TITLE: Probability in a circle
QUESTION [0 upvotes]: Suppose a circle with the radius R. We randomly select a point within or on the circle. What is the probability of the chosen point distance from the center of the circle be smaller or equal to R/2?
I have two approaches as below:
Approach 1: The chosen point distance from the center is smaller or equal to R/2, if the point is within or on the concentric circle with the radius of R/2. The probability of randomly chosen point a point been in that area is $\pi*(R/2)^2\;\div\;\pi*R^2=1/4$
Approach 2: We consider a random radius and then pick a point on this radius. Now the chance of picking a point located on the inner half of the radius (between the center and R/2) is 1/2.
Could someone explain this?

REPLY [1 votes]: Let $R$ be the radius of the original circle. If randomly selecting a point from the circle implies that every point in the circle has the same (relative) probability of being picked, then the first approach is correct.
That is, the probability of a random point being within the circle of radius $R/2$ is
$$\frac{\pi (R/2)^2}{\pi R^2} = \frac{1}{4}.$$
The reason the second approach doesn't work is that there is a higher likelihood of picking points outside of the smaller circle with radius $R/2$ because of the difference in area. The concept of uniformly picking a random point $p \in (0,R)$ and then extending the result to an area doesn't work because area depends on the radius squared. This is why if you double the radius of the circle, the area does not double, but rather quadruples.
Now, we can 'fix' your second approach as follows:
We need a density function in the interval $[0,R]$ that is not uniform, but is rather an increasing function of the radius. We can start by considering the circumference of a circle or radius $r$:
$$f(r) = 2 \pi r.$$
Now notice
$$\int_{- \infty}^\infty f(r) dr = \int_{0}^R 2 \pi r dr =  \pi R^2$$
so that the area $A= \pi R^2$of the bigger circle is the normalizing constant for the density. That is,
$$f(r) = \frac{2 \pi r}{\pi R^2} = \frac{2r}{R^2} \tag{1}$$
corresponds to a valid density function defined for $r \in [0,R].$ Using $(1)$ as the new density, we obtain that the probability of a random point falling within the circle of radius $R/2$ corresponds to
$$\int_0^{R/2} f(r) dr =  \frac{2}{R^2} \int_0^{R/2} r dr =  \frac{2}{R^2}\cdot \frac{R^2/4}{2} = \frac{1}{4}.$$
This last result is essentially the same as the first approach.


TITLE: Please help with the following questions or at least one of them.
QUESTION [0 upvotes]: Problem 1. Let $f \colon X\to Y$ be a function from the set $X$ to the set $Y$. For a subset $A\subset X$, let $f_*(A)=\{ y\in Y | \exists x\in A\text{ such that }f(x)=y\}$. Prove the following:
  (i) The function $f \colon X\to Y$ is one-to-one if and only if $f_*(A\cap B)=f_*(A) \cap f_*(B)$ for all subsets $A$, $B$ of $X$.
  (ii) The function $f \colon X\to Y$ is onto if and only if $Y\setminus f_*(A) \subseteq f_*(X\setminus A)$ for all subsets $A$ of $X$.

I've managed to prove the first part of problem 1. Stuck on the second. And completely clueless about problem 2.
My attempt:
Problem 1
i) Suppose $y\in f_*(A \cap B)$. Then $f_*(x)=y$ such that $x \in A \cap B$. Then, $x \in A$ and $x \in B$.
                Therefore, $y \in f_*(A)$ and $y \in f_*(B)$.
              This implies $y \in f_*(A) \cap f_*(B)$.
              Therefore, $f_*(A\cap B)=f_*(A) \cap f_*(B)$.

REPLY [0 votes]: In your attempt you prove that $f_{*}\left(A\cap B\right)\subseteq f_{*}\left(A\right)\cap f_{*}\left(B\right)$
but not that $f_{*}\left(A\cap B\right)=f_{*}\left(A\right)\cap f_{*}\left(B\right)$
as you claim.
(i)$\Rightarrow$ From $A\cap B\subseteq A\wedge A\cap B\subseteq B$
it follows immediately that $f_{*}\left(A\cap B\right)\subseteq f_{*}\left(A\right)\cap f_{*}\left(B\right)$ (as you show in your attempt).
If $y\in f_{*}\left(A\right)\cap f_{*}\left(B\right)$ then $y=f\left(a\right)=f\left(b\right)$
for some $a\in A$ and some $b\in B$. Then $a=b\in A\cap B$ (because $f$ is one-to-one) so that
$y\in f_{*}\left(A\cap B\right)$.
(i) $\Leftarrow$ Let $u,v\in X$ with $y=f\left(u\right)=f\left(v\right)$.
For $A=\left\{ u\right\} $ and $B=\left\{ v\right\} $ we find $y\in f_{*}\left(A\right)\cap f_{*}\left(B\right)=f_{*}\left(A\cap B\right)$.
This implies that $A\cap B\neq\emptyset$ or equivalently that $u=v$.
(ii) $\Rightarrow$ Let $y\in Y-f_{*}\left(A\right)$. Then $y=f\left(x\right)$
for some $x\in X$. However, if $x\in A$ then $y=f\left(x\right)\in f_{*}\left(A\right)$
contradicting that $y\in Y\backslash f_{*}\left(A\right)$. We conclude
that $x\in X-A$ and consequently $y=f\left(x\right)\in f_{*}\left(X-A\right)$.
(ii) $\Leftarrow$Substituting $A=X$ leads to $Y-f_{*}\left(X\right)\subseteq f_{*}\left(X-X\right)=f_{*}\left(\emptyset\right)=\emptyset$.
So $Y=f_{*}\left(X\right)$, i.e. $f$ is onto.


text:
\begin{document}

\title[Hankel operators on the Fock-Sobolev spaces]{Hankel operators on the Fock-Sobolev spaces}

\author[\textbf{Anuradha Gupta} and \textbf{Bhawna Gupta}]
{\textbf{Anuradha gupta} \and \textbf{Bhawna gupta}\\}

\address{Anuradha gupta,
Delhi college of arts and commerce,
Department of mathematics,
University of Delhi,
Delhi,
India.}
\email{dishna2@yahoo.in}
\address{Bhawna Gupta,
Department of mathematics,
University of Delhi,
Delhi,
India.}
\email{swastik.bhawna26@gmail.com}

\begin{abstract} In this paper, we study operator-theoretic properties (boundedness and compactness) of Hankel operators on the Fock-sobolev spaces $ \mathscr{F}^{p,m} $ in terms of $ \mathcal{BMO}_r^p $ and $ \mathcal{VMO}_r^p $ spaces, respectively, for a non-negative integers $ m $, $ 1 \leq p < \infty $ and $ r > 0 $. Along the way, we also study Berezin transform of Hankel operators on $ \mathscr{F}^{p,m} $. The results in this article are analogous to Zhu's characterization and Per\"al\"a's characterization of bounded and compact Hankel operators on the Bergman spaces of unit disc and the weighted Fock spaces, respectively.
\end{abstract}

\subjclass[2010]{Primary 47B35; Secondary 30H20, 30H35.}

\keywords{Fock-Sobolev spaces, Hankel operators, Berezin transform,  $\mathcal{BMO}_r^p$ spaces, $\mathcal{VMO}_r^p$ spaces.}

\maketitle

\section{Introduction}

 The investigation of Hankel operators on several spaces like Hardy spaces, Bergman spaces, Bergman spaces on a cartain domains, Fock spaces, Fock-type spaces etc., has a long history in mathematics and mathematical physics. We refer \cite{georghankel, karelhankel, zhubmo, zhuoperator, zhuanalysis} for the study on these spaces. Zhu \cite{zhubmo} obtained a characterization of bounded and compact Hankel operators on the Bergman space by defining the spaces of bounded mean oscillation and vanishing mean oscillation with respect to the Bergman metric and analogous results were obtained by Perala, Schuster and Virtanen \cite{peralahankel} on the weighted Fock spaces. After that Cho and Zhu \cite{chofock1, chofock} gave a very useful Fourier characterization of a new class of spaces known as Fock-Sobolev spaces on $ \mathbb{C}^n $ ($ n \geq 1 $). Our aim is to study the basic properties of Hankel operators on the Fock-Sobolev spaces. In particular, we examine the boundedness and compactness of these operators in terms of $ \mathcal{BMO}_r^p $ and $ \mathcal{VMO}_r^p $ spaces for the generating symbols.

