate along with Lemma 3.1 we can apply Hölder's inequality as in (2.1) to get Also, using Lemma 3.1 we have So combining this along with (3.1) we get Using (3.2) we get that Since the first term is of size X 1−ε we must have that the second term is We are now ready to prove Theorem 2.
Proof of Theorem 2. Let y = X 6ε and Λ = y 1/2 X ε . By Lemma 3.2 we have x≤8n≤x+y 2n is -free on a subset of cardinality at least X 1−3ε/2 . Therefore the two subsets intersect on at least ≫ X 1−3ε/2 values of x, and therefore give rise to at least X 1−15ε/2 sign changes in [X, 2X].
The proof of Theorem 3 is completely elementary and does not depend on any of the other techniques developed here.
Proof of Theorem 3. Let g ∈ S + k+1/2 be a Hecke cusp form and f denote the weight 2k Hecke cusp form of level 1 that corresponds to g. For d a fundamental discriminant with (−1) k d > 0 where λ f (n) denotes the nth Hecke eigenvalue of f (see equation (2) of [17]). In particular if p is prime this becomes Since there exists p which depends at most on k such that λ f (p) < −2/ √ p it follows that if c(|d|) = 0 then c(|d|) and c(|d|p 2 ) have opposite signs. By Ono and Skinner's result there are ≫ X/(p 2 log(X/p 2 )) ≫ X/ log X fundamental discriminants |d| ≤ X/p 2 such that c(|d|) = 0. So by considering the signs of the Fourier coefficients c(n) at n = |d| ≤ X/p 2 with c(|d|) = 0 and at m = |d|p 2 ≤ X we arrive at the claimed result.

Upper bounds for mollified moments
Let f be a level 1, Hecke cusp form of weight 2k. The aim of this section is to compute an upper bound for mollified moments of L( 1 2 , f ⊗ χ d ), conditionally under GRH. This gives an upper bound for the mean square of the mollified Fourier coefficients of g ∈ S + k+1/2 . The second moment L( 1 2 , f ⊗χ d ) has been asymptotically computed assuming GRH by Soundararajan and Young [29]. However, a direct adaptation of their method cannot handle the introduction of a mollifier of length ≫ |d| ε . For this reason we use a different approach, which is based on the refinement of Soundararajan's [28] method for upper bounds on moments due to Harper [6].
where the sum is over fundamental discriminants and the implied constant depends on f, l, κ.
Using the proposition above we can easily deduce Proposition 1.1.
Proof of Proposition 1.1. Using (1.1) it follows that 1≤n≤x 2n is −free where the sum is over all fundamental discriminants.

Preliminary results.
In this section we introduce our mollifier for the Fourier coefficients of g at fundamental discriminants. The shape of the mollifier is motivated by Harper's refinement [6] of Soundararajan's [28] bounds for moments.
Notation. Let λ(n) = (−1) Ω(n) denote the Liouville function (which should not be confused with λ f (n)). Denote by ν(n) the multiplicative function with ν(p a ) = 1 a! and write ν j (n) = (ν * · · · * ν)(n) for the j-fold convolution of ν. A useful observation is that which may be proved by induction or otherwise. Also for an interval I and m ∈ Z let An L-function inequality. We now prove an inequality in the spirit of Harper's work on sharp upper bounds for moments of L-functions, in the context of quadratic twists of L-functions attached to Hecke cusp forms. The upshot of the inequality is it essentially allows us to almost always bound the L-function by a very short Dirichlet polynomial.
