Positive lower density of large prime gaps

Proof. The displayed inequality is equivalent to p_{n+1}-p_n>p_n/n. The prime number theorem gives p_n/n\sim\log p_n. Apply Theorem 1.1 with C=2 and discard the finitely many indices for which p_n/n\ge2\log p_n.□

Proof. For each such m, let p(m) be the last prime in (m,m+h]. There is no prime in (p(m),m+2h], so p(m)^+>m+2h\ge p(m)+h. For a fixed selected prime p, the condition m<p\le m+h implies p-h\le m\le p-1. This interval contains exactly h integers. Thus each prime is selected at most h times, proving the claim.□

Proof of Theorem 1.1. Fix C>0 and choose \lambda>\max\{C,1\}. Take \varepsilon=\lambda^2/(2K_\lambda) in Proposition 2.1, and fix the resulting weight family. There is a finite constant M>0 such that \mathbb E_XW_X^2\le M for every sufficiently large X. Let A_X=\{V_J>0\} and B_X=\{V_I>0\}. Weighted Cauchy–Schwarz gives

for all sufficiently large X; its denominator is positive because the numerator tends to \lambda^2>0. By (1)–(2),

Since \mathbf1_{B_X}\le V_I and W_X\ge0, it follows that

Another use of Cauchy–Schwarz yields, for every sufficiently large X,

This is an ordinary proportion of integer starting points. Lemma 2.2 therefore gives at least

distinct prime starts p\in(X,2X+h] with consecutive gap exceeding h. Uniformly on this range, \log p=\log X+O(1) and h=\lambda\log X+O(1). The strict inequality \lambda>C consequently gives h>C\log p for every sufficiently large X.

For an arbitrary sufficiently large integer N, choose X=\lfloor p_N/3\rfloor. Then 2X+h<p_N, so all these prime starts have indices at most N. The prime number theorem gives

In particular X/\log X\ge N/4 for every sufficiently large N. The required count is at least \delta N/(4\lambda). Thus c(C)=\delta/(4\lambda)>0 is admissible. All choices preceding X depend only on the fixed threshold.□

Proof. For R\ge2, choose a smooth \phi_R on (0,\infty), equal to 1/u on [1,R], with 0\le\phi_R(u)\le1/u and support in [1/2,2R]. Put

Then A_R\ge\log R, B_R\le\log(4R), and 1/2\le N_R^2\le2. Set g_0=\phi_R/N_R, c=\lambda/(\int g_0)^2, and g(u)=c^{-1/2}g_0(u/c). Changes of variable give exactly

Choose R sufficiently large and then hold it fixed.□

Proof. Fix 0<\beta<\gamma<\tau and choose g from Lemma 4.1. Write d=\beta-\mu_g>0. Let B be an upper endpoint of the support of g, and let s_g^2 be the variance of the probability density g^2. All these quantities are fixed and finite; they need not be bounded uniformly in \lambda.

For a large integer k and 0\le j\le k, define

with Q_0=1 and S_0=0. Before imposing the simplex cutoff, these products satisfy

Since \alpha_j/\alpha_{j+1}=(j+1)/\lambda, alternating products (-1)^jQ_j/\sqrt{\alpha_j} at neighboring levels therefore cancel up to the factor 1-\sqrt{(j+1)/k}. We use r neighboring levels near k: the condition r/k\to0 makes these factors small, while r\to\infty makes the two endpoint contributions small after normalizing the squared mass equally across the levels.

Take r=\lfloor\sqrt k\rfloor and a=k-r+1. Choose once and for all a smooth cutoff \chi with 0\le\chi\le1, equal to one on [0,\beta] and zero on [\gamma,\infty). At the r indices a\le j\le k, set

and set all other f_j, including f_0, to zero. Each selected finite family has compact interior support: every coordinate is bounded away from zero by a positive support endpoint of g divided by k, and the cutoff forces S_j\le\gamma<\tau.

The density Q_j^2 has total mass one. Under it the variables kt_i are independent with density g^2, so, for j\le k,

Chebyshev’s inequality therefore gives

This proves w\to1.