 Let $ dA $ be the Lebesgue area measure on $ \mathbb{C} $. For $ 1 \leq p \leq \infty $, let $ \mathscr{F}^p $ be the space of all entire functions $ f $ on the complex plane $ \mathbb{C} $ such that $ g(v)  e^{- \frac{1}{2} |v|^2} \in L^p (\mathbb{C}, dA(v)) $ with norm
\begin{equation*}
\| g \|_p = \left \{ \frac{p}{2\pi} \int_{\mathbb{C}} |g(v)|^p e^{- \frac{p}{2} |v|^2} dA(v)  \right\}^{\frac{1}{p}},
\end{equation*} for $ 1 \leq p < \infty $ and
\begin{equation*}
\| g \|_\infty = \underset{v \in \mathbb{C}}{\text{ ess sup }} \{ |g(v)| e^{- \frac{1}{2} |v|^2} \},
\end{equation*} for $  p = \infty. $

Throughout the paper, $ m $ is a fixed non-negative integer. The Fock-Sobolev space $ \mathscr{F}^{p,m} $ is the space of all entire functions $ g $ on $ \mathbb{C} $ such that
\begin{equation*}
 \| g\|_{p,m} = \sum_{0 \leq k \leq m} \| f^{(k)} \|_p < \infty.
\end{equation*}

Cho and Zhu \cite{chofock} proved that $ g \in \mathscr{F}^{p,m} $ if and only if $ z^m g \in \mathscr{F}^p $ and $ \| g \|_{p,m} $ can be taken as
\begin{equation*}
\| g \|_{p,m} = \left \{ \omega_{p,m}  \int_{\mathbb{C}} | g(v)|^p |v|^{mp} e^{- \frac{p}{2} |v|^2} dA(v) \right \}^\frac{1}{p} ; 1 \leq p < \infty
\end{equation*} and
\begin{equation*}
\| g \|_\infty = \underset{v \in \mathbb{C}}{\text{  sup }} { |g(v) v^m e^{- \frac{1}{2} |v|^2} |}  ; p = \infty,
\end{equation*}
where  $ \omega_{p,m} = (\frac{p}{2})^{\frac{mp}{2}+1} \frac{1}{\pi \Gamma (\frac{mp}{2}+1)} $.

Denote by $ L^{p,m} $, the space of all Lebesgue measurable functions $ g $ such that $ g(v) |v|^m e^{- \frac{1}{2} |v|^2} \in L^p(\mathbb{C}, dA(v))$ and $ \mathscr{F}^{p,m} $ is  a closed subspace of $ L^{p,m} $.

The space $ \mathscr{F}^{2,m} $, the closed subspace of the Hilbert space $ L^{2,m} $, is a Hilbert space with inner product
\[ \left \langle f, g \right \rangle_{p,m} = \frac{1}{\pi} \int_\mathbb{C} f(v) \overline{g(v)} |v|^{2m} e^{- |v|^2} dA(v) \text{ for all } f, g \in   \mathscr{F}^{2,m},
\]and  having reproducing kernel
\[ K^m(v,z)=  K_z^m(v) = \sum_{k=0}^\infty \frac{m!}{(k+m)!}( \bar{z} v)^k = m! \frac{( e^{\bar{z} v} - Q_m(\bar{z} v))}{(\bar{z} v)^m },
\]
where $ Q_m(w) $ is the Taylor polynomial of $ e^w $ of order $ (m-1) $ that is, $ Q_m(w) = \displaystyle \sum_{k = 0 }^{ m-1} w^k $. Let \[ k_z^m(v)  = \frac{K_z^m(v)}{\sqrt{K_z^m(z)}} = \frac{ ( e^{\bar{z}v} - Q_m(\bar{z}v))}{(\bar{z}v)^m} \left \{\frac{m!|z|^{2m}}{(e^{|z|^2} - Q_m (|z|^2))}\right \}^\frac{1}{2}\] denote the normalized reproducing kernel of $ \mathscr{F}^{2,m} $. Also, the sequence $ { b_k(v) }_{k=0}^\infty $ forms an orthonormal basis of $ \mathscr{F}^{2,m} $, where
\[
b_k(v) = \sqrt{\frac{m!}{(k+m)!}} v^k .
\]
Cho and Zhu \cite{chofock} showed that the orthogonal projection $ P^m : L^{2,m} \rightarrow \mathscr{F}^{2,m} $ given by
\[ P^m g(z)= \left \langle g, K_z^m \right \rangle_{2,m}  = \frac{1}{\pi} \int_\mathbb{C} g(v) \overline{K_z^m(v)} |v|^{2m} e^{- |v|^2} dA(v)
\] is a bounded projection from $ L^{p,m} $ onto $ \mathscr{F}^{p,m} $ for $ 1 \leq p \leq \infty $.

\section{\bf $\mathcal{BMO}_r^p $ spaces and boundedness of Hankel operators on $ \mathscr{F}^{p,m}$}
Let $ \Omega_m $ denote the space of all Lebesgue measurable functions $ g $ on $ \mathbb{C} $ such that $ g k_v^m  \in \bigcup_{p' \geq 1}L^{p',m} $ for each $ v \in  \mathbb{C} $. Let $ I $ denotes the identity operator on $ L^{p,m} $.

The following result can be found in \cite{chofock1}, following which it is clear that for each $ v \in \mathbb{C} $, the reproducing kernel $ \|K_v^m\|_{p', m} $ is finite for all possible $ p' \geq 1 $.
\begin{lemma}  Suppose $ m $ is a fixed non-negative integers and $Q_m(z) $ is the Taylor polynomial of $ e^z $ of order $ m-1 $(with the convention that $Q_0 = 0$). For any parameter $ p'> 0 $, $ \sigma > 0 $, $c> 0 $ and $d > -mp'-2$, we can find a positive constant $ C_0 $ such that
\[ \int_{\mathbb{C}} |e^{\bar{z}w} - Q_m{\bar{z}w} |^{p'} e^{-c|w|^2} |w|^{d}dA(w) \leq C_0 |z|^d e^{\frac{{p'}^2}{4c}|z|^2},
\] for all $ |z| \geq \sigma$. Furthermore, this holds for all $z$ if $ d \leq p'm $ as well.
\end{lemma}

Therefore, it follows that if $ g \in \Omega_m $, then the Hankel operator $ H_g^p : \mathscr{F}^{p,m} \rightarrow L^{p,m} $  with symbol $ g $, defined by $ H_g^p f = (I - P^m) gf $ for all $ f \in \mathscr{F}^{p,m} $, is densely defined on $ \mathscr{F}^{p,m} $, since the set of linear span of all kernel functions $ \{ k_v^m: v\in \mathbb{C} \} $ is dense in the space $ \mathscr{F}^{p,m} $.
 By using the definition of $ P^m $,
 we write
 \begin{equation}\label{seq}
  H_g^p f(z) = \frac{1}{\pi} \int_\mathbb{C} (g(z) - g(v) ) f(v) \overline{K_z^m(v)} |v|^{2m} e^{- |v|^2} dA(v).
\end{equation}
 Henceforth, for the convergence of integral in \eqref{seq}, we will assume that the symbol $ g  $ is in $ \Omega_m $.

For some $ z \in \mathbb{C} $, $ 1 \leq p < \infty $ and $ 0 < r < \infty $, let $ \mathscr{B}(z; r) = \{ v \in \mathbb{C}: | v - z | \leq r \} $ be the Euclidean disk centred at $ z $ and of radius $r$. Denote by $ L_{Loc}^p $, the space of all Lebesgue measurable functions $ g $ on $ \mathbb{C} $ such that $ g(v) \in L^p(K, dA(v)) $ for each compact subset $ K $ of $ \mathbb{C} $. Let $\mathcal{BA}_r $ be the set of all $ L_{Loc}^1$ integrable functions $ g $ on $ \mathbb{C} $ such that $ \tilde{g}_r $ defined by
\[ \tilde{g}_r (z) = \frac{1}{ \pi r^2} \int_{\mathscr{B}(z; r)} g(v) dA(v) \] is bounded on $ \mathbb{C}$. For finite $ p \geq 1$ and $ g \in L_{Loc}^p $, denote
\[ \tilde{g}_r ^p (z) = \frac{1}{ \pi r^2} \int_{\mathscr{B}(z; r)} |g(v)|^p  dA(v).\]
Let $ \mathcal{BA}_r^p $ be the set of all $ L_{Loc}^p $  integrable functions $ g $ on $ \mathbb{C} $ such that $\tilde{g}_r ^p $ is bounded on $ \mathbb{C}$.
Let $ \mathcal{BMO}_r^p $ denote the set of all $ L_{Loc}^p $  integrable functions $ g $ such that
\[ \| g \|_{\mathcal{BMO}_r^p} = \underset{z \in \mathbb{C}}{Sup} \left \{ \frac{1}{ \pi r^2} \int_{\mathscr{B}(z; r)} |g(v) - \tilde{g}_r  (z)|^p  dA(v)\right \}^{\frac{1}{p}} \] is finite.
Let $ \mathcal{BO}_r $ be the set of all continuous functions $ g $ on $ \mathbb{C} $ such that
\[ \| g \|_{\mathcal{BO}_r} = \underset{z \in \mathbb{C}}{Sup} \left \{ \underset{v \in \mathscr{B}(z; r)}{Sup} | g(v) - g(z) |\right \} < \infty.
\]

The following lemmas will be instrumental in further study of Hankel operators on $ \mathscr{F}^{p,m} $.

