Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 150 / 2

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.

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The symmetric form of the Functional equation of the Riemann zeta function isThe complex conjugation identity $\zeta (\overline{s})=\overline{\zeta (s)}$ and the functional equation show that every Nontrivial zero of the Riemann zeta function $\rho$ is accompanied by $\overline{\rho }$, $1-\rho$, and $1-\overline{\rho }$.

Let $\rho =\beta +i\gamma$ be the given nonreal zero. If $\beta >1/2$, use $\rho$ itself; if $\beta <1/2$, use $1-\rho$, whose real part is $1-\beta >1/2$. The resulting zero cannot have real part greater than one, by the stated zero-free half-plane. It therefore has real part in $(1/2,1]$.

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Write $M(x)={\sum }_{n\le x}\mu (n)$ for the Mertens function. Suppose, to the contrary, that for some $\epsilon >0$ the quotient $|M(x)|/x^{1/2-\epsilon }$ were bounded. Partial summation would then makeconverge and define a holomorphic function throughout $\Re s>1/2-\epsilon$. In $\Re s>1$ the Euler product identifies this function with $1/\zeta (s)$, so analytic continuation would make $1/\zeta (s)$ holomorphic in that larger half-plane.

By assumption, $\zeta$ has a nontrivial zero. The Functional equation of the Riemann zeta function reflects one of that zero and its partner into $\Re s\ge 1/2$, where $1/\zeta$ must have a pole, a contradiction. Thus $|M(x)|/x^{1/2-\epsilon }$ is unbounded, which gives an $x\ge 1$ exceeding any prescribed constant $C$.

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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 $n^{it}$.

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Put $g=f\mu$. At every prime, $f(p{)}^2=1$, so the triangle inequality for pretentious distance givesThe standard strong aperiodicity of the Möbius function states, for example with $T=(\log x{)}^{1/10}$, thatIndeed, its left side is controlled by the prime sum ${\sum }_{p\le x}(1+\cos (v\log p))/p$, uniformly in this range.

Choose $t$ and $u$ minimizing the two distances in Halász theorem. The displayed triangle inequality implies that at least one of $M(f;x,T)$ and $M(g;x,T)$ tends to infinity. Halász's bound, and $T\to \infty$, then show that at least one oftends to zero. Their minimum is consequently $o(1)$, which is the claimed $o(x)$ estimate before normalization.

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