TheoremComplete

Selberg's Upper Bound Sieve

Theorem7.2Selberg's Upper Bound

With the standard sieve notation, let A\mathcal{A} have density ω(d)/d\omega(d)/d with 0≀ω(p)<p0 \leq \omega(p) < p for all primes pp, and let D=zsD = z^s for s≄2s \geq 2. Then: S(A,P,z)≀XG(z)(1+O(e−s/2))+∑d1,d2≀D∣λd1λd2∣∣r[d1,d2]∣S(\mathcal{A}, \mathcal{P}, z) \leq \frac{X}{G(z)}\left(1 + O(e^{-s/2})\right) + \sum_{\substack{d_1, d_2 \leq D}} |\lambda_{d_1}\lambda_{d_2}| |r_{[d_1, d_2]}| where G(z)=∑d∣P(z)d≀DÎŒ2(d)∏p∣dω(p)p−ω(p)∌CÎș(log⁥z)ÎșG(z) = \sum_{\substack{d | P(z) \\ d \leq D}} \mu^2(d) \prod_{p | d} \frac{\omega(p)}{p - \omega(p)} \sim C_\kappa (\log z)^\kappa with Îș=∑pω(p)p/log⁥log⁥z\kappa = \sum_{p} \frac{\omega(p)}{p}/ \log\log z the sieve dimension, and λd\lambda_d are the optimal Selberg weights.


Proof

Proof

Step 1: Upper bound by squaring. For any real weights λd\lambda_d with λ1=1\lambda_1 = 1 and λd=0\lambda_d = 0 for d>Dd > D: (∑d∣n,d∣P(z)λd)2≄1\left(\sum_{d|n, d|P(z)} \lambda_d\right)^2 \geq 1 when (n,P(z))=1(n, P(z)) = 1 and λ1=1\lambda_1 = 1. Therefore: S(A,z)≀∑a∈A(∑d∣(a,P(z))d≀Dλd)2=∑d1,d2≀Dλd1λd2∣A[d1,d2]∣S(\mathcal{A}, z) \leq \sum_{a \in \mathcal{A}} \left(\sum_{\substack{d | (a, P(z)) \\ d \leq D}} \lambda_d\right)^2 = \sum_{d_1, d_2 \leq D} \lambda_{d_1}\lambda_{d_2} |\mathcal{A}_{[d_1, d_2]}|

Step 2: Separate main term and error. Substituting ∣Ae∣=ω(e)eX+re|\mathcal{A}_e| = \frac{\omega(e)}{e}X + r_e with e=[d1,d2]e = [d_1, d_2]: S≀X∑d1,d2λd1λd2ω([d1,d2])[d1,d2]+∑d1,d2∣λd1λd2∣∣r[d1,d2]∣S \leq X \sum_{d_1, d_2} \lambda_{d_1}\lambda_{d_2} \frac{\omega([d_1,d_2])}{[d_1,d_2]} + \sum_{d_1, d_2} |\lambda_{d_1}\lambda_{d_2}| |r_{[d_1,d_2]}|

Step 3: Optimize the quadratic form. The main term is a quadratic form Q(λ)=∑d1,d2λd1λd2h(d1,d2)Q(\lambda) = \sum_{d_1, d_2} \lambda_{d_1}\lambda_{d_2} h(d_1, d_2) where h(d1,d2)=ω([d1,d2])/[d1,d2]h(d_1, d_2) = \omega([d_1,d_2])/[d_1,d_2]. Using the multiplicativity of ω\omega and Mobius factoring, this diagonalizes: Q=∑e∣P(z)1g(e)(∑e∣dλd)2Q = \sum_{e | P(z)} \frac{1}{g(e)} \left(\sum_{e | d} \lambda_d\right)^2 where g(d)=∏p∣dp−ω(p)ω(p)g(d) = \prod_{p|d} \frac{p - \omega(p)}{\omega(p)}.

Subject to λ1=1\lambda_1 = 1, the minimum of QQ (by Lagrange multipliers) is 1/G(z)1/G(z) where G(z)=∑d≀D,d∣P(z)ÎŒ2(d)/g(d)G(z) = \sum_{d \leq D, d|P(z)} \mu^2(d)/g(d).

Step 4: Evaluate G(z)G(z). By multiplicativity and the Mertens-type estimate: G(z)=∏p<z(1+ω(p)p−ω(p))⋅(1+O(D−c))≍(log⁥z)ÎșG(z) = \prod_{p < z}\left(1 + \frac{\omega(p)}{p - \omega(p)}\right) \cdot (1 + O(D^{-c})) \asymp (\log z)^\kappa where Îș\kappa is the sieve dimension. The error O(e−s/2)O(e^{-s/2}) comes from the truncation d≀D=zsd \leq D = z^s.

Step 5: Conclusion. S≀X/G(z)(1+O(e−s/2))+RS \leq X/G(z)(1 + O(e^{-s/2})) + R where RR is the remainder depending on the individual error terms rdr_d. □\square

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ExampleSelberg Sieve for Goldbach-Type Problems

For the Goldbach problem (even N=p+pâ€ČN = p + p'): A={N−p:p≀N}\mathcal{A} = \{N - p : p \leq N\}, ω(p)=p/(p−1)\omega(p) = p/(p-1) for p∀Np \nmid N and ω(p)=1/(p−1)\omega(p) = 1/(p-1) for p∣Np | N. The Selberg sieve gives the upper bound ∣{p:N−p prime}âˆŁâ‰€8S(N)N/(log⁥N)2|\{p : N - p \text{ prime}\}| \leq 8 \mathfrak{S}(N) N/(\log N)^2 where S(N)=∏p∣N,p>2(p−1)/(p−2)∏p>2(1−1/(p−1)2)\mathfrak{S}(N) = \prod_{p|N, p>2}(p-1)/(p-2) \prod_{p>2}(1 - 1/(p-1)^2).