\begin{lemma}  \cite{peralahankel} \label{lem1} Let $ p \geq 1 $. Then the following conditions hold:
\begin{enumerate}
\item Let $ g \in L_{Loc}^p $ then $ g \in \mathcal{BMO}_r^p $ if and only if there is a constant $ C > 0 $ such that for every $ z \in \mathbb{C} $ there exists a constant $ \mu_z $ such that
\begin{equation*}
 \int_{\mathscr{B}(z; r)} |g(v) - \mu_z |^p  dA(v) \leq C.
\end{equation*}
\item For $ 0 < r < R $, $ \mathcal{BMO}_R^p \subset \mathcal{BMO}_r^p $.
\item $ \mathcal{BO}_r $ is independent of $r$. Moreover, for any continous function $ g $ on $ \mathbb{C} $, $ g \in \mathcal{BO} $ if and only if there exists a constant $ C_0 > 0 $ such that
 \begin{equation*}
 | g(z) - g(v) | \leq C_0(|z-v| +1)
\end{equation*}  for all $ z, v \in \mathbb{C}$.
\item If $ g \in \mathcal{BMO}_{2r}^p $, then $ \tilde{g}_r \in \mathcal{BO}_r $.
\item If $ g \in \mathcal{BMO}_{2r}^p $, then $ g - \tilde{g}_r \in \mathcal{BA}_r^p $.
\item $ \mathcal{BMO}_r^p \subset \mathcal{BO}_r + \mathcal{BA}_r^p$ for $ 0 < r < \infty $.
\end{enumerate}
\end{lemma}

\begin{lemma} \label{lem2} \cite{chofock1}  Suppose $ t \in \mathbb{R} $ and $ M > 0 $ be a fixed real number.
\begin{enumerate} \item Then there exists a constant $ C_0 > 0 $  such that
\begin{equation*}
\sum_{k=0}^\infty \left(\frac{y}{k+1}\right)^t \frac{y^k}{k!} \leq C_0 e^y
\end{equation*} for all real $ y \geq M $. Futhermore, this holds for all $ y \geq 0 $ if $ t \geq 0 $.
\item Then there exists a constant $ C_0 > 0 $  such that
\begin{equation*}
\sum_{k=0}^\infty \left(\frac{y}{k+1}\right)^t \frac{y^k}{k!} \geq C_0 e^y
\end{equation*} for all real $ y \geq M $. Futhermore, this holds for all $ y \geq 0 $ if $ t \leq 0 $.
\end{enumerate}
\end{lemma}
For any two points $ u $ and $v$ such that $ u $ and $ v $ do not lie on the same ray emanating from the origin, the lattice generated by $ u $ and $ v $ is the set $ \{ au+bv| a,b \in \mathbb{Z} \}$.

\begin{lemma} \label{lem3} \cite{wangtoeplitz}
Suppose $ \lambda $ is a locally integrable positive measure, $ p >0 $, $ r > 0 $,  $ m $ is a non- negative integer and $ \{ b_n \} $ is the lattice in $ \mathbb{C} $ generated by $ r $ and $ r i$. Then the following conditions are equivalent.
\begin{enumerate}
\item There exists a constant $ C_0 $ such that
\begin{equation*}
\int_{\mathbb{C}} |g(v) v^m e^{- \frac{|v|^2}{2}}|^p d \lambda \leq C_0 \| g \|_{p,m}^p
\end{equation*} for all entire functions $ g $.
\item There exists a constant $ C_0 > 0$ such that $ \lambda(\mathfrak{B}(z;r)) < C_0 $ for all $ z \in \mathbb{C} $.
\item There exists a constant $ C_0 > 0 $ such that $ \lambda(\mathfrak{B}(b_n ;r)) < C_0 $ for all positive integers $ n $.
\end{enumerate}
\end{lemma}

The Berezin transform of a function $ g $ is given by
\begin{align*}
\mathfrak{B_m} (g)(z) &= \left \langle g k_z^m, k_z^m \right \rangle_{2, m} = \frac{1}{\pi m!} \int_{\mathbb{C}} g(v) |k_z^m(v) |^2 |v|^{2m} e^{-|v|^2} dA(v)\\
& = \frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathbb{C}} g(v) | e^{\bar{z}v} - Q_m(\bar{z}v)|^2 e^{-|v|^2} dA(v),
\end{align*}where $ k_z^m $ denotes the normalized reproducing kernel of $ \mathscr{F}^{2,m} $.

\begin{proposition}
\label{lem4}
Let $ g \in \Omega_m $. For $ 1 \leq p \leq \infty $, the following conditions are equivalent:
\begin{enumerate}
\item $ g \in \mathcal{BA}_r^p$;
\item There exists a positive constant $ C $ such that \begin{equation*}
\frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathbb{C}}| g(v)|^p | e^{\bar{z}v} - Q_m(\bar{z}v)|^2 e^{-|v|^2} dA(v) \leq C
\end{equation*} for all $ z \in \mathbb{C}$;
\item  The multiplication operator $ L_g: \mathscr{F}^{p,m} \rightarrow L^{p,m} $ is bounded.
\end{enumerate}
\end{proposition}

\begin{proof} $ (1) \Leftrightarrow (2) $
Let $ g \in \mathcal{BA}_r^p $ then $ \int_{\mathscr{B}(z; r)} |g(v)|^p  dA(v) $ is bounded on $ \mathbb{C} $. Then lemma \ref{lem3} gives
\begin{equation*}
\int_{\mathbb{C}} |h(v) v^m e^{- \frac{|v|^2}{2}}|^p d \lambda \leq C_0 \| h \|_{p,m}^p
\end{equation*} for all entire functions $ h $ where $ d\lambda(v) = |g(v)|^p dA(v) $  if and only if $ g \in \mathcal{BA}_r^p $ and hence, it follows that  $ g \in \mathcal{BA}_r^p $ if and only if $ \mathfrak{B_m}|g|^p $ is bounded on $ \mathbb{C}$ where
\begin{equation*}
\mathfrak{B_m}| g(z)|^p = \frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathbb{C}}| g(v)|^p | e^{\bar{z}v} - Q_m(\bar{z}v)|^2 e^{-|v|^2} dA(v).
\end{equation*}
$(1) \Leftrightarrow (3) $ Let $ g \in \mathcal{BA}_r^p $ then by definition $ \tilde{g}_r^p $ is bounded. Define  a non-negative measure $ d \lambda (z) = |g(z)|^p  dA(z) $ on $ \mathbb{C} $ then $ \lambda(\mathscr{B}(z;r)) = \int_{\mathscr{B}(z;r)} d\lambda(v) = \int_{\mathscr{B}(z;r)} |g(v)|^p  dA(v) $. Therefore, from lemma \ref{lem3}, it follows that \begin{equation*}
\int_{\mathbb{C}} |h(v) v^m e^{- \frac{|v|^2}{2}}|^p d \lambda \leq C_0 \| h \|_{p,m}^p
\end{equation*} for all entire functions $ h $ if and only if  $ g \in \mathcal{BA}_r^p $. Thus, for all $ h \in \mathscr{F}^{p,m} $, we have
\begin{align*}
\| L_g(h)\|_{p,m}^p &= \| hg\|_{p,m}^p = \omega_{p,m} \int_{\mathbb{C}} |h(v)|^p |g(v)|^p |v|^{mp} e^{-\frac{p}{2} |v|^2 } dA(v)\\
& = \omega_{p,m} \int_{\mathbb{C}} |h(v)v^m e^{-\frac{1}{2} |v|^2 }|^p d\lambda (v) \leq C \| h \|_{p,m}^p
\end{align*} for some constant $ C > 0 $.
\end{proof}

Thus, from lemmas \ref{lem1} and \ref{lem4}, it can be observed that $ \mathcal{BO}_r $ and $ \mathcal{BA}_r^p $ are independent of $ r $ and hence, we will denote them by $ \mathcal{BO} $ and $ \mathcal{BA}^p $, respectively.

\begin{lemma} \label{lem5}
Let $ g \in \mathcal{BMO}_r^p $. Then
\begin{equation*}
\frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathbb{C}} |g(v)- \mathfrak{B_m}g(z)|^p | e^{\bar{z}v} - Q_m(\bar{z}v)|^2 e^{-|v|^2} dA(v)
\end{equation*} is bounded for  $ |z| > M $, for some positive constant $ M $.
\end{lemma}

