Prime Number Theorem And ...

2 First approach: Wiener-Ikehara Tauberian theorem.

2.1 A Fourier-analytic proof of the Wiener-Ikehara theorem

The Fourier transform of an absolutely integrable function \(\psi : \mathbb {R}\to \mathbb {C}\) is defined by the formula

\[ \hat\psi (u) := \int _\mathbb {R}e(-tu) \psi (t)\ dt \]

where \(e(\theta ) := e^{2\pi i \theta }\).

Let \(f: \mathbb {N}\to \mathbb {C}\) be an arithmetic function such that \(\sum _{n=1}^\infty \frac{|f(n)|}{n^\sigma } {\lt} \infty \) for all \(\sigma {\gt}1\). Then the Dirichlet series

\[ F(s) := \sum _{n=1}^\infty \frac{f(n)}{n^s} \]

is absolutely convergent for \(\sigma {\gt}1\).

Lemma 1 first_fourier
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If \(\psi : \mathbb {R}\to \mathbb {C}\) is integrable and \(x {\gt} 0\), then for any \(\sigma {\gt}1\)

\[ \sum _{n=1}^\infty \frac{f(n)}{n^\sigma } \hat\psi ( \frac{1}{2\pi } \log \frac{n}{x} ) = \int _\mathbb {R}F(\sigma + it) \psi (t) x^{it}\ dt. \]
Proof

By the definition of the Fourier transform, the left-hand side expands as

\[ \sum _{n=1}^\infty \int _\mathbb {R}\frac{f(n)}{n^\sigma } \psi (t) e( - \frac{1}{2\pi } t \log \frac{n}{x})\ dt \]

while the right-hand side expands as

\[ \int _\mathbb {R}\sum _{n=1}^\infty \frac{f(n)}{n^{\sigma +it}} \psi (t) x^{it}\ dt. \]

Since

\[ \frac{f(n)}{n^\sigma } \psi (t) e( - \frac{1}{2\pi } t \log \frac{n}{x}) = \frac{f(n)}{n^{\sigma +it}} \psi (t) x^{it} \]

the claim then follows from Fubini’s theorem.

Lemma 2 second_fourier
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If \(\psi : \mathbb {R}\to \mathbb {C}\) is continuous and compactly supported and \(x {\gt} 0\), then for any \(\sigma {\gt}1\)

\[ \int _{-\log x}^\infty e^{-u(\sigma -1)} \hat\psi (\frac{u}{2\pi })\ du = x^{\sigma - 1} \int _\mathbb {R}\frac{1}{\sigma +it-1} \psi (t) x^{it}\ dt. \]
Proof

The left-hand side expands as

\[ \int _{-\log x}^\infty \int _\mathbb {R}e^{-u(\sigma -1)} \psi (t) e(-\frac{tu}{2\pi })\ dt\ du \atop {?}= x^{\sigma - 1} \int _\mathbb {R}\frac{1}{\sigma +it-1} \psi (t) x^{it}\ dt \]

so by Fubini’s theorem it suffices to verify the identity

\begin{align*} \int _{-\log x}^\infty e^{-u(\sigma -1)} e(-\frac{tu}{2\pi })\ du & = \int _{-\log x}^\infty e^{(it - \sigma + 1)u}\ du \\ & = \frac{1}{it - \sigma + 1} e^{(it - \sigma + 1)u}\ \Big|_{-\log x}^\infty \\ & = x^{\sigma - 1} \frac{1}{\sigma +it-1} x^{it} \end{align*}

Now let \(A \in \mathbb {C}\), and suppose that there is a continuous function \(G(s)\) defined on \(\mathrm{Re} s \geq 1\) such that \(G(s) = F(s) - \frac{A}{s-1}\) whenever \(\mathrm{Re} s {\gt} 1\). We also make the Chebyshev-type hypothesis

\begin{equation} \label{cheby} \sum _{n \leq x} |f(n)| \ll x \end{equation}
1

for all \(x \geq 1\) (this hypothesis is not strictly necessary, but simplifies the arguments and can be obtained fairly easily in applications).

Lemma 3 Preliminary decay bound I

If \(\psi :\mathbb {R}\to \mathbb {C}\) is absolutely integrable then

\[ |\hat\psi (u)| \leq \| \psi \| _1 \]

for all \(u \in \mathbb {R}\). where \(C\) is an absolute constant.

