text:
\begin{document}

\setcounter{page}{1}

\title[Dual continuous $K$-Frames in Hilbert spaces]{Dual continuous $K$-Frames in Hilbert spaces}

\author[Mohamed Rossafi, Brahim Moalige, Hamid Faraj, Abdeslam Touri and Samir Kabbaj]{Mohamed Rossafi$^1$$^{\ast}$, Brahim Moalige$^2$, Hamid Faraj$^2$, Abdeslam Touri$^2$ \MakeLowercase{and} Samir Kabbaj$^2$}

\address{$^{1}$Department of Mathematics, Faculty of Sciences Dhar El Mahraz, University Sidi Mohamed Ben Abdellah, Fes, Morocco}
\email{\textcolor[rgb]{0.00,0.00,0.84}{rossafimohamed@gmail.com}}

\address{$^{2}$Department of Mathematics, University of Ibn Tofail, B.P. 133, Kenitra, Morocco}
\email{\textcolor[rgb]{0.00,0.00,0.84}{bmoalige@hotmail.com}}
\email{\textcolor[rgb]{0.00,0.00,0.84}{farajham19@gmail.com}}
\email{\textcolor[rgb]{0.00,0.00,0.84}{touri.abdo68@gmail.com}}
\email{\textcolor[rgb]{0.00,0.00,0.84}{samkabbaj@yahoo.fr}}

\subjclass[2010]{Primary 42C15; Secondary 42C40, 41A58.}

\keywords{Continuous frame, Continuous $K$-frame, dual continuous $K$-frame.}

\date{Received:;
\newline \indent $^{*}$Corresponding author}

\begin{abstract}
Frame theory is recently an active research area in mathematics, computer science and engineering with many exciting applications in a variety of different fields. This theory has been generalized rapidly and various generalizations of frames in Hilbert spaces. In this papers we study the notion of dual continuous $K$-frames in Hilbert spaces. Also we etablish some new properties.
\end{abstract} \maketitle

\section{Introduction}
A frame is a set of vectors in a Hilbert space that can be used to reconstruct each vector in the space from its inner products with the frame vectors. These inner products are generally called the frame coefficients of the vector. But unlike an
orthonormal basis each vector may have infinitely many different representations in terms of its frame coefficients.

Introduced by Duffin and Schaeffer in 1952 \cite{1} to study some deep problems in nonharmonic Fourier series, theory of frame in Hilbert space has grown rapidly. After the fundamental paper \cite{2} by Daubechies, Grossman and Meyer, frames
theory began to be widely used, particularly in the more specialized context of wavelet frames and Gabor frames \cite{3}.

A discrete frame in a separable Hilbert space $\mathcal{H}$ is a sequence $\{f_{i}\}_{i\in I}$  for which there exist positive constants $A, B > 0$ called frame bounds such that
\begin{equation*}
	A\|x\|^{2}\leq\sum_{i\in I}|\langle x,f_{i}\rangle|^{2}\leq B\|x\|^{2}, \;\forall x\in\mathcal{H}.
\end{equation*}

The continuous frames has been defined by Ali, Antoine and Gazeau \cite{4}, called these kinds frames, frames associated with measurable space. For more details, the reader can refer \cite{5}. The concept of continuous $K$-frame in Hilbert space have been introduced in \cite{c-k-frame}.

Many generalizations of the concept of frame have been defined in Hilbert Spaces and Hilbert $C^{\ast}$-modules \cite{r1, r11, r2, r3, r4, r5, r6, r7, r10, r8, r9}.

In this papers, we characterize the concept of dual continuous $K$-frames in Hilbert spaces and we give some new properties.
\section{Preliminaries}
Let $X$ be a Banach space, $(\Omega,\mu)$ a measure space, and function $f:\Omega\to X$ a measurable function. Integral of the Banach-valued function $f$ has defined Bochner and others. Most properties of this integral are similar to those of the integral of real-valued functions.

Let $(\Omega,\mu)$ be a measure space, let $H$ and $K$ be two separables Hilbert Spaces, we denote $B(H,K)$ the collection of all bounded linear operators from $H$ to $K$, as well $B(H,H)$ is abbreaviated to $B(H)$.\\
For $T \in B(H,K)$, we use the notation $\mathcal{R}(T)$ and $\mathcal{N}(T)$ to denote respectively the range and the null space of $T$.

\begin{definition} \cite{5} \label{d1}
	Let $H$ be a complex Hilbert space, and $(\Omega,\mu)$ be a measure space with positive measure $\mu$.\\
	A map $ F : \Omega \longrightarrow H$ is called a continuous frame with respect to $(\Omega,\mu)$ if :
	\begin{itemize}
		\item [1]- $F$ is weakly measurable, ie: $ \forall f\in H, \quad w \longrightarrow \langle f,F(\omega)\rangle$ is a measurable fonction on $\Omega$.
		\item [2]- There exists two constants $A,B > 0$ such that :
		\begin{equation}\label{d1eq1}
		A\|f\|^{2}\leq \int_{\Omega}|\langle f,F(\omega)\rangle |^{2}d\mu(\omega)\leq B\|f\|^{2} \quad \forall f\in H.
		\end{equation}
	\end{itemize}
\end{definition}
Let $F$ be a continuous frame, the analysis operator $T$ is defined by :
\begin{align*}
T: &H \longrightarrow L^{2}(\Omega)\\
& f \longrightarrow \{\langle f,F(\omega)\rangle \}_{\omega \in \Omega}
\end{align*}
The adjoint operator of $T$, called synthesis operator, is defined by :
\begin{align*}
T^{\ast}: &L^{2}(\Omega) \longrightarrow H\\
& x \longrightarrow \int_{\Omega}x(\omega)F(\omega)d\mu(\omega)
\end{align*}
The frame operator of the continuous frame $F$ is defined by : $S=T^{\ast}T$ such that, is bounded and invertible.\\
Recall that a continuous Bessel sequence $G$ is a dual continuous frame of $F$ if :
\begin{equation*}
f=\int_{\Omega}\langle f,G(\omega)\rangle F(\omega)d\mu (\omega) \qquad \forall f\in H
\end{equation*}
We have :
\begin{equation*}
f=\int_{\Omega}\langle f,S^{-1}_{F}F(\omega)\rangle F(\omega)d\mu (\omega) \qquad \forall f\in H
\end{equation*}
This show that $S^{-1}_{F}F$ is a dual continuous frame of $F$, called the canonical dual continuous frame of $F$.
\begin{definition}\cite{c-k-frame} \label{d2}
Let $K\in B(H)$, a map $F:\Omega \longrightarrow H$ is said to be a continuous K-frame, if there exists a constants $0<A<B<\infty$ such that :
\begin{equation*}
A\|K^{\ast}f\|^{2}\leq \int_{\Omega}|\langle f,F(\omega)\rangle |^{2}d\mu(\omega)\leq B\|f\|^{2} \quad \forall f\in H.
\end{equation*}
The constants $A$ and $B$ are called the lower and upper continuous K-frame bounds.\\
If $A=B$, $F$ is called a  tight continuous K-frame.\\
If $A=B=1$, $F$ is called the parseval continuous K-frame.\\
If \eqref{d1eq1} holds right, $F$ is called continuous K-Bessel sequence.
\end{definition}
\begin{definition}\cite{Tousi}
	A Bessel mapping $F$ is said to be $L^{2}$-independent if $\int_{\Omega}\phi(\omega)F(\omega)d\mu(\omega)=0$ for $\phi\in L^{2}(\Omega)$, implies that $\phi=0$ a. e.
\end{definition}
\begin{lemma}\cite{10}\label{l1}
Let $\Lambda \in B(H,K)$ has a closed range, then there exists a unique operator $\Lambda^{\dagger}\in B(K,H)$, called the pseudo-inverse of $\Lambda$, satisfying :
\begin{align*}
&\Lambda\Lambda^{\dagger}\Lambda=\Lambda \qquad (\Lambda\Lambda^{\dagger})^{\ast}=\Lambda\Lambda^{\dagger} \\
&\Lambda^{\dagger}\Lambda\Lambda^{\dagger}=\Lambda^{\dagger} \qquad (\Lambda^{\dagger}\Lambda)^{\ast}=\Lambda^{\dagger}\Lambda \qquad (\Lambda^{\ast})^{\dagger}=(\Lambda^{\dagger})^{\ast}\\
&\mathcal{N}(\Lambda^{\dagger})=(\mathcal{R}(\Lambda))^{\perp} \qquad \mathcal{R}(\Lambda^{\dagger})=(\mathcal{N}(\Lambda))^{\perp}
\end{align*}
\end{lemma}
\begin{lemma}\cite{11}\label{l2}
	Let $H, H_{1}$ and $H_{2}$ be three Hilbert Spaces, also let $S\in B(H_{1},H)$ and $T\in B(H_{2},H)$. The following statements are equivalents :
	\begin{itemize}
		\item [1 -] $\mathcal{R}(S) \subset \mathcal{R}(T)$.
		\item [2 -] There exist $\lambda > 0 $ such that $SS^{\ast} \leq \lambda TT^{\ast}$.
		\item [3 -] There exist $\theta \in B(H_{1},H_{2})$ such that $S=T\theta$.
	\end{itemize}
	Moreover, if $(1)$, $(2)$ and $(3)$ are valide, then there exist a unique operator $\theta$ such that :
	\begin{itemize}
\item	[a -] $\|\theta\|^{2}=inf\{\mu : SS^{\ast}\leq \mu TT^{\ast}\}$.
\item	[b -] $\mathcal{N}(S)=\mathcal{N}(\theta)$.
\item	[c -] $\mathcal{R}(\theta)\subset \overline{\mathcal{R}(T^{\ast})}$.
	\end{itemize}
\end{lemma}

\section{Main result}
Before giving our main results, we will first demonstrate the following lemmas.
\begin{lemma}\label{l3}
	Let $K\in B(H)$ and $F$ be a continuous Bessel sequence of $H$ with analysis operator $T$. Then $F$ is a continuous $K$-frame of $H$ if and only if:

	 $\mathcal{R}(K) \subset \mathcal{R}(T^{\ast})$.
\end{lemma}
\begin{proof}
	It is an immediate consequence of lemma \ref{l2}.
\end{proof}
\begin{lemma}\label{l4}
Suppose that $K\in B(H)$ has closed range and $F$ is a parseval continuous $K$-frame of $H$, then $K^{\dagger}F$ is a dual continuous $K$-Bessel sequence of $F$.
\end{lemma}
\begin{proof}
$F$ is a parseval continuous $K$-frame of $H$, then :
\begin{equation}\label{1}
\|K^{\ast}f\|^{2}=\int_{\Omega}|\langle f,F(\omega)\rangle |^{2}d\mu(\omega) \qquad \forall f\in H.
\end{equation}
Let $g\in {\mathcal{R}(K^{\ast})}$, we have : $g=K^{\ast}(K^{\ast})^{\dagger}g=K^{\ast}(K^{\dagger})^{\ast}g$.  Replace $f$ by $(K^{\ast})^{\dagger}g$ in \eqref{1}, then :
\begin{equation*}
\|K^{\ast}(K^{\ast})^{\dagger}g\|^{2}=\int_{\Omega}|\langle (K^{\ast})^{\dagger}g, F(\omega)\rangle |^{2}d\mu(\omega)
\end{equation*}
so,
\begin{equation*}
\|g\|^{2}=\int_{\Omega}|\langle g,K^{\dagger}F(\omega)\rangle |^{2}d\mu(\omega).
\end{equation*}
Hence, $K^{\dagger}F$ is a continuous Bessel sequence.

Since $F$ is a parseval continuous $K$-frame, one has
\begin{equation*}
KK^{\ast}f=\int_{\Omega}\langle f,F(\omega)\rangle F(\omega)d\mu(\omega).
\end{equation*}
Then we have : $g=K^{\ast}(K^{\ast})^{\dagger}g=K^{\ast}(K^{\dagger})^{\ast}g$,
\begin{align*}
Kg&=KK^{\ast}(K^{\dagger})^{\ast}g\\
&=\int_{\Omega}\langle (K^{\dagger})^{\ast}g,F(\omega)\rangle F(\omega)d\mu(\omega)\\
&=\int_{\Omega}\langle g,K^{\dagger}F(\omega)\rangle F(\omega)d\mu(\omega).
\end{align*}
If $h\in (\mathcal{R}(K^{\ast}))^{\perp}=\mathcal{N}(K)$\\
lemma \ref{l1} $\Longrightarrow h\in (\mathcal{N}((K^{\ast})^{\dagger})=\mathcal{N}((K^{\dagger})^{\ast})$,\\
then :
\begin{equation*}
\int_{\Omega}\langle h,K^{\dagger}F(\omega)\rangle F(\omega)d\mu(\omega)=\int_{\Omega}\langle (K^{\dagger})^{\ast}h,F(\omega)\rangle F(\omega)d\mu(\omega)=0=Kh.
\end{equation*}
So, for all $f\in H$, we have :
\begin{equation*}
Kf=\int_{\Omega}\langle g,K^{\dagger}F(\omega)\rangle F(\omega)d\mu(\omega).
\end{equation*}
\end{proof}
\begin{lemma}\label{l5}
Suppose that $K\in B(H)$ has closed range and $F$ is a parseval continuous $K$-frame of $H$ with analysis operator $T$, then $G$ is a dual continuous $K$-Bessel sequence of $F$ if and only if there exists $\varphi \in B(H,L^{2}(\Omega))$ such that : $T^{\ast}\varphi = 0$ and $(\varphi f)_{\omega}=\langle f, G(\omega)-K^{\dagger}F(\omega)\rangle \quad \forall f\in H, \quad \forall \omega \in \Omega$.
\end{lemma}
\begin{proof}
Let $G$ be a dual continuous $K$-Bessel sequence of $F$, \begin{align*}
\varphi : &H\longrightarrow L^{2}(\Omega)\\
&f\longrightarrow \varphi f
\end{align*}
wich is defined by: $(\varphi f)_{\omega}=\langle f,G(\omega)-K^{\dagger}F(\omega)\rangle$.
One has
\begin{align*}
T^{\ast}(\varphi f)
&=\int_{\Omega}\langle f,G(\omega)-K^{\dagger}F(\omega)\rangle F(\omega) d\mu(\omega)\\
&=\int_{\Omega}\langle f,G(\omega)\rangle F(\omega) d\mu(\omega) - \int_{\Omega}\langle f,K^{\dagger}F(\omega)\rangle F(\omega) d\mu(\omega)\\
&=Kf-Kf=0.
\end{align*}
Conversely, suppose that exist $\varphi \in B(H,L^{2}(\Omega))$ such that : $T^{\ast}\varphi = 0$ and
\begin{equation*}
(\varphi f)_{\omega}=\langle f,G(\omega)-K^{\dagger}F(\omega)\rangle \quad \forall f\in H, \quad \forall \omega \in \Omega.
\end{equation*}
We have :
\begin{align*}
\int_{\Omega}\langle f,G(\omega)\rangle F(\omega)d\mu (\omega)&=\int_{\Omega}\langle f, K^{\dagger}F(\omega)\rangle F(\omega)d\mu(\omega)\\
&=Kf.
\end{align*}
\end{proof}
\begin{theorem}\label{l6}
Suppose that $K\in B(H)$ has closed range and $F$ is a parseval continuous $K$-frame of $H$ with analysis operator $T_{F}$, then $K^{\dagger}F$ is the canonical dual continuous $K$-Bessel sequence of $F$.
\end{theorem}
\begin{proof}
From lemma \ref{l4}, we know that $K^{\dagger}F$ is a dual continuous $K$-Bessel sequence of $F$. To complete the proof, it only needs to prove: $\|T_{F^{'}}\|\leq \|T_{G}\|$ for any dual continuous $K$-Bessel sequence $G$ of $F$, where $T_{F^{'}}$ is the analysis operator of $K^{\dagger}F$.\\
By lemma \ref{l5}, there exist $\varphi \in B(H,L^{2}(\Omega))$ such that $T^{\ast}_{F}\varphi= 0$ and
\begin{equation*}
(\varphi f)_{\omega}=\langle f,G(\omega)-K^{\dagger}F(\omega)\rangle \quad \forall f\in H, \quad \forall \omega \in \Omega.
\end{equation*}
On the other hand : $T_{G}^{\ast}=T_{F^{'}}^{\ast} + \varphi$,
\begin{align*}
\|T_{G}^{\ast}f\|^{2}&=\langle T_{G}^{\ast}f,T_{G}^{\ast}f\rangle\\
&=\langle T_{F^{'}}^{\ast}f + \varphi f,T_{F^{'}}^{\ast}f + \varphi f\rangle \\
&=\|T_{F^{'}}^{\ast}f\|^{2} + \langle T_{F^{'}}^{\ast}f,\varphi f\rangle + \langle \varphi f, T_{F^{'}}^{\ast}f \rangle + \|\varphi f\|^{2}\\
&=\|T_{F^{'}}^{\ast}f\|^{2}  +  \|\varphi f\|^{2}\geq \|T_{F^{'}}^{\ast}f\|^{2}
\end{align*}
Hence, $\|T_{F^{'}}\|\leq \|T_{G}\|$ as desired.
\end{proof}
\begin{lemma}
\begin{itemize}
\item [1 -] The canonical continuous dual $K$-Bessel sequence of a parseval continuous $K$-frame $F$, wich will be denoted by $\tilde{F}$ later, is actually a parseval continuous frame on $(\mathcal{N}(K))^{\perp}$.