Let us now introduce the following notation, which will be used throughout this section. Let l, κ > 0 be such that l · κ ∈ N and suppose that 1 ≤ l · κ ≤ C. Also, let η 1 , η 2 > 0 be sufficiently small in terms of C. For j = 0, . . . , J let 5 and Note that the choice of parameters θ j , ℓ j depends on C and is independent of l, κ satisfying 1 ≤ l · κ ≤ C. For j = 0, . . . , J and t > 0 let and define the completely multiplicative function a(·; j) by The smooth weights w(·; j) appear for technical reasons and their effect is mild. Let ℓ be an positive even integer. For t ∈ R, let > 0 for any t ∈ R since ℓ is even; the latter inequality may be seen using the Taylor expansion for e t . Moreover, using the Taylor expansion it follows for t ≤ ℓ/e 2 that For each j = 0, . . . , J let and note D j (m; l) > 0, since each term in the product is > 0. A useful observation is that for a positive integer s P I (m; a(·)) s = (4.6) Additionally, for any real number l = 0 We are now ready to state our main inequality for L( 1 2 , f ⊗ χ d ).
Proposition 4.2. Assume GRH. Suppose that 1 ≤ |d| ≤ x is a fundamental discriminant and l > 0 is a real number. Also let Let ℓ j , θ j , I j and a(·, j) be as in (4.2) and (4.3) (resp.). Then there exists a sufficiently large absolute constant C 1 = C 1 (l) > 0 such that for each |d| ≤ x either: for any even integers s 1 , . . . , s J , where the implied constant depends on f and l.
Proof. Applying Theorem 2.1 of Chandee [4] gives, for any X ≥ 10, that where α p , β p are the Satake parameters (note α n p + β n p is real valued see Remark 2 of [20]). In the sum over primes, the contribution from the prime powers with n ≥ 3 is bounded so that it may be included in the O(1) term. Also, noting that where the implied constant depends on f . For r = 0, . . . , J, let E r be the set of fundamental discriminants d such that max r≤u≤J |P Ir (d; a(·; u))| < ℓr le 2 . For each d we must have one of the following: Hence, we get for each fundamental discriminant with |d| ≤ x either: Here (4.12), (4.13) correspond to possibilities (i) and (ii) above (resp.), while (4.14) and (4.15) corresponds to (iii).
If (4.12) holds for d then we are done. If (4.13) holds, we apply (4.10) with X = x θ J and so the term 3 · log |d| log X is ≪ 1. Also, the contribution from (4.11) is Hence, applying these estimates along with (4.4) we get that Finally, if (4.14) and (4.15) hold we apply (4.10) with X = x θ j and for each 0 ≤ r ≤ j we argue as above and bound the term exp(lP Ir (d; a(·; j))) by (1 + e −ℓr/2 )E ℓ j (lP Ir (d; a(·, j))). In (4.10) we use the inequality 3 · log |d| log X ≤ 3 θ j and note that the contribution from the squares of primes (4.11) is So that applying these estimates, we get that for any non-negative, even integer s j+1 where we have included the extra factor max j+1≤u≤J e 2 lP I j+1 (d; a(·, u)) ℓ j+1 Combining (4.16) and (4.17) and using the inequality max(a, b) ≤ a+ b for a, b > 0 gives (4.9).
The definition of the mollifier. With Proposition 4.2 in mind we now choose our mollifier so that it will counteract the large values of D J (d; l). For j = 0, . . . , J let where ℓ j , θ j and I j are as in (4.2). Let and note that in Propositions 1.1 and 1.2 we specialize I 0 = (c, x θ 0 ] so that c is sufficiently large, yet fixed. A useful observation is that M(m; 1 κ ) > 0, which can be seen by using (4.7) along with the fact that E ℓ (t) > 0 for even ℓ and t ∈ R. Also, let δ 0 = j≤J ℓ j θ j . Observe that by construction δ 0 > 0 is fixed, sufficiently small and the length of the Dirichlet polynomial M(m; 1 κ ) is x δ 0 . Also, and for n, r, ℓ ∈ N (4.21) ν r (n; ℓ) = n 1 ···nr=n Ω(n 1 )≤ℓ,...,Ω(nr)≤ℓ ν(n 1 ) · · · ν(n r ).