Proof

Immediate from the triangle inequality.

Lemma 4 Preliminary decay bound II

If \(\psi :\mathbb {R}\to \mathbb {C}\) is absolutely integrable and of bounded variation, and \(\psi '\) is bounded variation, then

\[ |\hat\psi (u)| \leq \| \psi \| _{TV} / 2\pi |u| \]

for all non-zero \(u \in \mathbb {R}\).

Proof

By integration by parts we will have

\[ 2\pi i u \hat\psi (u) = \int _\mathbb {R}e(-tu) \psi '(t)\ dt \]

and the claim then follows from the triangle inequality.

Lemma 5 Preliminary decay bound III

If \(\psi :\mathbb {R}\to \mathbb {C}\) is absolutely integrable, absolutely continuous, and \(\psi '\) is of bounded variation, then

\[ |\hat\psi (u)| \leq \| \psi ' \| _{TV} / (2\pi |u|)^2 \]

for all non-zero \(u \in \mathbb {R}\).

Proof

Should follow from previous lemma.

Lemma 6 Decay bound, alternate form

If \(\psi :\mathbb {R}\to \mathbb {C}\) is absolutely integrable, absolutely continuous, and \(\psi '\) is of bounded variation, then

\[ |\hat\psi (u)| \leq ( \| \psi \| _1 + \| \psi ' \| _{TV} / (2\pi )^2) / (1+|u|^2) \]

for all \(u \in \mathbb {R}\).

Proof

Should follow from previous lemmas.

It should be possible to refactor the lemma below to follow from Lemma 6 instead.

Lemma 7 Decay bounds
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If \(\psi :\mathbb {R}\to \mathbb {C}\) is \(C^2\) and obeys the bounds

\[ |\psi (t)|, |\psi ''(t)| \leq A / (1 + |t|^2) \]

for all \(t \in \mathbb {R}\), then

\[ |\hat\psi (u)| \leq C A / (1+|u|^2) \]

for all \(u \in \mathbb {R}\), where \(C\) is an absolute constant.

Proof

From two integration by parts we obtain the identity

\[ (1+u^2) \hat\psi (u) = \int _{\bf R} (\psi (t) - \frac{u}{4\pi ^2} \psi ''(t)) e(-tu)\ dt. \]

Now apply the triangle inequality and the identity \(\int _{\bf R} \frac{dt}{1+t^2}\ dt = \pi \) to obtain the claim with \(C = \pi + 1 / 4 \pi \).

Lemma 8 Limiting Fourier identity
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If \(\psi : \mathbb {R}\to \mathbb {C}\) is \(C^2\) and compactly supported and \(x \geq 1\), then

\[ \sum _{n=1}^\infty \frac{f(n)}{n} \hat\psi ( \frac{1}{2\pi } \log \frac{n}{x} ) - A \int _{-\log x}^\infty \hat\psi (\frac{u}{2\pi })\ du = \int _\mathbb {R}G(1+it) \psi (t) x^{it}\ dt. \]
Proof

By Lemma 1 and Lemma 2, we know that for any \(\sigma {\gt}1\), we have

\[ \sum _{n=1}^\infty \frac{f(n)}{n^\sigma } \hat\psi ( \frac{1}{2\pi } \log \frac{n}{x} ) - A x^{1-\sigma } \int _{-\log x}^\infty e^{-u(\sigma -1)} \hat\psi (\frac{u}{2\pi })\ du = \int _\mathbb {R}G(\sigma +it) \psi (t) x^{it}\ dt. \]

Now take limits as \(\sigma \to 1\) using dominated convergence together with 1 and Lemma 7 to obtain the result.

Corollary 1 Corollary of limiting identity
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With the hypotheses as above, we have

\[ \sum _{n=1}^\infty \frac{f(n)}{n} \hat\psi ( \frac{1}{2\pi } \log \frac{n}{x} ) = A \int _{-\infty }^\infty \hat\psi (\frac{u}{2\pi })\ du + o(1) \]

as \(x \to \infty \).

Proof

Immediate from the Riemann-Lebesgue lemma, and also noting that \(\int _{-\infty }^{-\log x} \hat\psi (\frac{u}{2\pi })\ du = o(1)\).

