text:
# Simple question on parallel transport in dually flat manifolds

I just started studying Information Geometry and its applications by Amari. Right in the first chapter, the author talks about parallel transport in Dually flat manifolds.

Just some quick notation: In the first chapter, the author introduces the definition of Manifold $$M$$ and one of the many coordinate systems on the manifold, say, $$\theta$$. The length of the curve from $$\boldsymbol {\theta}$$ to $$\boldsymbol{ \theta + d\theta}$$ is given by $$d s^{2}=2 D_{\psi}[\theta : \theta+d \theta]=\sum g_{i j} d \theta^{i} d \theta^{j}$$

A tangent vector can be expressed as $$d \boldsymbol{\theta}=\sum d \theta^{i} \boldsymbol{e}_{i}$$ where $$\left\{\boldsymbol{e}_{i}\right\}$$, $$i \in \{1,.. ,n\}$$ are the basis of the tangent space of M at point $$\boldsymbol \theta$$. Similarly, the author introduces a dual affine coordinate system whose corresponding basis is $$\left\{e^{* i}\right\}$$. Therefore,we can write

$$d \boldsymbol \theta^{*}=\sum d \theta_{i}^{*} e^{* i}$$

Now, one can also write the length of the small line vector as $$d s^{2}=\langle d \boldsymbol{\theta}, d \boldsymbol{\theta}\rangle= g_{i j} d \theta^{i} d \theta^{j}$$, which is rewritten as $$d s^{2}=\left\langle d \theta^{i} e_{i}, d \theta^{j} e_{j}\right\rangle=\left\langle e_{i}, e_{j}\right\rangle d \theta^{i} d \theta^{j}$$ Hence, it is clear that $$g_{i j}(\boldsymbol{\theta})=\left\langle\boldsymbol{e}_{i}, \boldsymbol{e}_{j}\right\rangle$$

Similarly, for the dual affine coordinate system $$\boldsymbol \theta^*$$, we have $$g^{* i j}(\boldsymbol \theta^*)=\left\langle e^{* i}, e^{* j}\right\rangle$$

If $$\bf G$$ is the Jacobian of the transformation from $$\boldsymbol \theta$$ to $$\boldsymbol \theta^*$$, then we can write $$\begin{array}{l}{d \boldsymbol{\theta}^{*}=\mathbf{G} d \boldsymbol{\theta}, \quad d \boldsymbol{\theta}=\mathbf{G}^{-1} d \boldsymbol{\theta}^{*}} \\ {d \theta_{i}^{*}=g_{i j} d \theta^{j}, \quad d \theta^{j}=g^{* j i} d \theta_{i}^{*}}\end{array}$$

Actual doubt:

If a tangent vector $$\mathbf{A} = A^i\mathbf{e_i}$$ is transported from a point $$\boldsymbol{\theta}$$ to $$\boldsymbol{\theta^{'}}$$, the components $$A^i$$ remain the same because $$\mathbf{e_i}$$ is same everywhere in a dually flat manifold. However, in the later paragraphs, it is stated that the length of the vector is not constant even in this dually flat manifold. If the components and basis remain the same, shouldn't it be the case where even the length of the vector is also the same when moved from one point to another point?

Also, why is dual parallel transport any different parallel transport? After all, they are just two different coordinate systems. Because, it is given that the manifold is dually flat, so both the coordinate systems remain the same across all the points. So, why does the parallel transport makes the vector invariant under the original basis but the same parallel transport changes the vector in the dual basis? Any intuitive explanations, illustrations or examples on how this happens?

P.S: Relevant page from the book.

• Offhand this sounds like a lot of nonsense to me. What connection is the author using? Not the Levi-Civita connection. And the vector doesn't live both in the vector space and in its dual. Apr 30 '19 at 18:44
• @TedShifrin The author has not introduced Levi-Civita connection yet. Can you please explain why you say that a vector cannot live in its vector and dual space? After all, here, dual space is nothing but a different coordinate system. Apr 30 '19 at 18:55

