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 identity

✓

If \(\psi : \mathbb {R}\to \mathbb {C}\) is continuous and 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 identity

✓

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}

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 Decay bounds

✓

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 4 Limiting Fourier identity

✓

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 the preceding two lemmas, 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 3 to obtain the result.

Corollary 1 Corollary of limiting identity

✓

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 5 Smooth Urysohn lemma

✓

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 6 Limiting identity for Schwartz functions

✓

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 5 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 3 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 7 Bijectivity of Fourier transform

✓

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

✓

If \(\Psi : (0,\infty ) \to \mathbb {C}\) is smooth and compactly supported away from the origin, then, 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 7, 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 6.

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

Proposition 1 Wiener-Ikehara in an interval

✓

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 5 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

✓

We have

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

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 Weak PNT

✓

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 8 Variant of limiting Fourier identity

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-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

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 Automatic Chebyshev bound

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 Wiener-Ikehara theorem, II

We have

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

Proof ▶

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

2.4 The prime number theorem in arithmetic progressions

Theorem 2 Weak PNT in AP

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 ▶

Applying Theorem 3 (or Theorem 6), one needs to show that 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 \(\mathrm{Re}(s)=1\). (The Chebyshev bound follows from the corresponding bound for \(\Lambda \).) By a Fourier expansion of Dirichlet characters, one can write

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

where the sum is over Dirichlet characters of modulus \(q\). 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.