For a\le j\le k-1, substituting u=kt_{j+1} and using \alpha_j/\alpha_{j+1}=(j+1)/\lambda gives

For k\ge2B/d, Chebyshev’s inequality gives, uniformly in j\le k,

On the complementary event every cutoff in (11) equals one. The bracket is then 1-\sqrt{(j+1)/k}, whose square is at most (r/k)^2. Everywhere, both terms in the bracket lie in [0,1], since g\ge0, \int g=\sqrt\lambda, and j+1\le k; its absolute value is at most one. After summing the r-1 interior terms, their total contribution to v/\lambda is at most

Here the probability estimate is with respect to Q_j^2, while the coordinate-deletion integral contains g(u)\,du. The passage between them uses no comparison of these measures: on the complementary event, S_j+u/k\le\beta holds for every u in the support of g. Thus the whole integral has its cutoff equal to one pointwise.

Exactly two further indices can contribute. At the top, j=k, the term is \alpha_k\|f_k\|_2^2\le1/r. At the bottom, j=a-1=k-r, only Tf_a occurs, and

Its squared norm is at most a/(kr)\le1/r. Consequently

Together with (10), this proves the proposition.□

Proof of Proposition 2.1. Fix \lambda>0 and one function G as in Proposition 3.1. Set

This choice is independent of \varepsilon. Given \varepsilon>0, choose \beta, \gamma, g, and \chi as in the preceding construction. By Proposition 4.2, one can select a single finite k and its family with w>0 and v/w\le\varepsilon. Freeze this family and define

All conclusions follow from Proposition 3.1 by division by w, or by w^2 for the second moment. Once the construction is chosen in terms of \lambda and \varepsilon, the fixed-family second-moment bound may be written O_{\lambda,\varepsilon}(1).

The parameter order is essential: the auxiliary limit k\to\infty produces one family, and only then does X\to\infty. No divisor-sum asymptotic uniform in k is required. For an arbitrary fixed gap threshold C>0, the application in Section 2 chooses a fixed \lambda>\max\{1,C\} before making these choices.□

Proof. At either endpoint x=X,2X, the range q\le X^\sigma is contained in the range of (15) for every fixed logarithmic saving once X is sufficiently large. Removing proper prime powers from a \psi-sum costs at most

in each residue class. Indeed there are O(X^{1/2}L) pairs p,j\ge2, and each summand is O(L). Translating the two endpoints through any |a|\le H_X costs O((H_X+1)L), uniformly in the residue class. Thus, writing Q=\lfloor X^\sigma\rfloor, for every A_0>0,

The last absorption uses the fixed strict inequality \sigma<1/2. The same bound controls the maximum over all shifts at once; no union bound over shifts is used.

An interval of length X contains at most X/q+1 integers in one residue class. Also

Since q\le Q<X for large X, these two facts give

For every fixed real T\ge1,

for a finite constant c_T. The elementary bound \sum_{p\le y}1/p=O(\log\log(3y)) suffices: for y\ge2, compare 1/p\le e p^{-1-1/\log y} with \log\zeta(1+1/\log y), and bound \zeta(1+u)\le1+1/u. No prime-number theorem is needed for this estimate.

As squarefree q satisfy d(q)=2^{\omega(q)}, (18)–(19) imply, for some finite c_K,

Cauchy–Schwarz, applied to the nonnegative numbers E_X(q), therefore gives

Choose A_0\ge c_K+2A. This proves (16).□

Proof. Expand the product of divisor sums. Möbius factors kill all nonsquarefree divisors. Whenever the product of the F_r’s is nonzero, the support hypothesis gives

This uses the budget of the sum of the coordinates in each factor, and does not impose a budget on the dimension.

For a squarefree tuple, the conditions d_\gamma\mid m+b_\gamma prescribe at prime p\mid q the residues m\equiv-b_\gamma\pmod p for all selected coordinates. These congruences are inconsistent unless the selected shifts agree modulo every p\mid q. When they agree, the Chinese remainder theorem prescribes exactly one class m\pmod q. In the unmarked case this is precisely compatibility, and the count is X/q+O(1), with an absolute constant in the O(1).

When M\ge1, the number of positive integer tuples whose product is at most Y\ge1 is at most

Indeed, fix the first M-1 entries and bound the number of possible last entries by Y divided by their product, then enlarge each preceding summation to d\le Y. Apply (24) with Y=X^\sigma, multiply the O(1) count error by \prod_r\|F_r\|_\infty, and divide by X. This proves (22), hence (21) in the unmarked case. If M=0, the unmarked average and (20) are both exactly \prod_rF_r.

For the marked case, put n=m+a_0. On a residue class satisfying these divisor congruences, the residue of n\pmod p is a_0-b_\gamma whenever p\mid d_\gamma. Consequently this class is reduced modulo q exactly when the additional marked compatibility condition holds. If it is not reduced, a prime n in that class would have to be a prime divisor of q, hence satisfy n\le q. But (23) and \sigma<1/2 give

for large X. The marked sum on a nonreduced class is therefore exactly zero, not an unestimated exceptional term. On a reduced class, the marked sum is

Since q is squarefree, \varphi(q)=\prod_{p\mid q}(p-1), proving the main density factor in (20).

