text:
\begin{document}
\author{Alia Hamieh}

\address[Alia Hamieh]{Department of Mathematics and Statistics \\
        University of Northern British Columbia \\
        Prince George, BC V2N4Z9 \\
        Canada}
\email{alia.hamieh@unbc.ca}

\author{Wissam Raji}
\address[Wissam Raji]{
Department of Mathematics\\
American University of Beirut\\
P.O. Box 11-0236\\
Riad El Solh\\
Beirut 1107 2020\\
Lebanon}

\email{wr07@aub.edu.lb}

\title[Non-vanishing of Derivatives of $L$-functions of Hilbert Modular Forms]{Non-vanishing of Derivatives of $L$-functions of Hilbert Modular Forms in the Critical Strip}

\begin{abstract}
In this paper, we show that, on average, the derivatives of $L$-functions of cuspidal Hilbert modular forms with sufficiently large weight $k$ do not vanish on the line segments $\Im(s)=t_{0}$, $\Re(s)\in(\frac{k-1}{2},\frac{k}{2}-\epsilon)\cup(\frac{k}{2}+\epsilon,\frac{k+1}{2})$. This is analogous to the case of classical modular forms.
 \end{abstract}

\maketitle

\section{Introduction}

In \cite{kohnen}, Kohnen proved that given any real number $t_0$ and $\epsilon>0$, then for $k$ large enough, the average of the normalized L-functions $L^*(f,s)$ with $f$ varying over a basis of Hecke eigenforms of weight $k$ on $SL_2(\mathbb{Z})$ does not vanish on the line segment $$ \Im(s)=t_0, \ \ (k-1)/2< \Re(s)<k/2-\epsilon, k/2+\epsilon<\Re(s)<(k+1)/2.$$  Recently in \cite{KSW}, Kohnen, Sengupta and Weigel extended their method and showed a non-vanishing result for the derivatives of $L$-functions associated to modular forms of integer weight on the full group.  In particular, they show that for $k$ large enough, \begin{equation}
\sum_{j=1}^{d}\frac{1}{\left<f_{k,j}, f_{k,j}\right>}\frac{d^n}{ds^n}[L^*(f_{k,j},s)]
\end{equation}
does not vanish on the line segment $\Im(s)=t_0$, $(k-1)/2< \Re(s)<k/2-\epsilon, k/2+\epsilon<\Re(s)<(k+1)/2$.

In \cite{HR}, we generalized the result of \cite{kohnen} to the context of cuspidal Hilbert modular forms. In order to describe our work, we introduce the following notation. Let $F$ be a totally real number field of  degree $n$ over $\mathbb{Q}$. Let $\ringO_F$ be the ring of integers of $F$, and assume its narrow class number is equal to 1.  For $\bk\in2\mathbb{N}^n$, we denote by $\mathcal{S}_{\bk}(\SL(\ringO_{F}))$ the space of Hilbert cusp forms of weight $\bk$ for $\SL(\ringO_{F})$.  Our main result in \cite{HR} is the following theorem.
\begin{theorem}\label{thm:old}
Let $\bk=(k,k,\dots,k)\in2\mathbb{N}^n$, and  let $\mathcal{B}_{\bk}(\ringO_{F})$ be a basis of normalized Hecke eigenforms of $\mathcal{S}_{\bk}(\SL(\ringO_{F}))$. Let $t_{0}\in\mathbb{R}$ and $\epsilon>0$. Then there exists a constant $C$ depending only on $t_{0}$, $\epsilon$ and $F$ such that for $k>C$ the average \[\sum_{f\in\mathcal{B}_{\bk}(\ringO_{F})}\frac{\Lambda(f,s)}{\left<f,f\right>}\] is non-vanishing for any $s=\sigma+it_{0}$ with $\sigma\in(\frac{k-1}{2},\frac{k}{2}-\epsilon)\cup(\frac{k}{2}+\epsilon,\frac{k+1}{2})$.
\end{theorem}

Here,  we extend the result in \cite{HR} by showing that the derivatives of $L$-functions of cuspidal Hilbert modular forms with sufficiently large weight $k$ do not vanish on the line segments
\begin{equation*}
\Im(s)=t_{0}, \ \ \ \  \Re(s)\in(\frac{k-1}{2},\frac{k}{2}-\epsilon)\cup(\frac{k}{2}+\epsilon,\frac{k+1}{2}).
\end{equation*}
More precisely, we prove the following theorem.

\begin{theorem}\label{thm:main}
 Let $\mathcal{B}_{\bk}(\ringO_{F})$ be a basis of normalized Hecke eigenforms of $\mathcal{S}_{\bk}(\SL(\ringO_{F}))$. Let $t_{0}\in\mathbb{R}$, $\epsilon>0$ and $\ell\in\mathbb{N}$. Then there exists a constant $C$ depending only on $t_{0}$, $\epsilon$ and $F$ such that for $k>C$ the average \[\sum_{f\in\mathcal{B}_{\bk}(\ringO_{F})}\frac{1}{\left<f,f\right>}\frac{d^{\ell}}{ds^{\ell}}\left(\Lambda(f,s)\right)\] is non-vanishing for any $s=\sigma+it_{0}$ with $\sigma\in(\frac{k-1}{2},\frac{k}{2}-\epsilon)\cup(\frac{k}{2}+\epsilon,\frac{k+1}{2})$.
\end{theorem}
We obtain the following corollary as a direct consequence

\begin{corollary}
Let $t_{0}$, $\epsilon$, $\ell$ and $C$ be as in Theorem \ref{thm:main}. Then for $k>C$ and any $s=\sigma+it_{0}$ with $\sigma\in(\frac{k-1}{2},\frac{k}{2}-\epsilon)\cup(\frac{k}{2}+\epsilon,\frac{k+1}{2})$, there exists a Hecke eigenform $f\in \mathcal{S}_{\bk}(\SL(\ringO_{F}))$ such that $\frac{d^{\ell}}{ds^{\ell}}\left(\Lambda(f,s)\right)\neq 0$.
\end{corollary}

\section{Setting and Preliminaries}
In this note, we work over a totally real number field $F$ of degree $n$ over $\mathbb{Q}$ with ring of integers $\ringO_{F}$. The group of units in $\ringO_{F}$ is denoted by $\ringO_{F}^{\times}$. For simplicity of exposition, we assume that the narrow class number of $F$ is 1.

The absolute norm of an ideal $\a\subset\ringO_{F}$ is given by $\norm(\a)= [\ringO_{F}:\a]$. The trace and the norm over $\mathbb{Q}$ of an element $x\in F$ are denoted by $\trace(x)$ and $\norm(x)$, respectively. We denote by $\Dif_{F}$ the different ideal of $F$ and  by $d_{F}$ its discriminant over $\mathbb{Q}$. We have the relation $\Dif_{F}=(d_{F})$ and $\norm(\Dif_{F})=|d_{F}|$.

The real embeddings of $F$ are denoted by $\sigma_{j}:x\mapsto x_{j}:=\sigma_{j}(x)$ for $j=1,\dots,n$. We say $x\in F$ is totally positive and write $x\gg0$ if $x_{j}>0$ for all $j$. Moreover, we use $X^{+}$ to denote the set of all totally positive elements in a subset $X$ of $F$.

To simplify exposition, we will often make use of the following notation. For $c,d\in F$, $z=(z_{1},\dots,z_{n})\in\mathbb{C}^{n}$ and $s\in\mathbb{C}$, we set \[\norm(cz+d)^{s}=\prod_{j=1}^n(c_jz_j+d_j)^s.\] Moreover, we set \[\trace(cz)=\sum_{j=1}^{n}c_{j}z_{j}.\]

Let us now recall the following results which are crucial for establishing Theorem \ref{thm:main} (see the proof of Lemma \ref{lem:bound} below).
\begin{lemma}\label{normlemmatrotabas}{$\mathrm{(Trotabas}$ \cite[Lemma~2.1]{trotabas}$\mathrm{)}$}
There exist constants $C_{1}$ and $C_{2}$ depending only on $F$ such that $$\forall \xi\in F,\exists\epsilon\in\ringO_{F}^{\times +},\forall j\in\{1,\dots,n\}: \quad C_{1}|\norm(\xi)|^{1/n}\leq|(\epsilon\xi)_{j}|\leq C_{2}|\norm(\xi)|^{1/n}.$$
\end{lemma}

\begin{lemma}\label{lem:luo}{$\mathrm{(Luo}$ \cite{luo}$\mathrm{)}$}
For $\lambda>0$, we have
\begin{equation}
\sum_{\eta\in\ringO_{F}^{\times+}}\prod_{|\eta_j|<1}|\eta_j|^{\lambda}<\infty.\end{equation}
\end{lemma}

\section{Fourier Expansion of the Kernel Function for Hilbert Modular Forms}
Let us now define the kernel function for Hilbert modular forms over $F$. Let $z=(z_1,z_2,\dots,z_{n})\in\mathbb{H}^{n}$, $k\in2\mathbb{N}$ and let $s\in\mathbb{C}$ with $1<\Re(s)<{k}-1$. We have
\begin{equation}\label{eqn:kernel-function}R_{s,k}(z)=\gamma_{k}(s)\sum_{ \begin{pmatrix}
a & b  \\
c & d  \end{pmatrix}\in T\backslash\SL(\mathcal{O}_{F})}\norm(cz+d)^{-k}\frac{\norm(az+b)^{-s}}{\norm(cz+d)^{-s}},\end{equation}
where $T=\left\{\begin{pmatrix}
\epsilon & 0 \\
0 & \epsilon^{-1}  \end{pmatrix}: \epsilon\in\ringO_{F}^{\times +}\right\}$ and $\displaystyle{\gamma_{k}(s)=\frac{\left(i^{s}\Gamma(s)\Gamma(k-s)\right)^n}{[\ringO_{F}^{\times}:\ringO_{F}^{\times+}]}}.$

In \cite{HR}, we computed the Fourier expansion of $R_{s,k}(z)$.
\begin{proposition}\label{lem:lemma2}
The function $R_{s,k}(z)$ has the following Fourier expansion:
\[R_{s,k}(z)=\sum_{\substack{\nu\in\Dif_{F}^{-1}\\\nu\gg0}}r_{s,k}(\nu)\exp\left(2\pi i\trace(\nu\bz)\right),\]
where \begin{align*}r_{s,k}(\nu)&=\frac{(2\pi)^{ns}\Gamma^n(k-s)}{\sqrt{|d_{F}|}}\norm(\nu)^{s-1}+(-1)^{n\frac{k}{2}}\frac{(2\pi )^{n(k-s)}\Gamma^n(s)}{\sqrt{|d_{F}|}}\norm(\nu)^{k-s-1}\\&\hspace{0.2in}+\gamma_{k}(s)\frac{(2\pi i)^{nk}\norm(\nu)^{k-1}}{\Gamma^n(k)\sqrt{|d_{F}|}}\sideset{}{^*}\sum_{\substack{(a,c)\in\ringO_{F}\times\ringO_{F} \\ac\neq0\\\gcd(a,c)=1}}\frac{\norm(c)^{s-k}}{\norm(a)^{s}}\exp\left(2\pi i\trace\left(\nu\frac{d_{0}}{c}\right)\right)\\&\hspace{2.5in}\times\prod_{j=1}^n{}_{1}F_{1}\left(s,k,-\frac{2\pi i\nu_{j}}{a_{j}c_{j}}\right).\end{align*}
\end{proposition}

We also showed in \cite{HR} that $R_{s,k}(z)$ can be expressed as follows: \begin{align}\label{eqn:kernel-expansion}
R_{\overline{s},k}&=(-1)^{n\frac{k}{2}}\pi^n2^{n(2-k)}\Gamma^n(k-1)\sum_{f\in\mathcal{B}_{\bk}(\ringO_{F})}\frac{\Lambda(f,s)}{\left<f,f\right>}f,
\end{align}
where $\mathcal{B}_{\bk}(\ringO_{F})$ is a basis of normalized Hecke eigenforms of $\mathcal{S}_{\bk}(\SL(\ringO_{F}))$,  and \[\Lambda(f,s)=|d_{F}|^{s}(2\pi)^{-ns}\Gamma^n(s)L(f,s).\] Recall that $\Lambda(f,s)$ satisfies the functional equation (see \cite[page~654]{shimura}) \begin{equation}\label{eqn:functional}\Lambda(k-s)=(-1)^{n\frac{k}{2}}\Lambda(s).\end{equation}

\section{Proof of Theorem \ref{thm:main}}\label{sec:proof}
In this section, we prove the main theorem of this paper following the recent work of Kohnen et al. \cite{KSW}. In view of the functional equation \eqref{eqn:functional}, it suffices to consider the left hand side of the critical strip. Hence, we take $s=\frac{k}{2}-\delta-it_{0}$ where $\epsilon<\delta<\frac12$ and $t_{0}\in\mathbb{R}$.

