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For $A_1$ in Lemma 3.1,

$$M(A_1)\le \left(1+O\!\left(\frac{\log_2x}{\log x}\right)\right) \frac{x}{\log x}. \tag{1}$$

Proof. First we record the weighted strengthening of the fibre formula needed for this exact error:

$$\sum_{\varphi(d)/d=q}\frac{\log d}{d}\le4 \qquad(q>0). \tag{2}$$

An empty fiber gives zero. For a nonempty fiber with support $P$, summing positive exponent series gives

$$\sum_{\varphi(d)/d=q}\frac{\log d}{d} =\left(\prod_{p\in P}\frac1{p-1}\right) \sum_{p\in P}\frac{p\log p}{p-1}.$$

This identity follows by writing $\log d=\sum_{p\in P}j_p\log p$, using $\sum_{j\ge1}j/p^j=p/(p-1)^2$, and summing each coordinate. The distributed term indexed by $p$ is

$$\frac{p\log p}{(p-1)^2} \prod_{\substack{r\in P\\r\ne p}}\frac1{r-1}.$$

Its first factor is at most two because $\log p\le p-1$. For $k=|P|\ge2$, at most one of the remaining primes is two, so the remaining product is at most $2^{-(k-2)}$. Summing bounds the total by $2k/2^{k-2}\le4$: the expression is four at $k=2$ and decreases thereafter. For $k=1$ the bound is two; for $k=0$ the sole $d$ is one and the sum is zero. This proves (2).

Let $B\subset A_1$ be monotone. Every $n\in B$ has a representation $n=dp$, $d\le D$, with

$$\frac{x}{DL}<p\le\frac{x}{d},\qquad p\ge D^3$$

for sufficiently large $x$. Thus $p$ is coprime to $d$ and

$$\varphi(n)=q(d)(1-1/p)n,\qquad q(d)=\varphi(d)/d. \tag{3}$$

List the distinct ratios for $d\in\mathbb N_{\le L}$, $d\le D$ as $0<q_1<\cdots<q_K\le1$. There are at most $D$ of them. Their reduced denominators are at most $D$, so

$$q_{k'}-q_k\ge D^{-2},\qquad q_{k'}\ge(1+D^{-2})q_k\quad(k'>k). \tag{4}$$

Partition $(x/L,x]$ into consecutive intervals $I_i$ of ratio $1+D^{-3}$, truncating the last. There are $O(D^3\log L)$ of them. Within $I_i$, let $H_{i,k}$ be the convex hull of points of $B$ with ratio $q_k$, with the empty and singleton conventions used in Proposition 3.3.

For two counted points $n,n'$ in $I_i$, with ratios $q_k<q_{k'}$, we have $n'/n=1+O(D^{-3})$. Equations (3)–(4) give $\varphi(n')>\varphi(n)$ once $D$ is large, because the relative $D^{-2}$ gap dominates the $O(D^{-3})$ errors. Hence $n'>n$. All the $H_{i,k}$ are consequently disjoint, and

$$\sum_{i,k}|H_{i,k}|\le x. \tag{5}$$

For a fiber $\mathcal D_k=\{d\le D:d\in\mathbb N_{\le L},q(d)=q_k\}$, the prime count for a fixed $d$ is, by Lemma 1.6,

$$\#\{p:dp\in H_{i,k}\} \le\frac1d\int_{H_{i,k}}\frac{dt}{\log(t/d)} +O\!\left(\frac{x}{d} e^{-c\sqrt{\log(x/D)}}\right). \tag{6}$$

This holds also for singleton hulls. Since $t>x/L$, put $v=\log(xd/t)$; then $0\le v\le\log d+\log L\le\log(DL)$. For sufficiently large $x$, $v\le\frac12\log x$. The elementary inequality $1/(a-v)\le1/a+2v/a^2$ therefore gives

$$\frac1{\log(t/d)} \le\frac1{\log x} +\frac{2(\log d+\log L)}{\log^2x}.$$

Sum (6) over $d\in\mathcal D_k$. Its reciprocal mass is at most one, and (2) bounds its logarithmic moment by four. Thus the integral contribution for this hull is at most

$$|H_{i,k}|\left(\frac1{\log x} +\frac{2(4+\log L)}{\log^2x}\right).$$

The error contribution is at most $Cx e^{-c\sqrt{\log(x/D)}}$. Summing over $i,k$ and using (5) bounds the total error by $CxD^4(\log L)e^{-c\sqrt{\log(x/D)}}$. Here $\log D=(\log_2x)^3=o(\sqrt{\log x})$, so this error is $o(x/\log^A x)$ for every fixed $A$. Since $\log L=10\log_2x$, the integral bound proves (1). $\square$

Source precision. The published last displayed bound replaces every $\log d$ by $\log D$, giving error $O((\log_2x)^3/\log x)$. That is enough for the main theorem but does not alone prove the stronger error in its Proposition 3.4. The explicit moment (2) completes that printed claim. This is a compilation expansion, not an author-issued erratum or a claim of a new bound.

Source. Tao, published paper, published pp.808–811, Proposition 3.4. This page uses that published version.

Bears on. Problem 49.