Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 29 (Prime Numbers)

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 $\mathrm{Re}s>1$, define the Riemann zeta function by the absolutely convergent Dirichlet seriesThe Euler product follows from unique prime factorization and absolute convergence. The counting-function integral givesBecause the fractional part is bounded, the last integral, and its derivatives with respect to $s$ on compact subsets, converge uniformly for $\mathrm{Re}s>0$. It is a holomorphic function there. This proves the Meromorphic continuation of the Riemann zeta function to the right half-plane, with a simple pole at one of residue one and no other singularity in that half-plane.

The Gamma function isThe Gamma function recurrence extends it meromorphically; its poles are at the nonpositive integers and it has no zeros. The functional equation of the Riemann zeta function can be stated asor, equivalently,These are identities of meromorphic functions; the second also extends the Riemann zeta function to the remaining half-plane. The critical strip is $0<\mathrm{Re}s<1$; its boundary lines will also be treated below, rather than included among possible exceptional zeros.

For $\mathrm{Re}s>1$, the absolutely convergent reciprocal Euler product gives $\zeta (s)\ne 0$. For $\mathrm{Re}s<0$, the factors $2^s$, ${\pi }^{s-1}$, $\Gamma (1-s)$ and $\zeta (1-s)$ in the second functional equation of the Riemann zeta function are finite and nonzero. Hence the only zeros there are the zeros of the sine factor:Each is simple. At zero, the sine zero cancels the pole of $\zeta (1-s)$; using its residue one gives $\zeta (0)=-1/2$, not zero. Nonvanishing on the rest of $\mathrm{Re}s=0$ follows from the boundary-line proof below and the functional equation of the Riemann zeta function. Together these facts prove that the trivial zeros of the Riemann zeta function are the only zeros outside the open critical strip.

Here is a Jensen disk proof of the zeta zero-count bound which avoids any unproved left-half-plane growth estimate. Set $F(s)=(s-1)\zeta (s)$, a holomorphic function throughout $\mathrm{Re}s>0$, including at one. The integral continuation formula givesFor each integer $j$, apply Jensen's formula in the disk with centre $2+ij$, outer radius $R=7/4$ and inner radius $r=8/5$. The outer disk stays in $\mathrm{Re}s\ge 1/4$, so its maximum modulus is $O((|j|+2{)}^2)$. At its centre,since $|1/\zeta (2+ij)|\le {\sum }_n|\mu (n)|n^{-2}\le \zeta (2)$. Therefore the number of zeros in its inner disk, counted with multiplicity, is at mostThe rectangle $1/2\le \mathrm{Re}s\le 1$, $|\mathrm{Im}s-j|\le 1/2$ fits in the inner disk because its farthest point has distance $\sqrt{(3/2{)}^2+(1/2{)}^2}=\sqrt{5/2}<8/5$. Summing over $O(T)$ such disks counts $O(T\log T)$ zeros in the right half of the critical strip up to height $T$. The first functional equation of the Riemann zeta function bijects zeros, with multiplicities, in the left half with their reflections $s\mapsto 1-s$ in the right half; its Gamma function factors are finite and nonzero in the strip. ThusThe pole of zeta function at one does not count as a zero: $F(1)=1$.

Finally, for real $\sigma >1$ and $t\ne 0$, the Euler-product logarithms and the nonnegative trigonometric polynomialgive the three-four-one product proof of zeta boundary nonvanishing:If $\zeta (1+it)$ had a zero of order $a\ge 1$, its factor would be $O((\sigma -1{)}^{4a})$ as $\sigma \downarrow 1$. The real zeta function factor has pole order three, and the factor at $1+2it$ stays bounded because $t\ne 0$. The product would tend to zero as $O((\sigma -1{)}^{4a-3})$, contradicting that it is at least one. ThereforeAt $t=0$ there is a pole, not a zero. Applying the sine-form functional equation of the Riemann zeta function at $s=it\ne 0$ now proves nonvanishing on the remaining imaginary axis, completing the earlier assertion about all zeros outside the critical strip.

A precise useful form of the principle is the type I–type II inverse principle for Möbius correlation. Normalize $|f(n)|\le 1$, and supposeFor any positive integer cutoffs $U,V$ with $U+V\le \delta X/2$, defineThen $|c_d|,|a_d|\le \tau (d)$, where $\tau$ is the divisor function, and at least one of the following two sums has modulus at least $\delta X/4$:The first is correlation against a controlled linear combination of indicators of multiples of small moduli, each a periodic function: this is the “somewhat periodic” branch. The second is a bilinear sum with independently weighted factors, both larger than the cutoffs: this is the “somewhat multiplicative” branch. The statement concerns these precise correlations, not an assertion that $f$ must itself be periodic or multiplicative.

For clarity, the Vaughan identity for the Möbius function proves this version immediately. Split $\mu ={\mu }_{\le U}+{\mu }_{>U}$ and likewise at $V$. Since $\mu \ast \mu \ast 1=\mu$,Multiply by $f(n)$ and sum. The first two terms contribute at most $U+V$ in modulus, and the remaining terms are $-T_{\mathrm{I}}+T_{\mathrm{II}}$. The triangle inequality gives the stated dichotomy. It is valid at finite $X$ and for any chosen cutoffs in the indicated range.

We now prove the required orthogonality without appealing to a stronger uniform exponential-sum theorem. Write $\alpha =\sqrt{2}$ and $e(u)=e^{2\pi iu}$. The Diophantine bound for the square root of two iswhere $\Vert u\Vert$ is distance to the nearest integer. If $a$ is that nearest integer, then $|2h^2-a^2|\ge 1$, while $h\sqrt{2}+a\le 4h$; dividing proves the bound.

Use the preceding Dirichlet convolution identity with $U=V=P=\lfloor X^{1/10}\rfloor$. The two short terms are $O(P)$. For the type I sum, the exponential geometric sum bound givesConsequentlyusing ${\sum }_{d\le Y}\tau (d)\ll Y\log (2Y)$ from counting factor pairs.

For the type II sum, perform a dyadic decomposition into $d\in (D,2D]$, $w\in (W,2W]$, keeping $dw\le X$. Nonempty blocks have $DW<X$, while $D,W\gg P$. A block $B$ satisfies, by the Cauchy-Schwarz inequality and the divisor-square summatory bound,To justify the cutoff, after expanding the square the permissible $d$ for a pair $w,w^{\prime }$ still form an interval, with upper endpoint $\min (2D,X/\max (w,w^{\prime }))$. Its exponential sum has the same geometric bound. Diagonal pairs supply $DW$; for each nonzero difference $h=w-w^{\prime }$ there are $O(W)$ pairs.

The points $0,\alpha ,2\alpha ,\dots ,\lfloor W\rfloor \alpha$ modulo one are separated by at least $1/(4W)$. Counting points in consecutive distance bands around zero therefore givesSubstitution yields the bilinear cancellation for badly approximable phases estimateThere are $O({\log }^{2}X)$ blocks, so $T_{\mathrm{II}}\ll X^{19/20}{\log }^{4}X$. The divisor-square bound itself can be obtained elementarily from $\tau (n{)}^2\le {\tau }_4(n)$ prime-power by prime-power and ${\sum }_{n\le Y}{\tau }_4(n)\le Y(1+\log Y{)}^3$; no uncontrolled coefficient bound is being suppressed.

Combining the short, type I and type II terms,Hence

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