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The constant-one arithmetic function has $1(n)=1$ for every positive integer $n$. Its Dirichlet convolution with $f$ is the divisor sum $(1\ast f)(n)={\sum }_{d\mid n}f(d)$.

For a prime $p$, the divisor sum defining the real weight is $\varphi (0)-\varphi (\log p/\log X)$. If $p>X^{1/3}$, the second term vanishes and the first is one. Thus $F(p)=1$, while $F(n)\ge 0$ for every integer because $\varphi$ is real and the sum is squared.

Put $h(x)=e^x\varphi (x)$. With the Fourier transform convention in the question, Fourier inversion givesThe function $h$ is smooth and has compact support. Using integration by parts $k$ times, with zero endpoint terms, provesChoose an integer $k\ge A$ to obtain $|\psi (t)|{\ll }_A|t{|}^{-A}$ for $|t|\ge 1$. In particular every polynomially weighted absolute integral of $\psi$ is finite.

For the double integral, useDifferentiating the second formula gives $\int \psi (t)(1+it)e^{-(1+it)u}dt=-{\varphi }^{\prime }(u)$. absolute convergence from the rapid decay permits Fubini's theorem. The derivative energy constant for a smooth sieve cutoff is thereforeThere is no complex conjugate in this integral: the two factors both become $-{\varphi }^{\prime }(u)$, which is real. The constant is positive; in fact the Cauchy-Schwarz inequality applied to $\varphi (1/3)-\varphi (0)=-1$ gives $c_{\varphi }\ge 3$.

Here is the basic structure of the smooth divisor-square sieve asymptotic, with enough uniform estimates to specify the error. Write $L=\log X$ and $R=X^{1/3}$. Expand the square and count multiples of the least common multiple $[d,e]$ in an arbitrary interval $I$ of length $X$:The sum is effectively restricted to $d,e<R$. Each count is $X/[d,e]+O(1)$, independent of the interval's position, so the total error is $O_{\varphi }(X^{2/3})$.

The Fourier representation of a smooth Selberg weight readsHence, with $z^{\prime }=(1+i{t}^{\prime })/L$,The four local terms correspond to the prime dividing neither divisor, only the first, only the second, or both. This is the Euler product for a smoothed divisor-square correlation. Factor it aswhereThe local factors of $H$ are $1+O(p^{-2+4\eta })$ when $|\mathrm{Re}z|,|\mathrm{Re}{z}^{\prime }|\le \eta$ for small fixed $\eta <1/4$. Thus $H$ is a holomorphic function near $(0,0)$, and evaluating each local factor at zero gives exactly one. Consequently $H(0,0)=1$ and $H(z,z^{\prime })=1+O(|z|+|z^{\prime }|)$ there.

The pole expansion $\zeta (1+w)=w^{-1}+O(1)$ now yieldswhen $|t|,|t^{\prime }|\le L^{1/4}$. Integrating the error against the rapidly decreasing transforms gives $O_{\varphi }(L^{-2})$. The complementary tails are harmless uniformly: absolute values in the original divisor series giveAny prescribed power decay of ${\int }_{|t|>L^{1/4}}|\psi (t)|dt$ is available, so these tails and the tails of the limiting kernel are $O_{\varphi }(L^{-2})$ after choosing sufficiently many applications of integration by parts. The same absolute bound justifies the original exchanges of infinite sums and integrals. Using the evaluated double integral,We conclude, uniformly over all interval locations,This proof does not assume that the interval starts near the origin.

Finally choose this one fixed smooth cutoff. primes above $R$ in $(X_0,X_0+X]$ contribute one each to the nonnegative weight, and at most $R$ primes lie below $R$. Thereforefor all sufficiently large $X$, uniformly in $X_0$. The bounded range $2\le X\le X_1$ follows by increasing the fixed constant and using the trivial interval bound $X+1$. Thus the short-interval prime upper bound from a smooth divisor weight is

The constant and first Fourier coefficients of ${\varphi }^8$ are $1$ and $-16$. In the weight-four Eisenstein basis at level two, these forceFor $n>0$, its $q^n$ coefficient is $-16{\sigma }_3(n)+256{\sigma }_3(n/2)$, where the second divisor sum is zero if $n$ is odd. The even divisors have cube sum $8{\sigma }_3(n/2)$, soOn the other hand, expanding the eighth power of the theta series of integer squares counts ordered integer eight-tuples of square sum $n$. Their sign is $(-1{)}^{{\sum }_j{m}_j}=(-1{)}^n$, since $m_j^2\equiv m_j(\mathrm{mod}2)$. If $r_8(n)$ denotes that count, thenSeparately $r_8(0)=1$, from the all-zero tuple. The positive-divisor formula is not a formula at zero. This is the eight-square representation formula.