\item [2 -] The canonical dual continuous $K$-Bessel sequence of parseval continuous $K$-frame $F$ is precisely a parseval continuous $K^{\dagger}K$-frame. But in general it is not a parseval continuous $K$-frame. It can naturally generate a new one in the form $K\tilde{F}$.

\end{itemize}
\end{lemma}
\begin{proof}
\begin{itemize}
	\item [1]-\begin{equation*}
	\int_{\Omega}|\langle f,\tilde{F}(\omega)\rangle|^{2}d\mu (\omega)=\|K^{\ast}(K^{\dagger})^{\ast}f\|^{2}=\|(K^{\dagger}K)^{\ast}f\|^{2}=\|K^{\dagger}Kf\|^{2}=\|f\|^{2}.
	\end{equation*}
	\item[2 -] \begin{align*}
	\int_{\Omega}|\langle f,\tilde{F}(\omega)\rangle|^{2}d\mu (\omega)&=\int_{\Omega}|\langle f,K^{\dagger}F(\omega)\rangle|^{2}d\mu (\omega)\\
	&=\|K^{\ast}(K^{\dagger})^{\ast}f\|^{2}\\
	&=\|(K^{\dagger}K)^{\ast}f\|^{2} \qquad \forall f\in H,
	\end{align*}
	\begin{align*}
	\int_{\Omega}|\langle f,K\tilde{F}(\omega)\rangle|^{2}d\mu (\omega)&=\int_{\Omega}|\langle f,KK^{\dagger}F(\omega)\rangle|^{2}d\mu (\omega)\\
	&=\int_{\Omega}|\langle (KK^{\dagger})^{\ast}f,F(\omega)\rangle|^{2}d\mu (\omega)\\
	&=\|K^{\ast}(KK^{\dagger})^{\ast}f\|^{2}=\|(KK^{\dagger}K)^{\ast}f\|^{2}\\
	&=\|K^{\ast}f\|^{2} \qquad \forall f\in H.
	\end{align*}

\end{itemize}
\end{proof}
\begin{theorem}
Suppose that $K\in B(H)$ has closed range and $F$ is a parseval continuous $K$-frame of $H$ with a dual continuous $K$-Bessel sequence $G$. Then $G$ is the canonical dual continuous $K$-Bessel sequence of $F$ if and only if $T^{\ast}_{G}T_{G}=T^{\ast}_{G}T_{H}$ for any dual continuous $K$-Bessel sequence $H$ of $F$, where $T_{G}$ and $T_{H}$ denote the analysis operators of $G$ and $H$ respectively.
\end{theorem}
\begin{proof}
Let us first assume that $G=\tilde{F}$.\\
If we denote by $T_{F}$ the analysis operator of $F$ then a direct calculation can show that $T_{G}=T_{F}(K^{\dagger})^{\ast}$.\\
From this fact and taking into account the fact that :
\begin{equation*}
T^{\ast}_{F}(T_{G}f-T_{H}f)=\int_{\Omega}\langle f,\tilde{F}(\omega)\rangle F(\omega) d\mu (\omega)-\int_{\Omega}\langle f,H(\omega)\rangle F(\omega)d\mu (\omega)=0.
\end{equation*}
We obtain for any $f,g \in H$:
\begin{equation*}
\langle (T_{G}-T_{H})f,T_{G}g\rangle = \langle (T_{G}-T_{H})f,T_{F}(K^{\dagger})^{\ast}g\rangle=\langle K^{\dagger}T^{\ast}_{F}(T_{G}-T_{H})f,g\rangle = 0.
\end{equation*}
Thus $T^{\ast}_{G}(T_{G}f-T_{H}f)=0$ then $T^{\ast}_{G}T_{G}=T^{\ast}_{G}T_{H}$.\\
For the converse, suppose that $T^{\ast}_{G}T_{G}=T^{\ast}_{G}T_{H}$, for any dual continuous $K$-Bessel sequence $H$ of $F$. Then :
\begin{equation*}
\|T_{G}\|^{2}=\|T^{\ast}_{G}T_{G}\|=\|T^{\ast}_{G}T_{H}\|\leq \|T_{G}\|\|T_{H}\|.
\end{equation*}
So, $\|T_{G}\|\leq \|T_{H}\|$ implying that $G$ is the canonical continous $K$-Bessel sequence of $F$.
\end{proof}
Now it is legitimate to pose the following question: Under what condition will a parseval continuous $K$-frame admit a unique dual continuous $K$-Bessel sequence?
\begin{theorem}\label{t1}
Suppose that $K\in B(H)$ has closed range and $F$ is a parseval continuous $K$-frame of $H$ with analysis operator $T_{F}$. Then $F$ has a unique dual continuous $K$-Bessel sequence if and only if $\mathcal{R}(T_{F})=L^{2}(\Omega)$.
\end{theorem}
\begin{proof}
Suppose that $\mathcal{R}(T_{F})=L^{2}(\Omega)$, then $T^{\ast}_{F}$ is injective. Let $G$ and $Q$ be two dual continuous $K$-Bessel sequences of $F$. Then: $\{\langle f, G(\omega)-Q(\omega)\rangle \}_{\omega \in \Omega} \in L^{2}(\Omega)$ and that :
\begin{align*}
0=Kf-Kf&=\int_{\Omega}\langle f,G(\omega)\rangle F(\omega) d\mu (\omega) - \int_{\Omega}\langle f,Q(\omega)\rangle F(\omega) d\mu (\omega)\\
&=\int_{\Omega}\langle f,G(\omega)-Q(\omega)\rangle F(\omega) d\mu (\omega) \\
&=T^{\ast}_{F}(\{\langle f, G(\omega)-Q(\omega)\rangle \}_{\omega \in \Omega}).
\end{align*}
Since, $T^{\ast}_{F}$ is injective, we have $$\langle f,G(\omega)-Q(\omega)\rangle = 0 \quad \forall \omega \in \Omega \quad and \quad \forall f\in H, $$
hence $$G(\omega)=Q(\omega) \quad \forall \omega \in \Omega,$$ so $G=Q$.\\
Conversely, assume contrarity that $\mathcal{R}(T_{F})\neq L^{2}(\Omega)$.\\
Since $F$ is a parseval continuous $K$-frame, it is easely seen that $KK^{\ast}=T^{\ast}_{F}T_{F}$.\\
Hence, $\mathcal{R}(T^{\ast}_{F})=\mathcal{R}(K)$, by lemma \ref{l2}, and $T^{\ast}_{F}$ has closed range as a consequence.\\
Let $S\in B(H,L^{2}(\Omega)) $ be an invertible operator and $0\neq \alpha \in (\mathcal{R}(T_{F}))^{\perp}$.\\
Taking $h=S^{-1}(\alpha)$ and $G(\omega)=\bar{\alpha(\omega)}h$,
for each $\omega \in \Omega$, for every $f\in H$, we have:
\begin{align*}
\int_{\Omega}|\langle f,G(\omega)\rangle|^{2}d\mu (\omega)&=\int_{\Omega}|\langle f,\bar{\alpha(\omega)}h\rangle|^{2}d\mu (\omega)\\
&=\int_{\Omega}|\langle f,h\rangle |^{2}|\alpha(\omega)|^{2}d\mu (\omega)\\
&=|\langle f,h\rangle |^{2}\|\alpha \|_{2}^{2}\\
&\leq \|\alpha \|_{2}^{2} \|h\|^{2} \|f\|^{2}.
\end{align*}
Hence, $G$ is a continuous $K$-Bessel sequence for $H$.\\
Now, let $Q(\omega)=\tilde{F}(\omega) + G(\omega)$ for every $\omega \in \Omega$, then it is easely seen that $Q$ is a continuous $K$-Bessel sequence for $H$.\\
Since $\alpha $ is orthogonal to $\mathcal{R}(T_{F})$,
\begin{align*}
\langle \int_{\Omega}\langle f,G(\omega)\rangle F(\omega)d\mu (\omega),e\rangle &= \int_{\Omega}\langle f,\bar{\alpha(\omega)}h\rangle\langle F(\omega),e\rangle d\mu (\omega)\\
&=\int_{\Omega}\langle f,h\rangle \alpha(\omega) \langle F(\omega),e\rangle d\mu (\omega)\\
&=\langle f,h\rangle \langle \alpha,\{\langle e,F(\omega)\rangle \}_{\omega \in \Omega}\rangle \\
&=\langle f,h\rangle \langle \alpha,T_{F}e\rangle = 0 \quad \forall e,f \in H.
\end{align*}
Then $\int_{\Omega}\langle f,G(\omega)\rangle F(\omega)d\mu (\omega)=0 \quad \forall f\in H$.\\
Which give:
\begin{align*}
\int_{\Omega}\langle f,Q(\omega)\rangle F(\omega)d\mu (\omega)&=\int_{\Omega}\langle f,\tilde{F}(\omega)\rangle F(\omega)d\mu (\omega) + \int_{\Omega}\langle f,G(\omega)\rangle F(\omega)d\mu (\omega)\\
&=\int_{\Omega}\langle f,\tilde{F}(\omega)\rangle F(\omega)d\mu (\omega)=Kf.
\end{align*}
Since $\alpha\neq 0$, there exists $w_{0} \in \Omega$ such that $\alpha(w_{0})\neq 0$, and thus $G(w_{0})\neq 0$. A simple calculation gives $(\frac{\alpha(w_{0})}{|\alpha(w_{0})|^{2}})S(G(w_{0}))=\alpha$.\\
	Hence $Q$ is a dual continuous $K$-Bessel sequence of $F$ and its different from $\tilde{F}$ wich is a contradiction.
\end{proof}
\begin{theorem}\label{t2}
Suppose that $K\in B(H)$ has closed range and $F$ is a parseval continuous $K$-frame of $H$ then the following results hold:
\begin{itemize}
	\item [1 -] $F$ is continuous $L^{2}$-independent if and only if $\tilde{F}$ is continuous $L^{2}$-independent.
	 \item [2 -] If $F$ admits a unique dual continuous $K$- Bessel sequence then $\tilde{F}$ admits a unique dual continuous $K^{\ast}$-Bessel sequence.
\end{itemize}
\end{theorem}
\begin{proof}
(1) One has
\begin{align*}
\int_{\Omega}\langle f,F(\omega)\rangle F(\omega)d\mu (\omega)&=T^{\ast}_{F}T_{F}f=KK^{\ast}f\\
&=\int_{\Omega}\langle K^{\ast}f,\tilde{F}(\omega)\rangle F(\omega)d\mu (\omega)\\
&=\int_{\Omega}\langle f,K\tilde{F}(\omega)\rangle F(\omega)d\mu (\omega) \qquad \forall f\in H.
\end{align*}
Hence,
\begin{align*}
0&=\int_{\Omega}\langle f,F(\omega)\rangle F(\omega)d\mu (\omega)-\int_{\Omega}\langle f,K\tilde{F}(\omega)\rangle F(\omega)d\mu (\omega)\\
&=\int_{\Omega}\langle f,F(\omega)-K\tilde{F}(\omega)\rangle F(\omega)d\mu (\omega).
\end{align*}
Since, $F$ is continuous $L^{2}$-independent, it is follows that:
\begin{equation*}
\langle f,F(\omega)-K\tilde{F}(\omega)\rangle = 0 \qquad \forall f\in H, \quad \forall \omega \in \Omega.
\end{equation*}
So $F=K\tilde{F}$.\\
Suppose now that $\int_{\Omega}c(\omega)\tilde{F}(\omega)d\mu (\omega)=0 \quad $ for some $c\in L^{2}(\Omega)$, then :
\begin{equation*}
0=K\int_{\Omega}c(\omega)\tilde{F}(\omega)d\mu (\omega)=\int_{\Omega}c(\omega)K\tilde{F}(\omega)d\mu (\omega)=\int_{\Omega}c(\omega)F(\omega)d\mu (\omega).
\end{equation*}
So, $c(\omega)=0 \qquad \forall \omega \in \Omega$, because $F$ is continuous $L^{2}(\Omega)$ independent.\\
For the converse, let $\int_{\Omega}c(\omega)F(\omega)d\mu (\omega)=0\quad $ for  $c \in L^{2}(\Omega)$, then:
\begin{equation*}
0=K^{\dagger}\int_{\Omega}c(\omega)F(\omega)d\mu (\omega)=\int_{\Omega}c(\omega)K^{\dagger}F(\omega)d\mu (\omega)=\int_{\Omega}c(\omega)\tilde{F}(\omega)d\mu (\omega)
\end{equation*}
Then $c(\omega) = 0 \qquad \forall \omega \in \Omega$.\\
(2) Since, $F$ has a unique dual continuous $K$-Bessel, by theorem \ref{t1} we know that its analysis operator $T_{F}$ is surjective and  thus $T^{\ast}_{F}$ is injective, which implies that $F$ is continuous $L^{2}$-independent.\\
Hence, by (1), $\tilde{F}$ is also continuous $L^{2}$-independent, from wich we conclude that $\tilde{F}$ has a unique dual continuous $K^{\ast}$-Bessel sequence.
\end{proof}

\begin{theorem}\label{t4}
Suppose that $K\in B(H)$ has closed range and $F$ is a parseval continuous $K$-frame of $H$. Then for any $c\in L^{2}(\Omega)$ satisfying the equation:\\
$Kf=\int_{\Omega}c(\omega)F(\omega)d\mu (\omega)$, we have:
\begin{equation*}
\int_{\Omega}|c(\omega)|^{2}d\mu (\omega)=\int_{\Omega}|c(\omega)-\langle f,\tilde{F}(\omega)\rangle |^{2}d\mu (\omega) + \int_{\Omega}|\langle f,\tilde{F}(\omega)\rangle|^{2}d\mu (\omega).
\end{equation*}
\end{theorem}
\begin{proof}
	We have :
	\begin{align*}
	\int_{\Omega}(c(\omega)-\langle f,\tilde{F}(\omega)\rangle) \langle \tilde{F}(\omega),f\rangle d\mu (\omega) &=\int_{\Omega} \langle (c(\omega)-\langle f,\tilde{F}(\omega)\rangle)\tilde{F}(\omega),f\rangle d\mu (\omega)\\
	&=\langle \int_{\Omega} (c(\omega)-\langle f,\tilde{F}(\omega)\rangle)\tilde{F}(\omega),f\rangle d\mu (\omega)\\
	&=\langle K^{\dagger}\int_{\Omega} (c(\omega)-\langle f,\tilde{F}(\omega)\rangle)\tilde{F}(\omega),f\rangle d\mu (\omega)\\
	&=\langle K^{\dagger}(Kf-Kf),f\rangle =0 \qquad \forall f\in H.
	\end{align*}
	Therefore
	\begin{align*}
	\int_{\Omega}|c(\omega)|^{2}d\mu (\omega)&=\int_{\Omega}c(\omega)\overline{c(\omega)}d\mu (\omega)\\
	&=\int_{\Omega}[(c(\omega)-\langle f,\tilde{F}(\omega)\rangle) +\langle \tilde{F}(\omega),f\rangle] \overline{[(c(\omega)-\langle f,\tilde{F}(\omega)\rangle) +\langle f,\tilde{F}(\omega)\rangle]}d\mu (\omega)\\
	&=\int_{\Omega}(((c(\omega)-\langle f,\tilde{F}(\omega)\rangle) \overline{(c(\omega)-\langle f,\tilde{F}(\omega)\rangle)} + (c(\omega)-\langle f,\tilde{F}(\omega)\rangle)\langle \tilde{F}(\omega),f\rangle  \\
	&+\langle f,\tilde{F}(\omega)\rangle \overline{(c(\omega)-\langle f,\tilde{F}(\omega)\rangle)} + \langle f,\tilde{F}(\omega)\rangle \langle \tilde{F}(\omega),f\rangle))d\mu (\omega)\\
	&=\int_{\Omega}(c(\omega)-\langle f,\tilde{F}(\omega)\rangle)\overline{(c(\omega)-\langle f,\tilde{F}(\omega)\rangle)}d\mu (\omega) + \int_{\Omega}\langle f,\tilde{F}(\omega)\rangle \langle \tilde{F}(\omega),f\rangle d\mu (\omega) \\
	&=\int_{\Omega}|c(\omega)-\langle f,\tilde{F}(\omega)\rangle|^{2} d\mu (\omega)+ \int_{\Omega}|\langle f,\tilde{F}(\omega)\rangle|^{2}d\mu (\omega).
	\end{align*}
\end{proof}
{\bf Conflict of interest}
On behalf of all authors, the corresponding author states that there is no conflict of interest.

