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An Euler product factors a Dirichlet series into local factors indexed by prime numbers. For a multiplicative arithmetic function $f$ and in a half-plane of absolute convergence,

Halász's theorem bounds the mean of a bounded multiplicative arithmetic function in terms of its least pretentious distance from an Archimedean character. A large mean can occur only when this distance is small.

Let $X\ge 3$, and let $f$ be a normalized multiplicative arithmetic function supported on squarefree integers, with $|f(p)|\le 1$. Form the finite measure $\nu ={\sum }_{n}f(n)n^{-1-1/\log X}{\delta }_{\log n}$ using Dirac measures. Under the transform convention $\hat{\nu }(t)=\int e^{-itu}d\nu (u)$, its Euler product is ${\prod }_p(1+f(p)p^{-1-1/\log X-it})$. Expanding its logarithmic modulus, and replacing the damped prime weights by $1/p$ up to $X$ with Mertens first theorem, gives the displayed absolute bound, uniformly in real $t$. Hence $|\hat{\nu }(t)|\ge \delta \log X$ forces $D(f,n^{it};X{)}^2\le \log (C/\delta )$. The distance expression still defines a nonnegative pretentious score for prime values bounded by one; its actual metric interpretation uses unit-modulus values.

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 $e^{-itu}$.

An arithmetic function $f$ is multiplicative if $f(mn)=f(m)f(n)$ whenever $m$ and $n$ are coprime integers. If $g(n)={\sum }_{d\mid n}f(d)$ and $(m,n)=1$, every positive integer divisor of $mn$ is uniquely $d_1{d}_2$ with $d_1\mid m$ and $d_2\mid n$. Henceso $g$ is multiplicative.

The Möbius function is defined by $\mu (1)=1$, $\mu (n)=0$ if the square of a prime number divides $n$, and $\mu (n)=(-1{)}^r$ if $n$ is a product of $r$ distinct prime numbers. Coprime integers have disjoint sets of prime factors, so this definition immediately gives $\mu (mn)=\mu (m)\mu (n)$ when $(m,n)=1$.

For $n=p_1^{a_1}\cdots p_r^{a_r}>1$, only the square-free divisors contribute, and thereforeFor $n=1$ the sum is $1$. Thus $1\ast \mu =\epsilon$ in the language of Dirichlet convolution. Convolving $g=1\ast f$ with $\mu$ gives the Möbius inversion formula

If $f(n)=n$, its divisor sum is the sum-of-divisors function. Conversely, if $g(n)=n$, Möbius inversion givesusing the standard divisor identity for the Euler totient function. Therefore the two requested answers are

Suppose the sieve distribution has the formwhere $g$ is a multiplicative arithmetic function on squarefree integers and $0\le g(p)<1$. DefineThen the Selberg upper-bound sieve statesfor $1<D\le z$, with immaterial endpoint changes under other level conventions.

To construct the weights, putand setThen ${\lambda }_1=1$. If $\gcd (n,P(z))=1$, the divisor sum ${\sum }_{d\mid n}{\lambda }_d$ equals one, and henceSumming against $a_n$ and expanding givesThe Selberg diagonalization of the positive quadratic form givesFinally, grouping the error by $m=[d,e]$ gives at most $3^{\omega (m)}$ pairs $(d,e)$ for each squarefree $m$; using $|{\lambda }_d|\le 1$ yields the stated remainder.

An arithmetic function $f$ is multiplicative when $f(1)=1$ and $f(mn)=f(m)f(n)$ whenever $m$ and $n$ are coprime. If $m$ and $n$ are coprime, every divisor $d$ of $mn$ has a unique factorization $d=d_1{d}_2$ with $d_1\mid m$ and $d_2\mid n$. Hence the Dirichlet convolution of multiplicative functions satisfiesand $(f\star g)(1)=1$, so it is multiplicative.

Let $\epsilon$ be the identity for Dirichlet convolution, with $\epsilon (1)=1$ and $\epsilon (n)=0$ for $n>1$. The Möbius function satisfies $\mu \star 1=\epsilon$. Therefore, using commutativity and associativity,This is the Möbius inversion formula.

The Möbius function is defined by $\mu (1)=1$ and, for $n>1$,It is a multiplicative arithmetic function: $\mu (mn)=\mu (m)\mu (n)$ whenever $m$ and $n$ are coprime.

The summand $\mu (d{)}^2/\varphi (d)$ is multiplicative, so its divisor sum is multiplicative. For a prime power $p^a$ with $a\ge 1$, only $d=1,p$ contribute andMultiplying these local identities over the prime divisors of $n$ gives

For the final claim, choose distinct primes $p_0,\dots ,p_k$. The moduli $p_j^2$ are pairwise coprime, so the Chinese remainder theorem gives an integer $a$ satisfyingEvery positive integer $n\equiv a(\mathrm{mod}{\prod }_j{p}_j^2)$ then has $p_j^2\mid n+j$. HenceThere are infinitely many positive representatives of this congruence class.

For a positive integer $d$, defineThe sequence $\mathcal{A}$ has sieve distribution $(X,P_z,g,(r_d))$ when $g$ is a multiplicative arithmetic function on the squarefree divisors of $P_z$, with $0\le g(p)<1$, andfor every $d\mid P_z$. The function $g(d)$ models the local density of divisibility by $d$, while $r_d$ is the remainder.

If $m$ and $n$ are coprime, every divisor of $mn$ is uniquely $ab$ with $a\mid m$ and $b\mid n$. Hence, for a multiplicative arithmetic function $f$,

The Möbius function is $\mu (n)=0$ if a prime square divides $n$, and $\mu (n)=(-1{)}^r$ if $n$ is a product of $r$ distinct primes. The Euler totient function $\varphi (n)$ counts residues modulo $n$ coprime to $n$. From the prime factorizations,Both sides of the second identity are multiplicative, and at a prime power $p^a$,Therefore

If $f$ is a multiplicative function with $|f(n)|\le 1$, then for $\Re s>1$ its Dirichlet series has the Euler productIndeed, expanding the product over a finite set of primes and using unique prime factorization gives the sum over integers having no other prime factors. Moreover,so absolute convergence permits rearrangement and passage to the limit over all primes. This proves the formula.

One quantitative form of Halász theorem is the following. If $f$ is multiplicative and $|f(n)|\le 1$, putThen, uniformly for $2\le T\le x$,Thus a bounded multiplicative arithmetic function can have a large mean only when it has small pretentious distance from some Archimedean character $n^{it}$.

Pretentious number theory studies multiplicative arithmetic functions by measuring how closely they imitate structured functions such as Dirichlet characters and Archimedean characters.

Put $L=\sqrt{D}$, $h(t)={\prod }_{p\mid t}(g(p{)}^{-1}-1)$ and $y_t={\sum }_{\begin{array}{c}m\le L\\ t\mid m\end{array}}g(m){\rho }_m$, where $g$ is a multiplicative arithmetic function supported on squarefree integers. For squarefree integers $m,n$,Thus the quadratic form is $\Sigma ={\sum }_{t\le L}h(t)y_t^2$. Möbius inversion gives $g(d){\rho }_d={\sum }_{d\mid t\le L}\mu (t/d)y_t$. On the permitted squarefree integers, the constraint ${\rho }_1=1$ becomes ${\sum }_t\mu (t)y_t=1$. The Cauchy-Schwarz inequality yields $\Sigma \ge 1/J$, where $J={\sum }_{t}1/h(t)$ over those integers; equality holds at $y_t=\mu (t)/(Jh(t))$.