For fixed squarefree q, each prime dividing q must be assigned to a nonempty subset of the M divisor coordinates. There are at most (2^M-1)^{\omega(q)}\le (2^M)^{\omega(q)} choices if M\ge1; support and compatibility only reduce this count. The empty tuple, if M=0, has q=1 and is handled directly by the same bound with K=1. Thus the total normalized error is at most

which is O_A(L^{-A}) by Lemma 5.2. This proves the marked case, including all scalar factors and the wholly zero-dimensional case.□

Proof of Proposition 5.1. Let M=|\Gamma|. By Lemma 5.3, it remains to evaluate \mathcal A_\delta from (20): the arithmetic error is O_N(L^{-N}) for every fixed N. If M=0, that sum is exactly \prod_rF_r, and the assertion follows. We henceforth assume M>0.

Choose a smooth compact extension \widetilde F_r of each positive-dimensional function and define its Fourier transform by

The function \Phi_r is Schwartz, and Fourier inversion gives exactly

For scalar factors use the scalar itself. Put \Phi=\prod_r\Phi_r, z_\gamma=(1+\mathrm i u_\gamma)/L, and z_S=\sum_{\gamma\in S}z_\gamma.

Insert (25) into the density sum (20), retaining the sum over all squarefree tuples. The following bound justifies moving that sum inside the integrals:

Indeed 1/(p-\delta)\le2/p, every nonempty selected subset contributes at most p^{-1/L}, and there are at most 2^M-1 subsets. Moreover \sum_pp^{-1-\varepsilon}\le\log\zeta(1+\varepsilon) \le\log(1+1/\varepsilon). Multiplication by the finite integral \int|\Phi| proves absolute convergence of the sum-integral interchange.

Call a nonempty subset S\subset\Gamma allowed at p if its shifts satisfy the same compatibility and mark-exclusion conditions. The exact Euler factor is then

By (26) and Schwartz decay, the part of this integral where \max_\gamma|u_\gamma|>\sqrt L is O_N(L^{-N}) for every fixed N. All subsequent comparisons are made on the complementary region.

The factors carrying the zeta poles are

For \Re z_\gamma\ge0, all denominators in this expression have absolute value at least 1/2, and |H_p(\mathbf z)|\le\exp(C_M/p). If distinct shifts in \mathcal H have distinct residues modulo p, the allowed subsets are exactly the nonempty subsets of the individual fibers \mathcal M_i. Matching the terms of order p^{-1} gives

uniformly for \Re z_\gamma\ge0.

At \mathbf z=0, summing signs over the nonempty subsets of each occupied allowed residue class gives

It follows that H_p(0) is the local factor of \mathfrak S(\mathcal H). This identity holds also when that factor is zero.

We now compare the product of the H_p’s additively with its value at zero. Take Y=\max(2,2DL^B). For p\le Y, differentiation along the segment from zero to \mathbf z gives

For example, |p^{-z_S}-1|\le|z_S|\log p on that segment, while the denominators stay bounded away from zero. Each mixed partial product in the additive telescoping identity is at most

Using even the elementary bound \sum_{p\le Y}(\log p)/p\le\sum_{n\le Y}(\log n)/n=O((\log Y)^2), we obtain an error O(L^{-1/2}(\log L)^A) for the finite product. For p>Y there are no collisions of distinct shifts, so (27) shows that both remaining tail products are 1+O_M(1/Y). Consequently

This argument never divides by a possibly zero local singular-series factor.

It remains to evaluate the zeta factors. Absolute Euler products for \Re z_\gamma>0 give

Write w_\gamma=1+\mathrm i u_\gamma and

The pole expansion \zeta(1+z)=z^{-1}(1+O(|z|)) and the identity \sum_{\varnothing\ne S\subset\mathcal M_i}(-1)^{|S|}=-1 imply

Every denominator in K_i has positive integer real part, so the product of the K_i has at most polynomial growth. Thus

is absolutely convergent. Integrating (28) and (29) against the Schwartz transform, then restoring the negligible frequency tails, proves (13). The constant is real by conjugation under \mathbf u\mapsto-\mathbf u.