Lemma 4.1. Let F be a Schwartz function such that its Fourier transform F has compact support in (−A, A), for some fixed A > 0. Also, let I 0 , . . . , are real numbers such that J j=0 ℓ j θ j < 1/2. Then for any arithmetic function g we have Proof. The LHS of (4.23) equals Using Poisson summation, see Remark 1 of [20], we get that for n ≤ X 1/2 if n = that and for n = the above sum equals zero. By assumption n 0 · · · n J ≤ x ℓ 0 θ 0 · · · x ℓ J θ J < x 1/2 . Hence, the inner sum in (4.24) equals zero unless n 0 · · · n J = and since (n j , n i ) = 1 for i = j in this case n 0 , . . . , n J = . We conclude that the inner sum in (4.24) equals n 0 · · · n J when n 0 , . . . , n J = and is zero otherwise, thereby giving the claim.
To prove Proposition 4.1 we will use Proposition 4.2 and then apply Lemma 4.1. This leaves us with the problem of bounding the resulting sum. This task will be accomplished in the next two lemmas.
Before stating the next lemma let us introduce the following notation. Given a statement S let 1 S equal one if S is true and zero otherwise.
Proof. In the sum over m, n on the LHS of (4.25) write m = g · m 1 , n = g · n 1 where g = (m, n). Since mn = it follows that m 1 = and n 1 = . We will proceed by estimating the sum in terms of an Euler product. To this end, observe that if max{Ω(n), Ω(m)} ≥ ℓ j then 2 Ω(m)+Ω(n) /2 ℓ j ≥ 1; so using this along with the remarks following (4.21) the sum on the LHS of (4.25) equals where we have trivially estimated ν, ν l·κ , ϕ, and λ in the second sum. Recall for any multiplicative function r(n), and m ∈ N whose prime factors all lie in an interval I (which also contains at least one prime) that It follows that the error term in (4.28) is To estimate the main term in (4.28) consider Note that σ(p; u, v) > 0 for p sufficiently large. Hence, evaluating the sum over g in the main term in (4.28) we see that it equals (4.30) Note f 1 (p u ), f 2 (p u ) ≪ l,κ 1 and recall (4.1). It follows that Using these estimates on the RHS of (4.30) we see that the main term in (4.28) is Proof. In the sum on the LHS of (4.32) write m = gm 1 , n = gn 1 where g = (m, n), so that mn = implies that m 1 = and n 1 = . Also recall that ν κ·l (m; ℓ j ) ≤ ν κ·l (m). Hence, this sum is where we have also used the estimates ν r (ab) ≤ ν r (a)ν r (b) and ν(a) ≤ 1, for a, b, r ∈ N, which follow from (4.1). The sum over m is (4.34) Hence, using (4.34) along with the inequality ν(n 2 ) ≤ ν(n)2 −Ω(n) it follows that (4.33) is by (4.6). Using the assumption that s ≥ 4(l · κ)ℓ j gives s − Ω(g) ≥ 3 4 s for g with Ω(g) ≤ (l · κ)ℓ j . This allows us to bound the bracketed term on the RHS of (4.35) by Applying this estimate in (4.35) it follows that there exists B > 0, which depends at most on l, κ, such that the RHS of (4.35) is bounded by In the next lemma we estimate averages of our mollifier M as defined in (4.18) against the terms which appear in Proposition 4.2.
Lemma 4.4. For j = 0, . . . , J let s j be an even integer with 4(l · κ)ℓ j ≤ s j ≤ 2 5θ j . Then we have the following estimates: for each j = 0, . . . , J − 1. The implied constants depend at most on f, κ, l (and not on j, u).