\begin{proof}
Let $ g \in \mathcal{BMO}_r^p \subset \mathcal{BO} + \mathcal{BA}_r^p $, therefore, there exist two functions $ g_+ , g_- $ on $\mathbb{C} $ such that $ g_+ \in \mathcal{BO}_r $ and $ g_- \in \mathcal{BA}_r^p $.
Since $ g_+ \in \mathcal{BO}_r $, therefore, by using lemma \ref{lem2} and the fact that $ \mathcal{BO}_r $ is independent of $ r $ and
\begin{equation*}
 \lim_{\substack{v \in \mathscr{B}(z; r)\\ |z| \rightarrow \infty}} ( 1 - e^{-\bar{z}v}Q_m(\bar{z}v)) = 1,
\end{equation*}
 it follows that
\begin{align*}
& \left \{ \frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathscr{B}(z;r)} |g_+(v)- \mathfrak{B_m}g_+(z)|^p | e^{\bar{z}v} - Q_m(\bar{z}v)|^2 e^{-|v|^2} dA(v) \right \}\\
=& \left \{ \frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathscr{B}(z;r)} |g_+(v)- \mathfrak{B_m}g_+(z)|^p |e^{\bar{z}v}|^2 | 1 - e^{-\bar{z}v}Q_m(\bar{z}v)|^2 e^{-|v|^2} dA(v)  \right \}\\
& \leq C \left \{  \int_{\mathscr{B}(z;r)} |g_+(v)- \mathfrak{B_m}g_+(z)|^p e^{-|z-v|^2} dA(v) \right \}\\
& \leq C \left \{  \int_{\mathbb{C}} |g_+(v)- \mathfrak{B_m}g_+(z)|^p e^{-|z-v|^2} dA(v) \right \}\\
& = C \left \{  \int_{\mathbb{C}} |g_+(z-v)- \mathfrak{B_m}g_+(z)|^p e^{-|v|^2} dA(v) \right \},
\end{align*} for all $|z| > M $ for some positive constants $C $ and $M $, where
\begin{align*}
&| g_+(z-v)- \mathfrak{B_m}g_+(z)|\\
& = |g_+(z-v)- \frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathbb{C}} g_+(u) | e^{\bar{z}u} - Q_m(\bar{z}u)|^2 e^{-|v|^2} dA(u)|\\
& = |\frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathbb{C}} (g_+(z-v)- g_+(u)) | e^{\bar{z}u} - Q_m(\bar{z}u)|^2 e^{-|v|^2} dA(u)|\\
& = | \displaystyle \lim_{r \rightarrow \infty } \frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathscr{B}(z;r)} ( g_+(z-v)- g_+(u)) | e^{\bar{z}u} - Q_m(\bar{z}u)|^2 e^{-|u|^2} dA(u)|\\
& \leq  \displaystyle  \lim_{r \rightarrow \infty } |\frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathscr{B}(z;r)}( g_+(z-v)- g_+(u)) | e^{\bar{z}u} |^2 | 1 - e^{-\bar{z}u} Q_m(\bar{z}u)|^2 e^{-|u|^2} dA(u)|\\
& \leq C \displaystyle  \lim_{r \rightarrow \infty }  |\int_{\mathscr{B}(z;r)}( g_+(z-v)- g_+(u)) |  e^{-|z -u|^2} dA(u)|\\
& \leq C | \int_{\mathbb{C}} ( g_+(z-v)-g_+(u))   e^{-|z-u|^2} dA(u)|\\
& = C  \int_{\mathbb{C}}| g_+(z-v)-g_+(z - u)|   e^{-|u|^2} dA(u)|
\end{align*} for all $|z| > M $.
Therefore,
\begin{align*}
& \left \{ \frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathscr{B}(z;r)} |g_+(v)- \mathfrak{B_m}g_+(z)|^p | e^{\bar{z}v} - Q_m(\bar{z}v)|^2 e^{-|v|^2} dA(v) \right \}\\
& \leq C^2 \iint_{\mathbb{C}} | g_+(z-v)-g_+(z - u)|^p   e^{-|u|^2} dA(u)e^{-|v|^2} dA(v)\\
& \leq \iint_{\mathbb{C}}(|u-v| +1)^p  e^{-|u|^2} dA(u)e^{-|v|^2} dA(v)\\
\end{align*} for all $|z| > M $.
Now, since $ g_- \in \mathcal{BA}_r^p $, therefore by lemma \ref{lem4}, there exists a positive constant $ C $ such that
\begin{equation*}
\frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathbb{C}}| g_-(v)|^p | e^{\bar{z}v} - Q_m(\bar{z}v)|^2 e^{-|v|^2} dA(v) \leq C
\end{equation*}
for all $ z \in \mathbb{C}$. Consider

\begin{align*}
& \left \{ \frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathbb{C}} |g_-(v)- \mathfrak{B_m}g_-(z)|^p | e^{\bar{z}v} - Q_m(\bar{z}v)|^2 e^{-|v|^2} dA(v) \right \}^\frac{1}{p} \\
& \leq \left \{ \frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathbb{C}} |g_-(v)|^p | e^{\bar{z}v} - Q_m(\bar{z}v)|^2 e^{-|v|^2} dA(v) \right \}^\frac{1}{p} + | \mathfrak{B_m}g_-(z)| \| k_z \|_{2,m}^\frac{2}{p}\\
& \leq \left \{ \frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathbb{C}} |g_-(v)|^p | e^{\bar{z}v} - Q_m(\bar{z}v)|^2 e^{-|v|^2} dA(v) \right \}^\frac{1}{p} + | \mathfrak{B_m}g_-(z)|\\
& \leq C + | \mathfrak{B_m}g_-(z)|,
\end{align*}
where
\begin{align*}
| \mathfrak{B_m}g_-(z)| & \leq  \frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathbb{C}} |g_-(v)| | e^{\bar{z}v} - Q_m(\bar{z}v)|^2 e^{-|v|^2} dA(v) \\
& \leq \left \{\frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathbb{C}} |g_-(v)|^p | e^{\bar{z}v} - Q_m(\bar{z}v)|^2 e^{-|v|^2} dA(v) \right \}^\frac{1}{p} \| k_z \|_{2,m}^\frac{2}{q}\\
& = \left \{\frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathbb{C}} |g_-(v)|^p | e^{\bar{z}v} - Q_m(\bar{z}v)|^2 e^{-|v|^2} dA(v) \right \}^\frac{1}{p}\\
& \leq C.
\end{align*}
Therefore,
\begin{equation*}
 \left \{ \frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathbb{C}} |g_-(v)- \mathfrak{B_m}h(z)|^p | e^{\bar{z}v} - Q_m(\bar{z}v)|^2 e^{-|v|^2} dA(v) \right \}^\frac{1}{p} \leq 2 C
\end{equation*}  and hence, we get the result.
\end{proof}

\begin{lemma} \label{lem6}
Suppose there exists a positive constant $ M $ such that
\begin{equation*}
\underset{|z |> M}{Sup} \left \{ \frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathbb{C}} |g(v)- \mathfrak{B_m}g(z)|^p | e^{\bar{z}v} - Q_m(\bar{z}v)|^2 e^{-|v|^2} dA(v) \right \},
\end{equation*} is bounded. Then, there exists a constant $ M' > 0 $ such that for each $ z \in \mathbb{C} $, there exists a constant $ \mu_z $ such that
\begin{equation*}
\underset{|z| > M'}{Sup} \left \{ \frac{1}{ \pi r^2} \int_{\mathscr{B}(z; r)} |g(v) - \mu_z|^p  dA(v)\right \}
\end{equation*} is bounded.
\end{lemma}

\begin{proof} By using the fact that $ e^{-|z-v|^2} \geq a  $ for $ v \in \mathscr{B}(z;r) $ and for some constant $ a > 0 $, it follows that
\begin{align}
&a C \left \{ \frac{1}{ \pi r^2} \int_{\mathscr{B}(z; r)} |g(v) - \mathfrak{B_m}g(z)|^p  dA(v)\right \}\nonumber \\
& \leq C \left \{ \frac{1}{ \pi r^2} \int_{\mathscr{B}(z; r)} |g(v) - \mathfrak{B_m}g(z)|^p e^{-|z-v|^2} dA(v)\right \}\nonumber \\
& = C \left \{ \frac{1}{ \pi r^2} \int_{\mathscr{B}(z; r)} |g(v) - \mathfrak{B_m}g(z)|^p  e^{-|z|^2} |e^{\bar{z}v}|^2 e^{-|v|^2}dA(v)\right \}\nonumber\\
\label{eq1} &\leq \left \{ \frac{1}{ \pi^2 r^2((e^{|z|^2} - Q_m (|z|^2))} \int_{\mathscr{B}(z; r)} |g(v) - \mathfrak{B_m}g(z)|^p | e^{\bar{z}v} - Q_m(\bar{z}v)|^2 e^{-|v|^2}dA(v)\right \}\\
& \leq \left \{ \frac{1}{ \pi^2 r^2((e^{|z|^2} - Q_m (|z|^2))} \int_{\mathbb{C}} |g(v) - \mathfrak{B_m}g(z)|^p | e^{\bar{z}v} - Q_m(\bar{z}v)|^2 e^{-|v|^2}dA(v)\right \}\nonumber
\end{align} for all $ |z| > M' $, where the Eq. \eqref{eq1} follows from lemma \ref{lem2} and \begin{equation*}
 \lim_{\substack{v \in \mathscr{B}(z; r)\\ |z| \rightarrow \infty}} ( 1 - e^{-\bar{z}v}Q_m(\bar{z}v)) = 1. \qedhere
\end{equation*}
\end{proof}

