An Elementary Proof of the Prime Number Theorem based on M\"obius Function

TL;DR

This paper provides an elementary proof of the prime number theorem by demonstrating that the M"obius function's summatory function divided by x tends to zero, offering a different approach from classical proofs.

Contribution

It presents a novel elementary proof of the prime number theorem using the M"obius function and Selberg's asymptotic formula, with unique treatment of key components.

Findings

Proves that M(x)/x approaches zero as x tends to infinity.

Establishes the prime number theorem through an elementary approach.

Uses Selberg's asymptotic formula with a different methodology.

Abstract

Let $\mu(n)$ denote the M\"obius function, define $M(x)= \sum_{n\leq x}^{}\mu (n)$. The main result of this paper is to prove that \begin{equation*} \displaystyle\lim_{x \to +\infty}\frac{M(x)}{x}=0 \end{equation*} which is equivalent to the prime number theorem. We also use Selberg's asymptotic formula, but the treatments of key parts are different from several classical proofs.

Equations

$$\underset{x\to +\infty }{\lim }\frac{M(x)}{x}=0$$

$$\mathcal{F}(x):=\sum _{n\le x}\mu (n)\log \frac{x}{n}={\int }_1^x\frac{M(y)}{y}\mathrm{d}y,\mathcal{H}(x):=\frac{\mathcal{F}\left(e^{\sqrt{x}}\right)}{e^{\sqrt{x}}}.$$

$$\overline{\lim }|\mathcal{H}(x)|=\ell ,\overline{\lim }\frac{1}{x}{\int }_0^x|\mathcal{H}(y)|\mathrm{d}y=L,$$

$$|\mathcal{H}(x)|\le \alpha +o(1),$$

$$\underset{x\to +\infty }{\lim }\frac{M(x)}{x}=0,$$

$$\ell \le \frac{1}{1+\lambda }\alpha ,$$

$$G(x)=\log x\sum _{n\le x}F\left(\frac{x}{n}\right).$$

$$F(x)\log x+\sum _{n\le x}F\left(\frac{x}{n}\right)\Lambda (n)=\sum _{d\le x}\mu (d)G\left(\frac{x}{d}\right).$$

$$\sum _{n\le x}{\Lambda }_2(n)=2x\log x+\mathcal{O}(x).$$

$$\sum _{n\le x}(\Lambda (n)+\Theta (n))=2x+\mathcal{O}(\frac{x}{\log x})$$

$$\sum_{n\leq x}\mathcal{F}\left(\frac{x}{n}\right)=\sum_{n\leq x}\mu(n)\biggl{\lfloor}\frac{x}{n}\biggr{\rfloor}\mathrm{log}\ \frac{x}{n}=\mathcal{O}(\mathrm{log}\ x).$$

$${\int }_1^x\mathcal{F}\left(\frac{x}{t}\right)\mathrm{d}t=\mathcal{O}(x),$$

$$\mathcal{F}\left(\frac{x}{n}\right)={\int }_n^{n+1}\mathcal{F}\left(\frac{x}{t}\right)\mathrm{d}t+\mathcal{O}\left(\sum _{x/(n+1)<k<x/n}1\right).$$

$$\left|{\int }_{x_1}^{x_2}\mathcal{H}(y)\mathrm{d}y\right|\le M,$$

$${\int }_a^{b}f(x)\mathrm{d}x\le \frac{1}{2}m(b-a{)}^2.$$

$$|\mathcal{H}(x)|\le \frac{1}{x}{\int }_0^x|\mathcal{H}(y)|\mathrm{d}y+\mathcal{O}\left(\frac{1}{\sqrt{x}}\right).$$

$$\mathcal{F}(x)\log x+\sum _{n\le x}\mathcal{F}\left(\frac{x}{n}\right)\Lambda (n)=\mathcal{O}(x)$$

$$\sum _{n\le x}{\left(\log \frac{x}{n}\right)}^2=2x+\mathcal{O}((\log x{)}^2).$$

$$\mathcal{F}\left(\frac{x}{n}\right)\log \frac{x}{n}+\sum _{m\le x/n}\mathcal{F}\left(\frac{x}{mn}\right)\Lambda (m)=\mathcal{O}\left(\frac{x}{n}\right).$$

