diff --git a/blueprint/Wiener.tex b/blueprint/Wiener.tex deleted file mode 100644 index 39873f63..00000000 --- a/blueprint/Wiener.tex +++ /dev/null @@ -1,141 +0,0 @@ - -The Fourier transform of an absolutely integrable function $\psi: \R \to \C$ is defined by the formula -$$ \hat \psi(u) := \int_\R e(-tu) \psi(t)\ dt$$ -where $e(\theta) := e^{2\pi i \theta}$. - -Let $f: \N \to \C$ be an arithmetic function such that $\sum_{n=1}^\infty \frac{|f(n)|}{n^\sigma} < \infty$ for all $\sigma>1$. Then the Dirichlet series -$$ F(s) := \sum_{n=1}^\infty \frac{f(n)}{n^s}$$ -is absolutely convergent for $\sigma>1$. - -\begin{lemma}[First Fourier identity]\label{first-fourier} If $\psi: \R \to \C$ is continuous and compactly supported and $x > 0$, then for any $\sigma>1$ - $$ \sum_{n=1}^\infty \frac{f(n)}{n^\sigma} \hat \psi( \frac{1}{2\pi} \log \frac{n}{x} ) = \int_\R F(\sigma + it) \psi(t) x^{it}\ dt.$$ -\end{lemma} - -\begin{proof} By the definition of the Fourier transform, the left-hand side expands as -$$ \sum_{n=1}^\infty \int_\R \frac{f(n)}{n^\sigma} \psi(t) e( - \frac{1}{2\pi} t \log \frac{n}{x})\ dt$$ -while the right-hand side expands as -$$ \int_\R \sum_{n=1}^\infty \frac{f(n)}{n^{\sigma+it}} \psi(t) x^{it}\ dt.$$ -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. -\end{proof} - -\begin{lemma}[Second Fourier identity]\label{second-fourier} If $\psi: \R \to \C$ is continuous and compactly supported and $x > 0$, then for any $\sigma>1$ -$$ \int_{-\log x}^\infty e^{-u(\sigma-1)} \hat \psi(\frac{u}{2\pi})\ du = x^{\sigma - 1} \int_\R \frac{1}{\sigma+it-1} \psi(t) x^{it}\ dt.$$ -\end{lemma} - -\begin{proof} -\uses{first-fourier} - The left-hand side expands as - $$ \int_{-\log x}^\infty \int_\R e^{-u(\sigma-1)} \psi(t) e(-\frac{tu}{2\pi})\ dt du = x^{\sigma - 1} \int_\R \frac{1}{\sigma+it-1} \psi(t) x^{it}\ dt$$ - so by Fubini's theorem it suffices to verify the identity -$$ \int_{-\log x}^\infty \int_\R e^{-u(\sigma-1)} e(-\frac{tu}{2\pi})\ du = x^{\sigma - 1} \frac{1}{\sigma+it-1} x^{it}$$ -which is a routine calculation. -\end{proof} - -Now let $A \in \C$, 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 $\mathrm{Re} s > 1$. 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). - -\begin{lemma}[Decay bounds]\label{decay} If $\psi:\R \to \C$ is $C^2$ and obeys the bounds - $$ |\psi(t)|, |\psi''(t)| \leq A / (1 + |t|^2)$$ - for all $t \in \R$, then -$$ |\hat \psi(u)| \leq C A / (1+|u|^2)$$ -for all $u \in \R$, where $C$ is an absolute constant. -\end{lemma} - -\begin{proof} This follows from a standard integration by parts argument. -\end{proof} - -\begin{lemma}[Limiting Fourier identity]\label{limiting} If $\psi: \R \to \C$ is $C^2$ and compactly supported and $x \geq 1$, then -$$ \sum_{n=1}^\infty \frac{f(n)}{n} \hat \psi( \frac{1}{2\pi} \log \frac{n}{x} ) - A \int_{-\log x}^\infty \hat \psi(\frac{u}{2\pi})\ du = \int_\R G(1+it) \psi(t) x^{it}\ dt.$$ -\end{lemma} - -\begin{proof} -\uses{first-fourier,second-fourier,decay} - By the preceding two lemmas, we know that for any $\sigma>1$, we have - $$ \sum_{n=1}^\infty \frac{f(n)}{n^\sigma} \hat \psi( \frac{1}{2\pi} \log \frac{n}{x} ) - A x^{1-\sigma} \int_{-\log x}^\infty e^{-u(\sigma-1)} \hat \psi(\frac{u}{2\pi})\ du = \int_\R G(\sigma+it) \psi(t) x^{it}\ dt.