{\bf Data availability statement:}  No data were used to support this study.

{\bf Funding statement:}  This study was not funded.
\bibliographystyle{amsplain}


TITLE: Solve this limit using equivalent infinitesimals
QUESTION [4 upvotes]: I am given the following limit, which I'm asked to solve.
$$L=\lim_{x\to 0}\dfrac{2(\tan x-\sin x)-x^3}{x^5}$$
I get confused with equivalent infinitesimals most of the time. I know the correct answer to this limit is $\boxed{1/4}$. However, this is what I did:
$$\left.
\begin{array}{l}
&\sin x \approx x\\
&\tan x \approx x
\end{array}
\right\}
\Rightarrow L=-\lim_{x \to 0} \dfrac{x^3}{x^5}=-\lim_{x \to 0} \dfrac{1}{x^2}=\boxed{-\infty}$$
I know this problem can indeed be solved by taking 3 terms in the MacLaurin Series of each function (i.e. for both the sine and tangent functions). But sometimes this is not always necessary. Sometimes it's sufficient to take the first non-zero term in its MacLaurin Series.
But as far as I was taught in school, the equivalent infinitesimal is defined as the first non-zero term in the MacLaurin Series of a function.
How many terms should I grab to go safe for every case? Why doesn't it suffice to take just the 1st non-zero term?

REPLY [3 votes]: The denominator is $x^5$, so you need to go up to degree $5$ in the numerator. There is no hard and fast rule like “take three terms just in case”.
Think to
$$
\lim_{x\to0}\frac{x-\sin x}{x^3}
$$
You can't stop at degree $1$, because this would just give
$$
\lim_{x\to0}\frac{o(x)}{x^3}
$$
that doesn't carry sufficient information for computing the limit. If you go to degree $3$, you get
$$
\lim_{x\to0}\frac{x-(x-x^3/6+o(x^3)}{x^3}=
\lim_{x\to0}\left(\frac{1}{6}+\frac{o(x^3)}{x^3}\right)
$$
which instead carries the requested information, because $\lim_{x\to0}\frac{o(x^3)}{x^3}=0$.
If the denominator is $x^k$ you generally have to go to degree $k$ also in the numerator.
In your case:

$\displaystyle\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}+o(x^5)$
$\displaystyle\tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}+o(x^5)$

and therefore
\begin{align}
2(\sin x-\tan x)-x^3
&=2x+\frac{2x^3}{3}+\frac{4x^5}{15}-2x+\frac{x^3}{3}-\frac{x^5}{60}-x^3+o(x^5)
\\[4px]
&=\frac{x^5}{4}+o(x^5)
\end{align}
Like before, your limit is $1/4$.
You can stop before degree $5$ if terms with lower degree don't cancel, so you can do it “incrementally”; in this case, terms with degree $1$ cancel as well as terms with degree $3$ (no even degree term).


text:
\section{Time Average Regular (TAR) Policies}
\label{sec:tar}

We specialize our study to the following class of policies, that satisfy a mild regularity condition.  As noted at the end of this subsection in Corollary \ref{cor:consistency}, all TAR policies make $\hat{\alpha}_n$ a consistent estimator of $\alpha$.

\begin{definition}
\label{def:tar}
Policy $A$ is \emph{time-average regular (TAR)} with (possibly random) {\em policy limits} $\gamma=(\gamma(\ell,x): x\in S, \ell=1,2)$ if:
\begin{align}
    \frac{1}{n} \Gamma_n(\ell,x)\xrightarrow{p} \gamma(\ell,x)
\end{align}
as $n\rightarrow \infty$ for each $x\in S, \ell=1,2$.
\end{definition}

In the sequel we typically require that $\gamma(\ell,x) > 0$ almost surely.  Note that in Definition \ref{def:tar}, in general the policy limits {\em will be dependent on the initial state $X_0$}.  We suppress this dependence in the notation, because this dependence on initial conditions will not play a significant role.  In particular, the policies we suggest for efficient experimentation will lead to deterministic policy limits, with no dependence on the initial state.

We now characterize the structure of policy limits; in particular, we show in Proposition \ref{prop: policy limits} below that policy limits almost surely lie in the set $\mathcal{K}$ defined next.  The proof is in the Appendix.

\begin{definition}\label{def: mathcal K}
Define the set $\mathcal{K}$ as follows:
\begin{align}
    \mathcal{K} & = \Big\{ \kappa = (\kappa(\ell, x) : x \in S, \ell = 1,2)\text{ such that: } \\
& \kappa(1,y)+\kappa(2,y)=\sum_{\ell=1}^{2}\sum_{x\in S} \kappa(\ell,x)P(\ell,x,y), \quad y\in S;\label{eq: kappa eq 1}\\
&    \sum_{\ell=1}^{2}\sum_{x\in S}\kappa(\ell,x)=1;\label{eq: kappa eq 2}\\
&    \kappa(\ell,x)\ge 0, \quad x\in S, \ell=1,2\Big\}.\label{eq: kappa eq 3}
\end{align}
\end{definition}

\begin{proposition}\label{prop: policy limits}
Let $A$ be a time average regular policy with policy limits $\gamma=(\gamma(\ell,x): x\in S, \ell=1,2)$.  Then almost surely, $\gamma \in \mathcal{K}$.

\end{proposition}

Although straightforward, the preceding proposition encodes a surprising benefit of estimation using both chains.  In particular, experimental designs in this setting can benefit from {\em cooperative exploration}: one chain can be used to drive the system into states for which we want samples for the {\em other} chain (cf.~the relation \eqref{eq: kappa eq 1}).  In Appendix \ref{sec:example_cooperative}, we illustrate that this possibility can yield substantial benefits in estimation variance.  Indeed, as shown in Appendix \ref{sec:example_cooperative}, examples can be constructed such that the variance of the MLE of the treatment effect after $n$ time steps is unboundedly lower for the optimal policy, relative to the variance of the difference in the MLEs of steady state rewards obtained by running each chain in isolation for $n$ time steps.

\begin{remark}
\label{rem:markov_TAR}
Given a TAR policy $A$, for any initial state, the law of the resulting policy limits $\gamma$ is a probability measure over $\mathcal{K}$, according to Proposition \ref{prop: policy limits}.

Conversely, suppose that $\kappa\in \mathcal{K}$ is  positive (i.e. $\kappa(\ell,x)>0$ for $x\in S, \ell=1,2$). We show that regardless of the initial state, $\kappa$ can be achieved as the (deterministic) policy limit of some TAR policy.  For example, define:
\begin{equation}
\label{eq:Markov_from_kappa}
    p(\ell,x)= \frac{\kappa(\ell,x)}{\kappa(1,x)+\kappa(2,x)}
\end{equation}
for $\ell=1,2$ and $x\in S$. Define $A_n$ to be the following {\em stationary Markov policy}:
\begin{align}
    P(A_n=\ell|\mathcal{G}_{n-1}, X_n)=p(\ell,X_n)
\end{align}
for $\ell=1,2$, $n\ge 0$.  This policy is Markov because it depends only on the current state $X_n$ (and the auxiliary randomness $U_n$) and stationary because the choice probabilities do not change with time.  Further, it is straightforward to check that since each $P(\ell)$ is irreducible for $\ell = 1,2$, this policy makes $X_n$ an irreducible Markov chain.  As a result, this chain therefore has a unique stationary distribution regardless of the initial state.  Elementary computation yields that the stationary distribution must be $\pi(x)=\kappa(0,x)+\kappa(1,x), x\in S$. Therefore, the policy limit of this policy is equal to $\kappa$, regardless of the initial state.
\end{remark}

The following proposition implies Corollary \ref{cor:consistency}: the MLE estimator $\hat{\alpha}_n$ is {\em consistent} under all TAR policies with  positive policy limits, i.e., $\hat{\alpha}_n$ converges in probability to $\alpha$.  Both proofs are relatively straightforward and provided in Appendix \ref{sec:tar_proofs}.

\begin{proposition}\label{prop: P convergence}
If $A$ is a TAR policy with policy limits that are almost surely  positive, then for $\ell = 1,2$, as $n \to \infty$, there holds
\begin{equation}
\label{eq:P_convergence}
    \hat{P}_n(\ell)\xrightarrow{p}P(\ell)
\end{equation}
and
\begin{equation}
\label{eq:pi_convergence}
    \hat{\pi}_n(\ell)\xrightarrow{p} \pi(\ell).
\end{equation}
\end{proposition}

\begin{corollary}
\label{cor:consistency}
If $A$ is a TAR policy with policy limits that are almost surely positive, then the MLE estimator $\hat{\alpha}_n$ is {\em consistent} under $A$, i.e., $\hat{\alpha}_n \xrightarrow{p} \alpha$ as $n \to \infty$.
\end{corollary}

config: arxiv


text:
\section{Annulus SLE martingale-observables} \label{SLE MOs ER}
In this section, we discuss the relation between CFT and SLE in a doubly connected domain. In the first subsection, we define the one-leg operator $\Psi_{\bfs{\beta}}$ associated with a given divisor $\bfs{\beta}$ satisfying certain neutrality condition and explain its  implementation as a boundary condition changing operator. In Subsection~\ref{Main Theorem_sub}, we show a version of BPZ-Cardy type equation to prove Theorem~\ref{main}. In the last subsection, we present an integral representation of (chordal type) annulus SLE partition functions introduced by Lawler and Zhan to construct the martingale-observables for $\SLE(\kappa,\Lambda)$ associated with them.

\subsection{One-leg operator $\Psi_{\bfs{\beta}}$} \label{one-leg subsec}
\hfill
\medskip

In this subsection we introduce the \textit{one-leg operator} $\Psi_{\bfs{\beta}}$, which is a specific form of the OPE exponential, set to satisfy level two degeneracy equation.

By definition, for $n \in \mathbb Z$, Virasoro and current generators $L_n$, $J_n$ are given as
$$
L_n(z):=\frac{1}{2\pi i}\oint_{(z)}(\zeta -z )^{n+1} T(\zeta) d\zeta, \quad
J_n(z):=\frac{1}{2\pi i}\oint_{(z)} (\zeta -z )^n J(\zeta) d\zeta,
$$
respectively.
First, let us recall the following proposition.
\begin{prop}[Cf. Proposition 7.5 \cite{MR3052311}] \label{V.P.F.}
	Let $X$ be a Fock space field. Then any two of the following assertions imply the third one:
	\begin{itemize}
		\item	$X$ belongs to $\FF(A,\bar{A})$;
		\item $X$ is a $[\lambda,\lambda_*]$-differential;
		\item $L_n X=0$ for all $n\ge 1,$ $L_0 X=\lambda X, L_{-1}X=\pa X,$ and similar equations hold for $\bar{X}.$
	\end{itemize}
\end{prop}

Any field satisfying all conditions in the previous proposition is called a \textit{(Virasoro) primary} field in $\FF(A,\bar{A}).$
By definition, a Virasoro primary field $X$ is called \textit{current primary} if
$ J_n X=J_n\bar{X} =0$ for all $n\ge 1$ and
$J_0 X=-iq X,$ $J_0 \bar{X}=i\overline{q}_* \bar{X}$ for some numbers $q$ and $q_*$, which are called charges. One of the main ingredients to connect CFT with SLE theory is the following \textit{level two degeneracy equation}.

\begin{prop}[Cf. Proposition 11.2 \cite{MR3052311}] \label{level2}
	For a current primary field $V$ with charges $q$,$q_*$ in $\FF_{(b)},$ we have
	\begin{align*}
L_{-2}V = \frac{1}{2q^2}L^2_{-1}V
	\end{align*}
	provided $2q(b+q)=1.$
\end{prop}
Suppose that the real parameters $a$ and $b$ are related to the SLE parameter $\kappa$ as
$$
a=\sqrt{2/\kappa}, \quad b=a(\kappa/4-1).
$$
In the sequel, we sometimes use the following Kac labeling of the conformal weights or dimensions:
$$
h_{r,s}=\frac{(r\kappa -4s)^2-(\kappa -4 )^2}{16\kappa}.
$$

For a divisor $\bfs{\beta}= \sum_{j=1}^N \beta_j \cdot z_j,$ we define the one-leg operator $\Psi_{\bfs{\beta}}$ associated with $\bfs\beta$ as
\begin{equation}\label{Psi}
 \Psi_{\bfs{\beta}} \equiv  \Psi_{   \bfs{\beta} }  (z,\bfs{z}) := \begin{cases} \OO \left[ a\cdot z +\dfrac{  \bfs{\beta}    } {2} , \dfrac{  \bfs{\beta}    }{2} \right] &\textrm{for ER b.c.,}
 \vspace{0.5em}
 \\
 \hspace{0.1em}\OO \bigg[a\cdot z + \bfs{\beta},\hspace{0.5em} \bfs{0} \bigg] &\textrm{for Dirichlet b.c.} \end{cases}
\end{equation}
For the well-definedness, we require that the divisor $\bfs{\beta}$ satisfies the neutrality condition \eqref{NC0_ER}, i.e.,
\begin{equation} \label{NC ins}
a+\int \bfs{\beta} =0.
\end{equation}
Note that by \eqref{c.d._ER}, conformal dimensions of $\Psi_{\bfs{\beta}}$ with respect to $z$ and $z_j$ are given by
\begin{equation} \label{oneleg dim}
 h_{z}=\dfrac{6-\kappa}{2\kappa}=h_{1,2},
\quad
h_{j}=
\begin{cases}
\dfrac{\beta_j^2}{8}-\dfrac{\beta_j b}{2} &\textrm{for ER b.c.,}
\vspace{0.5em}
\\
\dfrac{\beta_j^2}{2}-\beta_j b &\textrm{for Dirichlet b.c.,}
\end{cases}
\end{equation}
and $h_{*j}=h_j$ for ER boundary condition, whereas $h_{*j}=0$ for Dirichlet one.
By \eqref{Coulomb ftn}, the evaluation of $\E \Psi_{\bfs{\beta}}$ in the identity chart of the cylinder $\CC_r$ is given as
\begin{align} \label{Partition ER}
\begin{split}
\E \Psi_{\bfs{\beta}}&= \Theta'(0)^{\frac{a^2}{2}+\sum \frac{\beta_j^2}{4}}
 \prod_{j} \Big( \Theta (z-z_j) \Theta (z-\bar{z}_j) \Big) ^{\frac{a \beta_j}{2}} \Theta(z_j-\bar{z}_j)^{\frac{\beta_j^2}{4}}
\\
&\times\prod_{j < k} \Big( \Theta(z_j-z_k)\Theta(\bar{z}_j-z_k) \Theta(z_j-\bar{z}_k)\Theta(\bar{z}_j-\bar{z}_k) \Big)^{\frac{\beta_j \beta_k}{4}}
\end{split}
\end{align}
for ER boundary condition and
\begin{align} \label{Partition Diri}
\begin{split}
\E  \Psi_{\bfs{\beta}}&=\Theta'(0)^{\frac{a^2}{2}+\sum \frac{\beta_j^2}{2}} \prod_{j} \tilde{\Theta}(z-z_j) ^{a \beta_j} \prod_{j < k} \tilde{\Theta}(z_j-z_k)^{\beta_j \beta_k}
\end{split}
\end{align}
for Dirichlet boundary condition.