Taking the first Fourier coefficients on both sides of \eqref{eqn:kernel-expansion} and using Proposition \ref{lem:lemma2}, we get
\begin{align}\label{eqn:main}&\frac{(2\pi)^{ns}\Gamma^n(k-s)}{\sqrt{|d_{F}|}}+(-1)^{n\frac{k}{2}}\frac{(2\pi)^{n(k-s)}\Gamma^n(s)}{\sqrt{|d_{F}|}}\nonumber\\&\hspace{0.3in}+\frac{(-1)^{n\frac{k}{2}}(2\pi )^{nk}i^{ns}}{\sqrt{|d_{F}|}[\ringO_{F}^{\times}:\ringO_{F}^{\times+}]}\nonumber\\&\hspace{0.5in}\times\sideset{}{^*}\sum_{\substack{(a,c)\in\ringO_{F}\times\ringO_{F} \\ac\neq0\\\gcd(a,c)=1}}\frac{\norm(c)^{s-k}}{\norm(a)^s}\exp\left(2\pi i\trace\left(\frac{d_{0}}{c}\right)\right)\prod_{j=1}^{n}{}_{1}f_{1}\left(s,k,-\frac{2\pi i}{a_{j}c_{j}}\right)\nonumber\\&=(-1)^{n\frac{k}{2}}\pi^n2^{n(2-k)}\Gamma^n(k-1)\sum_{f\in\mathcal{B}_{\bk}(\ringO_{F})}\frac{\Lambda(f,s)}{\left<f,f\right>},\end{align}

where \[{}_{1}f_{1}\left(s,k,-\frac{2\pi i}{a_{j}c_{j}}\right)=\frac{\Gamma(k-s)\Gamma(s)}{\Gamma(k)}{}_{1}F_{1}\left(s,k,-\frac{2\pi i}{a_{j}c_{j}}\right).\]

Taking the $\ell$-th derivative of both sides with respect to $s$ gives

\begin{align}\label{eqn:main}&\frac{1}{\sqrt{|d_F|}}\frac{d^{\ell}}{ds^{\ell}}\left[(2\pi)^{ns}\Gamma^n(k-s)\right]+\frac{(-1)^{n\frac{k}{2}}}{\sqrt{|d_{F}|}}\frac{d^{\ell}}{ds^{\ell}}\left[(2\pi)^{n(k-s)}\Gamma^n(s)\right]\nonumber\\&\hspace{0.3in}+\frac{(-1)^{n\frac{k}{2}}(2\pi )^{nk}}{\sqrt{|d_{F}|}[\ringO_{F}^{\times}:\ringO_{F}^{\times+}]}\nonumber\\&\hspace{0.5in}\times\sideset{}{^*}\sum_{\substack{(a,c)\in\ringO_{F}\times\ringO_{F} \\ac\neq0\\\gcd(a,c)=1}}\frac{d^{\ell}}{ds^{\ell}}\left[\frac{\norm(c)^{s-k}}{\norm(a)^s}\exp\left(\frac{\pi}{2} i ns\right)\exp\left(2\pi i\trace\left(\frac{d_{0}}{c}\right)\right)\prod_{j=1}^{n}{}_{1}f_{1}\left(s,k,-\frac{2\pi i}{a_{j}c_{j}}\right)\right]\nonumber\\&=(-1)^{n\frac{k}{2}}\pi^n2^{n(2-k)}\Gamma^n(k-1)\sum_{f\in\mathcal{B}_{\bk}(\ringO_{F})}\frac{1}{\left<f,f\right>}\frac{d^{\ell}}{ds^{\ell}}\left(\Lambda(f,s)\right),\end{align}

Let us consider first the expression
\[I_1=\frac{1}{\sqrt{|d_F|}}\frac{d^{\ell}}{ds^{\ell}}\left[(2\pi)^{ns}\Gamma^n(k-s)\right].\] We have
\begin{align*}
I_1&=\frac{1}{\sqrt{|d_F|}}\sum_{j=0}^{\ell}\binom{\ell}{j}\frac{d^{j}}{ds^{j}}\left[(2\pi)^{ns}\right]\frac{d^{\ell-j}}{ds^{\ell-j}}\left[\Gamma^n(k-s)\right]\\&=\frac{1}{\sqrt{|d_F|}}\sum_{j=0}^{\ell}\binom{\ell}{j}\left(n\log\left(2\pi\right)\right)^{j}(2\pi)^{ns}\sum_{\nu_1+\nu_2+\cdots+\nu_n=\ell-j}\frac{(\ell-j)!}{\nu_1!\nu_2!\cdots\nu_n!}\prod_{1\leq t\leq n}(-1)^{\nu_t}\Gamma^{(\nu_t)}(k-s)\\&=\frac{(2\pi)^{ns}}{\sqrt{|d_F|}}\sum_{j=0}^{\ell}(-1)^{\ell-j}\binom{\ell}{j}\left(n\log\left(2\pi\right)\right)^{j}\sum_{\nu_1+\nu_2+\cdots+\nu_n=\ell-j}\frac{(\ell-j)!}{\nu_1!\nu_2!\cdots\nu_n!}\prod_{1\leq t\leq n}\Gamma^{(\nu_t)}(k-s)\\&=\frac{(2\pi)^{ns}}{\sqrt{|d_F|}}\left(n\log(2\pi)\right)^{\ell}\Gamma^{n}(k-s)\\&\hspace{2em}+ \frac{(2\pi)^{ns}}{\sqrt{|d_F|}}\sum_{j=0}^{\ell-1}(-1)^{\ell-j}\binom{\ell}{j}\left(n\log\left(2\pi\right)\right)^{j}\sum_{\nu_1+\nu_2+\cdots+\nu_n=\ell-j}\frac{(\ell-j)!}{\nu_1!\nu_2!\cdots\nu_n!}\prod_{1\leq t\leq n}\Gamma^{(\nu_t)}(k-s).
\end{align*}

Hence,
\begin{align}\label{eqn:I1}
\frac{\sqrt{|d_F|}I_1}{(2\pi)^{ns}\Gamma^n(k-s)}&=\left(n\log(2\pi)\right)^{\ell}\\&\hspace{2em}+ \sum_{j=0}^{\ell-1}(-1)^{\ell-j}\binom{\ell}{j}\left(n\log\left(2\pi\right)\right)^{j}\sum_{\nu_1+\nu_2+\cdots+\nu_n=\ell-j}\frac{(\ell-j)!}{\nu_1!\nu_2!\cdots\nu_n!}\prod_{1\leq t\leq n}\frac{\Gamma^{(\nu_t)}(k-s)}{\Gamma(k-s)}.
\end{align}
In \cite{KSW}, it is observed that $\frac{\Gamma^{(m)(z)}}{\Gamma(z)}$ can be expressed as a polynomial $P\in\mathbb{Z}[\psi,\psi^{(1)},\dots,\psi^{(m)}]$, where $\psi=\frac{\Gamma'}{\Gamma}$ is the digamma function. It is important to note that the highest power of $\psi$ appearing in the polynomial $P$ is $\psi^m$.

We also recall the following important estimates of the digamma function:

\[\psi(z)\sim\log(z)-\frac{1}{2z}-\sum_{k=1}^{\infty}\frac{B_{2k}}{2kz^{2k}}\]
\[\psi^{(m)}(z)\sim(-1)^{m-1}\left(\frac{(m-1)!}{z^m}+\frac{m!}{2z^{m+1}}+\sum_{k=1}^{\infty}B_{2k}\frac{(2k+m-1)!}{(2k)!z^{2k+m}}\right),\]
for $z\to\infty$ in $|\arg(z)|<\pi$, where $B_n$ is the $n$th Bernoulli number  (see \cite[6.3.18~\&~6.4.11]{stegun}).

For $s=\frac{k}{2}-\delta+it_0$ with $\epsilon<\delta<\frac12$, see that $\frac{\Gamma^{(\nu_t)}(k-s)}{\Gamma(k-s)}=P_t\left(\log(k-s)\right)+ o(1)$ where $P_t$ is a polynomial with integer coefficients and  degree $\nu_t$. It follows that
\[\prod_{1\leq t\leq n}\frac{\Gamma^{(\nu_t)}(k-s)}{\Gamma(k-s)}=\prod_{1\leq t\leq n}P_t\left(\log(k-s)\right)+ o(1).\]
\begin{align*}
\frac{\sqrt{|d_F|}I_1}{(2\pi)^{ns}\Gamma^n(k-s)}&=\left(n\log(2\pi)\right)^{\ell}\\&\hspace{2em}+ \sum_{j=0}^{\ell-1}(-1)^{\ell-j}\binom{\ell}{j}\left(n\log\left(2\pi\right)\right)^{j}\sum_{\nu_1+\nu_2+\cdots+\nu_n=\ell-j}\frac{(\ell-j)!}{\nu_1!\nu_2!\cdots\nu_n!}\prod_{1\leq t\leq n}P_t\left(\log(k-s)\right) +o(1)\\&=\tilde{P}\left(\log(k-s)\right)+o(1),
\end{align*}
where $\tilde{P}$ is a polynomial of degree $\ell$ and with leading coefficient $(-1)^{\ell}\sum_{\nu_1+\cdots+\nu_n=\ell}\frac{\ell!}{\nu_1!\cdots\nu_n!}$.

Next, we deal with the sum
\[I_2=\frac{(-1)^{n\frac{k}{2}}}{\sqrt{|d_{F}|}}\frac{d^{\ell}}{ds^{\ell}}\left[(2\pi)^{n(k-s)}\Gamma^n(s)\right].\]

We have \begin{align*}
I_2&=\frac{(-1)^{n\frac{k}{2}}(2\pi)^{nk}}{\sqrt{|d_F|}}\sum_{j=0}^{\ell}\binom{\ell}{j}\frac{d^{j}}{ds^{j}}\left[(2\pi)^{-ns}\right]\frac{d^{\ell-j}}{ds^{\ell-j}}\left[\Gamma^n(s)\right]\\&=\frac{(-1)^{n\frac{k}{2}}(2\pi)^{n(k-s)}}{\sqrt{|d_F|}}\sum_{j=0}^{\ell}\binom{\ell}{j}(-1)^j\left(n\log\left(2\pi\right)\right)^{j}\sum_{\nu_1+\nu_2+\cdots+\nu_n=\ell-j}\frac{(\ell-j)!}{\nu_1!\nu_2!\cdots\nu_n!}\prod_{1\leq t\leq n}\Gamma^{(\nu_t)}(s).
\end{align*}

It follows that

\begin{align*}
\frac{\sqrt{|d_F|}I_2}{(2\pi)^{ns}\Gamma^n(k-s)}&=(-1)^{n\frac{k}{2}}(2\pi)^{n(k-2s)}\frac{\Gamma^n(s)}{\Gamma^n(k-s)}\sum_{j=0}^{\ell}\binom{\ell}{j}(-1)^j\left(n\log\left(2\pi\right)\right)^{j}\\&\hspace{1in}\times\sum_{\nu_1+\nu_2+\cdots+\nu_n=\ell-j}\frac{(\ell-j)!}{\nu_1!\nu_2!\cdots\nu_n!}\prod_{1\leq t\leq n}\frac{\Gamma^{(\nu_t)}(s)}{\Gamma(s)}.
\end{align*}
If $s=\frac{k}{2}-\delta+it_0$ with $\epsilon<\delta<\frac12$, we have

\begin{align*}\frac{\sqrt{|d_F|}I_2}{(2\pi)^{ns}\Gamma^n(k-s)}&=(-1)^{n\frac{k}{2}+\ell}(2\pi)^{n(k-2s)}(n\log(2\pi))^{\ell}\frac{\Gamma^n(s)}{\Gamma^n(k-s)}\\&+(-1)^{n\frac{k}{2}}(2\pi)^{n(k-2s)}\frac{\Gamma^n(s)}{\Gamma^n(k-s)}\left(\tilde{Q}\left(\log(k-s)\right)+o(1)\right),\end{align*}
where $\tilde{Q}$ is a polynomial of degree $\ell$ and with leading coefficient $\sum_{\nu_1+\cdots+\nu_n=\ell}\frac{\ell!}{\nu_1!\cdots\nu_n!}$.

Observe that (see \cite[6.1.23~\&~6.1.47]{stegun})
 \[\left|\frac{\Gamma^n(s)}{\Gamma^n(k-s)}\right|=\left|\frac{k}{2}+it_0\right|^{-2n\delta}\cdot\left|1+O\left(\frac{1}{\left|\frac{k}{2}+it_0\right|^n}\right)\right|,\] uniformly in $\epsilon<\delta<\frac12$. It follows that $\frac{\sqrt{|d_F|}I_2}{(2\pi)^{ns}\Gamma^n(k-s)}$ tends to $0$ as $k\to\infty$.