Proof. We will first prove (4.38), which is the most complicated of the bounds, and at the end of the proof we will indicate how to modify the argument to establish and (4.18)). So for any Schwartz function F which majorizes 1 [−1,1] , such that F has compact support in (−A, A) the LHS of (4.38) is The Dirichlet Polynomial above is supported on integers n ≤ x θ where θ = l · κ 0≤r≤J θ r ℓ r + 0≤r≤j θ r ℓ r + s j+1 θ j+1 < 1/2. Hence, using (4.5), (4.6), (4.7), (4.19), (4.20) and applying Lemma 4.1 the above equation is bounded by Let α(·) be as in (4.26) and take a(·) = a(·; J), b(·) = a(·; j). Applying Lemma 4.2, the contribution to (4.39) from the product over 0 ≤ r ≤ j is bounded in absolute value by (4.40) where in the last step we applied (4.27). Applying Lemma 4.3 with a(·) = a(·; J) and b(·) = a(·; u) the term in (4.39) with r = j + 1 contributes It remains to handle the factors in the product in (4.39) with j + 1 < r ≤ J. For such factors, m is a square so these terms are bounded by where we used (4.1) and the bound |λ f (p)| ≤ 2, in the last steps.
Using (4.40), (4.41) and (4.42) in (4.39) completes the proof of (4.38). To establish (4.36) we repeat the same argument, the only differences are that in (4.40) the product is over all 0 ≤ r ≤ J and the terms estimated in (4.41) and (4.42) do not appear. To establish (4.37) note that the term estimated in (4.40) does not appear and in the bound (4.42) the only difference is that j = 0, so we bound this term by O((log x) O(1) ). Finally in place of (4.41) we get by using Lemma 4.3 the bound From these estimates (4.37) follows.
Proof of Proposition 4.1. Let A(d) be as in the statement of Proposition 4.2. In particular, using |λ f (p 2 )| ≤ d(p 2 ) = 3 and λ f ( Using this observation we note that it suffices to prove that for l > 0 (note the definition of M is independent of l, for 1 ≤ l · κ ≤ C). Since by Cauchy-Schwarz (4.43) implies We will now establish (4.43). For j = 0, . . . , J, let s j = 2⌊ 1 5θ j ⌋. We first divide the sum over |d| ≤ x into two sums depending on whether (4.44) for some 0 ≤ u ≤ J. To bound the contribution to the LHS of (4.43) of the terms with |d| ≤ x for which (4.44) holds, we use the bound A(d) ≪ (log log x) 2 , Chebyshev's inequality and then apply Cauchy-Schwarz to see that ♭ |d|≤x |P I 0 (d;a(·;u))|≥ℓ 0 /(le 2 ) for any A ≥ 1. Hence, the contribution to (4.43) from |d| ≤ x satisfying (4.44) for some u ≤ J is ≪ J x (log x) A ≪ x (log x) A log log x for any A ≥ 1. For the remaining fundamental discriminants |d| ≤ x we apply (4.9) to see that their contribution to (4.43) is bounded by (4.45) To complete the proof it suffices to show that the expression above is ≪ x. Applying (4.36) the first term in (4.45) is

Using (4.38) the second term in (4.45) is bounded by
Applying Stirling's formula and estimating the inner sum over primes trivially as ≤ 5 log( log x θ j+1 log x θ j ) = 5, we see that the above is (4.47) .

By construction s
Hence, there exists c > 0 (which may depend on l, κ) such that (4.47) is where we used that θ J /θ j = e J−j in the last step. Combining (4.48) with (4.46), gives that (4.45) is ≪ x, which completes the proof.

The Proof of Proposition 1.2
The main input into the proof of Proposition 1.2 is a twisted first moment of L( 1 2 , f ⊗ χ d ). Similar moment estimates were established in [27], [29] and [24] and our proof closely follows the methods developed in those papers. We will include a proof of the following result for completeness. Proposition 5.1. Let φ be a Schwartz function with compact support in the positive reals. Also, let u = u 1 u 2 2 , be odd where u 1 is square-free. Then where C > 0 depends only on f and ϑ(·) is a multiplicative function with ϑ(p 2j+1 ) = λ f (p) + O(1/p) and ϑ(p 2j ) = 1 + O(1/p).
The constant C is explicitly given in the proof below, see (5.30). Before proving Proposition 5.1, we will use the result to deduce Proposition 1.2.