Lemmas \ref{lem6} and \ref{lem7} jointly lead to the following result:
\begin{theorem}\label{thm1}
Let $ g \in \mathcal{BMO}_r^p$. Then the following conditions are equivalent:
\begin{enumerate}
\item There exists a constant $ M > 0 $ such that \begin{equation*}
\underset{|z| > M}{Sup} \left\{ \frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathbb{C}} |g(v)- \mathfrak{B_m}g(z)|^p | e^{\bar{z}v} - Q_m(\bar{z}v)|^2 e^{-|v|^2} dA(v)\right \} < \infty;
\end{equation*}
\item There exists a constant $ M > 0 $ such that for each $ z \in \mathbb{C}  $, there exists a constant $ \mu_z $ such that
\begin{equation*}
\underset{|z| > M}{Sup} \left \{ \frac{1}{ \pi r^2} \int_{\mathscr{B}(z; r)} |g(v) - \mu_z|^p  dA(v)\right \} < \infty;
\end{equation*}
\item There exists a constant $ M > 0 $ such that for each $ z \in \mathbb{C}  $, there exists a constant $ \mu_z $ such that
 \begin{equation*}
\underset{|z| > M}{Sup}\left \{ \frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathbb{C}} |g(v)- \mu_z |^p | e^{\bar{z}v} - Q_m(\bar{z}v)|^2 e^{-|v|^2} dA(v)\right\} < \infty.
\end{equation*}
\end{enumerate}
\end{theorem}

\begin{proof} $ (1) \Leftrightarrow (3) $ This follows on the proof of lemma $ 3.1 $ of \cite{peralahankel}.
\end{proof}

\begin{proposition}\label{lem7}  Let $ g \in \mathcal{BMO}_{2r}^p $. Then there exists a positive constant $ M $ such that the followings hold:

(1)
$
\underset{|z |>M +r}{Sup}\left\{ \underset{v \in \mathscr{B}(z; r)}{Sup} | \mathfrak{B_m}g(v) - \mathfrak{B_m}g(z) |\right\} < \infty
$

(2)
$
\underset{|z |>M +r }{Sup} \left\{\frac{1}{ \pi r^2} \int_{\mathscr{B}(z; r)} |(g - \mathfrak{B_m}g)(v)|^p  dA(v) \right\}< \infty.
$
\end{proposition}

\begin{proof}
Let $ g \in \mathcal{BMO}_{2r}^p \subset \mathcal{BMO}_{r}^p $.
Consider
\begin{align*}
&| \mathfrak{B_m}g(z) - \tilde{g}_r(z) | \\
& = |\frac{1}{ \pi r^2} \int_{\mathscr{B}(z; r)} \mathfrak{B_m}g(z)  dA(v)- \frac{1}{ \pi r^2} \int_{\mathscr{B}(z; r)} g (v)  dA(v)|\\
& \leq  \frac{1}{ \pi r^2} \int_{\mathscr{B}(z; r)} |g(v)- \mathfrak{B_m}g(z)|  dA(v)\\
& \leq  \left \{ \frac{1}{ \pi r^2} \int_{\mathscr{B}(z; r)} |g(v)- \mathfrak{B_m}g(z)|^p  dA(v)\right\}^\frac{1}{p}\\
& \leq C \left \{ \frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathbb{C}} |g(v)- \mathfrak{B_m}g(z)|^p | e^{\bar{z}v} - Q_m(\bar{z}v)|^2 e^{-|v|^2} dA(v)\right \}^\frac{1}{p}\\
& \leq C_0,
\end{align*} for all $ |z| > M >0 $ and for some constant $C_0>0 $. Thus, $$ \underset{|z |>M +r}{Sup} \left\{\underset{v \in \mathscr{B}(z; r)}{Sup} |( \mathfrak{B_m}g -\tilde{g}_r) (v) - (\mathfrak{B_m} - \tilde{g}_r)g(z) | \right\} <\infty $$ and $$ \underset{|z |> M +r}{Sup} \left\{\frac{1}{ \pi r^2} \int_{\mathscr{B}(z; r)} |( \mathfrak{B_m}g - \tilde{g}_r)(v)|^p  dA(v)\right\}<\infty.
$$  Since $ g \in \mathcal{BMO}_{2r}^p \subset \mathcal{BMO}_{r}^p $, therefore by lemma \ref{lem1}, it follows that $ \tilde{g}_r \in \mathcal{BO} $ and $ g -  \tilde{g}_r \in \mathcal{BA}^p $, so  $ \mathfrak{B_m}g = \mathfrak{B_m}g - \tilde{g}_r + \tilde{g}_r $, $ g - \mathfrak{B_m}g = g - \tilde{g}_r + \tilde{g}_r - \mathfrak{B_m}g $, we get the desired result.
\end{proof}

\begin{lemma} \label{lem8}
If $ g \in \mathcal{BA}^p $, then $ H_g^p $ is bounded on $ \mathscr{F}^{p,m}$ for finite $ p \geq 1$.
\end{lemma}

\begin{proof}
By using lemma \ref{lem4}, $ g \in \mathcal{BA}^p $ if and only if $ L_g $ is bounded on $ \mathscr{F}^{p,m} $ if and only if $ H_g $ is bounded on $ \mathscr{F}^{p,m}$.
\end{proof}

\begin{lemma} \label{lem9}
If $ g \in \mathcal{BO} $, then $ H_g^p $ is bounded on $ \mathscr{F}^{p,m} $ for all $ 1 \leq p \leq \infty$.
\end{lemma}

\begin{proof}
Let $ h \in \mathscr{F}^{p,m} $ and $1 < p <\infty  $. Since $ g \in \mathcal{BO} $, therefore, by lemma \ref{lem1}, we obtain that
\begin{align*}
|H_g(h)(z)|^p  &\leq \left \{\omega_{p,m} \int_{\mathbb{C}} |g(z) - g(v))||h(v)| |\frac{e^{z \bar{v}} - Q_m(z \bar{v})}{(z \bar{v})^m }| e^{-|v|^2} |v|^{2m} dA(v)\right \}^p\\
& \leq C \left \{ \omega_{p,m} \int_{\mathbb{C}} (|z-v|+1)|h(v)| |\frac{e^{z \bar{v}} - Q_m(z \bar{v})}{(z \bar{v})^m }| e^{-|v|^2} |v|^{2m} dA(v) \right \}^p.
\end{align*}
 By Using (see \cite{chofock})\[ |\frac{e^{z \bar{v}} - Q_m(z \bar{v})}{(z \bar{v})^m }| \lesssim \frac{e^{\frac{1}{2}|z|^2 + \frac{1}{2} |v|^2 - \frac{1}{8}|z-v|^2}}{(1 + |z||v|)^m} \leq \frac{e^{\frac{1}{2}|z|^2 + \frac{1}{2} |v|^2 - \frac{1}{8}|z-v|^2}}{ (|z||v|)^m},\]
  it follows that
\begin{align*}
&|H_g(h)(z)|^p e^{-\frac{p}{2}|z|^2} |z|^{pm} \\
& \leq C  e^{-\frac{p}{2}|z|^2} |z|^{pm} \left \{ \omega_{p,m} \int_{\mathbb{C}} (|z-v|+1)|h(v)| |\frac{e^{z \bar{v}} - Q_m(z \bar{v})}{(z \bar{v})^m }| e^{-|v|^2} |v|^{2m} dA(v) \right \}^p\\
& \leq C  \left \{ \omega_{p,m} \int_{\mathbb{C}} (|z-v|+1)|h(v)| e^{-\frac{1}{2}|v|^2} e^{- \frac{1}{8}|z-v|^2} |v|^{m} dA(v) \right \}^p\\
& \leq C  \left \{ \omega_{p,m} \int_{\mathbb{C}} |h(v)|^p e^{-\frac{p}{2}|v|^2} |v|^{pm} dA(v) \right \}  \left \{ \omega_{p,m} \int_{\mathbb{C}} (|z-v|+1)^q  e^{- \frac{q}{8}|z-v|^2} dA(v) \right \}^\frac{p}{q}\\
& = C \| h\|_{p,m}^p  \left \{ \omega_{p,m} \int_{\mathbb{C}} (|z-v|+1)^q  e^{- \frac{q}{8}|z-v|^2} dA(v) \right \}^\frac{p}{q},
\end{align*} for some constant $ C >0$. Therefore, $\| H_f(h) \|_{p,m}^p \leq C_0 \| h\|_{p,m}^p $ and hence, $ H_f $ is bounded on $ \mathscr{F}^{p,m} $ for $1 < p < \infty$.
For $ p=1 $, we can conclude by using Fubini's theorem  that
\begin{align*}
 \| H_g(h)\|_{1,m} & = \omega_{1,m}\int_\mathbb{C} |H_g(h)(z)| e^{-\frac{1}{2}|z|^2} |z|^{m} dA(z)\\
 & \leq  C _0 \left \{  \int_{\mathbb{C}} |h(v)| e^{-\frac{1}{2}|v|^2} |v|^{m} dA(v) \right \}  \left \{  \int_{\mathbb{C}} (|z-v|+1)  e^{- \frac{1}{8}|z-v|^2} dA(z) \right \} \\
 & \leq C_0 \| h\|_{1,m},
\end{align*}where $ C_0 $ is a constant. For $ p = \infty $,
\begin{align*}
\| H_g(h)\|_{\infty ,m} & = |H_g(h)(z)| e^{-\frac{1}{2}|z|^2} |z|^{m}\\
& \leq \| h\|_{\infty,m} \left \{  \int_{\mathbb{C}} (|z-v|+1)  e^{- \frac{1}{8}|z-v|^2} dA(z) \right \} \leq C_1,
\end{align*} where $ C_1 >0 $ is a constant and hence, the result follows for all $ 1 \leq p \leq \infty $.
\end{proof}

\begin{theorem} \label{thm2} Let $ g \in \mathcal{BMO}_r^p $. Then the operators $ H_g^p $ and $ H_{\bar{g}}^p $ are bounded for all $ 1 \leq p < \infty $.
\end{theorem}
\begin{proof}
By using lemmas \ref{lem1}, \ref{lem8} and \ref{lem9} and the fact that if $ g \in \mathcal{BMO}_r^p $ then so is $ \bar{g} $, the result follows.
\end{proof}