We still have to estimate the following sum:
\begin{align*}I_3&=\frac{(-1)^{n\frac{k}{2}}(2\pi )^{nk}}{\sqrt{|d_{F}|}[\ringO_{F}^{\times}:\ringO_{F}^{\times+}]}\nonumber\\&\hspace{0.5in}\times\sideset{}{^*}\sum_{\substack{(a,c)\in\ringO_{F}\times\ringO_{F} \\ac\neq0\\\gcd(a,c)=1}}\frac{d^{\ell}}{ds^{\ell}}\left[\frac{\norm(c)^{s-k}}{\norm(a)^s}\exp\left(\frac{\pi}{2} i ns\right)\exp\left(2\pi i\trace\left(\frac{d_{0}}{c}\right)\right)\prod_{t=1}^{n}{}_{1}f_{1}\left(s,k,-\frac{2\pi i}{a_{t}c_{t}}\right)\right].\end{align*}

We have
\begin{align*}
I_3&=\frac{(-1)^{n\frac{k}{2}}(2\pi )^{nk}}{\sqrt{|d_{F}|}[\ringO_{F}^{\times}:\ringO_{F}^{\times+}]}\sideset{}{^*}\sum_{\substack{(a,c)\in\ringO_{F}\times\ringO_{F} \\ac\neq0\\\gcd(a,c)=1}}\norm(c)^{-k}\sum_{j=0}^{\ell}\frac{\norm(c)^{s}}{\norm(a)^s}\log\left(\frac{\norm(c)}{\norm(a)}\right)^{j}\\&\hspace{9em}\times\frac{d^{\ell-j}}{ds^{\ell-j}}\left[\exp\left(\frac{\pi}{2} i ns\right)\exp\left(2\pi i\trace\left(\frac{d_{0}}{c}\right)\right)\prod_{t=1}^{n}{}_{1}f_{1}\left(s,k,-\frac{2\pi i}{a_{t}c_{t}}\right)\right].\end{align*}

For $1\leq j\leq \ell-1$, we have
\begin{align*}
&\frac{d^{\ell-j}}{ds^{\ell-j}}\left[\exp\left(\frac{\pi}{2} i ns\right)\exp\left(2\pi i\trace\left(\frac{d_{0}}{c}\right)\right)\prod_{t=1}^{n}{}_{1}f_{1}\left(s,k,-\frac{2\pi i}{a_{t}c_{t}}\right)\right]\\&=\sum_{m=0}^{\ell-j}\binom{\ell-j}{m}\exp\left(2\pi i\trace\left(\frac{d_{0}}{c}\right)\right)\left(\frac{\pi}{2}in\right)^{m}\exp\left(\frac{\pi}{2} i ns\right)\frac{d^{\ell-j-m}}{ds^{\ell-j-m}}\left[\prod_{t=1}^{n}{}_{1}f_{1}\left(s,k,-\frac{2\pi i}{a_{t}c_{t}}\right)\right]\\&=\sum_{m=0}^{\ell-j}\binom{\ell-j}{m}\exp\left(2\pi i\trace\left(\frac{d_{0}}{c}\right)\right)\left(\frac{\pi}{2}in\right)^{m}\exp\left(\frac{\pi}{2} i ns\right)\\&\hspace{5em}\times \sum_{\nu_1+\nu_2+\cdots+\nu_n=\ell-j-m}\frac{(\ell-j-m)!}{\nu_1!\nu_2!\cdots\nu_n!}\prod_{t=1}^{n}(-1)^{\nu_t}\frac{d^{\nu_t}}{ds^{\nu_t}}{}_{1}f_{1}\left(s,k,-\frac{2\pi i}{a_{t}c_{t}}\right).
\end{align*}

Therefore,

\begin{align*}
\frac{\sqrt{|d_F|}I_3}{(2\pi)^{ns}\Gamma^n(k-s)}&=\frac{(-1)^{n\frac{k}{2}}(2\pi )^{n(k-s)}}{\Gamma^{n}(k-s)[\ringO_{F}^{\times}:\ringO_{F}^{\times+}]}\sideset{}{^*}\sum_{\substack{(a,c)\in\ringO_{F}\times\ringO_{F} \\ac\neq0\\\gcd(a,c)=1}}\frac{\norm(c)^{s-k}}{\norm(a)^s}\exp\left(\frac{\pi}{2} i ns\right)\exp\left(2\pi i\trace\left(\frac{d_{0}}{c}\right)\right)\\&\hspace{2em}\times\Big[\log\left(\frac{\norm(c)}{\norm(a)}\right)^{\ell}\prod_{t=1}^{n}{}_{1}f_{1}\left(s,k,-\frac{2\pi i}{a_{t}c_{t}}\right)\\&\hspace{5em}+\sum_{j=0}^{\ell-1}\log\left(\frac{\norm(c)}{\norm(a)}\right)^{j}\sum_{m=0}^{\ell-j}\binom{\ell-j}{m}\left(\frac{\pi}{2}in\right)^{m}\\&\hspace{5em}\times \sum_{\nu_1+\nu_2+\cdots+\nu_n=\ell-j-m}\frac{(\ell-j-m)!}{\nu_1!\nu_2!\cdots\nu_n!}\prod_{t=1}^{n}(-1)^{\nu_t}\frac{d^{\nu_t}}{ds^{\nu_t}}{}_{1}f_{1}\left(s,k,-\frac{2\pi i}{a_{t}c_{t}}\right)\Big].\end{align*}

Let us now consider the sums
\begin{align*}
E_{1,\ell}(s,k)&=\sideset{}{^*}\sum_{\substack{(a,c)\in\ringO_{F}\times\ringO_{F} \\ac\neq0\\\gcd(a,c)=1}}\frac{\norm(c)^{s-k}}{\norm(a)^s}\exp\left(\frac{\pi}{2} i ns\right)\exp\left(2\pi i\trace\left(\frac{d_{0}}{c}\right)\right)\\&\hspace{2em}\times\log\left(\frac{\norm(c)}{\norm(a)}\right)^{\ell}\prod_{t=1}^{n}{}_{1}f_{1}\left(s,k,-\frac{2\pi i}{a_{t}c_{t}}\right)
\end{align*}
and
\begin{align*}
E_{2,\ell}(s,k)&=\sideset{}{^*}\sum_{\substack{(a,c)\in\ringO_{F}\times\ringO_{F} \\ac\neq0\\\gcd(a,c)=1}}\frac{\norm(c)^{s-k}}{\norm(a)^s}\exp\left(2\pi i\trace\left(\frac{d_{0}}{c}\right)\right)\exp\left(\frac{\pi}{2} i ns\right)\sum_{j=0}^{\ell-1}\log\left(\frac{\norm(c)}{\norm(a)}\right)^{j}\\&\hspace{2em}\times\sum_{m=0}^{\ell-j}\binom{\ell-j}{m}\left(\frac{\pi}{2}in\right)^{m} \sum_{\nu_1+\nu_2+\cdots+\nu_n=\ell-j-m}\frac{(\ell-j-m)!}{\nu_1!\nu_2!\cdots\nu_n!}\prod_{t=1}^{n}(-1)^{\nu_t}\frac{d^{\nu_t}}{ds^{\nu_t}}{}_{1}f_{1}\left(s,k,-\frac{2\pi i}{a_{t}c_{t}}\right).
\end{align*}
We have
\begin{align*}
E_{1,\ell}(s,k)&=\sum_{\substack{\eta\in\ringO_{F}^{\times+}\\c\in\ringO_{F}/\ringO_{F}^{\times+},\\c\neq0}}\sum_{\substack{a\in\ringO_{F}/\ringO_{F}^{\times+}\\\gcd(a,c)=1}}\frac{\norm(c)^{s-k}}{\norm(a)^s}\exp\left(\frac{\pi}{2} i ns\right)\exp\left(2\pi i\trace\left(\frac{d_{0}}{\eta c}\right)\right)\\&\hspace{2em}\times\log\left(\frac{\norm(c)}{\norm(a)}\right)^{\ell}\prod_{t=1}^{n}{}_{1}f_{1}\left(s,k,-\frac{2\pi i}{a_{t}\eta_t c_{t}}\right),
\end{align*}
and
\begin{align*}
E_{2,\ell}(s,k)&=\sum_{\substack{\eta\in\ringO_{F}^{\times+}\\c\in\ringO_{F}/\ringO_{F}^{\times+},\\c\neq0}}\sum_{\substack{a\in\ringO_{F}/\ringO_{F}^{\times+}\\\gcd(a,c)=1}}\frac{\norm(c)^{s-k}}{\norm(a)^s}\exp\left(2\pi i\trace\left(\frac{d_{0}}{\eta c}\right)\right)\exp\left(\frac{\pi}{2} i ns\right)\\&\hspace{5em}\times\sum_{j=0}^{\ell-1}\log\left(\frac{\norm(c)}{\norm(a)}\right)^{j}\sum_{m=0}^{\ell-j}\binom{\ell-j}{m}\left(\frac{\pi}{2}in\right)^{m}\\&\hspace{7em}\times \sum_{\nu_1+\nu_2+\cdots+\nu_n=\ell-j-m}\frac{(\ell-j-m)!}{\nu_1!\nu_2!\cdots\nu_n!}\prod_{t=1}^{n}(-1)^{\nu_t}\frac{d^{\nu_t}}{ds^{\nu_t}}{}_{1}f_{1}\left(s,k,-\frac{2\pi i}{a_{t}\eta_{t}c_{t}}\right).
\end{align*}
To deal with $E_{1,\ell}(s,k)$ and $E_{2,\ell}(s,k)$, we need the following lemma.

\begin{lemma}\label{lem:hyper-bound}
 Let $s=\frac{k}{2}-\delta+it_{0}$ where $\epsilon<\delta<\frac12$ and $t_{0}\in\mathbb{R}$. Let $\ell$ be a non-negative integer. For all $x\in\mathbb{R}$ and all sufficiently large $k$, we have \[\frac{d^{\ell}}{ds^{\ell}}\left({}_{1}f_{1}\left(s,k,ix\right)\right)\ll_{\ell} \min\left\{1,|x|^{-1}\left(|s-1|+|k-s-1|\right)\right\}.\]
\end{lemma}
\begin{proof}
By \cite[13.2.1]{stegun}, we have \[{}_{1}f_{1}\left(s,k,ix\right)=\int_{0}^{1}\exp(ixu)u^{s-1}(1-u)^{k-s-1}\;du.\]  It follows that $\left|{}_{1}f_{1}\left(s,k,ix\right)\right|\leq 1$ whenever $\Re(s)>1$ and $\Re(k-s)>1$. Moreover, by differentiating both sides with respect to $s$, we get

\begin{equation}\label{eqn:derivativebound1}\frac{d^{\ell}}{ds^{\ell}}\left({}_{1}f_{1}\left(s,k,ix\right)\right)\ll_{\ell}1\end{equation} for any non-negative integer $\ell$ (see \cite[page 326]{KSW}). On the other hand, integration by parts yields
\begin{align*}
{}_{1}f_{1}\left(s,k,ix\right)&=\int_{0}^{1}\exp(ixu)u^{s-1}(1-u)^{k-s-1}\;du\\&=-\frac{s-1}{ix}\int_{0}^{1}\exp(ixu)u^{s-2}(1-u)^{k-s-1}\;du\\&\hspace{0.5in}+\frac{k-s-1}{ix}\int_{0}^{1}\exp(ixu)u^{s-1}(1-u)^{k-s-2}\;du.
\end{align*}
Taking the $\ell$-th derivative of both sides with respect to $s$ yields

\begin{align*}
\frac{d^{\ell}}{ds^{\ell}}\left({}_{1}f_{1}\left(s,k,ix\right)\right)
&=-\frac{1}{ix}\int_{0}^{1}\exp(ixu)u^{s-2}(1-u)^{k-s-2}\;du\\&\hspace{0.1in}-\frac{s-1}{ix}\int_{0}^{1}\exp(ixu)\sum_{j=0}^{\ell}(-1)^{\ell-j}\binom{\ell}{j}(\log u)^ju^{s-2}(\log(1-u))^{\ell-j}(1-u)^{k-s-1}\;du\\&\hspace{0.1in}+\frac{k-s-1}{ix}\int_{0}^{1}\exp(ixu)\sum_{j=0}^{\ell}(-1)^{\ell-j}\binom{\ell}{j}(\log u)^ju^{s-1}(\log(1-u))^{\ell-j}(1-u)^{k-s-2}\;du.
\end{align*}