$$\mathcal{O}\left(x\right)\log x-\sum _{n\le x}\Lambda (n)\mathcal{O}\left(\frac{x}{n}\right)=\mathcal{O}(x\log x),$$

$$\sum _{n\le x}\frac{\Lambda (n)}{n}=\log x+\mathcal{O}(1).$$

$$\left\{\mathcal{F}(x)\log x+\sum _{n\le x}\mathcal{F}\left(\frac{x}{n}\right)\Lambda (n)\right\}\log x-$$

$$\sum _{n\le x}\Lambda (n)\left\{\mathcal{F}\left(\frac{x}{n}\right)\log \frac{x}{n}+\sum _{m\le x/n}\mathcal{F}\left(\frac{x}{mn}\right)\Lambda (m)\right\}=\mathcal{O}(x\log x).$$

$$\mathcal{F}(x){\left(\log x\right)}^2=\sum _{n\le x}{\Lambda }_2^{-}\mathcal{F}\left(\frac{x}{n}\right)+\mathcal{O}(x\log x),$$

$$|\mathcal{F}(x)|{\left(\log x\right)}^2\le \sum _{n\le x}{\Lambda }_2\left|\mathcal{F}\left(\frac{x}{n}\right)\right|+\mathcal{O}(x\log x),$$

$${\Lambda }_2^{-}:=\Lambda \ast \Lambda -\Lambda \log .$$

$$\sum _{n\le x}\left\{{\Lambda }_2^{-}(n)-2\Lambda (n)\log \frac{x}{n}\right\}=\mathcal{O}(x\log x).$$

$${\Lambda }_2^{-}(n)-2\Lambda (n)\log \frac{x}{n}=\mathcal{O}(\log n),$$

$$2\Lambda (n)\log \frac{x}{n}=\left|{\Lambda }_2^{-}(n)\right|+\mathcal{O}(\log n).$$

$$2\log n={\Lambda }_2(n)+\mathcal{O}(1).$$

Full text

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which is equivalent to the prime number theorem. We also use Selberg’s asymptotic formula, but the treatments of key parts are different from several classical proofs.

1 Introduction

The prime number theorem, one of the most well-known and significant theorems in number theory, was first proved by Hadamard and de la Vallée Poussin in 1896 using analytic methods. The elementary proof was discovered in 1949 by A. Selberg [1] and P. Erdős [2]. Their proof makes no use of complex analysis nor of the Riemann zeta function but is intricate. In addition, there are more classic elementary proofs such as [3, 4, 5, 6].

Let

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Then we can assume that exist positive constant $\ell,L,\alpha$ such that

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by noticing that $|\mathcal{H}(x)|\leq 1$. To derive

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we shall prove $\ell=0$ in two cases, which can be proved to be equivalent to (2). In one case, we verified that $\ell=0$ directly by some simple skills in calculus; in another case, we got an inequality

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where $\lambda\in\left(0,1\right)$ is a constant. Then we can easily obtain $\ell=0$ through iteration.

In this paper, we omit many computational details because we believe that such processing can enable readers to grasp the framework and ideas of the proof faster. In our proof, Lemma 2 is based on N. Levinson [4, part 3]. The whole method, of course, still belongs to Selberg in essence.

2 Preliminaries

We list some well-known or trivial results without proof.

Proposition 1 (Tatuzawa, Iseki). Let F be a real- or complex-valued function defined on $(1,\infty)$, $\Lambda$ be Mangoldt function, *and let *

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Then

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**Proposition 2 **(Selberg). Let $\Lambda_{2}:=\mu\ast(\mathrm{log})^{2}=\Lambda\ast\Lambda+\Lambda\mathrm{log}$, thus

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Defining that $\Theta=(\Lambda\ast\Lambda)/\mathrm{log}$, then (3) can be written as

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by partial summation.

Proposition 3. We have

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Then

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since

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It gives

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where $M$ is a constant, $x_{1}$, $x_{2}$ are arbitrary non-negative numbers.

Proposition 4. Assume $f(x)\geq 0,\ f(a)=0,\ |f^{\prime}(x)|\leq m$. Then

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3 Lemmas

Lemma 1. For $x>0$ we have

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Proof.