\section{\bf $\mathcal{VMO}_r^p $ spaces and Compactness of Hankel operators on $ \mathscr{F}^{p,m}$}
Define $ \mathcal{VA}_r $ be the set of all $ L_{Loc}^1 $  integrable functions $ g $ on $ \mathbb{C} $ such that $ \displaystyle \lim_{|z| \rightarrow \infty}\tilde{g}_r   = 0 $.
 For finite $ p \geq 1$, let $ \mathcal{VMO}_r^p $ denote the set of all $ L_{Loc}^p $  integrable functions $ g $ such that
\[  \displaystyle \lim_{|z| \rightarrow \infty} \left \{ \frac{1}{ \pi r^2} \int_{\mathscr{B}(z; r)} |g(v) - \tilde{g}_r  (z)|^p  dA(v)\right \}^{\frac{1}{p}} =0. \]
Let $ \mathcal{VO}_r \subset \mathcal{BO}_r $ be the set of all continuous functions $ g $ on $ \mathbb{C} $ such that
\[  \displaystyle \lim_{|z| \rightarrow \infty} \underset{v \in \mathscr{B}(z; r)}{Sup} | g(v) - g(z) | =0.
\]
Let $ \mathcal{VA}_r^p $ be the set of all $ L_{Loc}^p $  integrable functions $ g $ on $ \mathbb{C} $ such that $ \displaystyle \lim_{|z| \rightarrow \infty}\tilde{g}_r^p   = 0 $.

The following lemma will be useful in the study of compact Hankel operators on $\mathscr{F}^{p,m} $ and the related results.

\begin{lemma} \label{lem10} \cite{wangtoeplitz} Let $ \lambda $ is a positive Borel measure, $ 0 < p < \infty $, $ r > 0 $, $ m $ is a non- negative integer and $ \{ b_n \} $ is the lattice in $ \mathbb{C} $ generated by $ r $ and $ ri$. Then the following conditions are equivalent.
\begin{enumerate}
\item $
 \displaystyle \lim_{n \rightarrow \infty} \int_{\mathbb{C}} |g_n (v) v^m e^{- \frac{|v|^2}{2}}|^p d \lambda =0
$ for all bounded sequence $ \{g_n\} $ in $ \mathscr{F}^{p,m} $ that converges to $ 0 $ uniformly on compact sets;
\item  $ \displaystyle \lim_{|z| \rightarrow \infty} \lambda(\mathscr{B}(z;r)) = 0$;
\item $ \displaystyle \lim_{n \rightarrow \infty} \lambda(\mathscr{B}(b_n;r)) = 0$.
\end{enumerate}
\end{lemma}

Similar to $ \mathcal{BO}_r $ and $ \mathcal{BA}_r^p $, it is easy to observe that $ \mathcal{VO}_r $ and $ \mathcal{VA}_r^p $ are independent of $r$, so we will denote them by $\mathcal{VO} $ and $ \mathcal{VA}^p $, respectively.

The following results are analogues to theorem \ref{thm1} and lemma \ref{lem1}.

\begin{theorem} \label{thm3} Let $ p $ is any natural number. Then the following conditions are equivalent:
\begin{enumerate}
\item $ g \in \mathcal{VMO}^p $;
\item $ g \in \mathcal{VO} + \mathcal{VA}^p $;
\item  \begin{equation*}
\displaystyle \lim_{|z| \rightarrow \infty} \left \{ \frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathbb{C}} |g(v)- \mathfrak{B_m}g(z)|^p | e^{\bar{z}v} - Q_m(\bar{z}v)|^2 e^{-|v|^2} dA(v)\right \} =0;
\end{equation*}
\item There exists a constant $ M > 0 $ such that for each $ z \in \mathbb{C}  $, there exists a constant $ \mu_z $ such that
\begin{equation*}
\displaystyle \lim_{|z| \rightarrow \infty} \left \{ \frac{1}{ \pi r^2} \int_{\mathscr{B}(z; r)} |g(v) - \mu_z|^p  dA(v)\right \} =0;
\end{equation*}
\item There exists a constant $ M > 0 $ such that for each $ z \in \mathbb{C}  $, there exists a constant $ \mu_z $ such that
 \begin{equation*}
\displaystyle \lim_{|z| \rightarrow \infty} \left \{ \frac{1}{\pi (e^{|z|^2} - Q_m (|z|^2)} \int_{\mathbb{C}} |g(v)- \mu_z |^p | e^{\bar{z}v} - Q_m(\bar{z}v)|^2 e^{-|v|^2} dA(v) \right \} =0.
\end{equation*}
\end{enumerate}
\end{theorem}
 From theorem \ref{thm3}, it follows that $ \mathcal{VMO}_r^p $ is independent of $ r $, so we will write $ \mathcal{VMO}^p $.

\begin{lemma}  \label{lem11}
\begin{enumerate}
 \item If $ g \in \mathcal{VMO}^p $, then $ \tilde{g}_r \in \mathcal{VO} $.
\item If $ g \in \mathcal{VMO}^p $, then $ g - \tilde{g}_r \in \mathcal{VA}^p $ for every $ r > 0$.
\item If $ g \in \mathcal{VMO}^p $, then $\mathfrak{B_m}(g) \in \mathcal{VO}$.
\item If $ g  \in \mathcal{VMO}^p $, then $ g - \mathfrak{B_m}(g) \in \mathcal{VA}^p$.
\item  The function $ g \in \mathcal{VO} $ if and only if for each constant $ C > 0 $, there exists $ r > 0 $ such that $ | g(z) - g(v) | \leq C (1 + | z-v |) $ for all $ z, v \in \mathbb{C} \backslash \mathscr{B}(0;r)$ (see \cite{bauermean}).
\end{enumerate}
\end{lemma}

\begin{lemma}\cite{bauermean} \label{lem12} For $ r > 0 $, consider a function $ g: \mathbb{C} \backslash \mathscr{B}(0;r) \rightarrow \mathbb{C} $ with
\[ | g(z) - g(v) | \leq C (1 + | z-v |) \text{ for all } z, v \in \mathbb{C} \backslash \mathscr{B}(0;r),
\] where $ C > 0 $ is independent of $ g $. Then there exists  a function $ G $ on $ \mathbb{C} $ such that $ g = G $ on $ \mathbb{C} \backslash \mathscr{B}(0;r) $  and $ |G(z) - G(v) | \leq 2C (1 + | z-v |) $ for all $ z, v \in \mathbb{C} $.
\end{lemma}

\begin{theorem} \label{thm4} Let $ 1 \leq p <\infty $. Then $ g \in \mathcal{VMO}^p $ then the Hankel operators $ H_g^p $ and $ H_{\bar{g}}^p $ are both compact.
\end{theorem}

\begin{proof} Let $ g \in \mathcal{VA}^p $. This gives the positive measure $ d \lambda = |g|^p d A $ satisfying $  \displaystyle \lim_{|z| \rightarrow \infty} \lambda(\mathscr{B}(z;r)) = 0$. So, by lemma \ref{lem10}, the multiplication operator $ L_g : \mathscr{F}^{p,m} \rightarrow L^{p,m} $ is compact and so is $ H_g^p $.

Let $ g \in \mathcal{VO} $. Let $ \epsilon > 0 $ be arbitrary. Using lemma \ref{lem11} and \ref{lem12}, it follows that there exists a function $ G $ on $ \mathbb{C} $ such that $ g = G $ on $ \mathbb{C} \backslash \mathscr{B}(0;r) $  and $ |G(z) - G(v) | \leq 2\epsilon (1 + | z-v |) $ for all $ z, v \in \mathbb{C} $. Then lemma \ref{lem1} and \ref{lem9} give $ H_G^p $ is bounded with $\| H_G^p\| \leq 2 \epsilon C_0 $ for some constant $ C_0 > 0$. Also, $ H_{g - G}^p $ is compact, since $ g-G $ has compact support, and $ \| H_G^p - H_{g-G}^p \| = \| H_g^p \| \leq 2\epsilon C_0 $. Since $ \epsilon >0 $ is arbitrary, therefore, the Hankel operators $ H_g^p $ is compact. Similarly, it can be proved that the Hankel operators $ H_{\bar{g}}^p $ is compact.
Hence, by using theorem \ref{thm3}, the result follows.
\end{proof}

\subsection*{Acknowledgment}
Support of UGC Research Grant to second author for carrying out the research work is gratefully acknowledged.

config: arxiv


TITLE: Kunen's result and $L(\mathbb{R})$
QUESTION [3 upvotes]: Recall Kunen's result: If $j:V \to M$ is a nontrivial elementary embedding then $V \ne M$.
Assume $AD$ is true in $L(\mathbb{R})$. Let $j : L(\mathbb{R}) \to M$ be a nontrivial elementary embedding. Must we have $L(\mathbb{R}) \ne M$? Thanks.