Hence, \begin{equation}\label{eqn:derivativebound2}\frac{d^{\ell}}{ds^{\ell}}\left({}_{1}f_{1}\left(s,k,ix\right)\right)\ll_{\ell} |x|^{-1}\left(|s-1|+|k-s-1|\right)\end{equation} whenever $\Re(s)>2$ and $\Re(k-s)>2$.
The desired result follows from \eqref{eqn:derivativebound1} and \eqref{eqn:derivativebound2}.
\end{proof}

\begin{lemma}\label{lem:bound}
Let $s=\frac{k}{2}-\delta+it_{0}$ where $\epsilon<\delta<\frac12$ and $t_{0}\in\mathbb{R}$. As $k\to\infty$, we have $E_{1,\ell}(s,k)=O(k^n)$ and $E_{2,\ell}(s,k)=O(k^n)$ where the implicit constants depend only on $\ell$, $t_{0}$, $\delta$ and the field $F$.
\end{lemma}
\begin{proof}
Upon taking absolute values, we get
\begin{align*}
\left|E_{2,\ell}(s,k)\right|&\leq \sum_{\eta\in\ringO_{F}^{\times +}}\sum_{\substack{c\in\ringO_{F}/\ringO_{F}^{\times +},\\c\neq0}}\sum_{\substack{a\in\ringO_{F}/\ringO_{F}^{\times +}\\\gcd(a,c)=1}}|\norm(c)|^{-\frac{k}{2}-\delta}|\norm(a)|^{\delta-\frac{k}{2}}\sum_{j=0}^{\ell-1}\left|\log\left(\frac{\norm(c)}{\norm(a)}\right)\right|^{j}\\&\hspace{5em}\times\sum_{m=0}^{\ell-j}\binom{\ell-j}{m}\left(\frac{\pi}{2}n\right)^{m}\exp\left(\frac{\pi}{2}  nt_0\right)\\&\hspace{5em}\times \sum_{\nu_1+\nu_2+\cdots+\nu_n=\ell-j-m}\frac{(\ell-j-m)!}{\nu_1!\nu_2!\cdots\nu_n!}\prod_{t=1}^{n}\left|\frac{d^{\nu_t}}{ds^{\nu_t}}{}_{1}f_{1}\left(s,k,-\frac{2\pi i}{a_{t}\eta_{t}c_{t}}\right)\right|.
\end{align*}
By Lemma \ref{lem:hyper-bound}, we know that  \[\frac{d^{\nu_t}}{ds^{\nu_t}}\left({}_{1}f_{1}\left(s,k,ix\right)\right)\ll_{\nu_t} \left(|s-1|+|k-s-1|\right) \left|\frac{2\pi}{\eta_{j}a_{j}c_{j}}\right|^{-\omega_j},\] where $\omega_j$ is either $0$ or $1$  depending on whether $|\eta_{j}|\geq1$ or $|\eta_{j}|<1$ respectively. Hence, we get \[\prod_{j=1}^n\left|\frac{d^{\nu_t}}{ds^{\nu_t}}\left({}_{1}f_{1}\left(s,k,ix\right)\right)\right|\ll_{\nu_t} \left(|s-1|+|k-s-1|\right)^{n}\prod_{|\eta_{j}|<1}\left|\frac{2\pi}{\eta_{j}a_{j}c_{j}}\right|^{-1}.\]
It follows that $E_{2,\ell}(s,k)$ is  \begin{equation}\label{eqn:S}\ll\left(|s-1|+|k-s-1|\right)^{n}\sum_{\eta\in\ringO_{F}^{\times +}}\sum_{\substack{c\in\ringO_{F}/\ringO_{F}^{\times +},\\c\neq0}}\sum_{\substack{a\in\ringO_{F}/\ringO_{F}^{\times +}\\\gcd(a,c)=1}}\frac{ \prod_{|\eta_{j}|<1}\left|\frac{2\pi}{\eta_{j}a_{j}c_{j}}\right|^{-1}}{|\norm(c)|^{\frac{k}{2}+\delta-\epsilon}|\norm(a)|^{\frac{k}{2}-\delta+\epsilon}}.\end{equation}
By Lemma~\ref{normlemmatrotabas}, we may assume that the representatives $a,c\in \ringO_{F}/\ringO_{F}^{\times +}$ in (\ref{eqn:S}) satisfy \begin{equation*}\label{eqn:norms}
\norm(a)^{1/n}\ll a_{j}\ll\norm(a)^{1/n}\quad \text{and} \quad |\norm(c)|^{1/n}\ll c_{j} \ll|\norm(c)|^{1/n},\end{equation*}
 for all  $j\in\{1,\cdots,n\}$ with implicit constants depending only on $F$. Therefore, we have \begin{equation*}E_{2,\ell}(s,k)\ll k^n\sum_{\eta\in\ringO_{F}^{\times +}}\prod_{|\eta_{j}|<1}|\eta_{j}|\sum_{\substack{c\in\ringO_{F}/\ringO_{F}^{\times +},\\c\neq0}}\sum_{\substack{a\in\ringO_{F}/\ringO_{F}^{\times +}\\\gcd(a,c)=1}}|\norm(c)|^{-\frac{k}{2}+1-\delta+\epsilon}|\norm(a)|^{-\frac{k}{2}+1+\delta-\epsilon}. \end{equation*}
Lemma \ref{lem:luo} allows us to factor out the sum over all $\eta\in\ringO_{F}^{\times +}$ since it is convergent and depends only on $F$. Thus, we get $E_{2,\ell}(s,k)=O(k^n)$ for sufficiently large $k$ as desired. We also emphasize that the implied constant in this estimate depends only on  $\delta$, $t_{0}$, $\ell$ and the field $F$. The sum $E_{1,\ell}(s,k)$ is treated similarly, and so we will not include the details here.
\end{proof}
It follows from Lemma \ref{lem:bound} that
\[\frac{\sqrt{|d_F|}I_3}{(2\pi)^{ns}\Gamma^n(k-s)}\ll\frac{k^n(2\pi)^{n\frac{k}{2}}}{\left|\Gamma^n(\frac{k}{2}+\delta-it_0)\right|},\] which tends to $0$
 as $k\to\infty$. Moreover, we have already established that as $k\to\infty$ we have
\[\frac{\sqrt{|d_F|}I_2}{(2\pi)^{ns}\Gamma^n(k-s)}\rightarrow0,\]  where as \[\frac{\sqrt{|d_F|}I_1}{(2\pi)^{ns}\Gamma^n(k-s)}\sim\tilde{P}(\log(\frac{k}{2}+\delta-it_0)),\] for some polynomial $\tilde{P}$ of degree $\ell$. Applying these estimates to \eqref{eqn:main}, we conclude that \[\sum_{f\in\mathcal{B}_{\bk}(\ringO_{F})}\frac{1}{\left<f,f\right>}\frac{d^{\ell}}{ds^{\ell}}\left(\Lambda(f,s)\right)\] is non-vanishing for any $s=\frac{k}{2}-\delta+it_{0}$ with $\epsilon<\delta<\frac12$ and $t_{0}\in\mathbb{R}$.

\normalsize


TITLE: Probability problem with random vectors
QUESTION [1 upvotes]: Problem
Suppose that $10$% of the american population smokes dark cigarettes, $35$% smokes white cigarettes, $3$% smokes pipe and the rest of the population doesn't smoke. A group of $35$ persons was surveyed and with the obtained results, we define the random variables:
$$Y_1=\{\text{number of people who smoke dark cigarettes}\}$$$$Y_2=\{\text{number of people who smoke light cigarettes}\}$$$$Y_3=\{\text{number of persons who smoke pipe}\}$$$$Y_4=\{\text{number of non smokers}\}$$
1) Find the probability mass of $(Y_1,Y_2,Y_3,Y_4)$ and the probability mass of $(Y_1,Y_2+Y_3,Y_4)$.
2) Find the probability mass of $Y_2+Y_3$.
This is what I could do:
1) I'll abbreviate $p_{Y_1,Y_2,Y_3,Y_4}(i,j,k,l)=p(i,j,k,l)$$$p(i,j,k,l)=\begin{cases} 0 & \mbox{if } i+j+k+l \neq 35 \\ {35 \choose i}{35-i \choose j}{35-i-j \choose k}{35-i-j-k \choose l}(\frac{1}{10})^i(\frac{7}{20})^j(\frac{3}{100})^k(\frac{13}{25})^l & \mbox{if } i+j+k+l=35 \end{cases}$$
I would like to know if what I've done is correct and I would appreciate help to calculate the other probability masses. Thanks in advance.

REPLY [0 votes]: a) Using the multinomial coefficient, we have
$$P(Y_1 = i, Y_2 = j, Y_3 = k, Y_4 = l) = \binom{35}{i,j,k,l}\left(\frac{10}{100}\right)^i\left(\frac{35}{100}\right)^j\left(\frac{3}{100}\right)^k\left(\frac{52}{100}\right)^l$$
b) Instead of considering four categories, we know consider three categories where categories two and three have been collapsed together. So you could say
\begin{align*}
X_1 &= \{\text{Number of people who smoke dark cigarettes}\}\\
X_2 &=\{\text{Number of people who light cigarettes or pipes}\}\\
X_3&=\{\text{Number of nonsmokers}\}
\end{align*}
Since you seemed to understand part a), then this should not very different.
c) Consider instead a fair six-sided die, and we wish to count the number of twos and threes in 35 rolls. Then it should be familiar that this follows a binomial distribution with $n = 35$, $p = 2/6$. Similarly, part c) also follows a binomial distribution with $n = 35,  p = 38/100$.


TITLE: Boolean algebra simplification with DeMorgan Laws
QUESTION [0 upvotes]: I can't simplify the following expression:
$$x·y'+z+(x'+y)·z'$$
I've tried to multiply the last term with the guys in the parentheses but I can't go any longer.
Thanks in advance.

REPLY [1 votes]: Since $xy'+z+(x'+y)z'$ is $1$ when $z=1$, the factor $z'$ in the last summand can be omitted, so it becomes
$$
xy'+z+x'+y.
$$
By the same reasoning, the factor $x$ in the first summand can now be omitted, so it becomes
$$
y'+z+x'+y.
$$
Now $y'+y=1$ so the expression is $1$ always, we conclude that
$$
xy'+z+(x'+y)z' = 1.
$$


TITLE: Prove that $M/N$ and $N$ are finitely generated $\Rightarrow M$ is finitely generated.
QUESTION [0 upvotes]: Let $M$ be an $R$-module, and $N$ be it's submodule.
I need to prove that if both $M/N$ and $N$ are finitely generated, then so is $M$.
How to prove it? Don't have any ideas at the moment.

REPLY [2 votes]: Suppose
$$N=\langle n_1,...,n_k\rangle\;,\;\;M/N=\langle m_1+N,...,m_r+N\rangle$$
Observe then that if we take any element $\;x\in M\;$ then exists $\;1\le i\le r\;$ and $\;n\in N\;$ with $\;x=m_i+n\;$ . But you can write that $\;n\;$ as $\;n=a_1n_1+\ldots+a_kn_k\;$ ....complete now.


TITLE: Any arbitrary closed smooth curve bounds a orientable surface?
QUESTION [4 upvotes]: I've got a question that, given an arbitrary closed smooth curve $C:[0,1]\rightarrow\mathbb{R}^3$, can you always find a orientable surface $\Omega$ which satisfy $\partial\Omega=C[0,1]$ ?
I have no idea on this question, and I suppose that the surface $\Omega$ has no restriction such as “smooth”. Thanks a lot for your help.

REPLY [2 votes]: Do we know that $C$ does not intersect itself, i.e. that $C$ restricted to the half-open interval $\left[ 0 , 1 \right[$ is injective?
Then the orientable surface exists, can be taken to be compact and connected, and is then called a Seifert surface of the knot.


text:
\begin{document}
\title{An ampleness criterion for rank 2 vector bundles on surfaces}
\author[Arnaud Beauville]{Arnaud Beauville}
\address{Universit\'e C\^ote d'Azur\\
CNRS -- Laboratoire J.-A. Dieudonn\'e\\

Parc Valrose\\
F-06108 Nice cedex 2, France}
\email{arnaud.beauville@unice.fr}

\maketitle
\section{Introduction}
We observe in this note that  the proof of the Bogomolov stable restriction theorem \cite{B} can be adapted to give
an  ampleness criterion for  globally generated rank 2 vector bundles on certain surfaces. This applies to the Lazarsfeld-Mukai bundles, to  congruences of lines in  $\P^3$, and possibly to the construction of surfaces with ample cotangent bundle.

\medskip
\section{Main result}
Throughout the note, $S$ will be a smooth projective surface over $\C$. We denote by $N^1(S)$ the group of divisors on $S$ modulo numerical equivalence; this is a free, finitely generated abelian group,  quotient of $\operatorname{NS}(S)=H^2(S,\Z)_{\mathrm{alg}} $ by its torsion subgroup.

\begin{prop}\label{fib}
Let $E$ be a globally generated rank 2 vector bundle on $S$, with $h^0(E)\geq 4$. Assume that $N^1(S)=\Z\cdot c_1(E)$. Then either $E$ is ample, or $E=\O_S\oplus  \det (E)$.
\end{prop}
We will need the following lemma:

\begin{lem}
Let $S$ be a smooth projective surface, and let $E$ be a globally generated rank 2 vector bundle on $S$, with $h^0(E)\geq 4$ and $H^1(S,\det(E)^{-1})=0$. Then $c_1^2(E)>c_2(E)$.
\end{lem}
\pr Let $V$ be a general $4$-dimensional subspace of $H^0(S,E)$. Then $V$ generates $E$ globally, giving rise to an exact sequence
\begin{equation}\label{se}
0\rightarrow N\rightarrow V\otimes _{\C}\O_S\rightarrow E\rightarrow 0\, .
\end{equation}
 Since $N^*$ is globally generated, the zero locus of a general section $s$ of $N^*$ is finite, of length  $c_2(N^*)=\allowbreak c_1^2(E)-c_2(E)$. Thus this number is $\geq 0$; if it is zero, $s$ does not vanish, so we have  an exact sequence
\[0\rightarrow \O_S \xrightarrow{\ s\ } N^* \rightarrow \det(E)\rightarrow 0\,.\]
Since $H^1(S,\det(E)^{-1})=0$, this sequence splits, so that $N\cong \O_S\oplus \det(E)^{-1}$. Thus the exact sequence (\ref{se}) reduces to
\[0\rightarrow \det(E)^{-1} \rightarrow \O_S^3\rightarrow E\rightarrow 0\,;\]
but using again $H^1(S,\det(E)^{-1})=0$ this implies  $h^0(E)\leq 3$, contradicting the hypothesis.\qed

\medskip
\noindent
\emph{Proof of the Proposition} : We denote  by $c_1$ and $c_2$ the Chern classes of $E$ in $H^*(S,\Z)$, and by $\Delta _E:=4c_2-c_1^2$ its discriminant. Assume that $E$ is not ample. By Gieseker's lemma \cite[Proposition 6.1.7]{L}, there exists an irreducible curve $C$ in $S$ and a surjective homomorphism $u:E\twoheadrightarrow \O_C$.   The kernel $F$ of $u$ is a vector bundle, with total Chern class
$c(F)=c(E)c(\O_C)^{-1}=\allowbreak (1+c_1+c_2)(1-[C])$, hence
\[c_1(F)=c_1-[C]\ ,\ c_2(F)=c_2-c_1\cdot [C] \ ,\ \mbox{and }\ \Delta _F=\Delta _E-2c_1\cdot [C]-[C]^2\,.\]

The curve $C$ is numerically equivalent to $rc_1$ for some integer  $r\geq 1$. Therefore  \[\Delta _F=4c_2 -(r+1)^2c_1^2\leq 4(c_2-c_1^2)\,.\]Because of our hypotheses   $\det(E)$ is ample, so $H^1(S,\det(E)^{-1}) =0$ and we can apply the Lemma, which gives $\Delta _F<0$. By Bogomolov's theorem (see \cite[Th\'eor\`eme 6.1]{Ra}),   we have an exact sequence
\[0\rightarrow L \rightarrow F \rightarrow \mathcal{I}_ZM\rightarrow 0\]
where $Z$ is a finite subscheme of $S$, $L$ and $M$ are line bundles on $S$, with $c_1(L)=ac_1$,   $c_1(M)=bc_1$ for some integers $a,b$ such that $a\geq b$.