We call
$$ Z_{\bfs{\beta}} \equiv Z_{\bfs{\beta}   } (z,\bfs{z}) :=\E \Psi_{   \bfs{\beta} }  (z, \bfs{z})$$
the \textit{partition function} associated with ${\bfs{\beta}}$ and set
$$ \Lambda_{\bfs{\beta}} \equiv \Lambda_{\bfs{\beta}  }(z,\bfs{z} ):= \kappa \pa_{z} \log Z_{\bfs{\beta}}.
$$
By \eqref{Partition ER}, and \eqref{Partition Diri}, $\Lambda_{\bfs{\beta}}$ is given as
\begin{equation} \label{Lambda bulk}
 \Lambda_{\bfs{\beta}}=
 \begin{cases}
 \displaystyle \sqrt{\dfrac{\kappa}{2}} \sum_{j=1}^N \dfrac{\beta_j}{2} \Big\{ H(r, z-z_j)+H(r,z-\bar{z}_j) \Big\} \hfill&\text{for ER b.c.;}
 \vspace{0.5em}
 \\
 \displaystyle \sqrt{\dfrac{\kappa}{2}}\sum_{j=1}^N \beta_j \ti{H}(r,z - z_j) \hfill&\text{for Dirichlet b.c.}
 \end{cases}
\end{equation}

Now we derive the level two degeneracy equation for $\Psi_{\bfs{\beta}}.$
Notice that $\Psi_{ \bfs{\beta} }$ is a Virasoro primary field.
\begin{lem} \label{psi current}
	The one-leg operator $\Psi(z) \equiv \Psi_{\bfs{\beta}}(z)=\Psi_{\bfs{\beta}}(z, \bfs{z})$ is a current primary field with charges $q = a, q_*=0$ at $z,$ i.e.,
	$$
	J_0 \Psi=-ia\Psi, \;\;\; J_0\overline{\Psi}=0,\;\;\; J_n\Psi=J_n \overline{\Psi}=0 \quad (n\ge1).
	$$
\end{lem}
\begin{proof}
We prove the ER case only. The Dirichlet case can be proved in a similar way.
 To show above equations, it is enough to show that as $\zeta \rightarrow z$,
	$$
	 	J_{(0)}(\zeta)\Psi \sim -ia\frac{\Psi}{\zeta-z},\;\;\; J_{(0)}(\zeta)\overline{\Psi}\sim 0
	$$
	holds in the identity chart. By Wick's formula, we have
	\begin{align*}
	 J_{(0)}(\zeta) \Psi &= J_{(0)}(\zeta) \odot \Psi + ia\E \left[J_{(0)}(\zeta)\Phiplus_{(0)}(z) \right]\Psi
	- \sum \frac{\beta_j}{2}\E \left[J_{(0)}(\zeta) \ti{\Phi}_{(0)}(z_j) \right]\Psi,
	\\
	J_{(0)}(\zeta) \overline{\Psi} &= J_{(0)}(\zeta) \odot \overline{\Psi} + ia\E\left[J_{(0)}(\zeta)\Phiminus_{(0)}(z)\right] \overline{\Psi}
	+ \sum \frac{\beta_j}{2}\E\left[J_{(0)}(\zeta) \ti{\Phi}_{(0)}(z_j) \right] \overline{\Psi}.
	\end{align*}
	Now, lemma follows from \eqref{formal ++ cor} and \eqref{h-zero}.
\end{proof}

By Proposition \ref{level2}, we immediately obtain the following level two degeneracy equation of $\Psi \equiv \Psi_{\bfs{\beta}}.$

\begin{prop} \label{level2_ER}
	If $2a(a+b)=1,$ then
	$$
	 	T\ast_{z} \Psi(z, \bfs{z})=\frac{1}{2a^2}\pa^2_{z}\Psi(z,\bfs{z}).
	$$
\end{prop}

 Now we show the following BPZ-type equation.

\begin{prop} \label{BPZ_ER}
	For any $\XX \in \FF_{(b)}$ with nodes outsides $\mathcal S_\Psi,$ we have
	$$
	 \frac{1}{a^2}\pa^2_{z}\E \left[ \Psi \XX \right] =\LL_{v_{z}}^+\E[ \Psi \XX]+\LL_{v_{\bar{z}}}^- \E[\Psi \mathcal X]+ \left( \frac{6}{\kappa} \frac{\zeta_r(\pi)}{\pi}+\pa_r \right) \E[\Psi \XX]
	$$
	for ER boundary condition, and
	$$
	 \frac{1}{a^2}\pa^2_{z}\E \left[ \Psi \XX \right] =\LL_{v_{z}}\E[ \Psi \XX]+ \left( \frac{6}{\kappa} \frac{\zeta_r(\pi)}{\pi}-\frac{1}{2r}+\pa_r \right) \E[\Psi \XX]
	$$
	for Dirichlet boundary conditions, where all fields are evaluated in the identity chart of $\mathcal C_r.$ Here, Lie derivative $\LL_{v_{z}}$ does not apply to $z.$
\end{prop}
\begin{proof}
By Proposition \ref{level2_ER}, we have
	$$
	 \frac{1}{a^2} \pa^2_{z} \E \Psi \XX =2 \E \left[ \frac{\pa_{z}^2 \Psi}{2a^2}\XX \right]= 2 \E \left[ \left( T\ast_{z}\Psi \right) \XX \right],
	$$
	which allows us to apply Corollary~\ref{Wequation_ER}. Thus, we obtain
	\begin{align*}
	2 \E \left[ \left( T\ast_{z}\Psi \right) \XX \right]
	&= 2 \E \left[ \left( A\ast_{z}\Psi \right) \XX \right]+ 2\E[T] \, \E[\Psi \XX]
	\\
	&= \LL_{v_{z}}^+\E[ \Psi \XX]+\LL_{v_{\bar{z}}}^- \E[\Psi \mathcal X]+ \left\{ 2h_{1,2} \frac{\zeta_r(\pi)}{\pi}+2\E[T]+\pa_r \right \} \E[\Psi \XX].
	\end{align*}
Now proposition follows from Proposition~\ref{Virasoro2_ER}.
\end{proof}

Given marked boundary points $p \in \R$ and $q_1, \cdots, q_N \in \pa \CC_r$, set $\bfs{\beta}= \sum \beta_j \cdot q_j$ satisfying \eqref{NC ins}. By abuse of notation, for $z_j \in \CC_r$, we sometimes use the same symbol $\bfs{\beta}$ for $\sum\beta_j \cdot z_j$.
As we explained in Remark~\ref{rmk Cou}, for ER boundary condition, a renormalization procedure is required to define $\Psi_{\bfs{\beta} }$. On the other hand, one can define $Z_{\bfs{\beta}  }=\E\Psi_{\bfs{\beta} }$ as a differential with conformal $h_{1,2}$ at $p$ and with conformal dimension $(\beta_j^2/4-\beta_j b/2, \hspace{0.5em}   \beta_j^2/4-\beta_j b/2  )$ at $q_j$. Its evaluation in the identity chart of $\CC_r$ is given by (up to multiplicative constant)
\begin{align*}
\begin{split}
Z_{\bfs{\beta}}(p,\bfs{q})&=
\prod_{j} \Big| \Theta (p-q_j) \Big| ^{a \beta_j}
\prod_{j < k} \Big|  \Theta(q_j-q_k) \Theta(q_j-\bar{q}_k) \Big|^{\frac{\beta_j \beta_k}{2}}
\end{split}.
\end{align*}
For Dirichlet boundary condition, $\Lambda_{\bfs{\beta}}(\xi,\bfs{q})$ is defined as
$$
\Lambda_{\bfs{\beta}}(\xi,\bfs{q}):=\kappa \frac{\pa_{\xi}Z_{\bfs{\beta}   } (\xi,\bfs{q})    }{  Z_{\bfs{\beta}   } (\xi,\bfs{q})   }.
$$
For ER boundary condition, we need a renormalization procedure.
In the cylinder $\CC_r$, $\Lambda_{\bfs{\beta}}(\xi,\bfs{q})$ is represented as
\begin{equation} \label{Lambda in gen}
\begin{cases}
\displaystyle\sqrt{\frac{\kappa}{2}} \left( \sum_{q_l \in \R} \beta_l H(r,\xi-q_l) + \sum_{q_m \in \R_r } \beta_m H_I(r,\xi-\Re\, q_m) \right) \hspace{0.5em}\hfill\text{\hfill for ER b.c.}&
\vspace{0.5em}
\\
\displaystyle \sqrt{\dfrac{\kappa}{2}}\sum_{j=1}^N \beta_j \ti{H}(r,\xi - q_j) \hfill\text{for Dirichlet b.c.}&
\end{cases}
\end{equation}
It is easy to see that $\Lambda_{\bfs{\beta}}(\xi,\bfs{q})$ is real-valued.

We now discuss how the one-leg operator $\Psi_{\bfs{\beta} }$ acts on Fock space fields in $D_r$ as a boundary condition changing operator.
From now on, we write $w_z=w(z)$, $w_p:=w(p)$ and $w_j:=w(q_j)$ for a conformal transformation $w: D_r \rightarrow \CC_r$.
The insertion of field $\Psi_{ \bfs{\beta} }/\E \Psi_{ \bfs{\beta} }$ produces an operator $\XX \mapsto \wh{\XX} $ on Fock space fields. It is given by
$$
\pa\XX \mapsto \pa\wh{\XX},\quad	\bar{\pa}\XX \mapsto \bar{\pa}\wh{\XX}, \quad \alpha\XX+\beta\YY \mapsto \alpha\wh{\XX}+\beta\wh{\YY},\quad	\XX \odot \YY \mapsto \wh{\XX}\odot \wh{\YY}
$$
and by the formula:
$$
\wh{\Phi}(z)=\Phi(z)+2a \arg \Theta(w_p-w_z)+\sum \beta_j \arg \Big\{ \Theta(w_{j}-w_z)\Theta(\bar{w}_{j}-w_z) \Big \}
$$
for ER boundary conditions and
$$
\wh{\Phi}(z)=\Phi(z)+2a\arg{\ti{\Theta}(w_p-w_z)}+2\sum\beta_{j}\arg\ti{\Theta}(w_{j}-w_z)
$$
for Dirichlet boundary conditions.
When $\kappa=4$, this insertion procedure produces specific height gap of GFF and gives rise to the coupling of GFF and SLE($4,\Lambda$).
For instance, in \cite{MR2651436}, Hagendorf, Bernard and Bauer studied coupling of GFF with Dirichlet boundary condition and SLE($4,\Lambda$) with one force point $q$ using the one-leg operator $\Psi \equiv \mathcal{O}[a\cdot p -a\cdot q,\bfs{0}]$.
In the case of multiple force points, the (multivalued) harmonic function $\wh{\varphi}:= \E \wh{\Phi}$ in $\bar{\CC}_r \sm \{p,q_1,\cdots, q_N\}$ has piecewise  Dirichlet boundary conditions having additional jump of $2\pi a$ at $p$ and $2\pi \beta_j$ at $q_j$'s. Due to the neutrality condition, we require that all jumps  should add up to $0$. This type of boundary condition of GFF was studied by Izyurov and Kyt\"{o}l\"{a}, see \cite{MR3010393}.

As in Subsection~\ref{Main result sub}, we let
$$
 \wh{\E}[\mathcal X]=
 \begin{cases}
 \displaystyle \E \Big[ e^{\odot ia \Phi_{(0)}^+(p)-\sum \frac{\beta_j}{2} \ti{\Phi}_{(0)}(q_j) } \mathcal X \Big]&\text{for ER b.c.}
 \vspace{0.5em}
 \\
 \displaystyle \E \Big[ e^{\odot ia \Phiplus_{(0)}(p)+i\sum \beta_j \Phiplus_{(0)}(q_j)} \mathcal X \Big]&\text{for Dirichlet b.c.}
 \end{cases}
$$
Then as in the simply connected case, it is easy to check that, for all $\XX\in \FF_{(b)},$ we have
$$
 			\wh{\E}[\XX]=\E[\wh{\XX}].
$$

\begin{egs*} Let $w$ be the conformal mapping from $D_r$ onto $\CC_r$.
\\
$\bullet$ The current field $\wh{J}$ is a PS-form of order $ib.$ We have
$$
\wh{J}(z)=J(z)+\frac{ia w'_z}{2} H(w_p-w_z) +\frac{iw'_z}{4}\sum \beta_j \Big \{ H(w_{j}-w_z)+H(\bar{w}_j-w_z) \Big \}
$$
for ER boundary conditions and
$$
\wh{J}(z)= J(z)+\frac{ia w'_z}{2} \ti{H}(w_p-w_z) +\frac{iw'_z}{2}\sum\beta_j\ti{H}(w_{j}-w_z)
$$
for Dirichlet boundary conditions.
\\
$\bullet$ The Virasoro field $\wh{T}$ is an S-form of order $c/12$.
We have
\begin{align*}
\wh{T}(z)&= -\frac{1}{2} \wh{J} \ast \wh{J}+ ib \pa \wh{J}
\\
&= T-(\wh{\jmath}-\jmath)J_{(0)}-\frac{1}{2} \left( \wh{\jmath}^2-\jmath^2 \right)+ib \pa \left(\wh{\jmath}-\jmath \right)
\\
&= A_{(0)}-\wh{\jmath}J_{(0)}+ib\pa J_{(0)}+\frac{c}{12}S_w+(w'_z)^2 \E T-\frac{1}{2} \left( \wh{\jmath}^2-\jmath^2 \right)+ib \pa(\wh{\jmath}-\jmath),
\end{align*}
where $\jmath:=\E J$ and $\wh{\jmath}:=\E \wh{J}$.
\end{egs*}
\subsection{BPZ-Cardy equations} \label{BPZ-Cardy_ER} \label{Main Theorem_sub}
\hfill
\medskip

In this subsection, we prove BPZ-Cardy type equation, which is a key ingredient to connect CFT with SLE theory and present the proof of Theorem~\ref{main}.

Fix marked boundary points $q_1, \cdots, q_N,$ and a marked point $p=\xi$ on the outer boundary. Recall the definitions
$$ Z_{\bfs{\beta}   } (z,\bfs{z}) :=\E \Psi_{   \bfs{\beta} }  (z, \bfs{z}), \quad \Lambda_{\bfs{\beta}  }(z,\bfs{z} ):= \kappa \pa_{z} \log Z_{\bfs{\beta}} (z,\bfs{z}).$$
For a Fock space functional $\mathcal X$ whose nodes do not intersect marked boundary points, set
$$
 R_{z}:=\wh{\E}_{z}[\XX]:=\frac{\E[\Psi_{   \bfs{\beta} }  (z, \bfs{z})\XX]}{ Z_{\bfs{\beta}   } (z,\bfs{z})  },
$$
and
\begin{align*}
 R_{\xi}&:=\wh{\E}_{\xi}[\XX]=
 \begin{cases}
 \displaystyle \E\Big[e^{\odot ia \Phiplus_{(0)}(\xi)-\sum \frac{\beta_j}{2} \ti{\Phi}_{(0)}(q_j)}\XX\Big], &\text{for ER b.c.}
 \vspace{0.5em}
 \\
 \displaystyle \E\Big[e^{\odot ia \Phiplus_{(0)}(\xi)+i\sum \beta_j \Phiplus_{(0)}(q_j)}\XX\Big], &\text{for Dirichlet b.c.}
 \end{cases}
\end{align*}
Let us denote
$ \check{\LL}_{v_z} := \LL_{v_z}(\bar{D}_r \sm ( \{ z \} \cup \mathcal S_\Psi ) )$, i.e., $\check{\LL}_{v_z} \Psi Y(z)X:= \Psi Y(z)\check{\LL}_{v_z}X$ if the nodes of $\XX$ do not intersect $z$ and $\mathcal S_\Psi$.