Finally, a singleton fiber contributes w_\gamma, and a double fiber contributes

After taking absolute values, the new variables contribute e^{-2\sum_i y_i}; the remaining factor is a Schwartz function times a fixed polynomial. Fubini therefore applies. Recombining the Fourier integrals by differentiating (25) proves (14).□

Proof. The cases s=0,1 are immediate. Otherwise let y=(1/4)\log h and Q=\prod_{p\le y}p. The prime number theorem gives Q\le h^{1/2} for large h. For all tuples, including those with equal entries, define

The exponent remains -s even on tuples with collisions. For independent uniform residues a_1,\ldots,a_s modulo a prime p, averaging over an additional uniform residue x gives

The Chinese remainder theorem therefore shows that \mathfrak S_y has exactly mean one on complete residue tuples modulo Q.

A uniform integer in a length-h interval has residue distribution modulo Q within total variation O(Q/h) of uniform. The elementary bound \sum_{p\le y}1/p=O(\log\log(3y)) used above gives

It follows that

There are O_s(h^{s-1}) tuples with equal entries, so deleting them changes this normalized mean by at most O_s(h^{-1}(\log y)^{A_s}).

For a distinct tuple put \Delta=\prod_{i<j}|a_i-a_j|, a positive integer. For y>2s every factor of the tail

is positive. A prime not dividing \Delta has \nu_p=s and a local factor 1+O_s(p^{-2}); the logarithm of a remaining factor is O_s(1/p). Hence

Since \log\Delta\le\binom s2\log(Kh), this gives, uniformly in the distinct tuples,

Now use the identity \mathfrak S(\{a_1,\ldots,a_s\})=\mathfrak S_y(\mathbf a)T_y(\mathbf a) and the nonnegative truncated mean. This identity is valid also if a small-prime factor vanishes; no quotient by \mathfrak S_y is taken. The stated error follows from Q\le h^{1/2} and \log y\asymp\log\log h.□

Proof of Proposition 3.1. First we check the smoothness and support needed by the moment formula. Extend f_j by zero to \mathbb R^j. Its cumulative integral is smooth on \mathbb R^j, and multiplication by smooth coordinate cutoffs that equal one near [0,\infty) and vanish below -1 makes it compactly supported without changing its orthant values or derivatives. Indeed, the cumulative integral already vanishes when any coordinate exceeds an upper support bound of f_j. On the nonnegative orthant, F_j vanishes if \sum_i v_i\ge\tau, and its signed complete mixed derivative is given by (4). This derivative is zero on every coordinate face. For j\ge1, Tf_{j+1} has the same smoothness, interior-support, and face-vanishing properties.

Expand Z^2 over pairs of subsets S,T\subset I. Any shift in S\mathbin\triangle T occurs once and sets a coordinate of one of the functions in (4) to zero. Its main coefficient is therefore zero. If S=T has size j, the coefficient is \|f_j\|_2^2. Averaging the singular series and using

proves the first limit in (5). With a prime mark in J, the same matching rule applies and

The two divisor factors have total support 2\tau<1/2, so the marked moment formula proves the first limit in (6).

For an internal mark a\in I, the exact deletion identity (8) replaces F_j by \widetilde F_j=F_j+F_{j+1}(\cdot,0) on the prime event. Deletion is a bijection, so no extra dimension factor occurs. The support radius of \widetilde F_j is still \tau, and its signed complete mixed derivative is f_j+Tf_{j+1}. It vanishes on coordinate faces when j\ge1. Apply the external marked formula to these functions. Only equal subsets U survive, and

This proves the formula for v, including the scalar term \lambda|f_0+\int f_1|^2.

The equality-pattern estimate preceding this subsection justifies summing the errors in these limits. Applied to Z^4, the same argument uses four divisor factors, total support 4\tau<1, and only finiteness of their pattern constants. Bounded averages of the singular series then give the second assertion in (5).

It remains to prove a detector estimate whose constant is independent of the chosen family. Write q_X=\log(2X+2h)/L=1+o(1). At a relevant prime m+b, we have D_G(m;b)=1. Consequently, pointwise,

The first inequality holds at composites as well, since its left side is zero. Because Z^2\ge0, these inequalities give

An off-diagonal summand contains six divisor factors, with total support 6\tau<1. The shifts b,c are paired and lie outside I. The two subsets from Z^2 must again be identical. For their common size j, the coefficient is c_G^2\|f_j\|_2^2. The pair (b,c) is ordered, and

Thus the off-diagonal sum tends to \lambda^2c_G^2w. Together with (6), this proves (7). Although the convergence rates may depend on the fixed family, the resulting C_* does not.□