From that exact sequence we get $c_1(F)=(a+b)c_1$, hence $a+b=1-r$, and $c_2(F)=\deg(Z)+ abc_1^2$, hence $\Delta _F=\allowbreak 4\deg(Z)-(a-b)^2c_1^2$. Comparing with the previous expression for $\Delta _F$ and using the Lemma  again we find
\[(a-b)^2c_1^2\geq -\Delta _F=(r+1)^2c_1^2-4c_2> (r^2+2r-3)c_1^2\geq (r^2-1)c_1^2\ ,\]
hence  $a-b\geq r$, and $a\geq 1$.

We have $H^0(E\otimes L^{-1})=H^0(E^*\otimes \det(E)\otimes L^{-1})\neq 0$. Since $E$ is globally generated, the natural map $E^*\rightarrow H^0(E)^*\otimes _{\C}\O_S$ is injective, hence
 $H^0(\det(E)\otimes L^{-1})\neq 0$. Since $c_1(L)=ac_1$ with $a\geq 1$, the only possibility is  $L\cong \det(E)$,   and therefore $H^0(E^*)\neq 0$. Using again that $E$ is globally generated, we obtain
 $E=\O_S\oplus \det(E)$.\qed

\begin{rem}
The condition $h^0(E)\geq 4$ is necessary: if $E$ is ample and globally generated, the rational map $\P(E)\rightarrow \P(H^0(E))$ associated to the linear system $\abs{\O_{\P(E)}(1)}$ is a finite morphism, hence $\dim \P(H^0(E))\geq 3$. On the other hand, the condition $N^1(S)=\Z\cdot c_1$ is quite restrictive, but it is not clear how it could be weakened. For instance, we will exhibit in Example 1 of \S4 a globally generated rank 2 vector bundle $E$ on $\P^2$ with $h^0(E)\geq 4$, $\det E=\O_{\P^2}(2)$, which is not ample.
\end{rem}

\section{Application 1: Lazarsfeld-Mukai bundles}

Let $C$ be an irreducible  curve in $S$,  $L$  a line bundle on $C$, and   $V$  a 2-dimensional subspace of $H^0(L)$ which generates $L$. The \emph{Lazarsfeld-Mukai bundle} $E_{C,V}$ is defined by the exact sequence
\[0\rightarrow E^*_{C,V}\rightarrow V\otimes_{\C} \O_S \rightarrow L\rightarrow 0\,.\]
Let $N_C:=\O_S(C)_{|C}$ be the normal of $C$ in $S$. By duality we get an exact sequence
\[0\rightarrow V^*\otimes_{\C} \O_S \rightarrow E_{C,V}\rightarrow N_C\otimes L^{-1}\rightarrow 0\,.\]
\begin{prop}
Assume   $H^1(S,\O_S)=0$,   $N^1(S)=\Z\cdot [C]$, and that the line bundle $N_C\otimes L^{-1}$ on $C$ is globally generated and nontrivial. Then $E_{C,V}$ is globally generated and ample.
\end{prop}

\pr We put $E:=E_{C,V}$. Since  $H^1(S,\O_S)=0$, we have a commutative diagram of exact sequences
\[\xymatrix{0\ar[r] & V^*\otimes _{\C}\O_S\ar[r]\ar@{=}[d] & H^0(S,E) \otimes_{\C}\O_S\ar[r]\ar[d] & H^0(C,N_C\otimes L^{-1})\otimes _{\C}\O_S\ar[r]\ar@{->>}[d] & 0\\
0\ar[r] & V^*\otimes _{\C}\O_S\ar[r]  & E \ar[r] & N_C\otimes L^{-1}\ar[r]& 0\,.
}\]This implies that $E$ is globally generated, with $h^0(E)=2+h^0(N_C\otimes L^{-1})\geq 4$. From the bottom exact sequence we get $c_1(E)=[C]$ and
 $c_2(E)= \deg(L) >0$. The conclusion follows from Proposition~\ref{fib}.\qed

\section{Application 2: congruences of lines}

\smallskip

Let $\G$ be the Grassmannian of lines in $\P^3$, which we view as a smooth quadric in $\P^5$; let $S\subset \G$ be a smooth surface. This defines a 2-dimensional family of lines in $\P^3$, classically called a \emph{congruence}. A point $p\in\P^3$ through which pass infinitely many lines of the congruence is called a \emph{fundamental point} (or, more classically, a singular point) of the congruence.

\begin{prop}
Assume that $S$ has degree $>1$ and that $N^1(S)$ is generated by the restriction of $\O_{\G}(1)$. Then $S$ has no fundamental point.
\end{prop}
\pr Let $E$ be the restriction to $S$ of the universal quotient bundle $Q$ on $\G$. The projective bundle $\P(E)$ on $S$ parametrizes pairs $(\ell,p)$ in $S\times \P^3$ with $p\in\ell$, and the second projection  $q:\P_S(E)\rightarrow \P^3$ satisfies $q^*\O_{\P^3}(1)=\O_{\P(E)}(1)$.
 Thus $q$ is finite (that is, $S$ has no fundamental point) if and only if $E$ is ample.

We have $h^0(Q)=4$, and a nonzero section of $Q$ vanishes along a linear plane; therefore $h^0(E)\geq 4$, and we can apply Proposition \ref{fib}. If $E=\O_S\oplus \O_S(1)$, we have $c_2(E)=0$, that is, $c_2(Q)\cdot [S]=0$; this can only happen if $S$ is a linear plane, which we have excluded. Therefore $E$ is ample. \qed

\begin{cor}
Let $d,e$ be two integers with  $d,e>1$, or $d=1$ and $e\geq 3$; let $S\subset \G$ be the complete intersection of two general hypersurfaces of degree $d$ and $e$. Then $S$ has no fundamental point.
\end{cor}
Indeed $\Pic(S)$ is generated by $\O_S(1)$ \cite[Th\'eor\`eme 1.2]{D}.

\medskip
\noindent\emph{Examples}$.-$ 1) Perhaps the simplest example of a nontrivial congruence is the surface $S$ of  lines bisecant to a twisted cubic $T\subset\P^3$; it is isomorphic to $\operatorname{Sym}^2T\cong \P^2 $, embedded in $\G\subset\P^5$ by the Veronese map. In that case $N^1=\Z\cdot [\O_{\P^2}(1)]$ but $\det E=\O_{\P^2}(2)$, and indeed the  fundamental locus of $S$ is $T$, so $E$ is not ample.

\smallskip

2) Let $A$ be an abelian surface such that $\operatorname{NS}(A)=\Z\cdot [L] $, where $L$ is a line bundle with $L^2=10$. The linear system $\abs{L}$ embeds $A$ into $\P^4$ \cite{R}, giving the famous Horrocks-Mumford abelian surface. The projection $\pi :\G\rightarrow \P^4$ from a general point of $\P^5$ is a double covering, and the surface   $S:=\pi ^{-1}(A)\subset\G$ is smooth. The line bundle $\pi ^*L$ is not divisible in $N^1(S)$: since
$(\pi ^*L)^2=20$, this could happen only if $\pi ^*L$ is divisible by 2; but $\pi ^*L=K_S$, so this would imply that $K_S^2$ is divisible by 8, a contradiction. It then follows from \cite{Bu} that $N^1(S)$ is generated by $\pi ^*L=\O_S(1)$, so Proposition 2 applies and $S$ has no fundamental point.

\bigskip
\section{Application 3 (virtual): surfaces with ample cotangent bundle}

\smallskip
The original motivation of this work was to obtain new examples of surfaces with ample cotangent bundle -- these surfaces have very interesting properties, but there are few concrete examples known. Applying Proposition \ref{fib} to $\Omega ^1_S$ we get the following result;  unfortunately we do not know any example of a surface satisfying the hypotheses
(help welcome!).
\begin{prop}
Assume that $\Omega ^1_S$ is globally generated (for instance that $S$ is a subvariety of an abelian variety), $q(S)\geq 4$, and $N^1(S)=\Z\cdot [K_S]$. Then $\Omega ^1_S$ is ample.
\end{prop}
\pr The hypotheses imply that $K_S$ is ample, hence $c_2(S)>0$; therefore $\Omega ^1_S$ is not isomorphic to $\O_S\oplus K_S$. The conclusion follows from
 Proposition \ref{fib}. \qed

 \bigskip

config: arxiv


TITLE: Calculus of Feynman diagrams
QUESTION [4 upvotes]: Feynman diagrams are wonderful tools to visualize particle interactions. They make deep connections/points of view in quantum mechanics intuitive (e. g. antiparticles = arrows against the direction of time) and give hints on how to decompose the integrals occuring by superposition. Often, one generates all possible Feynman diagrams to calculate probability amplitudes.
What I am missing is the ability to calculate with Feynman diagrams directly. In other words:

Take a complicated problem
Translate it into Feynman diagrams
Do some cool graphtheoretical stuff directly on the Feynman diagrams to obtain a more simple diagram
Retranslate them back into more simple/fewer formulae which otherwise would have been hard to derive.

Note that Feynman's rules "only" give assertions about how a diagram must look, not procedures on how to simplify it.
Is there such a Feynman calculus? Or is it somehow proven that such a calculus cannot exist? Perhaps, one could see commutative diagrams in Homological Algebra as having such a more direct calculus. (For example, the Snake lemma or results relating to the logical completeness of diagram chasing)
Disclaimer: Coming from mathematics, I read the Wikipedia article about Feynman diagrams some days ago and was instantly fascinated. Therefore, my physics background may not be the strongest one...

REPLY [0 votes]: Yes, there is such a thing, and Feynman diagrams were not invented by Feynman, but by mathematicians in the 19-th century, starting with Arthur Cayley. See, for example, my MO answer:
https://mathoverflow.net/questions/168888/who-invented-diagrammatic-algebra/260016#260016
A Feynman diagram is just an extremely compact yet precise notation for a contraction of tensors, say something like
$$
G_{i,r}=\sum_{j,\ell, p,q=1}^N C_{i,j}V_{j,\ell, p, q}C_{\ell, p}C_{q ,r}\ .
$$
If $C$ is a lattice propagator in position space and $V_{j,\ell,p,q}=\delta_{j,\ell}\delta_{j,p}\delta_{j,q}$, the above is just the tadpole diagram in the two-point function of the $\phi^4$ QFT. In the physics context, one wants to take $N\rightarrow\infty$ so summation over indices become integrals over continuous variables of position or momentum.
One can certainly do computation entirely in terms of diagrams, especially in the context of representation theory and invariant theory. This is the spirit of the book by Cvitanović, for instance.
You can also see examples of diagrammatic computations related to Pascal's Theorem in my article with Chipalkatti
"On the Reconstruction Problem for Pascal Lines",
Discrete & Computational Geometry volume 60, pages 381–405 (2018). Preprint version is here.


TITLE: Can black holes evaporate into Neutron stars?
QUESTION [5 upvotes]: If adding mass to a neutron star eventually makes it a black hole, why do black holes after losing mass through hawking radiation not evaporate into Neutron stars?

REPLY [2 votes]: Entropy argument:
A black hole has the maximum possible entropy for a given volume (or, by extension, mass - see here https://en.wikipedia.org/wiki/Black_hole_thermodynamics )
This holds true for any size of a black hole, including the mass interval where the neutron stars are stable.
In order to decay into a neutron star, a black hole has to dispose off some entropy and it has no imaginable means to do so.