We will now estimate the inner sum on the RHS of (5.4). First note that ϑ(mn 2 ) ≪ d(m)( mn 2 ϕ(m)ϕ(n 2 ) ) O(1) and recall the remarks after (4.21), so we get that the sum equals p|mn 2 ⇒p∈I j a(mn 2 ; J)λ(m)ϑ(mn 2 )µ(m) 2 2 Ω(m)+2Ω(n) mn ν 2 (mn 2 ) The error term is To estimate the main term in ( (5.7) Using (5.6), (5.7) along with the estimates a(p; we get that the RHS of (5.4) equals To esimate the second product above note that (5.9) since η 2 is sufficiently small. Also, there exists some constant C ′′ > 0 such that the Euler product over c < p ≤ x θ J in (5.8) is since c is sufficiently large (so each term in the Euler product is positive). Hence, using this and (5.9) we get that (5.8) is ≫ x(log x) 1/2 · (log x) −1/2 = x, which completes the proof.

5.2.
The proof of Proposition 5.1. Let f be a weight 2k, level 1, Hecke cusp form. For a fundamental discriminant d, let The functional equation for Λ(s, f ⊗ χ d ) is given by Note that the central value vanishes when (−1) k d < 0. For c > 0 let Our starting point in the proof of Proposition 5.1 is the following approximate functional equation for L(s, f ⊗ χ d ). Also, define W = W 1/2 .
Since we will sum over fundamental discriminants we introduce a new parameter Y with 1 ≤ Y ≤ x to be chosen later. Also, write F (ξ; x, y) = φ ξ x W y ξ . Applying Lemma 5.1 the LHS of (5.1) equals (5.10) The terms with a > Y . Write m = r 2 e where e is square-free, and note that since in the sum in (5.10) a 2 |m it follows a|r. So the second sum in (5.10) equals where in the second line we have used the rapid decay of W . Using the definition of W and for Re(s) > 1, writing we get for c > 1/2 that The integrand is holomorphic for 1/ log x ≤ Re(s) ≤ 2 and in this region bounded by (note r 2 e ≤ x 3 ) Hence, shifting contours on the RHS of (5.12) to Re(s) = 1/ log x and applying the above estimate we get that the LHS of (5.12) is bounded by Also note, that by Cauchy-Schwarz and Corollary 2.5 of [29] (which follows from Heath-Brown's result [7]) for Re(s) ≥ 0. Applying this estimate in (5.11) it follows that the second sum in (5.10) is bounded by The terms with a < Y : preliminary lemmas. It remains to estimate the first sum on the LHS of (5.10). This will be done by applying Poisson summation to the character sum, as developed in [27].
a n e aℓ n .
In Lemma 2.3 of [27] it is shown that G ℓ is a multiplicative function. Moreover, G 0 (n) = ϕ(n) if n is a square and is identically zero otherwise. Also, for p α ||ℓ, ℓ = 0 (5.14) Lemma 5.2. Let F be a Schwartz function. Then for any odd integer n Proof. This is established in the proof of Lemma 2.6 of [27], in particular see the last equation of the proof.
Using Lemma 5.1 it follows that F (·; x, n) is a Schwartz function, since W and φ and their derivatives decay rapidly. Hence, applying Lemma 5.2 the first sum on the RHS of (5.10) equals (note u is odd) Note that for odd k after applying Lemma 5.2 in (5.15) we also made the change of variables ξ → −ξ. Before proceeding we require several estimates for F . The first such result is a basic estimate on the rate of decay of F .
Proof. Using the rapid decay of W it follows that Next, write W(ξ) = W y xξ φ(ξ) and note that for any A, B, C ≥ 0, with A ≥ B, and |ξ| ≫ 1 so using this and the fact that φ has compact support it follows that Hence, integrating by parts and using the above bound it follows that We also require the following information about the Mellin transform of F . Let Then for Re(s) > 0 (5.16) Moreover, the functionF (s; ℓ, u, a 2 ) extends to an entire function in the half-plane Re(s) ≥ −1 + ε and in this region for any A ≥ 1.