$$ - Now take limits as $\sigma \to 1$ using dominated convergence together with \eqref{cheby} and Lemma \ref{decay} to obtain the result. -\end{proof} - -\begin{corollary}\label{limiting-cor} 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 $x \to \infty$. -\end{corollary} - -\begin{proof} -\uses{limiting} - Immediate from the Riemann-Lebesgue lemma, and also noting that $\int_{-\infty}^{-\log x} \hat \psi(\frac{u}{2\pi})\ du = o(1)$. -\end{proof} - -\begin{lemma}\label{schwarz-id} 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. -\end{lemma} - -\begin{proof} -\uses{limiting-cor} -For any $R>1$, one can use a smooth cutoff function to write $\psi = \psi_{\leq R} + \psi_{>R}$, where $\psi_{\leq R}$ is $C^2$ (in fact smooth) and compactly supported (on $[-R,R]$), and $\psi_{>R}$ obeys bounds of the form -$$ |\psi_{>R}(t)|, |\psi''_{>R}(t)| \ll R^{-1} / (1 + |t|^2) $$ -where the implied constants depend on $\psi$. By Lemma \ref{decay} we then have -$$ \hat \psi_{>R}(u) \ll R^{-1} / (1+|u|^2).$$ -Using this and \eqref{cheby} one can show that -$$ \sum_{n=1}^\infty \frac{f(n)}{n} \hat \psi_{>R}( \frac{1}{2\pi} \log \frac{n}{x} ), A \int_{-\infty}^\infty \hat \psi_{>R} (\frac{u}{2\pi})\ du \ll R^{-1} $$ -(with implied constants also depending on $A$), while from Lemma \ref{limiting-cor} 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. -\end{proof} - -\begin{lemma}\label{bij} The Fourier transform is a bijection on the Schwartz class. -\end{lemma} - -\begin{proof} This is a standard result in Fourier analysis. -\end{proof} - -\begin{corollary}\label{WienerIkeharaSmooth} - If $\Psi: (0,\infty) \to \C$ 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 $u \to \infty$. -\end{corollary} - -\begin{proof} -\uses{bij,schwarz-id} - By Lemma \ref{bij}, we can write -$$ y \Psi(y) = \hat \psi( \frac{1}{2\pi} \log y )$$ -for all $y>0$ 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 \ref{schwarz-id}. -\end{proof} - -\begin{lemma}[Smooth Urysohn lemma]\label{smooth-ury} If $I$ is a closed interval contained in an open interval $J$, then there exists a smooth function $\Psi: \R \to \R$ with $1_I \leq \Psi \leq 1_J$. -\end{lemma} - -\begin{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 $J$. -\end{proof} - -Now we add the hypothesis that $f(n) \geq 0$ for all $n$. - -\begin{proposition} -\label{WienerIkeharaInterval} - For any closed interval $I \subset (0,+\infty)$, we have - $$ \sum_{n=1}^\infty f(n) 1_I( \frac{n}{x} ) = A x |I| + o(x).$$ -\end{proposition} - -\begin{proof} -\uses{smooth-ury} - Use Lemma \ref{smooth-ury} 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 $f$. -\end{proof} - -\begin{corollary}\label{WienerIkehara} - We have -$$ \sum_{n\leq x} f(n) = A x |I| + o(x).