TITLE: Brave New Number Theory
QUESTION [24 upvotes]: I suppose this is an extremely general question, so I apologize, and perhaps it should be deleted.  On the other hand it's an awesome question.
Is it clear exactly how much (assumedly algebraic) number theory can written down diagrammatically, and if so, has there been any effort to write such problems in the category of spectra (whichever category you like) and solve problems there?  It seems that some problems may become easier to solve, if only because there are in some sense "more" spectra to work with than there are regular algebraic objects (i.e. we have the Eilenberg-MacLane spectrum for whatever ring or field of whatever, but we also have things that don't come from any algebraic object).
I have heard about Rognes' work on Galois extensions in this sense, and that there are lots of connections to things like Morava K-theory (and the associated spectra), and plan on at least attempting to pursue such ideas.

REPLY [12 votes]: This is probably not the answer you are looking for but, since there has been no other answers, it may be of some use. Recently, I have enjoyed finding out how wonderful and mysterious the relationship between (algebraic) number theory and low-dimensional topology is through the links between Galois groups and homotopy/fundamental groups. In particular the framing of properties of prime numbers in class field theory and arithmetic geometry having duality correlations with knots, links and three-dimensional manifolds! This subject of Arithmetic Topology was hinted time ago by several people, like D. Mumford and B. Mazur, which has very recently got its first MARVELOUS book:

Morishita, M. - Knots and Primes: An Introduction to Arithmetic Topology, Springer 2011.

Some of the first explorations on this were:

Mazur, B. - Remarks on the Alexander Polynomial,
Mazur, B. - Notes on Étale Cohomology of Number Fields,
Morishita, M. - Analogies between Knots and Primes, 3-Manifolds and Number Rings,
Kohno T.; Morishita M. (eds.) - Primes and Knots, AMS 2006.

The whole point is the study of the fundamental groups of topology but from a number-theoretic perspective, that is, working with the (étale) algebraic fundamental group of varieties and schemes as originally thought by A. Grothendieck in his anabelian geometry. For this and as a prerequisite to Morishita's book, along with standard arithmetic geometry, one should get the specific background on the relationship of Galois gropus and fundamental groups of schemes, provided by:

Szamuely, T. - Galois Groups and Fundamental Groups, CUP 2009.
Schneps, L. (ed.) - Galois Groups and Fundamental Groups, CUP 2011.

These new dualities between apparently so different worlds may open the door to a deep understanding of the mysterious arithmetic-topologic relationship.


TITLE: The composition of a convex function with an integrable function is integrable
QUESTION [5 upvotes]: If $f$ is (Riemann) integrable on $[0,1]$ and $\varphi$ is convex on $\mathbb{R}$, are there any simple or elegant proofs that $\varphi\circ f$ is also integrable on $[0,1]$?
This paper by J. Lu answers the question (and other similar questions) in a bit more generality--basically whenever $\varphi$ is merely continuous rather than convex--but I'm wondering if there's a simpler method here.

REPLY [5 votes]: This is not much of answer, but I just want to state that in my opinion the simplest proof of the result is the one the OP has already alluded to:

Every convex function $\varphi$ is continuous.
If $f: [a,b] \rightarrow \mathbb{R}$ is Riemann integrable and $\varphi: \mathbb{R} \rightarrow \mathbb{R}$ is continuous, then $\varphi \circ f: [a,b] \rightarrow \mathbb{R}$ is Riemann integrable.

The proof of 1) can be found in most "honors calculus" texts (e.g. Spivak).  If anyone wants a link to a proof, please let me know.  The proof of 2) given in the (nicely written, very elementary, expository) article by Jitan Lu cited by the OP works from the "Riemann notion of Riemann integrability", i.e., convergence of the Riemann sums uniformly in the mesh of the partition.  A lot of discussions of the Riemann integral (e.g. Spivak, Rudin) introduce the "Darboux notion of Riemann integrability" -- i.e., the one with the upper integral and the lower integral -- and with this notion the proof is shorter: closer to half a page than a full page.  See for instance the proof given on page 7 of these notes (Wayback Machine), which is taken almost verbatim from Russell Gordon's analysis text.
I suppose it would be of some interest if there were an even shorter proof that took advantage of the convexity, but in practice if you are at all interested in theorems like this you are going to want to know that $\varphi \circ f$ is Riemann integrable even for not necessarily convex continuous $\varphi$.

REPLY [2 votes]: Jensen's theorem says
$$
\varphi\left(\int_If(x)\;\mathrm{d}x\right)\le\int_I\varphi(f(x))\;\mathrm{d}x
$$
where the measure of $I=1$. However, what you are asking is in the opposite direction and is not true. Consider the functions
$$
\begin{array}{}f(x)=x^{-\frac{1}{2}}&\text{and}&\varphi(x)=x^2\end{array}
$$
$f$ is integrable on $[0,1]$ and $\varphi$ is convex on $\mathbb{R}$, but $\varphi\circ f(x)=x^{-1}$ is not integrable on $[0,1]$.
On the other hand, if $\varphi$ were concave and non-negative, then Jensen's inequality insures that $\varphi\circ f$ is integrable.
Addendumb: As Pete Clark points out, and as has been shown here, the question must pertain only to proper Riemann integrals not converging improper integrals.


text:
\begin{definition}[Definition:Order Embedding/Definition 2]
$\phi$ is an '''[[Definition:Order Embedding|order embedding]] of $S$ into $T$''' {{iff}} both of the following conditions hold:
:$(1): \quad \phi$ is an [[Definition:Injection|injection]]
:$(2): \quad \forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
\end{definition}

config: wiki


TITLE: $\lim_{x \to + \infty} \left \{ x-x^2 \ln (1+ \sin \frac{1}{x}) \right \}$
QUESTION [0 upvotes]: I've tried to solve this limit:
$\lim_{x \to + \infty} \left \{ x-x^2 \ln (1+ \sin \frac{1}{x}) \right \} $
$x-x^2 \ln (1+ \sin \frac{1}{x}) \sim x-x^2 \sin \frac{1}{x} \sim x-x = 0 $
but the final result should be $\frac{1}{2}$

REPLY [0 votes]: In the following proof, I won't be using L'hopital or series expansion.
Let $ x\in\mathbb{R}^{*} $, we have :
\begin{aligned}x-x^{2}\ln{\left(1+\sin{\left(\frac{1}{x}\right)}\right)}=\frac{\frac{1}{x}-\sin{\left(\frac{1}{x}\right)}}{\frac{1}{x^{2}}}+\left(x\sin{\left(\frac{1}{x}\right)}\right)^{2}\left(\frac{\sin{\left(\frac{1}{x}\right)}-\ln{\left(1+\sin{\left(\frac{1}{x}\right)}\right)}}{\sin^{2}{\left(\frac{1}{x}\right)}}\right)\end{aligned}
We ended with well known limits, that I'm sure have been computed many times on this site.
$ \textbf{Let's proof first that : $ \displaystyle\lim_{x\to 0}{\frac{x-\sin{x}}{x^{2}}}=0$ : } $
Note that for any $ x\in\mathbb{R}^{*} $, we have : $$ \frac{x-\sin{x}}{x^{2}}=\int_{0}^{1}{\left(1-y\right)\sin{\left(xy\right)}\,\mathrm{d}y} $$
Thus : \begin{aligned} \left|\frac{x-\sin{x}}{x^{2}}\right|&\leq\int_{0}^{1}{\left(1-y\right)\left|\sin{\left(xy\right)}\right|\mathrm{d}y}\\ &\leq\int_{0}^{1}{\left(1-y\right)\left|xy\right|\mathrm{d}y}\\ \left|\frac{x-\sin{x}}{x^{2}}\right|&\leq\frac{\left|x\right|}{6} \end{aligned}
Hence : $$ \lim_{x\to 0}{\frac{x-\sin{x}}{x^{2}}}=0 $$
$\textbf{Let's proof then that : $ \displaystyle\lim_{x\to 0}{\frac{x-\ln{\left(1+x\right)}}{x^{2}}}=\frac{1}{2} $ : }$
Note that for any $ x\in\left(-1,1\right)\setminus\left\lbrace 0\right\rbrace $, we have : $$ \frac{x-\ln{\left(1+x\right)}}{x^{2}}=\int_{0}^{1}{\frac{1-y}{\left(1+xy\right)^{2}}\,\mathrm{d}y}=\frac{1}{2}-x\int_{0}^{1}{\frac{y\left(2+xy\right)}{\left(1+xy\right)^{2}}\,\mathrm{d}y} $$
Since : \begin{aligned} \left|x\int_{0}^{1}{\frac{y\left(2+xy\right)}{\left(1+xy\right)^{2}}\,\mathrm{d}y}\right|&\leq\left|x\right|\int_{0}^{1}{\frac{y\left(2+\left|x\right|y\right)}{\left(1-\left|x\right|y\right)^{2}}\,\mathrm{d}y}\\ &\leq\frac{\left|x\right|\left(2+\left|x\right|\right)}{2\left(1-\left|x\right|\right)^{2}}\underset{x\to 0}{\longrightarrow}0 \end{aligned}
we get that : $$ \lim_{x\to 0}{\frac{x-\ln{\left(1+x\right)}}{x^{2}}}=\frac{1}{2} $$
$\textbf{Now let's apply this to our expression : }$
\begin{aligned}\lim_{x\to +\infty}{\left(x-x^{2}\ln{\left(1+\sin{\left(\frac{1}{x}\right)}\right)}\right)}&=\lim_{x\to +\infty}{\left(\frac{\frac{1}{x}-\sin{\left(\frac{1}{x}\right)}}{\frac{1}{x^{2}}}+\left(x\sin{\left(\frac{1}{x}\right)}\right)^{2}\left(\frac{\sin{\left(\frac{1}{x}\right)}-\ln{\left(1+\sin{\left(\frac{1}{x}\right)}\right)}}{\sin^{2}{\left(\frac{1}{x}\right)}}\right)\right)}\\ &=0 + 1^{2}\times\frac{1}{2}\\ &=\frac{1}{2}\end{aligned}


TITLE: what is a basis for $\ell^p$?
QUESTION [1 upvotes]: For the familiar vector space $\mathbb{R}^n$ we have the following standard orthonormal basis: $\{e_1, e_2, \ldots, e_n\}$.  Where $e_1 = (1, 0, \ldots, 0)$ and so on...
For $\ell^p$ would we have $\{x_1 = (1, 0, 0, \ldots), x_2 = (0, 1, 0, \ldots), \ldots, x_n = (0,\ldots,0,1,0,\ldots), \ldots\}$?
Would this be orthonormal as well?

REPLY [0 votes]: It is no basis, because every basis of an infinite-dimensional Banach space is uncountable as a consequence of the Baire category Theorem, See https://en.wikipedia.org/wiki/Baire_category_theorem


TITLE: How to find the value of the integral $\int_{-N}^{-2}\frac{1}{1+x}\text{d}x$
QUESTION [0 upvotes]: The below integral:
$$\int_{-N}^{-2}\frac{1}{1+x}\text{d}x$$
where $N \in \Bbb{Z}$ and $N>2$, represent the area surrounded by:$x=-N,x=-2,y=0,y=\frac{1}{1+x}$. how to find it?

REPLY [1 votes]: The antiderivative of $\frac{1}{1+x}$ is just $\ln(|1+x|)$ and so the integral is
$$
  \int_{-N}^{-2} \frac{1}{1+x} dx
= \left. \ln(|1+x|) \right|^{-2}_{-N}
= \ln(N-1)
$$

REPLY [1 votes]: Let $u = 1 + x$; then $du = dx$.
Then we have $$\int_{-N}^{-2} \frac{1}{1+x}dx = \int_{-N+1}^{-1} \frac{1}{u}du = (\ln|u|)|^{-1}_{-N+1} = (\ln(1) - \ln(N-1)) = -\ln(N-1)$$


TITLE: Is the sequence of functions $f_n=\chi_{[n,n+1]}$ uniformly integrable?
QUESTION [3 upvotes]: I wish to prove or disprove that the sequence of functions $f_n=\chi_{[n,n+1]}$ is uniformly integrable?
At a glance my judgement is YES, it is uniformly integrable.
From the definition of Uniform integrability, that's

A sequence ${f_n}$ is called uniformly integrable if $\forall \epsilon >0 \exists \delta > 0 $ such that if $E \subset X$, $E$ measurable and  $\mu (E)< \delta $ then $\forall n$ $\int_E |f_n| d\mu < \epsilon$.

So I let $E \subset R$ such that $\mu (E)<\delta$
then $\int_E|f_n|=\int|f_n|\chi_E \leq \mu (E)<\delta$.
So in this case $\epsilon =\delta$.
Does this make sense?

REPLY [2 votes]: The approach is correct with this definition of uniform integrability.
A concern about this one is that a uniform integrable family with this definition may not contain an integrable function. For example, the family consisting of the constant function equal to $1$ would be uniformly integrable.
A common definition is that a family $\mathcal F$ of function on a measure space $\left(X,\mathcal A,\mu\right)$ is uniformly integrable if for each positive $\varepsilon$, there exists an integrable function $g$ such that
$$
\sup_{f\in\mathcal F}\int_{\left\{\left\lvert f\right\rvert\gt g\right\}}\left\lvert f\right\rvert\mathrm d\lambda\lt \varepsilon.
$$
With this definition, the family $(f_n)$ defined by $f_n=\chi_{[n,n+1]}$ is not uniformly integrable. Indeed, let $\varepsilon=1/2$. If a function $g$ is such that $$\sup_{n\geqslant 1}\int_{\left\{\left\lvert f_n\right\rvert\gt g\right\}} f_n\mathrm d\lambda\lt 1/2$$
then $\lambda\left(\{g\lt 1\}\cap [n,n+1]\right)\lt 1/2$ hence $\lambda\left(\{g\geqslant 1\}\cap [n,n+1]\right)\geqslant 1/2$ for all $n$ hence $g$ cannot be integrable.


