I'm working my way through Daboussi's proof of the prime number theorem [C. R. Acad. Sc. Paris, t. 298, Serie I, n$^{\circ}$ 8, 1984]. He writes, for all $y\geq 2$,

$v_y(n)=\left\{\begin{array}{cl} 1 & \text{if all prime factors of }n\text{ are }\leq y \\ 0 & \text{otherwise} \end{array}\right.$ ,

$u_y(n)=\left\{\begin{array}{cl} 1 & \text{if all prime factors of }n\text{ are }>y \\ 0 & \text{otherwise} \end{array}\right.$ ,

$V_y(t)=\sum_{n\leq t}v_y(n)\mu(n)$ and

$M(t)=\sum_{n\leq t}\mu(n)$

One of the steps is to prove

$\displaystyle{ (1)\ \ \ \ \limsup_{x\to\infty}|M(x)/x|\leq\bigg(\prod_{p\leq y}\big(1-\frac{1}{p}\big)\bigg)\int_{1}^{\infty}\frac{|V_y(t)|}{t^2}\ dt }$

I've followed his reasoning to $V_y(x/n)=V_y(d_j)$ for $x/d_{j+1}<n\leq x/d_j$ and $\displaystyle{ (2)\ \ \ \ \limsup_{x\to\infty}|M(x)/x|\leq\sum_{j=1}^{q-1}|V_y(d_j)|\lim_{x\to\infty}(1/x)\sum_{x/d_{j+1}<n\leq x/d_j}u_y(n)+|V_y(d_q)|\lim_{x\to\infty}(1/x)\sum_{n\leq x/d_q}u_y(n)\ , }$ where $1=d_1<d_2<\cdots<d_q\leq y$ is the complete sequence of square-free integers between 1 and $y$.

Daboussi now claims that (2) gives (1) without further detail. Can anyone help me understand how? Many thanks.

  • $\begingroup$ It looks like the word "where" in line (2) should be erased. For layout purposes I think it would be better to write $$\lim \sup_{x\to \infty}|M(x)/x|=A+B $$ $$\text {where }\quad A=....$$ $$\text {and }\quad B=...$$ (where $A,B$ are the summations on the RHS of (2)...)... Sorry I can't help you with the math itself. $\endgroup$ – DanielWainfleet Sep 2 '17 at 19:39

Denote $P:=\prod_{p\le y}p$ and $Q:=\prod_{p\le y}(p-1)$. For any half-open interval $I$ of length $P$, we have $$\sum_{n\in I}u_y(n)=\#\{n\in I:\gcd(n,P)=1\}=\#\{n\in (0,P]:\gcd(n,P)=1\}=Q.$$ On the other hand, for any interval $I$ of length smaller than $P$, we have $$0\leqslant \sum_{n\in I}u_y(n)\leqslant Q.$$

Given $b>a>0$ and $x>0$, we may estimate the sum of $u_y(n)$ over $(ax,bx]$ by dividing it into intervals of length $P$ and the last interval of length smaller than $P$, which implies $$\Bigl\lfloor\frac{bx-ax}{P}\Bigr\rfloor Q\leqslant\sum_{ax<n\le bx}u_y(n)\leqslant\biggl(\Bigl\lfloor\frac{bx-ax}{P}\Bigr\rfloor+1\biggr)Q.$$ It is easy to see that both sides are equivalent to $(b-a)xQ/P$ as $x\to+\infty$, hence $$\lim_{x\to+\infty}\frac{1}{x}\sum_{ax<n\le bx}u_y(n)=(b-a)\frac{Q}{P}=(b-a)\prod_{p\le y}\biggl(1-\frac{1}{p}\biggr).$$ So \begin{align} |V_y(d_j)|\lim_{x\to+\infty}\frac{1}{x}\sum_{x/d_{j+1}<n\le x/d_j}u_y(n)&=|V_y(d_j)|\biggl(\frac{1}{d_j}-\frac{1}{d_{j+1}}\biggr)\prod_{p\le y}\biggl(1-\frac{1}{p}\biggr)\\ &=\prod_{p\le y}\biggl(1-\frac{1}{p}\biggr)\cdot\int_{d_j}^{d_{j+1}}\frac{|V_y(d_j)|}{t^2}\,\mathrm{d}t\\ &=\prod_{p\le y}\biggl(1-\frac{1}{p}\biggr)\cdot\int_{d_j}^{d_{j+1}}\frac{|V_y(t)|}{t^2}\,\mathrm{d}t \end{align} for $j=1,\ldots,q-1$, while \begin{align} |V_y(d_q)|\lim_{x\to+\infty}\frac{1}{x}\sum_{n\le x/d_q}u_y(n)&=|V_y(d_q)|\frac{1}{d_q}\prod_{p\le y}\biggl(1-\frac{1}{p}\biggr)\\ &=\prod_{p\le y}\biggl(1-\frac{1}{p}\biggr)\cdot\int_{d_q}^{+\infty}\frac{|V_y(d_q)|}{t^2}\,\mathrm{d}t \end{align} Assume $V_y(t)=V_y(d_q)$ for all $t\geqslant d_q$, then we may replace $V_y(d_q)$ by $V_y(t)$ in the integration and conclude that \begin{align} &\phantom{=}\sum_{j=1}^{q-1}|V_y(d_j)|\lim_{x\to+\infty}\frac{1}{x}\sum_{x/d_{j+1}<n\le x/d_j}u_y(n)+|V_y(d_q)|\lim_{x\to+\infty}\frac{1}{x}\sum_{n\le x/d_q}u_y(n)\\ &=\prod_{p\le y}\biggl(1-\frac{1}{p}\biggr)\cdot\Biggl(\sum_{j=1}^{q-1}\int_{d_j}^{d_{j+1}}\frac{|V_y(t)|}{t^2}\,\mathrm{d}t+\int_{d_q}^{+\infty}\frac{|V_y(t)|}{t^2}\,\mathrm{d}t\Biggr)\\ &=\prod_{p\le y}\biggl(1-\frac{1}{p}\biggr)\cdot\int_{d_1=1}^{+\infty}\frac{|V_y(t)|}{t^2}\,\mathrm{d}t. \end{align} In other words, the right hand sides of (1) and (2) are just equal, so (2) certainly gives (1).

However, the assumption $V_y(t)=V_y(d_q)$ for $t\geqslant d_q$ can not be derived from your clarification that $d_q$ is the largest square-free integer not greater than $y$. I think the correct statement should be that $d_q$ is the largest square-free integer without any prime factors greater than $y$, which clearly implies the assumption. Please check it in the paper.

  • $\begingroup$ Thank you very much. I don't think I could have got this by myself. $\endgroup$ – Teddy38 Sep 3 '17 at 17:04

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