REPLY [3 votes]: The existence of a nontrivial elementary
embedding $j: L(\mathbb{R}) \to L(\mathbb{R})$, or equivalently, the existence of $\mathbb{R}^\sharp$, is consistent with the statement that AD holds in $L(\mathbb{R})$.  Both are consequences of the existence of large cardinals, such as $\omega$ many Woodin cardinals with a measurable cardinal above them.  Also, both are conseqences of stronger determinacy axioms, for example AD$_\mathbb{R}$, the axiom of determinacy for games where the moves are reals rather than integers.


TITLE: Analytic continuation of $\frac{\log(1-z^2)}{z^2}$
QUESTION [0 upvotes]: I want to find the largest open set where I can analytically continue  $\frac{\log(1-z^2)}{z^2}$.
Attempt
The numerator is analytic in $\mathbb{C}\setminus[-\infty,-1]\cup[1,\infty]$ (by linking the branch points $\pm 1$ and $\infty$)and the denumerator non-zero outside zero, so by intersecting:
$$U=\mathbb{C}\setminus[-\infty,-1]\cup[1,\infty].$$
However, taking them together I think there is a larger set i.e.: $\mathbb{C}\setminus [-1,1]$
How to show that $\infty$ is not a branch point for $\dfrac{\log(1-z^2)}{z^2}$? Any hints.
As $z\to \infty$ we set $\frac{1}{z}$ and take $z\to 0$ then:
$$r^{2}e^{i2\theta}(log(z-1)+log(z+1)-2log(z))$$
so by making a small circle around 0, the first two terms are fixed but the last one is shifted. So infinity is actually a branch point.
Any mistakes?

REPLY [0 votes]: To analyze behaviour at $\infty$, set $z = 1/\zeta$.  Then you are studying the neighborhood of $0$ of \begin{align*}
    \zeta^2 \log(1-\zeta^{-2}) &= \zeta^2 \log(\zeta^{-2}(\zeta^{2}-1)) \\
        &= \zeta^2 \left( \log(\zeta + 1) + \log(\zeta - 1) - 2 \log(\zeta) \right) \text{.}
\end{align*}  Put a small loop around $0$.  The $\zeta^2$, $\log(\zeta + 1)$, and $\log(\zeta - 1)$ do not cause any branch difficulties.  However, $-2 \log(\zeta)$ moves by $-4\pi \mathrm{i}$, so $\infty$ is a branch point and there is a branch cut terminating at $\infty$.  (It's other end is one of the other two branch points or is a point on your cut, $[-1,1]$.)
Page 11 here has much nicer diagrams than I can make.  (The document is Dr. R. Rosales's Branch Points and Branch Cuts (18.04, MIT), from 11 October 1999.)
In slightly more detail, $\oint -2 \zeta^2 \log \zeta \,\mathrm{d}\zeta$ where the path is a small circle of radius $r \in (0,1)$ centered at $0$ is $\frac{4\pi\mathrm{i}}{3} \mathrm{e}^{3r} \rightarrow \frac{4\pi\mathrm{i}}{3}$ as $r \rightarrow 0$, so $\zeta = 0$ is a branch point.  Therefore $z = \infty$ is a branch point.  (I did that integral "by memory" and I don't recall if I've dropped a minus sign.  Either way, you get a branch point.)


TITLE: Inequality involving first and second moments
QUESTION [1 upvotes]: Knowing that $X$ is a positive random variable, and that for some $m>0$ and $M>0$ we have $\mathbb{P}\left( X \leq m \right) \leq M$, how can one show the following result ?
$$\forall \alpha > 0 \text{ s.t } \mathbb{E}\left[X^{2\alpha}\right] < +\infty \text{ and } \frac{\sqrt{M \mathbb{E}\left[X^{2 \alpha}\right]}}{\mathbb{E}\left[X^{\alpha}\right]} < 1, \quad \mathbb{E}\left[X^{\alpha}\right] \leq \frac{m^{\alpha}}{1-\frac{\sqrt{M \mathbb{E}\left[X^{2 \alpha}\right]}}{\mathbb{E}\left[X^{\alpha}\right]}}$$
Applying the Paley-Zygmund bound on $X^{\alpha}$ and choosing $\theta = 1-\frac{\sqrt{M \mathbb{E}\left[X^{2 \alpha}\right]}}{\mathbb{E}\left[X^{\alpha}\right]}$ yields
$$\mathbb{P} \left\{ X^{\alpha} > \left(1-\frac{\sqrt{M \mathbb{E}\left[X^{2 \alpha}\right]}}{\mathbb{E}\left[X^{\alpha}\right]}\right) \mathbb{E}\left[ X^{\alpha} \right] \right\} \geq M$$
Then applying the Markov inequality to get rid of the probability gives :
$$\frac{1}{1-\frac{\sqrt{M \mathbb{E}\left[X^{2 \alpha}\right]}}{\mathbb{E}\left[X^{\alpha}\right]}} \geq M$$
It would help if $\mathbb{E}\left[ X^{\alpha} \right] \leq M m^{\alpha}$, but sadly I can't prove it, nor do I think it is even true.
One can however show a weaker result :
$$\frac{m^{\alpha}}{1-\frac{\sqrt{M \mathbb{E}\left[X^{2 \alpha}\right]}}{\mathbb{E}\left[X^{\alpha}\right]}} \geq \mathbb{E}\left[ X^{\alpha} \left| X \leq m \right. \right] \mathbb{P}\left( X \leq m \right)$$

REPLY [0 votes]: Okay found it, using a Markov inequality where we condition by $X \leq m$ instead of the "vanilla" version of the Markov inequality in my previous attempt yields the result.
EDIT : false flag sorry, I made a mistake and a direct application of Markov doesn't yield the result. I will update this answer if I find something.


TITLE: The preimage $\phi^{-1}\{H\}$ of a subgroup $H$ of a factor group $G\big /N$ is a subgroup of $G$
QUESTION [0 upvotes]: Let $N$ be a normal subgroup of a finite group $G$, furthermore, let $H$ be a subgroup of the factor group $G\big /N$. Denote by $\phi:G \rightarrow G\big/N$ the canonical homomorphism: $g\mapsto gN$. The problem is to show that $\phi^
{-1}\{H\} = \{g\in G: \phi(g)\in H\}$ is a subgroup of $G$.
We need to show the following properties of the set $\phi^{-1}\{H\}$:

$e \in \phi^{-1}\{H\}$
If $x_1, x_2\in \phi^{-1}\{H\}$ then $x_1x_2 \in \phi^{-1}\{H\}$
$\forall x\in \phi^{-1}\{H\} \exists x^{-1}\in\phi^{-1}\{H\}: xx^{-1}=e=x^{-1}x$

Since $H$ is a subgroup of $G\big/N$ it is immediately clear that: $N=\phi(e)\in H$ so $e \in \phi^{-1}\{H\}$.
Furthermore, if $x_1$ and $x_2$ are in $\phi^{-1}\{H\}$ then both $\phi(x_1)$ and $\phi(x_2)$ are in $H$, since $H$ is already a subgroup we have:
$\phi(x_1)\phi(x_2) = \phi(x_1x_2)\in H$, so $x_1x_2\in \phi^{-1}\{H\}$.
It is the last part that has me confused; suppose $x\in \phi^{-1}\{H\}$, then since $H$ is a subgroup we have:
$\phi(e) = \phi(x)[\phi(x)]^{-1}$, let $g\in G$ be such that $\phi(g) = [\phi(x)]^{-1}$, then
$\phi(e) = \phi(x)\phi(g) = \phi(xg)$, so the inverse coset of $xN$ is $gN$, can I immediately draw the conclusion that $g = x^{-1}$ ?
My idea was the following:
Since $xNgN = xgN = eN$ then for any $n\in N$ we have: $xgn = n \Leftrightarrow xgnn^{-1} = nn^{-1} \Leftrightarrow xg = e$ so $g$ must be $x^{-1}$.

REPLY [1 votes]: Note that you always have to show that the subgroup is closed under the group operation. Then the easiest way to show that $G' \neq \emptyset$ is a subgroup of $G$ is to show $a,b \in G'\Rightarrow ab^{-1} \in G'$, which is an equivalent criterion of being a subgroup. Clearly $\phi$ is a homomorphism of groups and the inverse image of a group under a homomorphism $\phi$ is always a group itself.  Concerning your question: You are right choosing $x^{-1}N$ as the inverse of $xN$. By definition of the factor group: $x^{-1}NxN=N$ and uniqueness of inverse proves the claim.