Now we prove the following BPZ-Cardy equation.
	\begin{prop}\label{BPZ-C_ER}
		For any $\XX \in \mathcal F_{(b)}$, in the $\mathcal C_r$-uniformization, we have
$$
	 	\frac{1}{a^2}\pa^2_{\xi} R_{\xi}+\Lambda_{ \bfs{\beta}  } (\xi,\q)\pa_{\xi} R_{\xi}= \check{\LL}_{v_{\xi}} R_{\xi}+ \Big( \sum_{j=1}^N v_{\xi}(q_j) \pa_{q_j}+\bar{v}_{\xi}(q_j)\bar{\pa}_{q_j} \Big) R_{\xi}+\pa_rR_{\xi},
	$$
	for ER boundary conditions and
		$$
	\displaystyle	\frac{1}{a^2}\pa^2_{\xi} R_{\xi}+\Lambda_{ \bfs{\beta}  } (\xi,\q)\pa_{\xi} R_{\xi}= \check{\LL}_{v_{\xi}} R_{\xi}+ \Big(\sum_{j=1}^N v_{\xi}(q_j) \pa_{q_j} \Big) R_{\xi}+ \pa_rR_{\xi},
	$$
	for Dirichlet boundary conditions. Here $\pa_{\xi}=\pa+\bar{\pa}$ is the operator of differentiation with respect to the real variable $\xi$ whereas $\pa_{q_j}$ denotes the one with respect to the complex variable $q_j$.
	\end{prop}
\begin{proof}
We prove the proposition for ER boundary condition only. The minor changes are needed for Dirichlet boundary condition.

Let $\bfs{\zeta}=S_{\XX}.$ Since $\Psi_{ \bfs{\beta} }(z, \bfs{z})$ is a holomorphic differential with respect to $z,$ we rewrite Proposition \ref{BPZ_ER} as
\begin{align*}
 \frac{1}{a^2}\pa^2_{z} \Big( R_{z} Z_{\bfs{\beta}   } (z,\bfs{z}) \Big) &= \left( \LL_{v_{z}}^+ +\LL_{v_{\bar{z}}}^- \right)(\bfs{\zeta,z}) \Big( R_{z} Z_{\bfs{\beta}   } (z,\bfs{z}) \Big)
\\
&+ \left(\frac{6}{\kappa} \frac{\zeta_r(\pi)}{\pi}+\pa_r \right) R_{z} Z_{\bfs{\beta}   } (z,\bfs{z}).
\end{align*}
By applying the trivial string $\mathcal X \equiv 1$ to above, we have
\begin{equation*}
 \frac{1}{a^2}\pa^2_{z} Z_{\bfs{\beta}   } (z,\bfs{z})=\left( \LL_{v_{z_0}}^+ +\LL_{v_{\bar{z}_0}}^- \right) Z_{\bfs{\beta}   } (z,\bfs{z})
 + \left(\frac{6}{\kappa} \frac{\zeta_r(\pi)}{\pi}+\pa_r \right) Z_{\bfs{\beta}   } (z,\bfs{z}).
\end{equation*}
Subtracting above two equations, we obtain
$$
 	\frac{1}{a^2}\pa^2_{z} R_{z}+\Lambda_{ \bfs{\beta} }(z,\bfs{z})\pa_{z} R_{z}=\left( \LL_{v_{z}}^+ +\LL_{v_{\bar{z}}}^- \right)(\bfs{\zeta,z})R_{z}+\pa_rR_{z}.
$$
Now let's take the limit $z \rightarrow \xi$ and $z_j \rightarrow q_j$. Then, by the facts that $\pa_{\xi}=\pa+\bar{\pa},$ we obtain
$$
 \frac{1}{a^2}\pa^2_{\xi} R_{\xi}+\Lambda_{\bfs{\beta}  }  (\xi,\bfs{q})\pa_{\xi} R_{\xi}=\LL_{v_{\xi}}(\bfs{\zeta})R_{\xi}+\LL_{v_{\xi}}(\bfs{q})R_{\xi}+\pa_rR_{\xi}.
$$
Now proposition follows from the fact $R_\xi$ is a scalar field with respect to each $q_j$.
\end{proof}

Now we prove one of our main theorems.
 \begin{proof}[Proof of Theorem~\ref{main}]
 Let us denote $R_{\xi}=\wh{\E}_{\xi}[\XX].$ Then
$$
 M_t=m \Big(r-t,\xi_t,\ti{g}_t(\q) \Big), \quad m(r,\xi,\q,t )=\Big( R(r,\xi,\q) \big\| \ti{g}_t^{-1}\Big).
 $$
 Hence, by It\^{o}'s formula, we have
 \begin{align*}
 dM_t &= -\pa_r mdt+ \pa_\xi m d\xi_t+ \pa^2_\xi m \frac{d\langle\xi\rangle_t}{2}
 + \left( L_t R\| \ti{g_t}^{-1}\right)dt
 \\
 &+ \sum_{j=1}^N \left[ \pa_{q_j}m \, dq_j(t) + \bar{\pa}_{q_j}m \, d \overline{q_j(t)}\right],
 \end{align*}
where
$$
 L_t:= \left. \frac{d}{ds} \right|_{s=0} \left( R \big(r-t, \xi_t, \ti{g}_t(\bfs{q})\big) \big\| \ti{g}_{t+s}^{-1} \right).
$$
Let $f_{s,t}= \ti{g}_{t+s}^{-1} \circ \ti{g}_t^{-1} .$ Then since the time-dependent flow $f_{s,t}$ satisfies the ODE
$$
 \frac{d}{ds} f_{s,t} = -v_{\xi_{t+s}}f_{s,t}, \quad v_{\xi}(z)=H(r,\xi-z),
$$
we obtain
$$
f_{s,t}=\text{id}-s \, v_{\xi_t}+o(s).
$$
From the fact that the fields in $\mathcal{F}_{(b)}$ depends smoothly on local charts, it follows
$$
L_t= - \left( \check{ \mathcal L}_{v_{\xi_t}} R_\xi \big \| \ti{g}_t^{-1} \right).
$$
By the Loewner's equation $dq_j(t)=-v_{\xi_t}(q_j(t)),$ the drift term of $dM_t$ is
$$
 -\pa_r m+ \Lambda \pa_\xi m + \frac{\kappa}{2} \pa^2_\xi m
- \left( \check{ \mathcal L}_{v_{\xi_t}}R_\xi \big\| \ti{g}_t^{-1}\right)- \sum_{j=1}^N \left[ v_{\xi_t}(q_j(t)) \pa_{q_j}m + \overline{v_{\xi_t}}(q_j(t)) \bar{\pa}_{q_j}m \right].
$$
Theorem now follows from Proposition~\ref{BPZ-C_ER}.
 \end{proof}

\begin{eg*}
	 The simplest examples of SLE martingale-observables are the 1-point function $M(z)$ and the 2-point function $N(z_1,z_2)$ of the bosonic field $\wh{\Phi}_{(b)}$.
	 Of course this is a direct consequence of Theorem~\ref{main}. One can directly show that $M_t$ and $N_t$ are local martingales by It\^{o} formula and the pseudo-addition theorem for the zeta function \eqref{add H}. In the ER boundary case, for all $\kappa$,
	 $$
	dM_t(z)=-\sqrt{2}\,\Im\, H \big(w_t(z)\big)+ \left( a+\sum \beta_j \right) \left( \frac{H^2}{2}+H' \right) \big(w_t(z) \big).
	 $$
	Due to the neutrality condition \eqref{NC ins}, $M_t$ is driftless.
	For the $2$-point function, $dN_t(z_1,z_2) $ is expressed as
	\begin{align*}
	&-\sqrt{2} \Big( M_t(z_1) \Im\,H(r-t, w_t(z_2))+ M_t(z_2) \Im\,H(r-t, w_t(z_1)) \Big) dB_t
	\\
	& +\left( a+\sum \beta_j \right) \left \{ M_t(z_1)\left( \frac{H^2}{2}+H' \right)(w_t(z_2))+ M_t(z_2)\left( \frac{H^2}{2}+H' \right)(w_t(z_1)) \right \}dt,
	\end{align*}
	so the drifts disappear due to the neutrality condition again. The Dirichlet boundary case can be shown similarly.
	In \cite{MR3010393}, Izyurov and Kyt\"{o}l\"{a} studied these types of martingale-observables with Dirichlet boundary condition.
\end{eg*}

\subsection{Lawler and Zhan's annulus SLE partition function} \label{Screening_sub}
\hfill
\medskip

In this subsection, we use the method of screening to find a solution to the null-vector equation
\begin{equation} 	\label{Lawler-Zhan PDE}
\displaystyle \pa_r Z = \frac{\kappa}{2} Z'' + H Z' + \left( \frac{3}{\kappa}-\frac{1}{2} \right)H' Z +C(r) Z
\end{equation}
for the chordal type annulus partition function and prove Theorem~\ref{main_SLE Z}.

\subsubsection{The null vector equations}

Throughout this subsection, we only consider the case that all marked boundary points $p, q_1,\cdots,q_N$ are on the outer boundary component.
In what follows, we use the symbol $\Psi_{\bfs{\beta}}$ for the one-leg operator of the form
$$
\Psi_{\bfs{\beta}} \equiv \Psi_{\bfs{\beta}}(p,\bfs{q}):= \OO [a\cdot p +\bfs{\beta},\hspace{0.5em} \bfs{0}],\quad \bfs{\beta}=\sum_{j=1}^{N} \beta_j \cdot q_j
$$
for both boundary conditions. Let $Z_{ \bfs{\beta} }\equiv Z_{ \bfs{\beta}  }(r,p,\bfs{q}):=\E \Psi_{\bfs{\beta}}$. By \eqref{Coulomb ftn}, in the identity chart of $\CC_r$, we have
\begin{equation} \label{Partition gen}
Z_{ \bfs{\beta} }=
\begin{cases}
\displaystyle C_{\bfs{\beta}  } \prod_j \Theta(r,p-q_j) ^{a \beta_j} \prod_{j < k} \Theta(r,q_j-q_k)^{\beta_j \beta_k}, &\text{for ER b.c.}
\vspace{0.5em}
\\
\displaystyle  C_{\bfs{\beta}  } \prod_{j} \tilde{\Theta}(r,p-q_j) ^{a \beta_j} \prod_{j < k} \tilde{\Theta}(r,q_j-q_k)^{\beta_j \beta_k}, &\text{for Dirichlet b.c.}
\end{cases}
\end{equation}
where
$$C_{ \bfs{\beta}  }=\Theta'(r,0)^{\frac{a^2}{2}+\sum \frac{\beta_j^2}{2}}.$$
We write $\lambda_p$, $\lambda_j$ the (holomorphic) conformal dimension of $\Psi_{ \bfs{\beta} } $ with respect to $p$ and $q_j$'s. By \eqref{c.d._ER}, we have
\begin{equation} \label{Partition gen dim}
\lambda_p=\frac{6-\kappa}{2\kappa},\quad \lambda_j=\frac{\beta_j^2}{2}-\beta_j b.
\end{equation}

We remark that as $r \rightarrow \infty$, the following holds:
$$
\lim\limits_{r\rightarrow \infty} e^{r/4} \Theta(r,x) =2\sin\frac{x}{2},  \quad \lim\limits_{r\rightarrow \infty}e^{r/4}\Theta'(r,0)=1,
$$
see e.g., \cite[pp.65-69]{MR808396}. Therefore by the neutrality condition \eqref{NC ins}, we have (up to a multiplicative constant)
\begin{equation} \label{degen gen}
Z_\infty(p,\bfs{q}):=\lim\limits_{r\rightarrow \infty} Z_{ \bfs{\beta}  }(r,p,\bfs{q})= \prod_{j} \sin ^{a \beta_j}\Big( \frac{p-q_j}{2} \Big) \prod_{j < k} \sin ^{\beta_j \beta_k} \Big( \frac{q_j-q_k}{2} \Big)
\end{equation}
for both boundary conditions.

First we obtain the null-vector equation for the partition function $Z_{\bfs{\beta}  }$ as a consequence of Proposition~\ref{BPZ_ER}. Here and subsequently, $\pa$ and $'$ stands for the differentiation with respect to $p$.
\begin{lem} \label{pre screen null}
	The partition function $Z_{ \bfs{\beta} } \equiv Z_{ \bfs{\beta}  }(r,p,\bfs{q})$ satisfies following partial differential equations:
	$$
	\pa_r Z_{ \bfs{\beta} } =\frac{\kappa}{2} \pa^2 Z_{ \bfs{\beta} }+\left(  \sum_{j=1}^N \lambda_j H'(p-q_j) \right) Z_{ \bfs{\beta} }-\sum_{j=1}^N H(p-q_j)\pa_{q_j}Z_{ \bfs{\beta} } +C(r) Z_{ \bfs{\beta} },
	$$
	where
	\begin{equation} \label{null vec const}
	C(r)=\begin{cases}
	\displaystyle -\frac{6}{\kappa} \frac{\zeta_r(\pi)}{\pi} &\text{for ER b.c.}
	\vspace{0.5em}
	\\
	\displaystyle -\frac{6}{\kappa} \frac{\zeta_r(\pi)}{\pi}+\frac{1}{2r} &\text{for Dirichlet b.c.}
	\end{cases}
	\end{equation}
\end{lem}
\begin{proof}
	By applying trivial string $\XX \equiv 1$ to Proposition~\ref{BPZ_ER}, we have
	$$
	\pa_r  Z_{ \bfs{\beta} }=\frac{\kappa}{2} \pa^2 Z_{ \bfs{\beta} }-\LL_{v_{p}}^+Z_{ \bfs{\beta} } +C(r) Z_{ \bfs{\beta} },
	$$
	where $v_p(z)=H(r,p-z)$. Now lemma follows from \eqref{Partition gen dim} and the fact that
	$\LL_{v_{p}}^+\tilde{Z}= \sum ( v_p(q_j)\pa_{q_j}+\lambda_j v'_p(q_j) ) \tilde{Z}.$
\end{proof}

\begin{eg*}
Let us consider the case that there are two marked points $p, q \in \R$. Set
$$
\Psi_*(p,q) \equiv \mathcal{O}[a\cdot p -a\cdot q,\bfs{0}].
$$
By \eqref{Partition gen}, we have
\begin{equation} \label{k=4}
Z_*(r,p,q)\equiv\E \Psi_*(p,q)=
\begin{cases}
\displaystyle \Theta'(r,0)^{\frac{1}{2}}\Theta(r,p-q) ^{-\frac{1}{2}}  &\text{for ER b.c.}
\vspace{0.5em}
\\
\displaystyle \Theta'(r,0)^{\frac{1}{2}}\tilde{\Theta}(r,p-q) ^{-\frac{1}{2}} , &\text{for Dirichlet b.c.}
\end{cases}
\end{equation}
Note that \eqref{k=4} implies $\pa_q Z_*= -Z'_*$. Also by \eqref{Partition gen dim}, the conformal dimension with respect to $q$ is given as
$$
\lambda_q=\frac{a^2}{2}+ab= \frac{1}{2}-\frac{1}{\kappa}.
$$
Therefore by Lemma~\ref{pre screen null}, $Z_*(r,p,q)$ satisfies
$$
\pa_r Z_*= \frac{\kappa}{2} Z_*''+H  Z_*'+\Big(\frac{1}{2}-\frac{1}{\kappa} \Big) H' Z_*+C(r) Z_*.
$$
We emphasize that these null-vector equations become \eqref{Lawler-Zhan PDE} if and only if $\kappa=4$.
\end{eg*}

\subsubsection{Screening} \label{sss: Screening}

We now find explicit solutions of \eqref{Lawler-Zhan PDE} for general $\kappa >0$ by means of screening, which is well-known method in CFT to find solutions of the Knizhnik-Zamolodchikov type equations, see e.g. \cite{MR2016311,MR1629472}. In general, this method is useful in the theory of multiple SLEs introduced by Dub\'{e}dat, see \cite{dubedat2007commutation}. For instance, in \cite{MR2253875}, Dub\'{e}dat found Euler type integral representation for the partition function of multiple SLEs in a simply connected domain. Lenells and Viklund utilized the idea of screening to find the explicit formula of Green's function for multiple SLEs, see \cite{LV17}.