Proof. Changing the order of integration and making a change of variables xξ/t → ξ in the integral over t it follows thatF (s; ℓ, u, a 2 ) equals which establishes the first claim.
The function is holomorphic in the region Re(s) ≥ ε. Write w(s) = I 1 (s) + I 2 (s) where I 1 is the portion of the integral over [0, 8au 2 |ℓ| ] and I 2 is the rest. Due to the rapid decay of W , I 1 (s) is holomorphic in the region Re(s) ≥ −1 and in this region The first integral on the RHS of (5.18) can be analytically continued to Re(s) > −1 by integrating by parts. This provides the analytic continuation of I 2 (s) to Re(s) ≥ −1 + ε, and shows that in this region Hence applying this estimate along with (5.17) in (5.16) and noting Φ decays rapidly establishes the claim.
The terms with a < Y : main term analysis i.e. ℓ = 0. Recall that G 0 (n) = ϕ(n) if n = and is zero otherwise. So using Lemma 5.3 the term with ℓ = 0 in (5.15) equals We now evaluate the inner sum on the RHS (5.19) (note Re(s) = c > 0). Since nu = write n = e 2 u 1 , (recall u = u 1 u 2 2 ) so The sum on the RHS can be expressed as an Euler product as follows. Let r(n) = . Also, let It follows that,  2s; 0, 0) .
The terms with a < Y : Off-diagonal analysis i.e. ℓ = 0. In (5.15) it remains to bound First, note that by Lemma 5.3 the contribution from the terms with |ℓ| ≥ Y 2 ux ε to (5.23) is bounded by where A has been chosen sufficiently large with respect to ε. It remains to estimate the terms in (5.23) with |ℓ| ≤ Y 2 ux ε . Let ψ 4ℓ (n) = 4ℓ n and note that ψ 4ℓ is a character of modulus at most 4|ℓ|. Using that G ℓ is a multiplicative function, and using (5.14) we can write It follows that for (p, 4auℓ) = 1 For p|auℓ and Re(s) ≥ 1 2 + ε, writing p θ ||u, we have that Also R 2 (s; ℓ, u, a) ≪ 1 for Re(s) ≥ 1 2 . Hence, for 0 < a, u, |ℓ| ≤ x 2 and Re(s) ≥ 1 2 + ε (5.26) R(s; ℓ, u, a) ≪ So R(·; ℓ, u, a) is an absolutely convergent Euler product and thereby defines a holomorphic function in the half-plane Re(s) ≥ 1 2 + ε. Hence, applying Mellin inversion and (5.25) we get for c > 0 that Completion of the proof of Proposition 5.1. Applying (5.22), (5.24), (5.29) in (5.15) and then using the resulting formula along with (5.13) in (5.10) it follows that m 2m is −free In order to establish Proposition 1.3 we will need the following variant of the shifted convolution problem for coefficients of half-integral weight forms.
6.1. The shifted convolution problem. We begin with a proof of Proposition 6.1. The proof is based on the by now standard combination of the circle method and modularity. Recall that a weight k + 1 2 modular form (with trivial character) transforms under Γ 0 (4) in the following way the quadratic residue symbol in the sense of Shimura [25] (see Notation 3) and Also, we record the following estimate for the Fourier coefficients of g (6.3) n≤X |c(n)| 2 ≪ X (see [9,Theorem 5.1] and use partial summation). This implies that (6.4) |c(n)| ≪ n 1/2+ε .
We record below a few standard lemmas.
In the notation of [5, Proposition 2] we specialize to δ = Q −2+η and notice that L ≍ ϕ(4∆) ∆ 2 · Q 2 log Q . Crucial to our analysis is the following consequence of the modularity of g. and (ℓ, r) = 1. Let ∆ ⋆ = ∆/(4v + b, ∆) and (4v + b) ⋆ = (4v + b)/(4v + b, ∆). Then, for any β ∈ R and any real X ≥ 1 where χ t (·) denotes a real Dirichlet character of modulus t. Finally whenever we write a q we denote by a an integer such that aa ≡ 1 (mod q). Remark 2. By inspection of the proof below, the conclusion of the lemma also holds when v = 0 and we will use this later.