TITLE: If the MH370 black box did sink to 15000 ft, how long would it have taken?
QUESTION [2 upvotes]: I have been following MH370 on the news just as everyone and latest reports seem to indicate that the black-box could be found. A recent info-graphic http://t.co/lyBBE9C2hF shows the insurmountable depth of the oceans and how the black-box could have sunk 15,000 ft! I wonder how long it would have taken for it to sunk to the bottom of the sea-bed? What is the equation of motion for a sinking object at sea, ignoring under-water currents?

REPLY [0 votes]: You say, ignore currents and I assume, other extraneous factors.
If that is the case, then, considering a hydrostatic balance in the water column where z is the vertical coordinate. Then, the motion of a water parcel with density, $\rho$, displaced upwards by a distance, $\Delta z$, in a fluid with a reference density, $\rho_0$, is governed by
$\rho_0 \frac{d^2\Delta z}{dt^2} = g \frac{d \bar{\rho}}{dt^2} \Delta z$
I am confident you will be able to deduce the rest.


TITLE: How to find minimal polynomial of primitive element (field theory)
QUESTION [2 upvotes]: I am given a primitive element $\alpha$ in the Galoisfield $F_{2^6}$. The question is to find the mimimal polynomial of $\alpha^7$. How to I find this?
My thoughts so far:
$$
\alpha^7 \rightarrow \alpha^2\\
\Rightarrow \alpha^7+ \alpha^2 = 0
$$
But it seems to simple to me...

REPLY [2 votes]: $x^9-1 = (x^3-1)(x^6+x^3+1)=(x-1)(x^2+x+1)(x^6+x^3+1)$.
$\beta = \alpha^7$ is an element of order $9$ and thus a root of $x^9-1$.
It cannot be a root of $x^3-1$ and so must be a root of $x^6+x^3+1$.
So, the minimal polynomial of $\beta$ is $x^6+x^3+1$ or a divisor
thereof. Now, the roots of the minimal polynomial of $\beta$
are $\beta$ and all its conjugates, that is,
$\beta, \beta^2, \beta^{2^2}, \beta^{2^3}, \cdots$ are the roots
of the minimal polynomial of $\beta$. (Also, the minimal
polynomial of $\beta$ is irreducible over $\mathbb F_2$).
How many (distinct) conjugates are there?
What does that tell you about the degree of the minimal polynomial of
$\beta$? Can you deduce the minimal polynomial of $\beta$ from all this
information?


TITLE: Why does the antipodal map on $S^2$ have degree $-1$?
QUESTION [2 upvotes]: I'm reading the post here by Arthur to explain that there is no smooth vector field on $S^2$. I don't understand it very well:

The simplest I can remember off the top of my head is this:
Assume there is such a vector field. Let $v_x$ denote the vector at the point $x$. Now, define the homotopy $H : S^2\times[0,1]\to S^2$ by the following: $H(x,t)$ is the point $t\pi$ radians away from $x$ along the great circle defined by $v_x$. This gives a homotopy between the identity and the antipodal map on $S^2$, which is impossible, since the antipodal map has degree $−1$. Hence there can be no such vector field.

I wonder how can we compute the degree of map defined here? As I imagine, there is probably a theorem saying that all antipodal mapping on $S^2$ has degree $-1$. I wonder why is that? No answers found online.

REPLY [2 votes]: As Michael answered, this is a general fact. If $f: S^n \to S^n$ is the antipodal, its degree is $(-1)^{n+1}$. Another way to prove it is to notice that $f$ is the composition of $n+1$ reflections (one for each coordinate). Since the degree of the composition is the product of the degrees and a reflection has degree $-1$, we obtain the result.


TITLE: Let $(x_n)$ be a sequence in a Banach space $X$. Which of the following conditions ensure(s) that $(x_n)$ is convergent in $X$?
QUESTION [3 upvotes]: Let $(x_n)$ be a sequence in a Banach space $X$. Which of the following conditions ensure(s) that $(x_n)$ is convergent in $X$?
(a) $\|x_n − x_{n+1}\| \to 0.$
(b) $\sum^{\infty}_{n=1} \|x_n − x_{n+1}\|<\infty$
I think that the first condition is insuffuicent.
What if
$x_n=\sum^{n}_{i=1}\frac{1}{i}?$.
What about the second one? I couldn't say anything about (b).
Edit 1
I think (b) works. we are supposed to see what about $\|x_m-x_n\|\leq\|x_m-x_{m+1}\|+\|x_{m+1}-x_{m+2}\|+ \dots +\|x_{n-1}-x_n\|$

REPLY [2 votes]: The second condition implies that the sequence is convergent since it is a Cauchy sequence:
$\|x_{n+p}-x_n\|\leq \|x_{n+p}-x_{n+p-1}\|+....+\|x_{n+1}-x_n\|$ and since $\sum \|x_{n+1}-x_n\|<\infty$, for every $c>0$ there exists an integer $N$, such that for every $n>N, \|x_{n+p}-x_{n+p-1}\|+...+\|x_{n+1}-x_n\|<c$.


TITLE: Evaluation of $\int_C{F.dr}$
QUESTION [1 upvotes]: For a given vector field and curve C, evaluate $\int_C \mathbf{F.dr}$
$\mathbf{F}(x,y)= -y \mathbf{i} +x \mathbf{j}$, C:Ellipse $\frac{x^2}{4}+\frac{y^2}{9}$=1
First we parametrise
$x=2\cos{t}$; $dx=-2\sin{t}$
$y=3\sin{t}$; $dy=3\cos{t}$
Then we write a proper integrand
The limits are found by replacing $x=2$ and $x=-2$ in $x=2\cos{t}$
$\int_{\pi}^{0}{6\sin^2{t}+6\cos^2{t}}$=
$\int_{\pi}^{0}{6}=-6\pi$
But the answer is $12\pi$

REPLY [3 votes]: Parametrize the ellipse :
$$\gamma(t)=(x(t),\,y(t))\;,\;\;\begin{cases}x=2\cos t\\{}\\y=3\sin t\end{cases}\;,\;\;0\le t\le2\pi\;\;\implies$$$${}$$
$$\int_CF\cdot dr=\int_0^{2\pi}F(\gamma(t))\cdot\gamma'(t)\;dt=\int_0^{2\pi}\left(-3\sin t,\,2\cos t\right)\cdot\left(-2\sin t,\,3\cos t\right)dt=$$
$$=\int_0^{2\pi}6(\cos^2t+\sin^2t)\;dt=6\int_0^{2\pi}dt=12\pi$$


TITLE: How tension force is shown going through a body in this problem even though tension is a pulling force?
QUESTION [1 upvotes]: I was solving some problems about tension forces in my textbook and in solution of one problem I saw that one of the tension forces is drawn through the body,which confused me because text book was always drawing tension forces out of body(because it always pulls the body).
The Problem

Textbook Solution

The solutions follows like this:
I.$$T_{2}<T_{3}$$ because its just a component of T3 obviously.
II.$$T_{1}=T_{5}$$ because purple tension forces cancel each other out.
III.$$T_{2}=T_{4}$$ because they are only horizontal and they cancel each other(green ones).
My Question
I am actually confused at more than one point.These are point that I am confused at:
I. How did we know that T2 and T1 is component of T3 and do the two opposite T3 forces cancel each other out?
II. I dont get how we can say that there is T5 thats going through the body even though tension is a pulling force.
III. I am ok with this step because it is obvious.
I will appreciate any help and would be really happy if you can end my confusion.

REPLY [0 votes]: Tension force arises from the tendency of elastic ropes to maintain their length. Hence, it acts both ways. Imagine a game of tug of war between you and your friend. You feel a force in the direction opposite to the direction you are trying to pull the rope, and so does your friend. This is a good way for getting an intuition for how tension forces work. They work only when the rope is taut and act on both ends, being opposite to each other in direction.
The system given in the problem is in equilibrium as none of the points has any acceleration. Therefore, forces must be balanced at each point.
Now, let me tackle your confusions:

$T_2$ and $T_1$ are not components of $T_3$, they are equal to the horizontal and vertical components of $T_3$ respectively. I have already mentioned the reason why.
I do not understand your confusion here. Perhaps you need to strengthen your understanding of the nature of tension forces.


TITLE: Formalizing a Parameter Selection Problem in Machine Learning
QUESTION [0 upvotes]: I have a simple problem,
I have an array (say of length 300), which gives me the fraction of importance of some value.
I plot it and see that the value becomes flat after some index.
So, I select the index after which the importance fraction just becomes flatter by just plotting the value.
So, my question is how should I formalize the problem of selecting this index.
The curve is like: Example Curve
How should I select the index? What mathematical condition is guaranteed to allow me to find that index after which the values become flat?

REPLY [0 votes]: Why not slope of the curve? $dy/dx$? Consider value at the consecutive indices then $dx=1$.
From figure
            @ index 1, value= 3.5
            @ index 2, value= 1.8
                     etc
            @ index 7, value= 0.5
            @ index 8, value= 0.3

search through the array and find out when the slope is less than the required...so that the curve is flat ...close to $dy/dx=0$
for i=1:300
       check=value(i)-value(i+1); %dy, dx=1
          if check<=0.2
             display('index value')
             i
             break
          end
end%end for loop

Isnt this you are asking for?