Lemma 9 Smooth Urysohn lemma
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If \(I\) is a closed interval contained in an open interval \(J\), then there exists a smooth function \(\Psi : \mathbb {R}\to \mathbb {R}\) with \(1_I \leq \Psi \leq 1_J\).

Proof

A standard analysis lemma, which can be proven by convolving \(1_K\) with a smooth approximation to the identity for some interval \(K\) between \(I\) and \(J\). Note that we have “SmoothBumpFunction”s on smooth manifolds in Mathlib, so this shouldn’t be too hard...

Lemma 10 Limiting identity for Schwartz functions
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The previous corollary also holds for functions \(\psi \) that are assumed to be in the Schwartz class, as opposed to being \(C^2\) and compactly supported.

Proof

For any \(R{\gt}1\), one can use a smooth cutoff function (provided by Lemma 9 to write \(\psi = \psi _{\leq R} + \psi _{{\gt}R}\), where \(\psi _{\leq R}\) is \(C^2\) (in fact smooth) and compactly supported (on \([-R,R]\)), and \(\psi _{{\gt}R}\) obeys bounds of the form

\[ |\psi _{{\gt}R}(t)|, |\psi ''_{{\gt}R}(t)| \ll R^{-1} / (1 + |t|^2) \]

where the implied constants depend on \(\psi \). By Lemma 7 we then have

\[ \hat\psi _{{\gt}R}(u) \ll R^{-1} / (1+|u|^2). \]

Using this and 1 one can show that

\[ \sum _{n=1}^\infty \frac{f(n)}{n} \hat\psi _{{\gt}R}( \frac{1}{2\pi } \log \frac{n}{x} ), A \int _{-\infty }^\infty \hat\psi _{{\gt}R} (\frac{u}{2\pi })\ du \ll R^{-1} \]

(with implied constants also depending on \(A\)), while from Lemma 1 one has

\[ \sum _{n=1}^\infty \frac{f(n)}{n} \hat\psi _{\leq R}( \frac{1}{2\pi } \log \frac{n}{x} ) = A \int _{-\infty }^\infty \hat\psi _{\leq R} (\frac{u}{2\pi })\ du + o(1). \]

Combining the two estimates and letting \(R\) be large, we obtain the claim.

Lemma 11 Bijectivity of Fourier transform
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The Fourier transform is a bijection on the Schwartz class. [Note: only surjectivity is actually used.]

Proof

This is a standard result in Fourier analysis. It can be proved here by appealing to Mellin inversion, Theorem 5. In particular, given \(f\) in the Schwartz class, let \(F : \mathbb {R}_+ \to \mathbb {C}: x \mapsto f(\log x)\) be a function in the “Mellin space”; then the Mellin transform of \(F\) on the imaginary axis \(s=it\) is the Fourier transform of \(f\). The Mellin inversion theorem gives Fourier inversion.

Corollary 2 Smoothed Wiener-Ikehara
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If \(\Psi : (0,\infty ) \to \mathbb {C}\) is smooth and compactly supported away from the origin, then,

\[ \sum _{n=1}^\infty f(n) \Psi ( \frac{n}{x} ) = A x \int _0^\infty \Psi (y)\ dy + o(x) \]

as \(x \to \infty \).

Proof

By Lemma 11, we can write

\[ y \Psi (y) = \hat\psi ( \frac{1}{2\pi } \log y ) \]

for all \(y{\gt}0\) and some Schwartz function \(\psi \). Making this substitution, the claim is then equivalent after standard manipulations to

\[ \sum _{n=1}^\infty \frac{f(n)}{n} \hat\psi ( \frac{1}{2\pi } \log \frac{n}{x} ) = A \int _{-\infty }^\infty \hat\psi (\frac{u}{2\pi })\ du + o(1) \]

and the claim follows from Lemma 10.

Now we add the hypothesis that \(f(n) \geq 0\) for all \(n\).

Proposition 1 Wiener-Ikehara in an interval
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For any closed interval \(I \subset (0,+\infty )\), we have

\[ \sum _{n=1}^\infty f(n) 1_I( \frac{n}{x} ) = A x |I| + o(x). \]
Proof

Use Lemma 9 to bound \(1_I\) above and below by smooth compactly supported functions whose integral is close to the measure of \(|I|\), and use the non-negativity of \(f\).