$$ -\end{corollary} - - - -\begin{proof} -\uses{WienerIkeharaInterval, cheby} - Apply the preceding proposition with $I = [\varepsilon,1]$ and then send $\varepsilon$ to zero (using \eqref{cheby} to control the error). -\end{proof} - - diff --git a/blueprint/wiener.tex b/blueprint/wiener.tex deleted file mode 100644 index 42a05aa3..00000000 --- a/blueprint/wiener.tex +++ /dev/null @@ -1,200 +0,0 @@ - -The Fourier transform of an absolutely integrable function $\psi: \R \to \C$ is defined by the formula -$$ \hat \psi(u) := \int_\R e(-tu) \psi(t)\ dt$$ -where $e(\theta) := e^{2\pi i \theta}$. - -Let $f: \N \to \C$ be an arithmetic function such that $\sum_{n=1}^\infty \frac{|f(n)|}{n^\sigma} < \infty$ for all $\sigma>1$. Then the Dirichlet series -$$ F(s) := \sum_{n=1}^\infty \frac{f(n)}{n^s}$$ -is absolutely convergent for $\sigma>1$. - - - -\begin{lemma}[First Fourier identity]\label{first-fourier} If $\psi: \R \to \C$ is continuous and compactly supported and $x > 0$, then for any $\sigma>1$ - $$ \sum_{n=1}^\infty \frac{f(n)}{n^\sigma} \hat \psi( \frac{1}{2\pi} \log \frac{n}{x} ) = \int_\R F(\sigma + it) \psi(t) x^{it}\ dt.$$ -\end{lemma} - - - -\begin{proof} By the definition of the Fourier transform, the left-hand side expands as -$$ \sum_{n=1}^\infty \int_\R \frac{f(n)}{n^\sigma} \psi(t) e( - \frac{1}{2\pi} t \log \frac{n}{x})\ dt$$ -while the right-hand side expands as -$$ \int_\R \sum_{n=1}^\infty \frac{f(n)}{n^{\sigma+it}} \psi(t) x^{it}\ dt.$$ -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. -\end{proof} - - - -\begin{lemma}[Second Fourier identity]\label{second-fourier} If $\psi: \R \to \C$ is continuous and compactly supported and $x > 0$, then for any $\sigma>1$ -$$ \int_{-\log x}^\infty e^{-u(\sigma-1)} \hat \psi(\frac{u}{2\pi})\ du = x^{\sigma - 1} \int_\R \frac{1}{\sigma+it-1} \psi(t) x^{it}\ dt.$$ -\end{lemma} - - - -\begin{proof} -\uses{first-fourier} - The left-hand side expands as - $$ \int_{-\log x}^\infty \int_\R e^{-u(\sigma-1)} \psi(t) e(-\frac{tu}{2\pi})\ dt du = x^{\sigma - 1} \int_\R \frac{1}{\sigma+it-1} \psi(t) x^{it}\ dt$$ - so by Fubini's theorem it suffices to verify the identity -$$ \int_{-\log x}^\infty \int_\R e^{-u(\sigma-1)} e(-\frac{tu}{2\pi})\ du = x^{\sigma - 1} \frac{1}{\sigma+it-1} x^{it}$$ -which is a routine calculation. -\end{proof} - - - -Now let $A \in \C$, 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 $\mathrm{Re} s > 1$. 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). - - - -\begin{lemma}[Decay bounds]\label{decay} If $\psi:\R \to \C$ is $C^2$ and obeys the bounds - $$ |\psi(t)|, |\psi''(t)| \leq A / (1 + |t|^2)$$ - for all $t \in \R$, then -$$ |\hat \psi(u)| \leq C A / (1+|u|^2)$$ -for all $u \in \R$, where $C$ is an absolute constant. -\end{lemma} - - - -\begin{proof} This follows from a standard integration by parts argument. -\end{proof} - - - -\begin{lemma}[Limiting Fourier identity]\label{limiting} If $\psi: \R \to \C$ is $C^2$ and compactly supported and $x \geq 1$, then -$$ \sum_{n=1}^\infty \frac{f(n)}{n} \hat \psi( \frac{1}{2\pi} \log \frac{n}{x} ) - A \int_{-\log x}^\infty \hat \psi(\frac{u}{2\pi})\ du = \int_\R G(1+it) \psi(t) x^{it}\ dt.$$ -\end{lemma} - - - -\begin{proof} -\uses{first-fourier,second-fourier,decay} - By the preceding two lemmas, we know that for any $\sigma>1$, we have - $$ \sum_{n=1}^\infty \frac{f(n)}{n^\sigma} \hat \psi( \frac{1}{2\pi} \log \frac{n}{x} ) - A x^{1-\sigma} \int_{-\log x}^\infty e^{-u(\sigma-1)} \hat \psi(\frac{u}{2\pi})\ du = \int_\R G(\sigma+it) \psi(t) x^{it}\ dt.