We start with Proposition 1. Let $F=\mathcal{F}$, (5) gives

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by noticing that

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Replacing $x$ with $x/n$, then (8) becomes

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Noticing that

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since

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So we combine (8), (10)

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It follows that

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or

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where

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From Proposition 2, we consider

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It follows that

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or

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Similarly, we have

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Then (5), (12), (14) yield

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Using (11), (13), (15) we obtain

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Similarly, we have

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just observe

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which is a corollary of (4).

Thus, by combining (16), (17) and using (4) again, we derive

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which follows by partial summation. Similar to the operation in Proposition 3, (18) can be written as follows

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Finally we have, on writing $e^{\sqrt{x}}$ for $x$

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This completes the proof. ∎

Lemma 2. Let $a,b$ be successive zeros of $\mathcal{H}(x)$, then there are positive constants $\kappa$, $\varepsilon$ such that

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where $a<\xi<b$, $\alpha$ is defined in Introduction.

Proof.

We take a point $a+\alpha/h$, where h is an arbitrary positive number such that

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Noticing that $b-a\geq 1$, it can always have achieved because of the arbitrariness of $h$. May as well let $h=h_{1}$ satisfy (20). Clearly, $\mathcal{H}^{\prime}(x)$ is bounded, so we get from (1) and Proposition 4

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We choose $h=h_{2}$ such that $m-2h<0$. By (6), existing $\xi$, $a<\xi<b$, such that

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Hence

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just assume $h=max\left\{h_{1},h_{2}\right\}$.

Marking $\kappa=(2h-m)\alpha/\left(2Mh^{2}\right),\varepsilon=\alpha/h$ then we complete the proof. ∎

4 Proof of (2)

Theorem 1. We have $\ell=0$.

Proof.

Now we discussing the zeros of $\mathcal{H}(x)$. If $\mathcal{H}(x)$ has finite zeros, it implies $\ell=0$ directly since (6) and Lemma 1. So we assume that $\mathcal{H}(x)$ has infinite zeros. Let $\delta(x)$ denote the number of zeros not exceeding $x$, $x_{i}\ (i=1,2,\dots,\delta(x),\dots)$ denote the ith zero, then

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and existing $\xi_{i}\ (i=1,2,\dots,\delta(x)-1)$ such that

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Thus, we divide it into two cases.

Case 1. $\displaystyle\lim_{n\to\infty}|\mathcal{H}(\xi_{n})|=0$.

Then when $x\to\infty$, we have

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This gives

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where

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Hence

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It can be written as

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By noticing that $x\geq x_{\delta(x)}$, we get from Lemma 1 and (22)

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which implies $\ell=0$.

Case 2. $\displaystyle\lim_{n\to\infty}|\mathcal{H}(\xi_{n})|>0$ or the limit does not exist.

Then the minimum number exists in $\left\{\left|\mathcal{H}(\xi_{i})\right|\mid i=1,2,\dots,x_{\delta(x)},\dots\right\}$. Let $\iota:=\underset{i}{min}\left|\mathcal{H}(\xi_{i})\right|$ , we have

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since Lemma 2. It follows that

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where $\lambda=\kappa\iota$, $0<\lambda<1$. By $0\leq\ell\leq\lambda$, we get

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Repeating the above operation n times, then (23) becomes

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where $\lambda_{n}=1+\lambda\lambda_{n-1}$, $\lambda_{0}=1$. It implies $\ell=0$ when we let $n\to\infty$.

To sum up, we complete the proof. ∎

Corollary 1. Let $H(x):=e^{-\sqrt{x}}M\left(e^{\sqrt{x}}\right)$, then

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Proof.

We have

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The proof is completely contained in the proof of Lemma 1. Finally, we get from Theorem 1 and (24)

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It implies the corollary and (2) at once. ∎

Final remark.

1. As one sees Lemma 1 is similar to [1, (2.14)], but if we use the method in [1], we cannot get the required accuracy (but the method can be used for the proof of (24)). In reality, similar to the method in [2], if we only use (15), we can also derive (2) by dealing with a double integral.
2. (24) is necessary. We cannot get (2) from Theorem 1 directly since $M(x)$ is not monotonic.