Corollary 3 Wiener-Ikehara theorem
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We have

\[ \sum _{n\leq x} f(n) = A x + o(x). \]
Proof

Apply the preceding proposition with \(I = [\varepsilon ,1]\) and then send \(\varepsilon \) to zero (using 1 to control the error).

2.2 Weak PNT

Theorem 1 WeakPNT
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We have

\[ \sum _{n \leq x} \Lambda (n) = x + o(x). \]
Proof

Already done by Stoll, assuming Wiener-Ikehara.

2.3 Removing the Chebyshev hypothesis

In this section we do *not* assume bound 1, but instead derive it from the other hypotheses.

Lemma 12 limiting_fourier_variant
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If \(\psi : \mathbb {R}\to \mathbb {C}\) is \(C^2\) and compactly supported with \(f\) and \(\hat\psi \) non-negative, and \(x \geq 1\), then

\[ \sum _{n=1}^\infty \frac{f(n)}{n} \hat\psi ( \frac{1}{2\pi } \log \frac{n}{x} ) - A \int _{-\log x}^\infty \hat\psi (\frac{u}{2\pi })\ du = \int _\mathbb {R}G(1+it) \psi (t) x^{it}\ dt. \]
Proof

Repeat the proof of Lemma reflimiting_fourier_variant, but use monotone convergence instead of dominated convergence. (The proof should be simpler, as one no longer needs to establish domination for the sum.)

Corollary 4 crude_upper_bound
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If \(\psi : \mathbb {R}\to \mathbb {C}\) is \(C^2\) and compactly supported with \(f\) and \(\hat\psi \) non-negative, then there exists a constant \(B\) such that

\[ |\sum _{n=1}^\infty \frac{f(n)}{n} \hat\psi ( \frac{1}{2\pi } \log \frac{n}{x} )| \leq B \]

for all \(x \geq 1\).

Proof

For \(x \geq 1\), this readily follows from the previous lemma and the triangle inequality. For \(x {\lt} 1\), only a bounded number of summands can contribute and the claim is trivial.

Corollary 5 auto_cheby
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One has

\[ \sum _{n \leq x} f(n) = O(x) \]

for all \(x \geq 1\).

Proof

For \(x \geq 1\) apply the previous corollary for all \(y {\lt} C x\) and \(\psi \) chosen be both nonnegative and have nonnegative Fourier transform, while being not identically zero, and \(C\) a large constant. This gives

\[ |\sum _{n=1}^\infty \frac{f(n)}{n} \int _0^{Cx} \hat\psi ( \frac{1}{2\pi } \log \frac{n}{y} )\ dy| \leq CB x. \]

But observe that the quantity \(\int _0^{Cx} \hat\psi ( \frac{1}{2\pi }\) is non-negative and equal to the positive constant \(\int _{{\bf R}} \hat\psi ( \frac{1}{2\pi } u ) e^u\ du\) if \(n \leq x\) and \(C\) is large enough. The claim follows.

Corollary 6 WienerIkeharaTheorem”
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We have

\[ \sum _{n\leq x} f(n) = A x + o(x). \]
Proof

Use Corollary 5 to remove the Chebyshev hypothesis in Theorem 3.

2.4 The prime number theorem in arithmetic progressions

Lemma 13 WeakPNT_character
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If \(q ≥ 1\) and \(a\) is coprime to \(q\), and \(\mathrm{Re} s {\gt} 1\), we have

\[ \sum _{n: n = a\ (q)} \frac{\Lambda (n)}{n^s} = - \frac{1}{\varphi (q)} \sum _{\chi \ (q)} \overline{\chi (a)} \frac{L'(s,\chi )}{L(s,\chi )}. \]
Proof

From the Fourier inversion formula on the multiplicative group \((\mathbb {Z}/q\mathbb {Z})^\times \), we have

\[ 1_{n=a\ (q)} = \frac{\varphi (q)}{q} \sum _{\chi \ (q)} \overline{\chi (a)} \chi (n). \]

On the other hand, from standard facts about L-series we have for each character \(\chi \) that

\[ \sum _{n} \frac{\Lambda (n) \chi (n)}{n^s} = - \frac{L'(s,\chi )}{L(s,\chi )}. \]

Combining these two facts, we obtain the claim.