Proof. Since q = 4∆r we can write 4∆r .
Notice that (d + 4vr, 4∆r) = (d + 4vr, ∆) since (d, 4r) = 1 by assumption. Therefore the above is equal to Throughout set w := (4v + b) ⋆ r + 4∆ ⋆ ℓ. We now consider A slightly tedious computation using (6.5) reveals that We also find that where · · denotes the extended quadratic residue symbol in the sense of Shimura. In particular since 4∆ ⋆ r is divisible by four, this extended quadratic residue symbol coincides with χ 4∆ ⋆ r (w), a real Dirichlet character of modulus 4∆ ⋆ r, so that χ 4∆ ⋆ r (w) = χ 4∆ ⋆ r (w). Moreover by multiplicativity of the Jacobi symbol Since 4∆ ⋆ is divisible by 4 and r is congruent to 1 (mod 4) both expressions are Dirichlet characters of modulus 4∆ ⋆ and modulus r respectively. In particular the above expression is equal to Using the modularity of g, (6.2), and combining this with (6.6), (6.7), (6.8) we conclude that .

So by the Chinese Remainder Theorem
and since g(x + iy) is 1-periodic in x the claim follows.
We are now ready to prove the main proposition of this section.
Proof of Proposition 6.1. Using the circle method we can re-write the sum on the LHS of (6.1) as Let Q = X 1/2+2η and η ∈ (0, 1] to be determined later (we will choose η = 1 13 ). Let 1 ≤ ∆ ≤ X η/4 ≤ Q η/2 . By Lemma 6.1 the above expression is equal to (6.9) 1 plus an error term of size ≪ XQ −η/4 ≤ X 1−η/8 , where in the estimation of the error term we have used the bound |g(α + i/X)| ≪ X k+ 1 2 2 which holds uniformly for all α ∈ R. This follows from the fact that y (k+ 1 2 )/2 |g(z)| is bounded on H since g is a cusp form.
6.2. Proof of Proposition 1.3. We are now ready to prove Proposition 1.3. Let h be the indicator function of the integers that can be written as 8n with n odd and square-free. We find that By the triangle inequality, Therefore (6.14) is bounded by This contributes ≪ X 1−δ+ε provided that Y ≥ X δ My. Therefore after an application of Cauchy-Schwarz, it remains to obtain a nontrivial upper bound for We introduce an auxiliary smoothing, and bound the above by We note that the implicit constant is allowed to depend on the weight k + 1 2 . Let α(n) = c(n)M((−1) k n)h ≤Y (n). Expanding the square we re-write the above expression as 1 Grouping terms together and using a Taylor expansion we can re-wite the above as where we used (6.3) in the estimation of the error term. The term h = 0 contributes Repeating the same argument as before we see that This gives a total contribution of ≪ X 1−δ+ε provided that Y > X δ M 2 y. It follows that (6.16) is bounded by We now focus on the terms with h = 0 in (6.15). Opening h ≤Y and M((−1) k n) we see that the RHS of (6.15) restricted to h = 0 is bounded by By the Chinese Remainder Theorem the condition 2 α 1 +3 d 2 1 |n and 2 α 2 +3 d 2 2 |n + h can be re-written as a single congruence condition to a modulus of size ≤ 4Y 4 . Moreover χ (−1) k n (m 1 ) is 4m 1 -periodic in n, where-as χ (−1) k (n+h) (m 2 ) is 4m 2 -periodic in n. Therefore fixing the congruence class of n modulo 4[m 1 , m 2 ] fixes the value of χ (−1) k n (m 1 )χ (−1) k (n+h) (m 2 ). Therefore we can bound the above supremum by (6.17) sup Also, let U m be the operator, which acts on power series as follows U m n≥1 a n q n = n≥1 a mn q n .
Theorem 4. Let k ≥ 2 be an integer and g be a Hecke cusp form of weight k + 1 2 on Γ 0 (4). Then it is possible to normalize g so that its nth Fourier coefficient is real for n ∈ N b .
