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Let $\mu$ be the Möbius function. Its Möbius divisor-sum identity isIndeed, for $n>1$ the sum is the expansion of ${\prod }_{p\mid n}(1-1)$; nonsquarefree divisors contribute zero. Thus, for arithmetic functions $a,b$,To prove Möbius inversion, substitute the first formula into the second and collect the coefficient of $a(e)$. It is ${\sum }_{d\mid n/e}\mu (d)$, which is one for $e=n$ and zero otherwise. Conversely the same divisor interchange recovers $b$ from the displayed formula for $a$. In Dirichlet convolution notation this is simply $\mu \ast 1=\epsilon$, where $\epsilon$ is supported at one.

The Prime number theorem with classical zero-free-region error states that some absolute $c>0$ satisfiesHere $\Lambda$ is the Von Mangoldt function. An equivalent prime-counting form, after decreasing $c$ if needed, is $\pi (x)=\mathrm{Li}(x)+O(x{e}^{-c\sqrt{\log x}})$. The contributions of proper prime powers are $O(\sqrt{x}{\log }^{2}x)$ and can be absorbed into this error. We use $\Psi$ to keep it distinct from the Fourier transform appearing in Question 5.

Here is a direct deduction of the requested Mertens bound from a log-integrable Chebyshev error from the stated Prime number theorem. First, the divisor identity givesSecond, the exact Dirichlet convolution identityfollows from $\log =1\ast \Lambda$: multiplication of a Dirichlet convolution by $\log n$ differentiates its two factors, so applying it to $\mu \ast 1=\epsilon$ and convolving again with $\mu$ gives this formula. Summing it yieldsThe last harmonic sum is $O(1)$. To see this without an endpoint approximation, split $x/d$ into dyadic ranges $[2^j,2^{j+1})$. In each range $\sum 1/d=O(1)$, and its exponential factor is at most $e^{-c\sqrt{j\log 2}}$. Their sum over $j\ge 0$ converges. The finitely many terms with $1\le x/d<2$ obey the same estimate after adjusting the constant in the prime-number-theorem error. ThereforeFor the Mertens function $M(x)={\sum }_{n\le x}\mu (n)$,The second sum has absolute value at most ${\sum }_{n\le x}\log (x/n)=O(x)$, by an integral comparison or the Stirling formula. ConsequentlyThe absolute value in this conclusion is present in the original PDF.

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 $e^{-itu}$.

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

Take a positive gradient coefficient ${\kappa }^{-1}$, a positive squared mass, a finite-volume regulator and an ultraviolet cutoff. Use the dimensionless statistical action convention $Z=\int \mathcal{D}\varphi e^{-H}$: inverse temperature is absorbed into the coefficients. If $H$ denotes dimensional energy instead, the exponent below gains a factor ${\beta }_{\mathrm{th}}$ and the covariance gains ${\beta }_{\mathrm{th}}^{-1}$. These conventions cannot be mixed.

Choose the Fourier transform conventionIntegration of the quadratic gradient and source terms givesComplete the square with ${\tilde{\varphi }}_{\mathrm{cl}}(p)=\tilde{\Delta }(p{)}^{-1}\tilde{J}(p)$. Translation of the regulated real Gaussian variables leaves their measure unchanged, so the Gaussian integral givesHere $Z_G=Z_0[0]$ is the source-free Gaussian partition function. For $n$ independent regulated real coordinates and a positive quadratic matrix $A$, it is $(2\pi {)}^{n/2}(\det A{)}^{-1/2}$ under Lebesgue normalization. The continuum determinant needs the stated regulator and a measure convention. Reality pairs the nonzero Fourier modes, so they must not be counted as two independent real fields.

A connected correlation function subtracts the product of the means. The logarithm $W[J]=\log Z_0[J]$ generates its cumulants: differentiating $W$ twice with respect to the source, using the measure $d^{D}p/(2\pi {)}^D$ in the functional derivative convention, yieldsThe source changes the mean but not the covariance; the stated zero-source limit therefore follows immediately. Momentum conservation uses $p+q$, appropriate for a real field without complex conjugation on either displayed factor.

For the continuum massive Gaussian field correlation tail, set $\mu =\sqrt{\kappa }m>0$ and $r=|x|$. Fourier inversion givesThe large-argument Modified Bessel function of the second kind has $K_{\rho }(z)\sim \sqrt{\pi /(2z)}{e}^{-z}$. Hence $G_0(r)$ is proportional to $r^{-(D-1)/2}{e}^{-\mu r}$ at large distance andAt zero mass the exponential length diverges. For $D>2$ the critical continuum tail is algebraic, $G_0(r)\propto r^{2-D}$; in lower dimensions infrared regulation needs additional care.

For a uniform source $h$, perform a momentum-shell renormalization group step: separate slow modes $|p|<\Lambda /b$ and fast modes $\Lambda /b<|p|<\Lambda$. In a Gaussian field theory they decouple; the uniform source couples only to the retained zero mode. Integrating fast modes contributes only a field-independent determinant to the free energy. Restore the cutoff by $x^{\prime }=x/b$, $p^{\prime }=bp$ andThe factors from the measure, gradient and two fields cancel in the gradient term. The mass term gains $b^2$, and the linear source gains $b^{(D+2)/2}$. ThereforeHere $m$ is the positive mass; for signed thermal coordinates it is the squared-mass relation that applies. The Gaussian fixed point has thermal scaling eigenvalue ${\lambda }_t=2$, source eigenvalue ${\lambda }_h=(D+2)/2$ and zero anomalous dimension.

