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Write N={1,2,…}\mathbb N=\{1,2,\ldots\} and [x]={n∈N:n≤x}[x]=\{n\in\mathbb N:n\le x\} for real x≥1x\ge1. For an arithmetic function ff and finite A⊂NA\subset\mathbb N, put

Mf(A)=max⁡{∣B∣:B⊂A,n<m, n,m∈B⟹f(n)≤f(m)}.M_f(A)=\max\{|B|:B\subset A,\quad n<m,\ n,m\in B\Longrightarrow f(n)\le f(m)\}.

The maximum exists, including Mf(∅)=0M_f(\varnothing)=0. Write Mf(x)=Mf([x])M_f(x)=M_f([x]), and omit the subscript for f=φf=\varphi. Here φ(1)=1\varphi(1)=1 and

φ(n)=n∏p∣n(1−1/p),σ(n)=∑d∣nd,ψ(n)=n∏p∣n(1+1/p).\varphi(n)=n\prod_{p\mid n}(1-1/p),\qquad \sigma(n)=\sum_{d\mid n}d,\qquad \psi(n)=n\prod_{p\mid n}(1+1/p).

Every product over an empty prime support is one. Throughout, pp denotes a prime, log⁡\log is natural logarithm, and log⁡2x=log⁡log⁡x\log_2x=\log\log x. An OO or ≪\ll constant is absolute unless a dependence is displayed. All asymptotic arguments first take xx sufficiently large; an all-x≥10x\ge10 conclusion includes an explicit bounded-range absorption.

The sets N≤y\mathbb N_{\le y} and N<y\mathbb N_{<y} contain positive integers all of whose prime factors are respectively at most yy and less than yy; both contain one. An almost prime in this source means a prime or the product of two primes, allowing repetition.

Exact external analytic inputs. The classical prime number theorem in the form used here states that some absolute C,c>0C,c>0 satisfy

∣π(t)−∫2tdulog⁡u∣≤Ctexp⁡(−clog⁡t)(t≥10).(1)\left|\pi(t)-\int_2^t\frac{du}{\log u}\right| \le Ct\exp(-c\sqrt{\log t})\qquad(t\ge10). \tag{1}

We also use the classical Mertens estimates, uniformly for y≥2y\ge2,

∑p≤y1p=log⁡log⁡y+O(1),∑p≤ylog⁡pp≪log⁡y,∏p≤y(1−1/p)−1≪log⁡(2y).(2)\sum_{p\le y}\frac1p=\log\log y+O(1),\qquad \sum_{p\le y}\frac{\log p}{p}\ll\log y,\qquad \prod_{p\le y}(1-1/p)^{-1}\ll\log(2y). \tag{2}

Their analytic proofs are external. The interval and smooth-number consequences actually needed are proved in Lemma 1.5, Lemma 1.6 and Lemma 1.7. Elementary partial summation and repeated integration by parts in (1) also give

π(t)=tlog⁡t+tlog⁡2t+O ⁣(tlog⁡3t).(3)\pi(t)=\frac{t}{\log t}+\frac{t}{\log^2t} +O\!\left(\frac{t}{\log^3t}\right). \tag{3}

For completeness, integrating by parts twice gives these first two terms and a remainder 2∫2tdu/log⁡3u+O(1)2\int_2^tdu/\log^3u+O(1). Split that integral at t\sqrt t to bound it by O(t/log⁡3t)O(t/\log^3t); the exponential error in (1) is smaller than that remainder.

Source precision. The published opening definition refers to the selected subset, correcting the phrase “both in AA” (arXiv v4 p.1). The main theorem concerns weak monotonicity. The strict question in Problem 49 is related by the separate strict transfer.

Source. Tao, published paper, published pp.793–799, Section 1.1 and Lemmas 1.5–1.7. This page uses that published version.

Bears on. Problem 49.