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.

A weight-$k$ modular form for $\Gamma =S{L}_2(\mathbb{Z})$ is a holomorphic function $f$ on the complex upper half-plane satisfyingand holomorphic at a cusp. Since the full modular group has one cusp class, this last condition means that the period-one function has a convergent expansion $f(\tau )={\sum }_{n\ge 0}{a}_n{q}^n$, $q=e^{2\pi i\tau }$, near infinity. A cusp form has $a_0=0$. The element $-I$ makes a nonzero weight odd form impossible.

For weight zero, $f$ descends to a holomorphic function on the quotient of the complex upper half-plane by the modular group. At an elliptic fixed point, invariance under its finite stabilizer makes the local Taylor series a series in the quotient coordinate. At the cusp, the $q$ expansion extends it over $q=0$. The standard fundamental domain of the modular group, with its boundary identified and its cusp added, is a compact Riemann surface, the compactified modular curve $X(1)$. The maximum modulus principle now provesHolomorphy at the cusp is essential to this conclusion.

For positive even $k\ge 4$, define the unnormalized Eisenstein seriesThe real-linear map $(m,n)\mapsto m\tau +n$ is invertible, and on a compact subset of the complex upper half-plane its norm is uniformly comparable to $\sqrt{m^2+n^2}$. Thus the sum converges absolutely and locally uniformly for $k>2$, proving holomorphy. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$,The map $(m,n)\mapsto (ma+nc,mb+nd)$ is a bijection on the nonzero integer pairs, giving $G_k(\gamma \tau )=(c\tau +d{)}^k{G}_k(\tau )$.

For the expansion, the cotangent partial-fraction Fourier kernel follows by differentiating the partial-fraction expansion of $\pi \cot \pi z$ and its geometric-series expansion $-\pi i(1+2{\sum }_{r\ge 1}{e}^{2\pi irz})$, valid for $\mathrm{Im}z>0$. It gives, for even $k\ge 2$,The $m=0$ part of $G_k$ is $2\zeta (k)$, where $\zeta$ is the Riemann zeta function; positive and negative $m$ contribute equally. Apply the kernel at $z=m\tau$ for $m>0$, and regroup the absolutely convergent sum according to $N=mr$. With the divisor sum ${\sigma }_{k-1}(N)={\sum }_{r\mid N}{r}^{k-1}$, this provesThe expansion has no negative powers and converges for $|q|<1$, so it also proves holomorphy at a cusp and completes the proof of modularity.

The supplied zeta values yield the Fourier expansion of a normalized Eisenstein seriesBoth $E_4^3$ and $E_6^2$ are modular forms of weight twelve. Their constant terms cancel, so their difference divided by $1728$ is a cusp form. Integrality requires an argument: dividing an integral series by $1728$ alone would not suffice. Expanding directly gives the integrality identity for the modular discriminantFor every integer $d$, $5d^3+7d^5$ is divisible by twelve. Modulo three, $d^3\equiv d^5\equiv d$; modulo four, an even $d$ makes both powers divisible by four, while an odd $d$ has $d^3\equiv d^5\equiv d(\mathrm{mod}4)$. Thus each coefficient $(5{\sigma }_3(n)+7{\sigma }_5(n))/12$ is integral. The other three series visibly have integral coefficients, provingThis proves integrality without assuming a product expansion for the modular discriminant.

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

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.

The elementary Selberg symmetry formula can be writtenUse the weights ${\lambda }_d=\mu (d)\log (x/d)/\log x$. Their divisor sum is $1_{n=1}+\Lambda (n)/\log x$. Apply them to the integrated Chebyshev sum, then use the Möbius harmonic logarithmic moments. The Chebyshev estimate shows this is equivalent, with an $O(x)$ error, to ${\sum }_{n\le x}\Lambda (n)\log n+{\sum }_{mn\le x}\Lambda (m)\Lambda (n)=2x\log x+O(x)$.

The sum-of-divisors function isIt is the divisor sum of the identity arithmetic function $n\mapsto n$.