Under the formal homogeneous Gaussian scaling hypothesis, substitution of those eigenvalues givesThe last two values are independently visible in $\xi \propto t^{-1/2}$ and $\chi ={\tilde{G}}_0(0)=1/m^2\propto t^{-1}$, agreeing with Landau-Ginzburg theory. The specific-heat power follows directly from the determinant: after two thermal derivatives its singular contribution is proportional toThe change of variables $p=\sqrt{\kappa t}q$ proves this power because the rescaled integral converges at large $q$ exactly when $D<4$. The two-dimensional free energy itself has a $t\log t$ resonance, while its second derivative still has the indicated $t^{-1}$ power.

The printed magnetization exponent is a formal Gaussian scaling index, not a proved ordered-phase law of the purely quadratic model. The ordered-phase obstruction in a Gaussian scalar model is explicit: the uniform field has energy density $m^2{M}^2/2-hM$. At $h=0,m^2<0$ it tends to $-\infty$ as $|M|\to \infty$, and the partition function diverges. For $m^2>0$, the mean is $M=h/m^2$ and tends to zero as $h\to 0$, so there is no spontaneously ordered branch. Thus no literal $M\sim |t{|}^{\beta }$ for $t<0$ can be established from this Statistical Hamiltonian alone. The value $(D-2)/4$ follows from the scaling dimension of the field if one assumes such a branch; for $D\le 2$ its nonpositive value further warns against that interpretation. Adding a stabilizing quartic interaction defines an ordered phase, but below four dimensions that interaction is relevant and generally leads to the Wilson-Fisher fixed point, not these Gaussian thermodynamic powers.

For the Gaussian specific-heat infrared threshold, at $D=4$ the same determinant integral grows like $\log ({\Lambda }^2/t)$, so $\alpha =0$ as a power exponent, with a logarithmic divergence. For $D>4$, it has a finite cutoff-dependent limit; the leading heat capacity is a regular constant and has no divergent power, again described by $\alpha =0$ in the requested convention. After subtraction of this background the Gaussian singular term can still have the negative power index $2-D/2$; it is not correct to erase that distinction. In a stable quartic Landau-Ginzburg theory above four dimensions, the quartic coefficient is a dangerously irrelevant coupling: it is needed for the ordered phase and contributes $f_s\propto -t^2/u$, giving the mean-field specific-heat jump and $\alpha =0$ while invalidating naive hyperscaling relation. At four dimensions, marginal interactions can add further logarithmic corrections. The agreement with Landau theory is in the leading heat-capacity power; it does not assert the absence of logarithms or supply a missing stable Gaussian ordered phase.

Set $\Delta =T/M$, $q_0=q$, and $q_M=q^{\prime }$. The Trotter product formula separates the kinetic and potential energy evolution to first order in $\Delta$. Inserting $M-1$ resolutions of the identity in the position operator basis and using the free short-time kernel givesThe exponent converges formally to the action along the path. The prefactors and intermediate integrations define the time-sliced configuration-space path integral measure, soA real-time path integral is an oscillatory limit with the Feynman i-epsilon prescription, rather than an ordinary probability integral. The operator derivation assumes the usual self-adjointness and product-formula hypotheses for the Hamiltonian.

For a quadratic potential energy write $q=q_c+\eta$, with $\eta (0)=\eta (T)=0$ and ${\ddot{q}}_c+V^{\prime }(q_c)=0$. The first variation vanishes by the Euler-Lagrange equation and the fixed endpoints. Because $V$ is quadratic, the remaining expansion is exact:Here integration by parts has no boundary contribution. Translation of the integration variables leaves the time-sliced measure unchanged. Thus the Gaussian path integral over $\eta$ is independent of the endpoint values, and the entire endpoint dependence lies in $e^{iS[q_c]}$.

Expanding $\eta$ in eigenfunctions satisfying Dirichlet boundary conditions, each real mode contributes an inverse square root of its quadratic eigenvalue, with the oscillatory phase fixed by continuation. Normalizing against $A_0={\partial }_t^2$ gives a precise determinant-ratio form:The determinant ratio is meaningful after a common regulator; the phase is part of the prescription. It is equivalent to writing $N=C(T)({\det }_{D}A{)}^{-1/2}$ with an endpoint-independent normalization.

For example, if $V^{\prime \prime }={\omega }^2$, the Dirichlet oscillator determinant ratio isA linear or constant term in $V$ changes the action but not this prefactor. The formula holds away from conjugate times, when the boundary problem and determinant are nonsingular. At $\sin \omega T=0$, it must be continued as a distribution with its Maslov index phase; the assertion of an ordinary finite prefactor cannot be used literally there. For the unshifted oscillator, at $T=r\pi /\omega$ the kernel is $e^{-ir\pi /2}\delta (q^{\prime }-(-1{)}^{r}q)$.

For the full-line source integral, impose vacuum boundary conditions by adiabatic damping. Take the source initially as a test function, so its pairings with the Green function are defined. After integration by parts, writeWith the Fourier transform convention $G(t)=\int d\omega e^{i\omega t}\tilde{G}(\omega )/(2\pi )$, the inverse isSet $q=\eta -GJ$. Completing the square givesand consequentlyThe Gaussian functional integral $Z[0]$ is source independent and contains the regulated determinant. This is most naturally read as a normalized vacuum generating functional. For $m>0$, closing the frequency contour gives $G(t)=-i{e}^{-im|t|}/(2m)$; the time-ordered oscillator two-point function is $iG(t)$. This checks the sign of the source exponent and identifies the required vacuum prescription.