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There is a constant $C>0$ such thatTaking logarithms reduces the result to the Mertens second theorem, because the terms of order $p^{-2}$ and smaller form an absolutely convergent series.

Let $A(x)$ count primes $p\le x$ for which $p+2$ is also prime. We prove $A(x)\ll x/{\log }^{2}x$ using a quadratic Selberg sieve, then apply Abel summation.

For $F(n)=n(n+2)$, each odd prime has exactly two roots modulo that prime. The Chinese remainder theorem gives the polynomial root density in a sievefor odd squarefree integers $d$. Put $R=x^{1/8}$ and choose the optimizing real Selberg sieve weights ${\lambda }_d$, supported on these $d\le R$, with ${\lambda }_1=1$. If $p,p+2$ are prime and $p>R$, their weight $({\sum }_{d\mid F(p)}{\lambda }_d{)}^2$ is one. It is nonnegative for every other integer. Expanding the square therefore givesThe Selberg sieve diagonalization makes the main quadratic form $1/J(R)$, whereThe optimal Selberg weights have modulus at most one. Since ${\rho }_F([d,e])\le [d,e]\le R^2$, the remainder is at most $O(R^4)$.

For completeness, the lower bound on $J$ is where the second logarithm enters. Restrict to primes $3\le p\le y=R^{1/8}$ and put $k(p)=2/(p-2)$. The full product $W={\prod }_{3\le p\le y}(1+k(p))$ is $\asymp {\log }^{2}y$, since its logarithm is $2{\sum }_{3\le p\le y}1/p+O(1)$ and the Mertens second theorem applies. Give the associated squarefree divisor $d$ probability $k(d)/W$. Prime inclusion probabilities are $2/p$, so the Mertens first theorem givesfor large $R$. The Markov inequality shows that at least half the weight has $d\le R$. This proves the truncated Euler-product lower bound $J(R)\gg W\gg {\log }^{2}R$. Consequently the twin-prime upper bound from a quadratic sieve isFinally, Abel summation givesThe tail is dominated by ${\int }_3^{\infty }dt/(t{\log }^{2}t)<\infty$. Thus reciprocal-sum convergence from a counting bound proves convergence of the first reciprocal sum. The second is smaller termwise. We have proved Brun's theorem:Counting each twin prime just once changes at most a finite repetition and gives the same conclusion.

For the unit-modulus prime values in this question, the Granville-Soundararajan distance isThe second identity uses $|f(p)|=|g(p)|=1$. Thus it is the Euclidean distance between the finite weighted vectors $(f(p)/\sqrt{2p}{)}_{p\le X}$ and $(g(p)/\sqrt{2p}{)}_{p\le X}$. Nonnegativity, symmetry and the triangle inequality follow from that norm representation, and it vanishes exactly when the two prime-value vectors agree. This proves the prime-restriction metric for pretentious distance.

On whole arithmetic functions at fixed $X$ it is a pseudometric. For example $f=1$ and $g={\mu }^2$ have identical values at every prime and zero distance for every $X$, but different values at four. Thus the literal identity-of-indiscernibles assertion needs the prime-restriction quotient.

Since $\mu (p)=-1$,For completeness, partial summation and the prime-counting form of the Prime number theorem giveTo verify the constant and the error, replace $\pi$ by $\mathrm{Li}+E$ in the first expression. Differentiating $\mathrm{Li}(u)/u$ shows its two contributions combine to $\log \log X$ plus a constant. The error integral converges absolutely because $|E(u)|/u^2\ll e^{-c\sqrt{\log u}}/u$, and $E(X)/X\to 0$. This proves the Mertens second theorem in the form needed here. In particular the pretentious distance between one and the Möbius function satisfies

For the last request, retain the earlier prime normalization $|f(p)|=1$; the proof also works with the weaker assumption $|f(p)|\le 1$. Without a bound on prime values, the original definition and the absolute convergence of the displayed measure need not apply to an unrestricted multiplicative arithmetic function. For example $f(n)=\mu (n{)}^{2}n$ is multiplicative and squarefree-supported, but for $X>e$ the displayed measure has infinite positive mass: its prime terms alone are ${\sum }_p{p}^{-1/\log X}$, which diverges by the Prime number theorem.