$$ - Now take limits as $\sigma \to 1$ using dominated convergence together with \eqref{cheby} and Lemma \ref{decay} to obtain the result. -\end{proof} - - - -\begin{corollary}\label{limiting-cor} 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 $x \to \infty$. -\end{corollary} - - - -\begin{proof} -\uses{limiting} - Immediate from the Riemann-Lebesgue lemma, and also noting that $\int_{-\infty}^{-\log x} \hat \psi(\frac{u}{2\pi})\ du = o(1)$. -\end{proof} - - - -\begin{lemma}\label{schwarz-id} 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. -\end{lemma} - - - -\begin{proof} -\uses{limiting-cor} -For any $R>1$, one can use a smooth cutoff function to write $\psi = \psi_{\leq R} + \psi_{>R}$, where $\psi_{\leq R}$ is $C^2$ (in fact smooth) and compactly supported (on $[-R,R]$), and $\psi_{>R}$ obeys bounds of the form -$$ |\psi_{>R}(t)|, |\psi''_{>R}(t)| \ll R^{-1} / (1 + |t|^2) $$ -where the implied constants depend on $\psi$. By Lemma \ref{decay} we then have -$$ \hat \psi_{>R}(u) \ll R^{-1} / (1+|u|^2).$$ -Using this and \eqref{cheby} one can show that -$$ \sum_{n=1}^\infty \frac{f(n)}{n} \hat \psi_{>R}( \frac{1}{2\pi} \log \frac{n}{x} ), A \int_{-\infty}^\infty \hat \psi_{>R} (\frac{u}{2\pi})\ du \ll R^{-1} $$ -(with implied constants also depending on $A$), while from Lemma \ref{limiting-cor} 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. -\end{proof} - - - -\begin{lemma}\label{bij} The Fourier transform is a bijection on the Schwartz class. -\end{lemma} - - - -\begin{proof} This is a standard result in Fourier analysis. -\end{proof} - - - -\begin{corollary}\label{WienerIkeharaSmooth} - If $\Psi: (0,\infty) \to \C$ 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 $u \to \infty$. -\end{corollary} - - - -\begin{proof} -\uses{bij,schwarz-id} - By Lemma \ref{bij}, we can write -$$ y \Psi(y) = \hat \psi( \frac{1}{2\pi} \log y )$$ -for all $y>0$ 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 \ref{schwarz-id}. -\end{proof} - - - -\begin{lemma}[Smooth Urysohn lemma]\label{smooth-ury} If $I$ is a closed interval contained in an open interval $J$, then there exists a smooth function $\Psi: \R \to \R$ with $1_I \leq \Psi \leq 1_J$. -\end{lemma} - - - -\begin{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 $J$. -\end{proof} - - - -Now we add the hypothesis that $f(n) \geq 0$ for all $n$. - -\begin{proposition} -\label{WienerIkeharaInterval} - For any closed interval $I \subset (0,+\infty)$, we have - $$ \sum_{n=1}^\infty f(n) 1_I( \frac{n}{x} ) = A x |I| + o(x).$$ -\end{proposition} - - - -\begin{proof} -\uses{smooth-ury, WienerIkeharaSmooth} - Use Lemma \ref{smooth-ury} 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 $f$. -\end{proof} - - - -\begin{corollary}\label{WienerIkehara} - We have -$$ \sum_{n\leq x} f(n) = A x |I| + o(x).$$ -\end{corollary} - - - -\begin{proof} -\uses{WienerIkeharaInterval, cheby} - Apply the preceding proposition with $I = [\varepsilon,1]$ and then send $\varepsilon$ to zero (using \eqref{cheby} to control the error). -\end{proof} - - - -\section{Weak PNT} - -\begin{theorem}[Weak PNT]\label{WeakPNT} We have -$$ \sum_{n \leq x} \Lambda(n) = x + o(x).$$ -\end{theorem} - - - -\begin{proof} -\uses{WienerIkehara, ChebyshevPsi} - Already done by Stoll, assuming Wiener-Ikehara. -\end{proof} - -