Proposition 2 WeakPNT_AP_prelim
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If \(q ≥ 1\) and \(a\) is coprime to \(q\), the Dirichlet series \(\sum _{n \leq x: n = a\ (q)} {\Lambda (n)}{n^s}\) converges for \(\mathrm{Re}(s) {\gt} 1\) to \(\frac{1}{\varphi (q)} \frac{1}{s-1} + G(s)\) where \(G\) has a continuous extension to \(\mathrm{Re}(s)=1\).

Proof

We expand out the left-hand side using Lemma 13. The contribution of the non-principal characters \(\chi \) extend continuously to \(\mathrm{Re}(s) = 1\) thanks to the non-vanishing of \(L(s,\chi )\) on this line (which should follow from another component of this project), so it suffices to show that for the principal character \(\chi _0\), that

\[ -\frac{L'(s,\chi _0)}{L(s,\chi _0)} - \frac{1}{s-1} \]

also extends continuously here. But we already know that

\[ -\frac{\zeta '(s)}{\zeta (s)} - \frac{1}{s-1} \]

extends, and from Euler product machinery one has the identity

\[ \frac{L'(s,\chi _0)}{L(s,\chi _0)} = \frac{\zeta '(s)}{\zeta (s)} + \sum _{p|q} \frac{\log p}{p^s-1}. \]

Since there are only finitely many primes dividing \(q\), and each summand \(\frac{\log p}{p^s-1}\) extends continuously, the claim follows.

Theorem 2 WeakPNT_AP
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If \(q ≥ 1\) and \(a\) is coprime to \(q\), we have

\[ \sum _{n \leq x: n = a\ (q)} \Lambda (n) = \frac{x}{\varphi (q)} + o(x). \]
Proof

Apply Theorem 3 (or Theorem 6) to Proposition 2. (The Chebyshev bound follows from the corresponding bound for \(\Lambda \).)

2.5 The Chebotarev density theorem: the case of cyclotomic extensions

In this section, \(K\) is a number field, \(L = K(\mu _m)\) for some natural number \(m\), and \(G = Gal(K/L)\).

The goal here is to prove the Chebotarev density theorem for the case of cyclotomic extensions.

Lemma 14 Dedekind_factor

We have

\[ \zeta _L(s) = \prod _{\chi } L(\chi ,s) \]

for \(\Re (s) {\gt} 1\), where \(\chi \) runs over homomorphisms from \(G\) to \(\mathbb {C}^\times \) and \(L\) is the Artin \(L\)-function.

Proof

See Propositions 7.1.16, 7.1.19 of https://www.math.ucla.edu/ sharifi/algnum.pdf .

Lemma 15 Simple pole

\(\zeta _L\) has a simple pole at \(s=1\).

Proof

See Theorem 7.1.12 of https://www.math.ucla.edu/ sharifi/algnum.pdf .

Lemma 16 Dedekind_nonvanishing

For any non-principal character \(\chi \) of \(Gal(K/L)\), \(L(\chi ,s)\) does not vanish for \(\Re (s)=1\).

Proof

For \(s=1\), this will follow from Lemmas 14, 15. For the rest of the line, one should be able to adapt the arguments for the Dirichet L-function.

2.6 The Chebotarev density theorem: the case of abelian extensions

(Use the arguments in Theorem 7.2.2 of https://www.math.ucla.edu/ sharifi/algnum.pdf to extend the previous results to abelian extensions (actually just cyclic extensions would suffice))

2.7 The Chebotarev density theorem: the general case

(Use the arguments in Theorem 7.2.2 of https://www.math.ucla.edu/ sharifi/algnum.pdf to extend the previous results to arbitrary extensions

Lemma 17 PNT for one character

For any non-principal character \(\chi \) of \(Gal(K/L)\),

\[ \sum _{N \mathfrak {p} \leq x} \chi (\mathfrak {p}) \log N \mathfrak {p} = o(x). \]
Proof

This should follow from Lemma ?? and the arguments for the Dirichlet L-function. (It may be more convenient to work with a von Mangoldt type function instead of \(\log N\mathfrak {p}\)).