Put $L=\log X$, $\sigma =1+1/L$, and use the Fourier transform convention $\hat{\nu }(t)=\int e^{-itu}d\nu (u)$. Since $f$ is supported on squarefree integers, the defining property of a multiplicative arithmetic function gives the absolutely convergent Euler productThe bound on prime values gives $|f(n)|\le 1$ for squarefree $n$, so the total variation norm of a measure is bounded by $\zeta (\sigma )$; thus $\nu$ is a finite measure. Expanding the logarithm of the modulus, with an absolute $O(1)$ remainder since ${\sum }_p{p}^{-2\sigma }\ll 1$, givesAll constants here are uniform in $t$. Moreoverusing the pole of the Riemann zeta function. The nonnegative deficits $1-\mathrm{Re}(f(p)p^{-it})$ giveIndeed, the replacement of $p^{-\sigma }$ by $1/p$ for $p\le X$ costs at mostby the Mertens first theorem, while discarding the $p>X$ terms costs nothing in this lower bound. Thus the large Fourier coefficient criterion for pretentiousness isIn particular,The comparison function is the Archimedean character $n^{it}$ with this sign because our Fourier transform uses $e^{-itu}$.

Write $f(n)=3^{\omega (n)}$, where $\omega$ is the prime omega function. If $p^a\Vert n$, its contribution to the Dirichlet convolution $(f\ast \Lambda )(n)$ isSince $3(a-1+1/3)\ge a$, summing over the prime factors proves the log-weighted convolution bound for three to the prime omega, $f(n)\log n\le 3(f\ast \Lambda )(n)$. On $\sqrt{D}\le n\le D$, $\log n\ge \frac{1}{2}\log D$, henceThis is the required inequality. The standard Chebyshev estimate $\psi (t)\ll t$ bounds its right side by $D(\log D{)}^{-1}{\sum }_{m\le D}f(m)/m$. To bound the latter sum, use multiplicativity and extend to all numbers with prime factors at most $D$:The Mertens second theorem states ${\sum }_{p\le D}1/p=\log \log D+O(1)$. Taking logarithms of the product, with a summable $O(p^{-2})$ remainder, gives $3\log \log D+O(1)$. Therefore the product is $O({\log }^{3}D)$, proving the summatory bound for three to the prime omega in the required range:

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 $1000\le z\le x$.

PutThe Abel summation formula with weight $1/\log t$ givesSubstituting $A(t)=\log t+E(t)$ yieldsfor a constant $c$. Both final terms are $O(1/\log x)$, soThis is the Mertens second theorem.

Taking the natural logarithm of the finite Euler product and using the Taylor seriesgivesThe double series converges absolutely, and the Mertens second theorem therefore makes the right sideforExponentiating provesThis is the Mertens third theorem.

Because $\mu (p)=-1$, the squared pretentious distance from one to the Möbius function is twice the reciprocal-prime sum. The Mertens second theorem therefore gives $D(1,\mu ;X{)}^2=2\log \log X+2B_1+o(1)$. In particular the distance diverges, illustrating that a multiplicative function can be far from the constant function even though both have unit-modulus prime values.

The progression version of the Mertens second theorem gives ${\sum }_{\begin{array}{c}p\le y\\ p\equiv 1(\mathrm{mod}4)\end{array}}1/p=\frac{1}{2}\log \log y+c+O(1/\log y)$. Subtracting from the all-prime sum shows that the forbidden primes $p\equiv 3(\mathrm{mod}4)$ contribute half the logarithmic density. This is the source of the square root in a half-dimensional interval sieve.

The Euler product and Mertens second theorem give ${\sum }_{n\le D}{3}^{\omega (n)}/n\ll {\log }^{3}D$. The log-weighted convolution bound for three to the prime omega, with the Chebyshev estimate $\psi (y)\ll y$, then gives ${\sum }_{\sqrt{D}\le n\le D}{3}^{\omega (n)}\ll D{\log }^{2}D$. A direct bound for the full range follows from $3^{\omega (n)}\le {\tau }_3(n)$, where ${\tau }_3(n)=\#\{(a,b,c):abc=n\}$:

For $F(n)=n(n+2)$, every odd prime has two forbidden residue classes. Use Selberg sieve weights supported on odd squarefree integers $d\le R=x^{1/8}$. Their main quadratic form is $1/J$, where $J={\sum }_{d\le R}{\prod }_{p\mid d}2/(p-2)$. The truncated Euler-product lower bound applied to primes up to $R^{1/8}$ gives $J\gg {\log }^{2}R$: the mean logarithmic divisor size is $2{\sum }_{3\le p\le R^{1/8}}(\log p)/p<\frac{1}{2}\log R$, by the Mertens first theorem, and the full product is $\asymp {\log }^{2}R$, by the Mertens second theorem. The optimal Selberg weights have modulus at most one, so the remainder is at most $O({\sum }_{d,e\le R}[d,e])=O(R^4)$. The finitely many possible pairs with $p\le R$ contribute $O(R)$. Thus the count is $O(x/J+R^4+R)$, proving the bound.