We now focus on the SLE$(\kappa,\Lambda)$ starting from $p$ with one force point $q$.
For $p,q \in \R$ and $\zeta \in \C$, set
\begin{equation} \label{R-screening m.v.f.}
\Psi^\sharp(p,q,\zeta):= \OO\left[a\cdot p +a\cdot q -2a \cdot \zeta,\,\bfs{0}\right], \quad Z^{\sharp}=\E \Psi^{\sharp}.
\end{equation}
Then $\Psi^{\sharp}$ has the same conformal dimensions both at $p,q$ and it is a $(1,0)$-differential with respect to the screening variable $\zeta$, i.e.,
\begin{equation}\label{pre screen dim}
\lambda_p=\lambda_q=\frac{6-\kappa}{2\kappa}, \quad \lambda_\zeta=1.
\end{equation}
By \eqref{Partition gen}, we have
\begin{equation}
Z^\sharp=\begin{cases} \label{pre Z}
\Theta'(0)^{\frac{6}{\kappa}} \Theta(p-q)^{\frac{2}{\kappa}}
\Theta(p-\zeta)^{-\frac{4}{\kappa}} \Theta(\zeta-q) ^{-\frac{4}{\kappa}} &\text{for ER b.c.}
\vspace{0.5em}
\\
\Theta'(0)^{\frac{6}{\kappa}} \tilde{\Theta}(p-q)^{\frac{2}{\kappa}}
\tilde{\Theta}(p-\zeta)^{-\frac{4}{\kappa}} \tilde{\Theta}(\zeta-q) ^{-\frac{4}{\kappa}} &\text{for Dirichlet b.c.}
\end{cases}
\end{equation}
For generic $\kappa$ ($4/\kappa\notin\mathbb{N}$), we need to specify a principal branch (as a function of $\zeta$) in the expression \eqref{pre Z}. In what follows, we use the branch cut $[-\infty,q] \cup [p,\infty]$.

For a closed contour $\gamma$, set
\begin{equation}
 \Psi \equiv \Psi(p,q):= C(\kappa) \oint_{\gamma} \Psi^\sharp(p,q,\zeta)\, d\zeta,
\end{equation}
where the normalization constant $C(\kappa)$ is given by
\begin{equation} \label{scr normal const}
C(\kappa)=\left( 2\sin \frac{4\pi}{\kappa}  \right)^{-2} \frac{1}{\Gamma(1-4/\kappa)}.
\end{equation}
Here $\Gamma$ is the Gamma function. It is easy to see that $C(\kappa)$ has a simple pole at $\kappa$ if and only if $4/\kappa$ is a positive integer.

We denote by $\Pi_1 \equiv \Pi_1(\C \sm \{p,q \}, \zeta_0)$ the fundamental group of $\C \sm \{ p,q \}$ with base point $\zeta_0$.
If $4/\kappa$ is not a positive integer, in order to make the partition function
\begin{equation} \label{scr Z}
Z \equiv Z(r,p-q):=\E \Psi = C(\kappa) \oint_{\gamma} Z^\sharp(p,q,\zeta)\,d\zeta,
\end{equation}
single-valued, the homotopy class $[\gamma]$ should reside in the commutator subgroup of the $\Pi_1$. Moreover, this specific choice of $C(\kappa)$ makes the expression \eqref{scr Z} non-trivial for all values of $\kappa >0$ after removing possible singularities. We will explain this in Proposition~\ref{screen null} below.

The simplest example of such $\gamma$ is the Pochhammer contour $\mathcal{P}(p,q)$ entwining $p$ and $q$.
See Figure~\ref{Pochh} below for the description of $\mathcal{P}(p,q)$, where we indicate the branch cut as a dotted line.
It is an elementary but remarkable property that the winding number of $\mathcal{P}(p,q)$ with respect to each point $p,q$ is zero. This property allows us to define $Z^{\sharp}(p,q,\cdot)$ as a single-valued analytic function on the contour.
\begin{figure}[h]
	\begin{center}
		\includegraphics[width=5.0in]{fig1}
	\end{center}
	\caption{Pochhammer contour} \label{Pochh}
\end{figure}

Indeed, the Pochhammer contour is the typical example of path contained in the class of \textit{first twisted homology} or \textit{loaded cycle}, see  \cite{yoshida2013hypergeometric,MR2016311}. In other words for any (single-valued) function $G(\zeta)$ defined on $\mathcal{P}(p,q)$, we have
\begin{equation} \label{boundary Poch}
\oint_{\mathcal{P}(p,q)} \pa_\zeta G(\zeta) \, d\zeta =0 .
\end{equation}

We now present the following version of null-vector equations, which is precisely the first assertion of Theorem~\ref{main_SLE Z}.
\begin{prop} \label{screen null}
For any $\kappa>0$, $Z(r,\cdot)$ is a (non-trivial) real-valued solution to
	\begin{equation} 	\label{Z PDE_cho}
\displaystyle \pa_r Z = \frac{\kappa}{2} Z'' + H Z' + \left( \frac{3}{\kappa}-\frac{1}{2} \right)H' Z +C(r) Z,
\end{equation}
where $C(r)$ is given as \eqref{null vec const}
\end{prop}
\begin{proof}
Recall that $Z^{\sharp}(p,q,\zeta):=\E \Psi^{\sharp}[p,q,\zeta]$. Combining Lemma~\ref{pre screen null} with \eqref{pre screen dim},
\begin{align*}
\pa_r  Z^\sharp&= \frac{\kappa}{2} \pa^2 Z^\sharp+\left[\left(\frac{3}{\kappa}-\frac{1}{2}\right) H'+H'(p-\zeta) \right]Z^\sharp
\\
&-\Big( H\pa_q \tilde{Z}+H(p-\zeta) \pa_{\zeta}Z^\sharp \Big)+C(r)Z^\sharp.
\end{align*}
Observe that by the evaluation \eqref{pre Z},
$$\left(\pa+\pa_{q}+\pa_{\zeta}\right) Z^\sharp(p,q,\zeta)=0.$$
Therefore the above equation is rewritten as
\begin{align}\label{pre screen null vec}
\begin{split}
\pa_r Z^\sharp &= \frac{\kappa}{2} \pa^2 Z^\sharp + H \pa Z^\sharp + \left( \frac{3}{\kappa}-\frac{1}{2} \right)H' Z^\sharp+C(r) Z^\sharp+\pa_{\zeta}F(\zeta),
\end{split}
\end{align}
where
$$
F(\zeta):= H(p-q) \,Z^\sharp- H(p-\zeta)Z^\sharp.
$$
By integrating \eqref{pre screen null vec} with respect to screening variable $\zeta$ along $\mathcal{P}(p,q)$, the desired null-vector equation \eqref{Z PDE_cho} follows from \eqref{boundary Poch}.

It remains to verify the real-valuedness and non-triviality of $Z(r,\cdot)$. For $\kappa >4$, the expression \eqref{scr Z} can be decomposed as
\begin{align} \label{Pochh int decom}
Z&= C(\kappa)\sum_{j=1}^{4} \int_{\gamma_j} Z^\sharp(p,q,\zeta)\,d\zeta,
\end{align}
where the path $\gamma_j$'s are given as in Figure~\ref{Decomp Pochh} below.
\begin{figure}[h]
\begin{center}
\includegraphics[width=5.0in]{fig2}
\end{center}
	\caption{A contour deformation}\label{Decomp Pochh}
\end{figure}

We denote by $f(\cdot)$ the principal branch of $Z^\sharp(p,q,\cdot)$ defined on $\C \setminus ( [-\infty,q] \cup [p,\infty] )$. Then one can express $Z^\sharp(p,q,\cdot)$ on $\gamma_j$ as
$$
Z^\sharp(p,q,\zeta)=
\begin{cases}
f(\zeta) & \textrm{if} \quad  \zeta \in \gamma_1
\\
\vspace{0.5em}
f(\zeta)e^{-2\pi i \frac{4}{\kappa}\pi} & \textrm{if} \quad  \zeta \in \gamma_2
\\
\vspace{0.5em}
f(\zeta) & \textrm{if} \quad  \zeta \in \gamma_3
\\
\vspace{0.5em}
f(\zeta)e^{2\pi i \frac{4}{\kappa}\pi} & \textrm{if} \quad  \zeta \in \gamma_4
\end{cases}
$$
Therefore the Pochhammer contour integration \eqref{Pochh int decom} coincides with the following Euler type integral
\begin{equation} \label{Euler integral}
Z(r,p-q)=\frac{1}{\Gamma(1-4/\kappa)} \int_{q}^{p} Z^\sharp(p,q,t) \, dt.
\end{equation}
The integrability condition is satisfied both at $p$ and $q$ if and only if $\kappa >4$. Note that the integrand $Z^\sharp(p,q,\cdot)$ is a positive function since $\Theta(r,\cdot)$ and $\tilde{\Theta}(r,\cdot)$ are non-negative functions in the interval $[0,2\pi]$ vanishing only at $0$,$2\pi$. Therefore the positivity of $Z(r,\cdot)$ for $\kappa>4$ follows.

We remark that as a function of $\kappa$, $Z$ can be interpreted as an analytic continuation of the right-hand side of \eqref{Euler integral}, which implies the real-valuedness of $Z$ for $\kappa \le 4$.
Now it remains to check the non-triviality of $Z$.
By \eqref{degen gen}, we have (up to a multiplicative constant)
\begin{align*}
\begin{split}
Z_\infty(p-q)&:=\lim\limits_{r\rightarrow \infty} Z(r,p-q)
\\
&=C(\kappa) \sin^{\frac{2}{\kappa}}\left( \frac{p}{2} \right) \oint_{\mathcal{P}(p,q)} \sin^{-\frac{4}{\kappa}} \left( \frac{p-\zeta}{2} \right) \sin^{-\frac{4}{\kappa}}\left( \frac{\zeta-q}{2} \right) d\zeta.
\end{split}
\end{align*}
Therefore for $\kappa >4$,
\begin{align*}
Z_\infty(p-q)&=\frac{1}{\Gamma(1-4/\kappa)} \sin^{\frac{2}{\kappa}}\left( \frac{p}{2} \right) \int_{q}^{p} \sin^{-\frac{4}{\kappa}}\left( \frac{p-t}{2} \right) \sin^{-\frac{4}{\kappa}}\left( \frac{t-q}{2} \right) dt.
\end{align*}
The regularized hypergeometric function
$$
{}_2\bfs{F}_1(a,b;c;z):=\frac{  {}_2F_1(a,b;c;z)   }{\Gamma(c)}
$$
is an entire function of $a,b,c$.
Then by \cite[Eq.(15.6.1)]{olver2010nist}, we have
\begin{equation} \label{Z deg}
Z_\infty(x)=\cos^{\frac{2}{\kappa}-1} \Big( \frac{x}{4} \Big)  \sin^{1-\frac{6}{\kappa}} \Big( \frac{x}{4} \Big) {}_2\bfs{F}_1\left(\frac{1}{2}, 1-\frac{4}{\kappa}; \frac{3}{2}-\frac{4}{\kappa}, -\tan\Big(\frac{x}{4}\Big)^2        \right).
\end{equation}
Therefore by the identity theorem, \eqref{Z deg} holds not only for $\kappa >4$ but also for $\kappa\le4$.
Therefore we conclude that $Z(r,\cdot)$ is a non-trivial function for all $\kappa >0$.
\end{proof}

\noindent \textbf{Example.} As a consequence of Proposition~\ref{screen null}, it is easy to observe that $Z_\infty$ solves following ordinary differential equation:
\begin{equation} \label{null vec deg}
0 = \frac{\kappa}{2} Z''_\infty + \cot\Big( \frac{x}{2}\Big) Z'_\infty + \left( \frac{3}{\kappa}-\frac{1}{2} \right)\cot\Big( \frac{x}{2}\Big)' Z_\infty -\frac{1}{2\kappa} Z_\infty.
\end{equation}
We remark that the ordinary differential equation \eqref{null vec deg} was also introduced by Zhan as a commutation relation for the radial SLE($\kappa,\Lambda$) process, see \cite[Section 4.5]{MR3334276}.
In particular for the values of $\kappa$ such that $4/\kappa$ is a positive integer, the expression \eqref{Z deg} is simplified and expressed in terms of trigonometric functions, see e.g., \cite[Section 15.4]{olver2010nist}. See for instance Table~\ref{table partition deg} below, where we write $\sin_2(x):=\sin(x/2)$ and $\cot_2(x)=\cot(x/2)$ to lighten notations.

\begin{table}[h] \renewcommand{\arraystretch}{2}
\caption{ Partition functions $Z_\infty$  } \label{table partition deg}
	\begin{center}
	\begin{tabular}{c|c}
	\toprule
			$\kappa$ &  $Z_\infty(\cdot)$ \\
	\midrule
			$4$  &$\sin_2^{-1/2}$ \\
			$2$  &$\sin_2^{-1} \cot_2$ \\
			$4/3$ & $\sin_2^{-3/2} \Big(3\cot_2^2-2\cot'_2+\dfrac{1}{3} \Big) $ \\
			$1$  & $\sin_2^{-2} \Big(4\cot_2^3-6\cot_2\cot'_2+\cot''_2+\cot_2 \Big) $ \\
	\bottomrule
	\end{tabular}
	\end{center}
\end{table}

\ms
\noindent \textbf{Example: Some particular solutions.} Observe that if $4/\kappa$ is a positive integer, $Z^\sharp(p,q,\cdot)$ is a well-defined meromorphic function in $\C$ having poles only at $p$ and $q$. Therefore one can choose sufficiently small circle around $p$ as an integration contour in \eqref{scr Z}. In this case, it is easy to calculate the annulus SLE partition functions by residue calculus. We present some particular solutions (up to multiplicative constant) in Table~\ref{table partition} for ER boundary condition.

\begin{table}[h] \renewcommand{\arraystretch}{2}
\caption{Partition functions $Z$} \label{table partition}
	\begin{center}
	\begin{tabular}{c|c}
	\toprule
		$\kappa$ &  $Z(r,\cdot)$ \\
	\midrule
		$4$ & $\Theta^{-1/2}$ \\
		$2$ & $\Theta^{-1} H$  \\
		$4/3$ & $\Theta^{-3/2} \Big(3H^2-2H'+4\dfrac{\zeta_r(\pi)}{\pi} \Big) $\\
		$1$ & $\Theta^{-2} \Big(4H^3-6HH'+H''+12 \dfrac{\zeta_r(\pi)}{\pi} H \Big) $ \\
	\bottomrule
	\end{tabular}
	\end{center}
\end{table}

Now we present the proof of Theorem~\ref{main_SLE Z}. Recall that
	$$\Psi \equiv \Psi(\xi,q):= C(\kappa) \oint_{\mathcal{P}(\xi,q)} \Psi^\sharp(\xi,q,\zeta)\, d\zeta. $$
	By definition, $\E\Psi=Z$ and $\Lambda=\kappa (\log Z)'$.

\begin{proof}[Proof of Theorem~\ref{main_SLE Z}]
	By Proposition~\ref{screen null}, it remains to prove the second assertion. For any string $\XX$ of fields in the OPE family $\FF_{(b)}$, let
	$$R_\xi=\wh{\E} \XX:= \frac{\E\Psi(\xi,q) \XX}{ \E\Psi(\xi,q) }.$$
	Note that all we need to show is the following version of BPZ-Cardy equation
	\begin{equation} \label{BPZ_C_scr}
	\frac{1	}{a^2}\pa_{\xi}^2R_\xi+ \Lambda \pa_{\xi}R_\xi
	= \LL^+_{v_{\xi}} R_\xi +\pa_r R_\xi.
	\end{equation}
	By Proposition \ref{BPZ_ER}, we have
	\begin{align}
	\begin{split}
	\label{scr BPZ}
	\frac{1	}{a^2	}\pa_{\xi}^2 \E[ \Psi^\sharp(\xi,q,\zeta) \XX] &
	=\LL_{v_{\xi} }^+(q, S_\XX )\E [\Psi^\sharp(\xi,q,\zeta) \XX]+\LL_{v_{\xi} }^+(\zeta )\E [\Psi^\sharp(\xi,q,\zeta) \XX]
	\\
	&+ \left(-C(r) +\pa_r \right) \E[\Psi^\sharp(\xi,q,\zeta) \XX],
	\end{split}
	\end{align}
	where $C(r)$ is given as \eqref{null vec const}. Note that since $\Psi^\sharp$ is a $(1,0)$-differential with respect to $\zeta$, we have
	\begin{equation*}
	\LL_{v_{\xi} }^+(\zeta )\E \left[\Psi^\sharp(\xi,q,\zeta) \XX \right]=\pa_\zeta \Big[		v_{\xi}(\zeta) \E [\Psi^\sharp(\xi,q,\zeta) \XX]\Big].
	\end{equation*}
   By \eqref{boundary Poch},
	$$ \oint_{\mathcal{P}(p,q)} \LL_{v_{p} }^+(\zeta )\E [\Psi^\sharp(p,q,\zeta) \XX]d\zeta=0.$$
	Integrating \eqref{scr BPZ} with respect to $\zeta$ along $\mathcal{P}(p,q)$, we obtain
	\begin{equation}\label{gen screen}
	\frac{1	}{a^2} \pa_{\xi}^2\Big(Z R_\xi \Big)=\LL^+_{v_{\xi}} \Big(Z R_\xi\Big)+ \Big( -C(r) +\pa_r \Big) \Big(Z R_\xi \Big).
	\end{equation}
	Combining \eqref{gen screen} with Proposition~\ref{screen null} we conclude \eqref{BPZ_C_scr}, which completes the proof.
\end{proof}