TITLE: Evaluate $ \int_0^3 \frac{x^3}{1-x^4}\, dx. $
QUESTION [4 upvotes]: Evaluate
$$
\int_0^3 \frac{x^3}{1-x^4}\, dx.
$$
I evaluated the integral and got $\left[\dfrac{-\ln(1-x^4)}{4}\right]_0^3$ which ended up diverging. Any help is appreciated!

REPLY [0 votes]: The integral gives up easily to direct attack, but with a little twist.
Use the obvious substitution $u=x^4$:
$$\int_0^3 \frac{x^3}{1-x^4}\, dx=\frac{1}{4} \int_0^{81} \frac{du}{1-u}=$$
Now it makes sense to separate the integral into two parts, so the denominator is positive for each part:
$$=\frac{1}{4} \int_0^1 \frac{du}{1-u}-\frac{1}{4} \int_1^{81} \frac{du}{u-1}=$$
Now in the first integral we use the substitution $t=1-u$ and in the second $p=u-1$:
$$=\frac{1}{4} \int_0^1 \frac{dt}{t}-\frac{1}{4} \int_0^{80} \frac{dp}{p}=\frac{1}{4} \int_0^1 \frac{dt}{t}-\frac{1}{4} \int_0^1 \frac{dp}{p}-\frac{1}{4} \int_1^{80} \frac{dp}{p}=$$
But the first two divergent integrals just cancel each other out (they are equal), so we are allowed to write:
$$\int_0^3 \frac{x^3}{1-x^4}\, dx=-\frac{1}{4} \int_1^{80} \frac{dp}{p}=-\frac{1}{4} \ln 80=-\frac{1}{4} (\ln 5+4 \ln 2)$$
This is the correct answer (formally). Although Wolfram Alpha writes that the original integral diverges, and the answer we obtained is just Cauchy principal value of the integral.

Edit. A little justification for the calcelling the divergent integrals. The problem is at $t \to 0$, thus:
$$\int_0^1 \frac{dt}{t}=\lim_{\epsilon \to 0} \int_\epsilon^1 \frac{dt}{t}$$
In this case the limit doesn't exist.
But for any positive real number $\epsilon>0$ we have formally:
$$\int_\epsilon^1 \frac{dt}{t} \equiv \int_\epsilon^1 \frac{dp}{p}$$
We just renamed the variable after all. Moreover, in our particular example we have $t=-p$. So setting:
$$\lim_{\epsilon \to 0} \int_\epsilon^1 \frac{dt}{t}-\lim_{\epsilon \to 0} \int_\epsilon^1 \frac{dp}{p}=\lim_{\epsilon \to 0} \left( \int_\epsilon^1 \frac{dt}{t}-\int_\epsilon^1 \frac{dp}{p} \right)=0$$
Makes perfect sense.
But the rigorous justification should involve contour integration in the complex plane.


TITLE: Munkres topology excercise show that the quotient space of a disk is homeomorphic to the unit sphere
QUESTION [3 upvotes]: Find a homeomorphism between a quotient space of disk and sphere. I was looking at this example from munkres quotient topology section:Let $X$ be the closed unit ball $\{(x,y)|x^2+y^2\leq1\}$ in $\mathbb{R}^2$ and let $X*$ be the partition of $X$ consisting of one point sets $\{(x,y)\}$ for which $x^2+y^2<1$ along with the set $S^1=\{(x,y)|x^2+y^2=1\}$. One can show that $X*$ is homeomorphic with the unit two sphere $S^2=\{(x,y,z)|x^2+y^2+z^2=1\}$, a subspace of $\mathbb{R}^3$.
Here was my attempt at this problem Let $X*=\{\bigcup\{(x,y)|x^2+y^2=1\} \} \cup \{\{(x,y)\}|x^2+y^2<1\}$
My question is would this be a valid homeomorphism between $X*$ and $S^2$?
Define the function $f:X* \rightarrow S^2$ given by $f(\{(x,y)\})=(\cos2\pi x \sin \pi y, \sin 2\pi x \sin  \pi y, \cos \pi y)$ where $0<x \leq 1$ and $0<y \leq 1$ and $f(\bigcup\{(x,y)|x^2+y^2=1\})=(0,0,1)$.
$f$ is injective since $f(\{(x,y)\}=f(\{w,z\}) \implies (\cos2\pi x \sin \pi y, \sin 2\pi x \sin  \pi y, \cos \pi y)=\cos2\pi w \sin \pi z, \sin 2\pi w \sin  \pi z, \cos \pi z)\implies x=w,y=z \implies \{x,y)\}=\{(w,z)\}$.
$f$ is surjective since for any $(\cos2\pi x \sin \pi y, \sin 2\pi x \sin  \pi y, \cos \pi y)\in S^2$ if $x=1,y=0$ then $f(\bigcup\{(x,y)|x^2+y^2=1\})=(0,0,1)$, otherwise $f(\{(x,y)\})=(\cos2\pi x \sin \pi y, \sin 2\pi x \sin  \pi y, \cos \pi y)$.
$X*$ is compact since it is the image of $S^1$ under the quotient map. $S^2$ is Hausdorff as a subspace of $\mathbb{R}^3$. Further more $f$ is continuous since if $g$ denotes the quotient map, $f \circ g:S^1 \rightarrow S^2$ is continuous since each component function in $(\cos2\pi x \sin \pi y, \sin 2\pi x \sin  \pi y, \cos \pi y)$ of $(0,0,1)$ is continuous. Thus $f$ is a homeomorphism being a continuous bijection from a compact space to a Hausdorff space.
I am fairly confident that this is correct but is my restriction of the function $f$ mapping the equivalence class $\{(x,y)\}$ to $(\cos2\pi x \sin \pi y, \sin 2\pi x \sin  \pi y, \cos \pi y)$ exactly correct, by restricting $x,y$ values to $0<x \leq 1$ and $0<y \leq 1$?

REPLY [2 votes]: Your definition does not work. You map $S^1$ to the north pole $N = (0,0,1)$ and for all other points $\{(x,y)\}$ of $X^*$ you define $f(\{(x,y)\})=(\cos2\pi x \sin \pi y, \sin 2\pi x \sin  \pi y, \cos \pi y)$.
These "other points" are characterized by
$$x^2 + y^2 < 1 \tag{1}$$
and not by $0 < x,y \le 1$. In fact $(1)$ allows that both $x,y$ can take the value $0$, but none of them can take the value $1$.
Anyway, $f$ is not injective. You have $f(\{(x,0)\}) = (0,0,-1)$ for $-1 < x < 1$.
Finally, whatever the definition of $f : X^* \to S^2$ looks like, you must verify the continuity with respect to the quotient topology on $X^*$. Your definition does not produce a continuous map. If it were, then $F = fq : D^2 \to S^2$ would be continuous, where $q : D^2 \to X^*$ is the quotient map But $\lim_{n \to \infty} F(1-\frac{1}{n},0) = (0,0,-1) \ne (0,0,1) = F(1,0)$.
So what can be done? Let us construct a map $F: D^2 \to S^2$ as follows:

Map $(0,0)$ to the south pole $S = (0,0,-1)$ and $S^1$ to $N$.

Map each circle $S^1_t = \{(x,y) \in D^2 \mid  \lVert (x,y) \rVert ^2 = x^2 + y^2 = t \}$ with $0 < t  < 1$ to the intersection circle $C(t)$ of $S^2$ with the plane $P(t) = \{(x,y,z) \mid z = 2t -1\}$ which has radius $r(t) = \sqrt{1 - (2t-1)^2}$.
That is, define
$$F(\xi)  = \begin{cases}\left(\frac{\sqrt{1 - (2\lVert \xi \rVert-1)^2}}{\lVert \xi \rVert} \xi,2\lVert \xi \rVert - 1\right) & \lVert \xi \rVert > 0 \\ S & \xi = (0,0) \end{cases}$$
Note that $F(\xi) = N$ for $\lVert \xi \rVert = 1$. The map $F$ is obviously continuous on $D^2 \setminus \{(0,0)\}$. For $\xi \ne (0,0)$ we have
$$\lVert F(\xi) - F(0,0) \rVert = \left\lVert \left(\frac{\sqrt{1 - (2\lVert \xi \rVert-1)^2}}{\lVert \xi \rVert} \xi,2\lVert \xi \rVert \right) \
\right\rVert = 4 \lVert \xi \rVert$$
which proves continuity in $(0,0)$.

Moreover $F$ is surjective. Clearly $S, N$ are in the image of $F$. Now let $(\eta,\tau) \in S^2 \setminus \{S,N\}$, where $\eta = (a,b)$ with $a,b \in \mathbb R$ and $-1 < \tau < 1$. Note that $\lVert \eta \rVert > 0$ and $\lVert \eta \rVert^2 + \tau^2 = 1$. Define $\xi =  \frac{1 +\tau}{2\sqrt{1 - \tau^2}}\eta$. Then $\lVert \xi \rVert =\frac{1+\tau}{2}$ ( so that $0  < \lVert \xi \rVert  < 1$) and we have $F(\xi) = (\eta,\tau)$.
The fibers $F^{-1}(\eta,\tau)$ are singletons for $(\eta,\tau) \ne N$ and $F^{-1}(N) = S^1$.
Thus $F$ gives us precisely your partition $X^*$.


TITLE: Why conjugate distributions can be identified up to a constant?
QUESTION [1 upvotes]: A family $\mathcal{F}$ of densities distributions and a family
  $\mathcal{P}$ of prior distributions are said conjugates if $\forall f
 \in \mathcal{F}, \pi(\theta) \in \mathcal{P},$ $\pi(\theta\mid x) \in
 \mathcal{P}$

In practice, to determine if two families are conjugates books just check if $f \pi \propto \Pi$ with $\Pi \in \mathcal{P}$. This happens for instance for the gamma and poisson distributions.
Why is it enough to check it up to a constant?
Edit
Perhaps, I should have given the whole explicit example beofre.
If $X_i \sim \mathcal{P}(\theta)$ one can show that
$$h(x,\theta) = f(x|\theta) \pi(\theta) = \frac{e^{-\theta (n+1/ \beta)}\theta^{(n\overline{x}+\alpha -1)}I_{(0,\infty)}(\theta)}{\Gamma(\alpha)\beta^\alpha \prod x_i!}$$ which is like a $Gamma(n\overline{x}+\alpha,(n+1/\beta)^{-1})$, up to a constant because $\prod x_i!$ does not appear in the expression of the gamma.
Then I think that one can say that $m(x) = h(x,\theta)/\pi(\theta|x)$ so that the marginal $m(x)$ compensates the assumption that $\pi(\theta|x)$ is a gamma distribution. However, is this not limiting on the form of the gamma?
Solution
This is more silly than I thought. The situation (for instance for the example above) is that I get $f(x|\theta)\pi(\theta) = B Gamma$. Then I ask why the marginal normalizes this and the answer is straight forward, because $\int f(x|\theta)\pi(\theta)  = \int B Gamma = B \int Gamma = B$

REPLY [0 votes]: Every density or probability function can be determined up to a constant, since that constant is in fact uniquely determined by the conditions
$$\int_{-\infty}^{\infty} f(x) \, dx=1,$$
for density functions and
$$\sum_{x\in R_X} p(x) =1,$$
for probability functions ($R_X$ being the range of the discrete r.v. $X$).
For instance, if I know that $X$ is a continuous r.v. with support in $[0,1]$ and its density is proportional to $x^2$ over the support, then we have
$$f_X(x)=kx^2 I_{[0,1]}(x),$$
but as its integral over the reals has to equal one, that is
$$\int_{-\infty}^{\infty} f_X(x) \, dx=k\int_0^1 x^2 \, dx=\frac k3=1$$
the only option is $k=3$.
So in particular when you multiply two densities or probabilities to obtain another one of them, as in $$f_{\Theta|\vec{X}=\vec{x}}(\theta)\ltimes f_{\vec{X}|\Theta=\theta}(\vec{x})\cdot \pi(\theta)$$
(whether you have conjugates or not) you can use any multiple of the prior and the likelihood (thinking as constant everything but $\theta$), and you will obtain a multiple of the posterior density (this happens even if you use the real $f_{\vec{X}}$ and the real $\pi$), so the posterior (the density or probability) will be determined up to a constant, as happens with any other density or probability function.
Two more comments:

Sometimes the formula shows clearly the distribution even if a constant is undetermined. For instance, if $f_Z(z)\ltimes z^3 e^{-2z}$ for $z>0$ and $0$ elsewhere, it is clear that $Z\sim \Gamma(\alpha=4,\lambda=2)$ (at least if you ar familiar enough with the Gamma distribution).
Some important algorithms for simulations used in estimation and other procedures need as input only a multiple of certain involved distributions, and not the exact formula; this happens for instance if a density $f$ only appears in the algorithm in an expression as $\frac {f(x_1)}{f(x_0)}$, in which case any additional constant in $f(x_1)$ and $f(x_0)$ can be cancelled out.


