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Let $L=\sqrt{D}$. Regroup the finite divisor sum using the least common multiple:This is nonnegative. If $n$ has no prime factor in $\mathcal{P}$, only $d=1$ can contribute to the inner sum, so it equals ${\rho }_1=1$. This proves the required pointwise upper-bound sieve inequality. Whenever a term defining ${\lambda }_d$ is nonzero, $u,v\le L$ and their prime factors lie in $\mathcal{P}$. Hence $d=[u,v]\le uv\le L^2=D$, and every prime factor of $d$ lies in $\mathcal{P}$. Thus these are upper-bound sieve weights of level $D$, as described by the Selberg least-common-multiple weights.

Take sieve level $D=x^{1/3}$ and coefficient cutoff $L=\sqrt{D}=x^{1/6}$. Let $\mathcal{P}$ be the finite set of odd primes at most $L$. Give $n=p+2$ one unit of weight for each prime $p\le x$. For an odd squarefree integer $d$ on these primes, the sieve distribution isThe residue $-2$ is coprime to every such $d$, and $g(p)=1/(p-1)\in (0,1)$ for all permitted primes. Values of $g$ outside the permitted set can be chosen arbitrarily in $(0,1)$ at primes and extended multiplicatively on squarefree integers.

For the Selberg sieve normalizing sum, $k(p)=g(p)/(1-g(p))=1/(p-2)$ andHere is a direct proof of this lower bound. Restrict the Euler product to $3\le p\le y=L^{1/4}$. The Mertens first theorem gives ${\sum }_{3\le p\le y}\log p/(p-1)=\log y+O(1)$, which is at most $\frac{1}{2}\log L$ for large $L$. The Mertens second theorem givesThe truncated Euler-product lower bound, with $\theta =1/2$, now gives $J\ge W/2\gg \log L$.

Use the minimizing Selberg sieve weights from Question 3. Their absolute values are at most one, so the Selberg least-common-multiple weights satisfy $|{\lambda }_d|\le 3^{\omega (d)}$. Parts (a) and (b) control the remainder:The first factor uses the Bombieri–Vinogradov theorem; $D=x^{1/3}$ is below $x^{1/2}/{\log }^{B}x$ for all sufficiently large $x$. The second uses the permitted ${\sum }_{q\le D}{9}^{\omega (q)}/\varphi (q)\ll {\log }^{9}D$. The PDF has the exponent nine, which the TeX transcribes incorrectly. Choose $A=14$ to obtain $O(x/{\log }^{3}x)$.

Every twin prime pair with $p+2>L$ survives the finite sieve, while the pairs with $p+2\le L$ number at most $L$. The Selberg upper-bound sieve thus givesTherefore the twin-prime count is $O(x/{\log }^{2}x)$. The finite cutoff on the forbidden primes is essential: sieving by all odd primes would also discard the large prime values $p+2$ that we are trying to count.

In the level-$D$ convention used here, the Selberg upper-bound sieve states that a nonnegative sequence with sieve distribution $A_d=Xg(d)+r(d)$ satisfiesIt follows by minimizing the quadratic form with the Selberg sieve weights and using the Selberg least-common-multiple weights bound $|{\lambda }_d|\le 3^{\omega (d)}$.

Sieve the interval $x<n\le x+z$ by two and by the primes congruent to three modulo four, up to $L=\sqrt{D}$. The interval divisor counts giveThus $X=z$, $g(p)=1/p$, and $r(d)=O(1)$ uniformly in the location of the interval. Take $D=z^{1/2}$ and $L=z^{1/4}$.

To bound $J$, let $y=L^{1/4}$ and form the Euler product over the forbidden primes at most $y$. The given reciprocal-prime sum in residue class one modulo four, subtracted from the Mertens second theorem, yieldsMoreover, the Mertens first theorem bounds the logarithmic mean by ${\sum }_{p\le y}(\log p)/p=\log y+O(1)\le \frac{1}{2}\log L$ for large $L$. The truncated Euler-product lower bound gives $J\ge W/2\gg \sqrt{\log z}$.

The full-range summatory bound for three to the prime omega bounds the error by $O(D{\log }^{2}D)=O(\sqrt{z}{\log }^{2}z)$, which is $O(z/\sqrt{\log z})$. Every integer whose prime factors are all congruent to one modulo four survives this finite sieve. Therefore the half-dimensional interval sieve givesThe constants are independent of $x$. Enlarging them covers the bounded range of $z$ before the asymptotic product estimates apply, so the result holds throughout $1000\le z\le x$.

Apply the Selberg upper-bound sieve to the numbers $p+2$ with $p\le x$ prime. For odd squarefree integers $d$, the local count is $\pi (x;d,-2)$ and the density is $1/\varphi (d)$. With sieve level $D=x^{1/3}$, the normalizing sum is $\gg \log x$ by the truncated Euler-product lower bound. The Selberg least-common-multiple weights have absolute values at most $3^{\omega (d)}$. The weighted arithmetic-progression error bound and Bombieri–Vinogradov theorem make their total error $O(x/{\log }^{3}x)$. Consequently the number of twin prime pairs with smaller member at most $x$ is $O(x/{\log }^{2}x)$.