In addition, suppose that c(n) = 0 for some n ∈ N ♭ . Then for every ε > 0 the sequence {c(n)} n∈N ♭ g (X) has ≫ X 1−ε sign changes. Assuming GRH the sequence {c(n)} n∈N ♭ g (X) has ≫ X sign changes. Remark 3. We will see that if g is of the form with G ∈ S + k+1/2 , and a, b ∈ C with a b = − 1 α 2 λ F (2), where F ∈ S 2k (1) is the Shimura lift of G, then c(8n)µ 2 (2n) = 0 for each n ∈ N. Otherwise, for a Hecke cusp form g ∈ S k+1/2 not of this form, the proof below and Lemma 3.1 imply c(8n)µ 2 (2n) = 0 for some n so that the conclusion of Theorem 4 holds for g.
Moreover, for g as in (7.1) with b = 0 the subsequent argument shows that c(2n)µ 2 (2n) = b α 2 c G (8n)µ 2 (2n), where c G (n) denotes the nth Fourier coefficient of G. Hence in this case we conclude that after suitable normalization, the sequence {c(2n)µ 2 (2n)} n∈N has ≫ X sign changes as n ranges over integers in [1, X] under GRH and ≫ X 1−ε such sign changes unconditionally.
Additionally, it is possible to extend our result to level 4N with N square-free and odd if we also restrict to fundamental discriminants that lie in a suitable progression. For each prime divisor of p|N we require that χ d (p) = w p where w p is the eigenvalue of the Atkin-Lehner operator W p . So by the Chinese Remainder Theorem there exists η (mod N) such that for d ≡ η (mod N) we have χ d (p) = w p . Let N ♭ N = {n ∈ N : n = 8m, µ 2 (2m) = 1, and (−1) k n ≡ η (mod N)} and N ♭ N,g (X) = {n ≤ X : n ∈ N ♭ N and c(n) = 0}. Note that for n ∈ N ♭ N by construction (N, n) = 1. Let S k+1/2 (4N) denote the space of weight k + 1/2 cusp forms of level 4N, with N odd and square-free. Also, let S − k+1/2 (4N) be as defined by Baruch-Purkait (see Section 6.3 of [2]), who showed that this space is isomorphic to S new 2k (2N). This complements Kohnen's result [15] that S new k+1/2 (4N) is isomorphic to S new 2k (N). We can also prove the following result.
Using the formulas (7.2) and (7.3) we have in each case above that To bound this mollified moment, the only modification needed in the proof of Proposition 4.1 is to change the definition of I 0 so that I 0 = (c, X θ 0 ] with c sufficiently large in terms of N. Repeating the argument (with no further modifications) we arrive at We also need to prove n∈N ♭ g,N (X) |c(8n)| 2 M((−1) k 8n; 1 2 ) 2 ≍ X.
This follows in the same way as before once we have established an analog of Proposition 5.1 for f ∈ S new 2k (M) with M = N or M = 2N, where we average over discriminants (−1) k d ∈ N ♭ N . The necessary modifications for this computation have already been worked out in the paper of Radziwi l l and Soundararajan [24]. Finally, we need to establish the estimate X≤x≤2X x≤8n≤x+y n odd (−1) k 8n≡γ (mod N ) µ 2 (n)c(8n)M((−1) k 8n) ≪ X √ y + X 1− 1 2148 +ε .
To do this we first need to modify the proof of Lemma 6.2 in a straightforward way. From here we arrive at the analog of Proposition 6.1, for any g ∈ S k+1/2 (4N). To establish the above bound we repeat the argument used in the proof of Proposition 1.3. The only modification necessary is that the range of ∆ in (6.17) will now be 1 ≤ ∆ ≤ 16NY 4 M 2 to account for the progression (−1) k 8n ≡ η (mod N).
Combining the three estimates above we argue as in the proof of Theorem 1 thereby finishing the proof.