TITLE: Points of a variety defined by Galois descent
QUESTION [8 upvotes]: Let k be a perfect field.  By a k-variety, I shall mean a geometrically reduced separated scheme of finite type over k.  I think that is enough conditions that the following data determine an affine k-variety:

A subset $X(\bar{k})$ of $\bar{k}^n$ which is defined by polynomials
A continuous action of $\mathop{\mathrm{Gal}}(\bar{k}/k)$ on $X(\bar{k})$, such that each $\sigma \in \mathop{\mathrm{Gal}}(\bar{k}/k)$ acts as $\sigma \circ f$ where f is a $\bar{k}$-regular map

When I say that these data determine an affine k-variety, I mean that there is a unique affine k-variety X whose $\bar{k}$-points are $X(\bar{k})$ with the correct Galois action.
Given these data, I want to work out the functor of points of X (which I consider to have domain the category of k-algebras).  You can do that by following through the proof that these data determine a k-variety: first construct the coordinate ring A of X, as the Galois-fixed points of the ring of regular functions $X(\bar{k}) \to \bar{k}$; then $X(R) = \mathop{\mathrm{Hom}}(A, R)$ for any k-algebra R.
But if L is an algebraic extension of k, then there is a much simpler way of working out the L-points of X: just take the subset of $X(\bar{k})$ fixed by $\mathop{\mathrm{Gal}}(\bar{k}/L)$.
If L is a transcendental extension of k (or even a k-algebra which is not a field), is there a direct way of writing down the L-points of X which does not require going through the coordinate ring (or essentially equivalently, going through defining equations for X)?

REPLY [12 votes]: The following seems to give a reasonable affirmative answer which avoids computing the coordinate ring directly, and replaces condition (2) with the more natural condition that the subset $\Sigma := X(\overline{k})$ in (1) is stable under the action of the Galois group on $\overline{k}^n$.
Let's be cleaner by working more generally over an arbitrary (not necessarily perfect) field $k$ and with geometrically reduced closed subschemes $X$ in a fixed separated $k$-scheme $Y$ locally of finite type.  (Note: now affine schemes are gone; can take $Y$ to be an affine space, but this is irrelevant.)  The
${\rm{Gal}}(k_s/k)$-stable set $\Sigma = X(k_s)$ in $Y(k_s)$ recovers $X$ as follows. For a $k$-algebra $A$, $X(A)$ is the ${\rm{Gal}}(k_s/k)$-invariants in $X(A_{k_s})$, so we just need to describe $X(A_{k_s})$ as a ${\rm{Gal}}(k_s/k)$-stable subset of $Y(A_{k_s})$. The description in this latter case will be in terms of $\Sigma$, and the ${\rm{Gal}}(k_s/k)$-stability of $\Sigma$ inside of $Y(k_s)$ will ensure that the description we give for  $X(A_{k_s})$ is ${\rm{Gal}}(k_s/k)$-stable inside of $Y(A_{k_s})$.  That being noted, we rename $k_s$ as $k$ so that $k$ is separably closed and $\Sigma$ is simply a set of $k$-rational points of $Y$ (so the notation is now marginally cleaner).
First assume $A$ is geometrically reduced in the sense that $A_K$ is reduced for any extension field $K/k$.   Since $X(A)$ is the direct limit (inside $Y(A)$) of the $X(A_i)$ as $A_i$ varies through $k$-subalgebras of finite type in $A$ (all of which are geometrically reduced), we may assume $A$ is finitely generated over $k$.
Then the $k$-points are Zariski-dense (as $k = k_s$) and so the condition on $y \in Y(A)$ that it lies in $X(A)$ is that $y(\xi) \in \Sigma$ for all $k$-points $\xi$ of $A$.  That describes $X(A)$ for any (possibly not finitely generated) $k$-algebra $A$ that is geometrically reduced. In general, to check if $y \in Y(A)$ lies in $X(A)$ amounts to the same for each local ring of $A$, so we can assume $A$ is local.  Then the condition for $y$ to be in $X(A)$ is exactly that there is a local map of local $k$-algebras $B \rightarrow A$ with $B$ geometrically reduced such that $y$ is in the image of $X(B)$ under the induced map $Y(B) \rightarrow Y(A)$.  I don't claim this formulation is the best way to think about it, but it "works".
Of course, one can apply this process to any ${\rm{Gal}}(k_s/k)$-stable subset $\Sigma$ of $Y(k_s)$ provided that we first replace $\Sigma$ with with the set of $k_s$-points of its Zariski-closure in $Y_{k_s}$. Then we just obtain the Galois descent $X$ of the Zariski closure in $Y_{k_s}$ of $\Sigma$.  In general $X(k_s)$ may be larger than $\Sigma$, but nonetheless $\Sigma$ is Zariski-dense in $X_{k_s}$.  This is perfectly interesting in practice, regardless of whether or not $\Sigma$ is equal to $X_{k_s}$, since it is what underlies the construction of derived groups, commutator subgroups, images, orbits, and related things in the theory of linear algebraic groups over a general field.  For example, the $k$-group ${\rm{PGL}}_n$ is its own derived group in the sense of algebraic groups, but the commutator subgroup of ${\rm{PGL}}_n(k_s)$ is a proper subgroup whenever $k$ is imperfect and ${\rm{char}}(k)|n$.
To give a nifty application, suppose one begins with an arbitrary closed subscheme $X'$ in $Y$ (such as $X' = Y$!), then forms the ${\rm{Gal}}(k_s/k)$-stable set $X'(k_s)$ (which could well be empty, or somehow really tiny), and then applies the above procedure to get a geometrically reduced closed subscheme $X$ in $X'$.  What is it?  It is the maximal geometrically reduced closed subscheme of $X'$, and one can check its formation is compatible with products (as well as separable extensions $K/k$, such as completions $k_v/k$ for a global field $k$).  If $k$ is perfect then $X = X'_{\rm{red}}$, so this is more interesting when $k$ is imperfect.  It is especially interesting in the special case when $X'$ is equipped with a structure of $k$-group scheme.  Then $X$ is its maximal smooth closed $k$-subgroup, since geometrically reduced $k$-groups locally of finite type are smooth.  So what?  If one is faced with the task of studying the Tate-Shararevich set for such an $X'$ (e.g., maybe $X'$ is a nasty automorphism scheme of something nice) then all that really intervenes is $X$ since it captures all of the local points, so for some purposes we can replace the possibly bad $X'$ with the smooth $X$. (This trick is used in the proof of finiteness of Tate-Shafarevich sets for arbitrary affine groups of finite type over global function fields.)  But beware: if the $k$-group $X'$ is connected (and $k$ is imperfect) then $X$ may be disconnected and have much smaller dimension; see Remark C.4.2 in the book "Pseudo-reductive groups" for an example.


TITLE: What is a Chain rule for gradient?
QUESTION [1 upvotes]: If $f:\mathbb R \to \mathbb R$ and $g:\mathbb R \to \mathbb R$ be defirrentiable funtion then we know that the composition funtion $h=g\circ f$ is diffentiable and $$\frac{d}{dx} h(x)= h'(x)= f'(g(x)) \cdot g'(x) $$ (Here $\cdot $ denotes the real number multiplication)
Suppose that $f:\mathbb R^{n} \to \mathbb C$ and $g:\mathbb C \to \mathbb C.$ Consider composition function $h:\mathbb R^{n} \to \mathbb C$ as $h(x)= g\circ f(x)$ for $x\in \mathbb R^n.$

My Question: (1) What is the anologe of the above formula in higher dimension?
   (2) What is a gradient of $h= g\circ f:\mathbb R^n \to \mathbb C$? In other words, what is, keeping  the above formula in mind,  $\nabla h (x)$, ($x\in \mathbb R^n$)? Can we say $\nabla h(x)= \nabla g (f(x)) \cdot \nabla u (x)$ and now what is a meaning of $\cdot $ here?

REPLY [2 votes]: It is easier to view $\mathbb{C}$ as $\mathbb{R}^2$ unless you assume analyticity. Looking at it this way, $g$ and $f$ both have Jacobian matrices $J_f : \mathbb{R}^n \to \mathbb{R}^2$ and $J_g : \mathbb{R}^2 \to \mathbb{R}^2$ and you have $J_{g \circ f}(x)=J_g(f(x)) J_f(x)$ (where concatenation denotes matrix multiplication or equivalently composition of linear maps). If in fact $g$ is a holomorphic function then $J_g$ is isomorphic to the complex-valued function $g'$ by the isomorphism $a+bi \sim \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$.


TITLE: What is the relationship between Elasticity and Compliance?
QUESTION [2 upvotes]: Compliance is like elasticity of hollow tube. Elasticity is less for instance for arteries so they are less compliant.
There seems to some sort of relationship between compliance and elasticity.
Based on Tyler's comments:

Compliance is the derivative of strain with respect to stress (or derivative of displacement with respect to force).
Elasticity is like opposite of stiffness.

about which I am not sure.
One figure about compliance for vein and artery:

where I think the slope of the curve (gradient) is the compliance of each curve.
What is an approximate relationship between the elasticity and compliance?

REPLY [1 votes]: But then more elastic the structure is more will be the difficulty in stretching it or trying to develop elastic energy in it having said that why cant we consider compliance to be the product of distentibility and the volume of the vessel and because of this even highly elastic vessels have higher compliance beacuse of the large volume of blood like the aorta which they can accomodate


TITLE: Why is $\alpha^n\beta^m+\alpha^m\beta^n\le\alpha^s\beta^t+\alpha^t\beta^s$ if $(n,m)\preceq(s,t)$?
QUESTION [3 upvotes]: I've recently come across the following statement:
$$\alpha^n\beta^m+\alpha^m\beta^n\le\alpha^s\beta^t+\alpha^t\beta^s$$
for $\alpha,\beta\ge0$, $n,m,s,t\in\mathbb N_+$, $n+m=s+t$, and $(n,m)\preceq(s,t)$.
If anyone is interested (though it's not particularly relevant for the question itself), I found this stated in a slightly different form in this paper, page 2, second column. They reference a book on majorization in which I wasn't able to find an easy answer to this particular point (both paper and book are paywalled, sorry about that).
I can prove the case $n=m$ (see below), but I'm at a loss for the more general statement. What is an easy way to prove this?
Also, can this sort of statement be generalized to products of more than two numbers? The direct generalization would seem to be
$$\sum_{\sigma\in\mathcal P_N}\prod_{i=1}^N \alpha_{\sigma(i)}^{n_i}\le\sum_{\sigma\in\mathcal P_N}\prod_{i=1}^N \alpha_{\sigma(i)}^{m_i}$$
when $(n_1,..,n_N)\preceq(m_1,...,m_N)$.
If this holds, can the same technique used for the first statement be applied to this more general case?

For $n=m$ we write $s=\mu+\Delta, t=\mu-\Delta$, where $$\mu\equiv\frac{s+t}{2}=\frac{n+m}{2}=n.$$
We thus have
\begin{align}
(\alpha^n\beta^m+\alpha^m\beta^n)-(\alpha^s\beta^t+\alpha^t\beta^s)
&=2(\alpha\beta)^n-(\alpha\beta)^{\mu-\Delta}(\alpha^{2\Delta}+\beta^{2\Delta}) \\
&=(\alpha\beta)^{\mu-\Delta}\left[2\alpha^\Delta\beta^\Delta-\alpha^{2\Delta}-\beta^{2\Delta}\right]\le0,
\end{align}
where the last step follows from
$$2\alpha^\Delta\beta^\Delta-\alpha^{2\Delta}-\beta^{2\Delta}
=-(\alpha^\Delta-\beta^\Delta)^2.$$
The sign of the inequality is consistent with the original statement because
$(n,n)\preceq(n+\Delta,n-\Delta)$ (provided $n-\Delta\ge0$).

REPLY [1 votes]: The inequality you are interested in is wrong. To see this, take $m=3, n=1, s=t=2$. Then it is easy to see that the following inequality $\alpha^3\beta + \alpha \beta^3 \leq 2 \alpha^2 \beta^2$ does not hold.
In fact, it must have an opposite sign. In this case it is a particular case of a powerful Muirhead's inequality.
Let $a = (a_1,\dots, a_n)$ and $x=(x_1,\dots, x_n)$ be real nonnegative numbers. Define a symmetric polynomial $a\mapsto\Phi(a)$ with variables $x_1,\dots, x_n$ to be an average of all possible terms $x_{\sigma_1}^{a_1}...x_{\sigma_n}^{a_n}$ over all possible permutations $\sigma $ of ${1,\dots, n}$. In other words, it is
$$\Phi (a) = \frac{1}{n!}\sum_{\sigma}x_{\sigma_1}^{a_1}...x_{\sigma_n}^{a_n}.$$
For example, $\Phi(3,2) = \frac 1 2 (x_1^3x_2^2 + x_1^2x_2^3)$, $\Phi(3,2,0) = \frac 1 6 (x_1^3x_2^2 + x_1^3 x_3^2 + x_1^2x_2^3+x_1^2x_3^3 + x_2^3x_3^2+x_2^2x_3^3)$.
Then Muirhead's inequality says that if $a$ majorizes $b$, then $\Phi(a)\geq \Phi(b)$.
In your case $\Phi(m,n) = \frac 1 2 (\alpha^m \beta^n +\alpha^n \beta^m)$ and similarly for $\Phi(s,t)$. Now since $(n,m)$ majorizes $(s,t)$, then $\Phi (n,m)\geq \Phi(s,t)$.
The generalization works in the same way.