TITLE: If $f:\omega_1 \to \mathbb{R}$ is continuous, then there is $a \in \omega_1$ such that if $x\geq a$, then $f(x) = f(a)$.
QUESTION [0 upvotes]: Show that if $f:\omega_1 \to \mathbb{R}$ is continuous, then there is $a \in \omega_1$ such that if $x\geq a$, then $f(x) = f(a)$.
I'm stuck on this question for a while, although it seems to me that the proof will be related to the fact that $f$ is bounded (I proved this as folows: $f$ is continuous and $w_1$ has the least element of the set of its limit point, say $a$, then $f(a)$ should be the upper bound of $f:\omega_1 \rightarrow \mathbb{R}$. Is this correct?). Can anyone please help on this problem?

REPLY [0 votes]: Say that a closed interval $[a, b]$ is recurrent if for cofinally many $\alpha\in\omega_1$, we have $f(\alpha)\in (a, b)$. Then we can show a few things:

If $[a, b]$ is recurrent, then $f$ eventually stays in $[a, b]$: for some $\alpha\in\omega_1$ we have $\forall  \beta>\alpha, f(\beta)\in [a, b]$
For any $\epsilon>0$, there is some closed interval of size $\epsilon$ which is recurrent.

Once we prove these two facts, we then (roughly) can argue as follows:

Get a sequence $[a_1, b_1], [a_2, b_2], [a_3, b_3], . . . $ of recurrent intervals, $[a_n, b_n]$ of size $2^{-n}$; along with them, get ordinals $\alpha_1, \alpha_2, \alpha_3, . . . $ such that $f(\beta)\in [a_n, b_n]$ for all $\beta>\alpha_n$.
Now take the intersection of all those closed intervals - this is some real number, $\{r\}$. Meanwhile, the $\alpha_i$s are bounded . . .


TITLE: Telescoping series: $\sum\limits_{n=1}^{∞}[\tan^{-1}(2n+1)-\tan^{-1}(2n-1)]$
QUESTION [1 upvotes]: In this question that was asked today the OP wrote that
$$\begin{align}\sum\limits_{n=1}^{∞}[\arctan(2n+1)-\arctan(2n-1)]&=\arctan\infty-\arctan 1\\
&=\frac{\pi}{2}-\frac{\pi}{4}\\&=\frac{\pi}{4}
\end{align}$$
I don't really understand why there is a $\arctan\infty$ term. Surely every single term other than $-\arctan 1$ has been cancelled out? Now I know that I am wrong, as obviously the value of the summation cannot be negative, but I'm not sure where I am wrong.
I have thought of considering the similar finite series, $$\sum\limits_{n=1}^k[\arctan(2n+1)-\arctan(2n-1)]=\arctan(2k+1)-\arctan1$$
and letting $k\to\infty$, so that
$$\lim_{k\to\infty}\sum\limits_{n=1}^k[\arctan(2n+1)-\arctan(2n-1)]=\arctan\infty-\arctan1$$ as required, but I still can't see why all the terms other than $-\arctan1$ don't cancel out, as the upper limit actually is $\infty$.
Thank you for your help.

REPLY [3 votes]: A few intermediate steps might be helpful.

We obtain
\begin{align*}
\color{blue}{\sum_{n=1}^{\infty}}&\color{blue}{\left[\arctan(2n+1)-\arctan(2n-1)\right]}\\
&=\lim_{N\to\infty}{\sum_{n=1}^{N}\left[\arctan(2n+1)-\arctan(2n-1)\right]}\\
&=\lim_{N\to\infty}\left[\arctan(2N+1)-\arctan(1)\right]\tag{1}\\
&=\lim_{N\to\infty}\arctan(2N+1)-\lim_{N\to\infty}\arctan(1)\tag{2}\\
&=\frac{\pi}{2}-\frac{\pi}{4}\\
&\,\,\color{blue}{=\frac{\pi}{4}}\\
\end{align*}

Comment:

In (1) we apply telescoping.

In (2) we use $\lim_{N\to\infty}\arctan(N)=\frac{\pi}{2}$ and $\lim_{N\to\infty} a=a$.


TITLE: Find locus of $\Delta ABC$ centroid with orthocentre at origin and side slopes 2, 3 and 5
QUESTION [2 upvotes]: Let $ABC$ be a triangle with slopes of the sides $AB$, $BC$, $CA$ are $2,3,5$ respectively. Given origin is the orthocentre of the triangle $ABC$. Then find the locus of the centroid of the triangle $ABC$.
Since sides have slopes $2,3,5$ then altitudes must have slopes $\frac{-1}{2}, \frac{-1}{3}$ and $\frac{-1}{5}$ respectively. Then equations of the altitudes are respectively  \begin{align*} & 2y+x=0 \\ & 3y+x=0 \\ & 5y+x=0. \end{align*} If $(\alpha, \beta)$ be the coordinates of the centroid, how can I find the locus of the point $(\alpha, \beta)$ from here?

REPLY [1 votes]: Respectively, the vertexes A, B and C are on the three altitude lines you obtained $3y+x=0$, $5y+x=0$ and $2y+x=0$. So, let their coordinates be $A(a,-\frac a3)$, $B(b,-\frac b5)$ and $C(c,-\frac c2)$. Then, use them to match the three side slopes
$$\frac{-\frac b5 + \frac a3 }{b-a}=2,\>\>\>\>\>\>\>
\frac{-\frac b5 + \frac c2 }{b-c}=3,\>\>\>\>\>\>\>
\frac{-\frac a3 + \frac c2 }{a-c}=5$$
which lead to the ratio $a:b:c = 33:35:32$ and the corresponding vertexes in terms of a single parameter $t$
$$A(33t,-\frac {33t}3),\>\>\>\>\>B(35t,-\frac {35t}5),\>\>\>\>\>C(32t,-\frac {32t}2)$$
Then, the coordinates of the centroid are
$$x = \frac{A_x+B_x+C_x}3=\frac{100t}3,\>\>\>\>\>y = \frac{A_y+B_y+C_y}3=-\frac{34t}3$$
Eliminate $t$ to obtain its locus
$$y=-\frac{17}{50}x$$


TITLE: Operators in Hilbert space ( orthogonal projection)
QUESTION [2 upvotes]: Some one help me to solve this problem i put some hints
Show that every projection $P \in \mathcal{B}(H)$ is and extreme point in the convex set
$$B_+ = \{T \in \mathcal{B}(H) : T \geq 0, \Vert T\Vert  \leq 1\}$$
Hints:
If $P = \lambda S + ( 1 – \lambda) T$ with $S$ and $T$ in $B_+$ , then $x = Sx = Tx$ for every $x \in P(H)$. Furthermore $\langle Ty,y\rangle = 0$ for every $y \in  P(H)^\perp$ and because of the positivity and the Cauchy Schwarz inequality this implies that $Sy  = Ty = 0$.

REPLY [1 votes]: Since $S,T\in B_+$, then $\Vert S\Vert\leq 1$ and $\Vert T\Vert\leq 1$. Hence
$$
\Vert Sx\Vert\leq \Vert x\Vert\qquad \Vert Tx\Vert\leq \Vert x\Vert$$
for all $x\in H$. Assume there exist $x_0\in P(H)$ such that $Sx_0\neq Tx_0$, then
$$
|\langle Sx_0,Tx_0\rangle|<\Vert Sx_0\Vert\Vert Tx_0\Vert\leq \Vert x\Vert^2
$$
Since $x_0\in P(H)$, then $x_0=Px_0$ so
$$
\begin{align}
\Vert x_0\Vert^2&=\Vert Px_0\Vert^2=\Vert\lambda Sx_0+(1-\lambda) Tx_0\Vert^2\\
&=\lambda^2\Vert Sx_0\Vert^2+2\lambda(1-\lambda)\mathrm{Re}\langle Sx_0,Tx_0\rangle+(1-\lambda)^2\Vert Tx_0\Vert^2\\
&\leq\lambda^2\Vert x_0\Vert^2+2\lambda(1-\lambda)|\langle Sx_0,Tx_0\rangle|+(1-\lambda)^2\Vert x_0\Vert^2\\
&<\lambda^2\Vert x_0\Vert^2+2\lambda(1-\lambda)\Vert x_0\Vert^2+(1-\lambda)^2\Vert x_0\Vert^2\\
&=\Vert x_0\Vert^2
\end{align}
$$
Thus $\Vert x_0\Vert^2<\Vert x_0\Vert^2$, contradiction. Therefore, for all $x\in P(H)$ we have
$$
Sx=Tx\\
x=Px=\lambda Sx+(1-\lambda)Tx=\lambda Sx+(1-\lambda)Sx=Sx=Tx
$$
Since $T\in B_+$ we have $T\geq 0$ and in particular for all $y\in P(H)^\perp$ holds $\langle Ty,y\rangle\geq 0$. Assume there exist $y_0\in P(H)^\perp$ such that $\langle Ty_0,y_0\rangle>0$. Since $y_0\in P(H)^\perp$ we have
$$
0=Py_0=\lambda Sy_0+(1-\lambda)T y_0
$$
hence $S y_0=-\lambda^{-1}(1-\lambda)t y_0$. Since $\lambda\in(0,1)$ then
$$
\langle S y_0,y_0\rangle=-\lambda^{-1}(1-\lambda)\langle Ty_0, y_0\rangle<0
$$
Hence $S\not{\geq}0$ and $S\notin B_+$. Contradiction, so
$$
\langle Ty,y\rangle=0
$$
for all $y\in P(H)^\perp$. Let $y_1,y_2\in P(H)^\perp$, then
$$
\langle T(y_1+y_2),y_1+y_2\rangle=0\\
\langle T(y_1+iy_2),y_1+iy_2\rangle=0
$$
Since $T\in B_+$, we have $T=T^*$. Using this fact previous two equations can be simplified to
$$
\langle Ty_1,y_1\rangle+2\mathrm{Re}\langle Ty_1,y_2\rangle+\langle Ty_2,y_2\rangle=0\\
\langle Ty_1,y_1\rangle+2\mathrm{Im}\langle Ty_1,y_2\rangle+\langle Ty_2,y_2\rangle=0
$$
Since $y_1,y_2\in P(H)^\perp$, then
$$
\langle Ty_1,y_1\rangle=\langle Ty_2,y_2\rangle=0
$$
so we get $\mathrm{Re}\langle Ty_1,y_2\rangle=\mathrm{Im}\langle Ty_1,y_2\rangle=0$ which is equivalent to
$$
\langle Ty_1,y_2\rangle=0
$$
Since $y_1\in P(H)^\perp$, then for all $x\in P(H)$ we have $\langle y_1,x\rangle=0$. Since $T=T^*$ we get $\langle Ty_1,x\rangle=\langle y_1,T^*x\rangle=\langle y_1,Tx\rangle=\langle y_1,x\rangle=0$ for all $x\in P(H)$. This means that $Ty_1\in P(H)^\perp$. Hence we may consider case $y_2=Ty_1$, then
$$
\Vert Ty_1\Vert^2=\langle Ty_1,Ty_1\rangle=\langle Ty_1,y_2\rangle=0
$$
so $Ty_1=0$. Thus for all $y_1\in P(H)^\perp$ we have $Ty_1=0$.
Now take arbitrary $z\in H$, and consider representation $z=Pz+(1_H-P)z$. Since $Pz\in P(H)$ and $(1_H-P)z\in P(H)^\perp$ we have
$$
Tz=T(Pz)+T((1_H-P)z)=T(Pz)=Pz
$$
Since $z\in H$ is arbitrary $T=P$. Similar argument shows that $S=P$. Since $S=T=P$ we see that $P$ is an extreme point of $B_+$.


TITLE: Finding an English version of Hjortnaes' paper on the series representation of $\zeta(3)$
QUESTION [4 upvotes]: Hjortnaes published a paper in 1953 establishing the following series representation for $\zeta(3)$, where $\zeta$ is the Riemann Zeta function:
$$
{\displaystyle {\begin{aligned}\zeta (3)&={\frac {5}{2}}\sum _{k=1}^{\infty }(-1)^{k-1}{\frac {k!^{2}}{(2k)!k^{3}}}\\&={\frac {5}{2}}\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{{\binom {2k}{k}}k^{3}}}\end{aligned}}}
$$
The paper is entitled "Overføring av rekken ${\displaystyle \sum _{k=1}^{\infty }\left({\frac {1}{k^{3}}}\right)}$  til et bestemt integral", and appeared in the Scandinavian Mathematical Society's Proc. 12th Scandinavian Mathematical Congress. Unfortunately the paper is in Norwegian, which I do not understand. A (quick) Google search did not yield any English translations for this paper. Does anyone know where I can find one?

REPLY [4 votes]: Unfortunately there is no English translation of Hjortnaes' paper. Moreover, after hours of searching I was not able to find Apéry's original paper proving the irrationality of $\zeta(3)$.
However, the following papers give various different proofs of the above identity:

Lemma 3.1.3 in this paper
Page 5 in this paper
A proof by Apéry in this small note
Van der Poorten's excellent paper, section 3.