TITLE: Domain of a Trigonometric Composite Function
QUESTION [2 upvotes]: I am struggling with a question based on finding the domain of a composite function. I have managed to complete questions such as in the following document at the bottom of page three:
http://www.sinclair.edu/centers/mathlab/pub/findyourcourse/worksheets/Algebra/CompositeFunctionsAndTheirDomains.pdf
However, this question feels different and I am unsure of how the answer was gained:
Let $f$ and $g$ be the functions:
$$ f : [0, 2\pi] \to \mathbb R, \quad f(x) = \cos (x), $$
and
$$g : [0, 20] \to\mathbb R, \quad g(x) = x^2$$
Determine the domain for each of the compositions
$f \circ g$ and $g \circ f$. "
The given answers are:

$$\operatorname{Dom}(f \circ g) = [0,\sqrt{2\pi} ]$$

and

$$\operatorname{Dom}(g \circ f) = \:\left[0,\frac{\pi }{2}\right]\:\bigcup \:\left[\frac{3\pi }{2},\:2\pi \right]$$

Could anybody help me in working out how to find these domains please?
Thank you,
Lewis

REPLY [1 votes]: Let's consider $f\circ g$; for $x$ in the still to find domain, we have
$$
f\circ g(x)=f(g(x))
$$
so in particular $g(x)\in[0,2\pi]$, because otherwise we can't compute $f$ on it. On the other hand, whenever $g(x)\in[0,2\pi]$, we can compute $f(g(x))$. So we need to ensure
$$
0\le x^2\le 2\pi
$$
that is,
$$
x\in[-\sqrt{2\pi},\sqrt{2\pi}]
$$
However, we must take into into account that $0\le x\le 20$. So we get
$$
x\in[-\sqrt{2\pi},\sqrt{2\pi}]\cap[0,20]=[0,\sqrt{2\pi}]
$$
For $g\circ f$ it's similar: we need $f(x)\in[0,20]$, that is
$$
0\le \cos x\le 20
$$
with the condition that $x\in[0,2\pi]$. The cosine is always at most $1$, so we just need $\cos x\ge0$, which happens for $x$ in the sets of the form
$$
[0+2k\pi,\pi/2+2k\pi]\cup[3\pi/2+2k\pi,2\pi+2k\pi]
$$
with $k$ integer. Intersecting with $[0,2\pi]$ gives
$$
[0,\pi/2]\cup[3\pi/2,2\pi]
$$


TITLE: Solve $x+yp=ap^2$ where $p=\frac {dy}{dx}$
QUESTION [2 upvotes]: Solve $x+yp=ap^2$ where $p=\frac {dy}{dx}$
My Attmept:
$$x+yp=ap^2$$
$$x=ap^2-yp$$
This is solvable for $x$ so differentiating both sides w.r.t $y$
$$\frac {dx}{dy} = 2ap\cdot \frac {dp}{dy} - y\cdot \frac {dp}{dy} -p$$
$$\frac {1}{p} + p = (2ap-y) \cdot \frac {dp}{dy}$$
$$\frac {1+p^2}{p} = (2ap-y) \cdot \frac {dp}{dy}$$
$$\frac {dy}{dp} = \frac {2ap^2}{1+p^2} - \frac {yp}{1+p^2}$$
$$\frac {dy}{dp} + \frac {p}{1+p^2} \cdot y = \frac {2ap^2}{1+p^2}$$
$$\textrm {This is Linear so}$$
$$\textrm {Integrating Factor}=e^{\int \frac {p}{1+p^2} dp}=\sqrt {1+p^2}$$
So we have
$$y\sqrt {1+p^2} = \int \frac {2ap^2}{1+p^2} \times \sqrt {1+p^2} dp + c$$
$$y\sqrt {1+p^2} = 2a \int \frac {p^2}{\sqrt {1+p^2}} dp + c $$
I could not solve further from here.

REPLY [1 votes]: Usually you get fastest to the relevant equations if you start with the derivative by $p$, using $\dot y(p)=p\dot x(p)$. Then
$$
\dot x+p\dot y+y=2ap
$$
Multiply with $p$ and eliminate $x$
$$
(1+p^2)\dot y+py=2ap^2\implies \sqrt{1+p^2}y(p)=\int \frac{2ap^2}{\sqrt{1+p^2}}dp
$$
Now substitute $p=\sinh(u)$ in the last integral,
$$
=\int a(\cosh(2u)-1)du
=a\sinh(u)\cosh(u)-au+C
=ap\sqrt{1+p^2}-a\sinh^{-1}(p)+C.
$$


TITLE: Why do you use the inverse matrix to find the image of a curve in the plane?
QUESTION [2 upvotes]: Given a $2\times2$ linear transformation matrix and the equation of a curve in the plane (e.g. $x^2+y^2=1$), why does one use the inverse matrix to find the equation of the image of the equation after the transformation?
I recently watched a video that explained how $2\times2$ matrices represent linear transformations in the plane.  For example, if I wanted to find the image of the point $(1,0)$ after the transformation given by the matrix
$$
\left[\begin{matrix}2&1\\-1&0\\\end{matrix}\right],
$$
I would simply multiply:
$$
\left[\begin{matrix}2&1\\-1&0\\\end{matrix}\right]\left[\begin{matrix}1\\0\\\end{matrix}\right]=\left[\begin{matrix}2\\-1\\\end{matrix}\right]\rightarrow(2,-1).
$$
To extend this idea, I wanted to see the image of the circle $x^2+y^2=1$ after the transformation given by the same matrix.  I figured that since
$$
\left[\begin{matrix}2&1\\-1&0\\\end{matrix}\right]\left[\begin{matrix}x\\y\\\end{matrix}\right]=\left[\begin{matrix}2x+y\\-x\\\end{matrix}\right],
$$
the equation of the transformed circle would be $(2x+y)^2+(-x)^2=1$.  But after testing a few points on the circle, I quickly realized that it was not the correct equation.  Instead, the correct equation turned out to be
$$
(-y)^2+(x+2y)^2=1,
$$
which is obtained by multiplying $x$ and $y$ by the inverse matrix:
$$
\left[\begin{matrix}2&1\\-1&0\\\end{matrix}\right]^{-1}\left[\begin{matrix}x\\y\\\end{matrix}\right]=\left[\begin{matrix}0&-1\\1&2\\\end{matrix}\right]\left[\begin{matrix}x\\y\\\end{matrix}\right]=\left[\begin{matrix}-y\\x+2y\\\end{matrix}\right]
$$
So why do we use the original matrix to find the coordinates of the image of individual points (i.e. $(1,0)\rightarrow(2,-1)$), but the inverse matrix to find the equation of the image of the equation of a curve?  Is there a geometric explanation for this (or how do I explain this to my students)?

REPLY [0 votes]: A nice way to think about it is to consider the transformed curve as a geometric locus. Fist of all let the starting curve be $S$ and the transformed curve be $C$ and $\alpha$ the transformation itself. What we need to obtain is a function $f(X,Y)=0$ defined as the set $C$ of points $(X,Y) \implies (x,y) \in S$, namely:  $(X,Y) \in (C) \iff (x,y) \in (S)$.
Generally, when we calculate a geometric locus, we choose an equation that implies that a point is on the curve, and then we write the variables in that equation in terms of $x$ and $y$.
For example: the condition for a point to be on the unit circle is $d=1$, where $d$ is the distance from the origin. At this point we start writing the distance from the origin in terms of $(x,y)$.
For the transformed curve it is exactly the same. The condition for a point to be on the transformed curve is: $(x,y) \in S$, which is the equation of the original curve. Now we just need to write $(x,y)$ in terms of $(X,Y)$, and the job is done :)


TITLE: a Maximum of Discrete Function
QUESTION [3 upvotes]: Define a set
$$X=\{(x_1,\ldots ,x_n)\mid x_i=\pm 1,1\leq i\leq n\}$$
Fix $a$, $b\in X$. Consider the discrete function
$$F(x_1,\ldots,x_n)=(x_1a_1+\cdots+x_na_n)^2+(x_1b_1+\cdots+x_nb_n)^2$$
$(x_1,\ldots,x_n)\in X$.
I want to find when the function reaches the maximum. Is it at $x=a$? And how to prove it? Thank you.

REPLY [3 votes]: Say $S$ is the set of $j$ with $a_j=b_j$ and $T$ is the set of $j$ with $a_j=-b_j$. Let $$A(x)=\sum_{j\in S}a_jx_j,\quad B(x)=\sum_{j\in T}a_jx_j.$$
Then $$F(x)=(A+B)^2+(A-B)^2=2A(x)^2+2B(x)^2.$$
Since none of the variables appear in both $A(x)$ and $B(x)$ we get the maximum by maximizing $A^2$ and $B^2$ separately. And so the maximum of $F$ is $$2|S|^2+2|T|^2,$$where $|S|$ denotes the number of elements of $S$ and I'll just let you guess what $|T|$ is.
Ah, you wanted to know when $F$ is maximized. Looks like the maximum is achieved at four points: $x=a,b,-a,-b$.
And we get the minimum by minimizing $A^2$ and $B^2$ separately. The minimum of $A^2$ is $0$ or $1$ depending on whether $|S|$ is even or odd. Similarly for $B^2$, so the minimum of $F$ is $0$, $2$, or $4$, depending on the parity of $|S|$ and $|T|$.


TITLE: A faster way than tedious multiplying
QUESTION [3 upvotes]: Show that $(a+b+c)(a+b\epsilon +c\epsilon^2)(a+b\epsilon^2 + c\epsilon) = a^3 + b^3 + c^3 - 3abc$ If $$\epsilon^2 + \epsilon + 1 =0$$

The solution in the back of the book is given as

 Proved by a direct check, taking into consideration that $\epsilon^2 = -\epsilon -1 \: \:$ and $\epsilon^3 = 1$

However the solutions still feels like tedious multiplication, doesn't it? Is there a faster, more elegant way to do this?

REPLY [1 votes]: One way would be by factorizing $a^3+b^3+c^3-3abc$.
$$a^3+b^3+c^3-3abc=(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)$$
where $\omega$ is the cube root of unity.
More on the factorization can be found here.
Then, clearly, $\epsilon=\omega$, and thus, $$\epsilon^3=1$$
Factorizing this gives,
$$(\epsilon-1)(\epsilon^2+\epsilon+1)=0$$
Thus,
$$\epsilon^2+\epsilon+1=0$$


text:
\begin{document}

\maketitle
\begin{abstract}
In this paper we identify the cotangent to the derived stack of representations of a quiver $Q$ with the derived moduli stack
of modules over the Ginzburg dg-algebra associated with $Q$. More generally, we extend this result to finite type dg-categories,
to a relative setting as well, and to deformations of these. It allows us to recover and generalize some results of Yeung, and
leads us to the discovery of seemingly new lagrangian subvarieties in the Hilbert scheme of points in the plane.
\end{abstract}

\setcounter{tocdepth}{2}
\tableofcontents

\section{Introduction}

\input{CritQuiv-Section1}

\section{Motivation: new lagrangian subvarieties}\label{sec:motiv}

\input{CritQuiv-Section2}

\section{Shifted symplectic and lagrangian structures}\label{sec:recap}

\input{CritQuiv-Section3}

\section{Derived and non-derived examples}\label{sec:examples}

\input{CritQuiv-Section4}

\section{Calabi--Yau structures}\label{sec:CY}

\input{CritQuiv-Section5}

\section{Comparison}\label{sec:comparison}

\input{CritQuiv-Section6}

\appendix\section{A Lemma}

\input{CritQuiv-Appendix}


TITLE: Finding the horizontal asymptotes $ f(x) = \frac{x}{\sqrt{x^2+1}}$
QUESTION [2 upvotes]: Find the horizontal asymptotes of the grpah of the function f defined by
$$ f(x) =  \frac{x}{\sqrt{x^2+1}}$$
Solution:
$$ \lim_{x\to +\infty} \frac{\sqrt{x^2}}{\sqrt{x^2 +1}}$$
$$ \lim_{x\to +\infty} \sqrt{\frac{1}{1 + 1/x^2}} = 1$$
Okay, the book says the answer is 1 (I skipped the last step).
I'm curious, why did x^2 turn into 1 on the numerator and denominator, and why did the whole fration get a big fraction?
I think I know this somewhere in my head, I need refresher!

REPLY [2 votes]: Since detail has already been given for positive $x$, we deal with negative $x$, which have not been mentioned. If $x$ is negative, then $x=-\sqrt{x^2}$. Then the same argument as the one given shows that
$$\lim_{x\to-\infty}\frac{x}{\sqrt{x^2+1}}=-1.$$
So $y=-1$ is also a horizontal asymptote.
Another way of seeing this is that $f(x)$ is an odd function, meaning that $f(-x)=-f(x)$ for all $x$, So the graph of $y=f(x)$ for negative $x$ is obtained by rotating the graph for positive $x$ around the origin through $180^\circ$, or equivalently by reflecting in the origin.


TITLE: Check my proof of the "Boundedness theorem"
QUESTION [1 upvotes]: Theorem: Let $f$ be continuous on a closed interval $[a, b]$. Then f is bounded on $[a, b]$.
Proof (sketch): Suppose $f$ is unbounded. Let's define the set $N$ containing those $x$ for which $f$ is bounded on [a, x). Since $f$ is continuous there is $\delta$ such that $f$ is bounded on $[a, a + \delta)$, so $N$ is not empty. Since $N$ is bounded by $b$, we take $sup = supN$ and appealing to continuity of $f$ show that $sup$ belongs to $N$ and if $sup < b$ there is some $\delta$ such that $f$ is bounded on $[a, sup + \delta)$, so $sup + \delta$ must be in $N$ (a contradiction).

REPLY [1 votes]: The sketch looks fine for now, of course you will need to elaborate on some points, but it looks OK and it should go through. Be careful, though, as the closedness of the interval is vital and you must use it (see $1/x$ on $[-1,0)$) (also see the comment by  Martin Argerami on this topic) 2 minor points, though:
1) You do not need to work with reductio ad absurdum here. Just drop the "suppose $f$ is unbounded" at the beginning, and your proof will still hold. Basically, you first suppose $\not A$, then prove, $A$, then say "$A$ is in contradiction with $\not A$, therefore the original proposition of "$\not A$" is false, which means $A$ is true". I know this is just a minor complaint, but basically, it's simpler to just take the set $N$ and prove $N=[a,b]$, therefore proving $f$ is bounded.
2) Another, maybe simpler proof (this one DOES use contradiction): Suppose $f$ is unbounded. Then there exists a sequence $x_n$ on $[a,b]$ so that $|f(x_n)|>n$ for all $n\in \mathbb N$. Because $[a,b]$ is compact, $x_n$ has a convergent subsequence $x_{n_i}$, $i\in \mathbb N$ with limit $c$. Now, since $f$ is continuous, you know that
$$\lim_{i\rightarrow\infty} f(x_{n_i})=f(\lim_{i\rightarrow\infty} x_{n_i})=f(c),$$
however you also know that the limit does not exist. This is a contradiction, therefore $f$ cannot be unbounded.


TITLE: How to solve a binomial inequality
QUESTION [1 upvotes]: How to solve the following binomial inequality, the variable in this inequality is k,
$$\sum\limits_{i=k+1}^{100}{\left( \begin{matrix}
   100  \\
   i  \\
\end{matrix} \right)}{{0.8}^{i}}{{0.2}^{100-i}}\le \alpha,\alpha>0 $$
I think that the following formula is used, but I can not get its result
$$(x+y)^{n}=\sum _{k=0}^{n}{n \choose k}x^{n-k}y^{k}=\sum _{k=0}^{n}{n \choose k}x^{k}y^{n-k}.$$
Please help to suggest me how to find the range of k to satisfy the inequality.

REPLY [0 votes]: After deeply thinking about this problem, I cannot find any analysis solution for this, because we cannot specify an explicit result of:
$$\sum\limits_{i=0}^k{\left( \begin{matrix}
   100  \\
   i  \\
\end{matrix} \right)}{{0.8}^{i}}{{0.2}^{100-i}}$$
I solved this problem by writing a Matlab function to determine k value. It means that I solved it by a numerical method.


TITLE: Easiest read to learn number theory?
QUESTION [2 upvotes]: Recently I bought a "so called" intro book on number theory "Elementary number theory by burton" after reading some online recommendations to self learn number theory; but it is too much dry for a beginner like me and it feels like the author already assuming that reader is expert in algebra, proofs, combinatorics, math induction etc.. which I am not, if i was, why would I want an introductory number theory book. It is not about what material is in this book but HOW it is explained, it is intentionally written that way. anyways, is there any book that would be easy to read and has in depth and detailed explanations, step by step and goes slow in progress, some visual example would be very helpful but not necessary, should not be a typical coffee table book either but an standard text with easiest examples. I am not an student neither am in that age or mind, I just developed interest in number theory after reading "Journey through genius". it was very interesting book though.
Thanks.

REPLY [1 votes]: I read a lot of Silverman's A Friendly Introduction to Number Theory and enjoyed it quite a bit.   It's written in a down to earth style, and was actually intended to sort of woo non-math majors.
Recently I became aware that Rosenlicht wrote a book with Andre Weil entitled A Beginner's Guide to Number Theory.  It's based on a series of lectures by the latter at the university of Chicago.   I have high hopes for it, based on the fact that I'm familiar with Rosenlicht's work,  and even took one of his classes.
I have heard good things about Baker's book.  He won a Field's medal.
I haven't seen Davenport' s  Higher Arithmetic, but love the title.
Do remember that any book at all will tend to assume a certain amount of mathematical maturity.
I am currently reading Vinogradov's Elements of Number Theory, and find there are not many prerequisites.   It's nicely written,  and includes various exercises with solutions.