In information geometry there are two flat connections $$\nabla^e$$ and $$\nabla^m$$ that are dual with respect to the Fisher-Rao metric $$g$$, which means that for any three vector fields $$X,Y,Z$$ we have $$X g(Y,Z)=g(\nabla^e_XY,Z)+g(Y,\nabla^m_XZ).$$ A consequence of this is that if $$Y$$ is e-parallel along $$X$$ and $$Z$$ is m-parallel along $$X$$, then the angle between $$Y$$ and $$Z$$ is constant. However, this is not true if both $$Y$$ and $$Z$$ are parallel along $$X$$ wrt. to one connection only. The duality above is different from the property enjoyed by the Levi-Civita connection $$\nabla^{\text{LC}}$$, which satisfies $$X g(Y,Z)=g(\nabla^{\text{LC}}_XY,Z)+g(Y,\nabla^{\text{LC}}_XZ)$$ for all vector fields $$X,Y,Z$$. This is known as self-dualilty or metricity of $$\nabla^{\text{LC}}$$ and can also be written as $$\nabla^{\text{LC}}g=0$$ since for any arbitrary connection $$\nabla$$ we have $$X g(Y,Z)=g(\nabla_XY,Z)+g(Y,\nabla_XZ)+(\nabla_X g)(Y,Z).$$ In contrast, $$\nabla^eg$$ and $$\nabla^mg$$ are both non-zero.
Since $$\nabla^{\text{LC}}$$ is not flat, parallel transport is complicated in general. On the other hand, $$\nabla^e$$ and $$\nabla^m$$ are flat, so parallel transport is easy. There are sets of dual charts $$\theta^i$$ and $$\eta_i$$ that are e- and m-affine, respectively. Dual means that if we denote $$e_i=\partial_i=\partial_{\theta^i}$$ and $$e^i=\partial^i=\partial_{\eta^i}$$ $$g(e_i, e^j)=\delta^j_i \Leftrightarrow e^i=g^{ij}e_j\Leftrightarrow e_i=g_{ij}e^j.$$ Affine means that all the Christoffel symbols vanish in that chart, i.e. all Christoffel symbols of $$\nabla^e$$ vanish in the $$\theta$$ chart and all Christoffel symbols of $$\nabla^m$$ vanish in the $$\eta$$ chart. This in turn implies that parallel transport corresponds to constant coefficients in the respective charts. I.e. $$\partial_i$$ are e-parallel and $$\partial^i$$ are m-parallel. The property $$g(\partial_i,\partial^j)=\delta^j_i$$ reflects the above self-duality condition.


text:
# Is $(0,1) \bigcap \mathbb{Q}$ an open set in the standard topology of $\mathbb{R}$?

From my point of view I think it isn't but I'm not sure. Here is what I thought:

$(0,1) \bigcap \mathbb{Q}$ is basically $$A = \{ \frac a b \in \mathbb{Q} \mid \frac a b \gt 0 \text{ and } \frac a b \lt 1 \}$$ and $\forall x \in A$ and $\forall R \gt 0$ there is a $q \in \mathbb{R} \setminus \mathbb{Q}$ such that $q \in \mathcal{B}_R(x)$ (i.e., open ball of center $x$ and radius $R$), so $\forall x \in A, \mathcal{B}_R(x) \nsubseteq \mathbb{Q}$ so $A \notin$ standard topology of $\mathbb{R}$. Conclusion, $A$ is not an open set in the standard topology of $\mathbb{R}$.

Is that correct ?

• Yes that's correct. Great! – Error 404 Jan 29 '17 at 15:06
• Oh for once I did it . Thank you ! – Eduard Valentin Jan 29 '17 at 15:07
• Also $q$ depends on $x$ and $R$. So for more clarity you can write $q(x,R).$ – Error 404 Jan 29 '17 at 15:08
• Yes, you are right,there are so many different notation for things. – Eduard Valentin Jan 29 '17 at 15:11
• Your question (in the header) asks if it is an open set in a topology of $\Bbb R$; the answer to this is "yes": in the discrete topology on $\Bbb R$, all sets are open. I assume that what you intended was "is it open in the standard topology on $\Bbb R$?", in which case your answer is correct. – John Hughes Jan 29 '17 at 15:41

format: plain


text:
# Journal Assignments Required for Exam 1 Spring 2017

Professor: Erika L.C. King
Email:eking@hws.edu
Office: Lansing 304
Phone: (315)781-3355

Our first exam is Tuesday (February 21) 7:15PM-9:15PM in Gulick 206A. You will turn in your journal at the beginning of the exam. I will randomly grade three to five of the entries completely. Then I will count to see how many total entries you have. Note that it is highly likely that the graded entries will be those things we did not specifically go over in class or have on a quiz. Remember to follow the journal guidelines (hightlighting exercise/problem numbers or the start of a question, etc.).

1. Vocabulary List
2. One Question you had followed by an answer for each journal assignment for a total of 10 questions with answers.
3. Section 1.2: problems 2, 4, 6 and 7
4. Check Yourself problems 1 and 2 at the end of Section 1.3
5. Check Yourself problems 1, 2 and 3 at the end of Section 1.4
6. The "nice version" of Example 1.4.7 as requested on page 13
7. Check Yourself problems 1 and 2 at the end of Section 1.5
8. Problem 17 in Section 1.7
9. On page 18, Prof. belcastro introduces existence proofs/theorems. In your journal, describe what they are and what makes them different from other proofs/theorems.
10. Here is an existence theorem: "There exists an integer whose cube equals its square." Try writing a nice proof of this theorem.
11. Check Yourself problems 1-6 at the end of Section 2.2
12. Problem 13 and 15 in Section 1.7
13. Check Yourself problems 1, 4, 5, and 8 at the end of Section 2.3
14. Richard Hammack's book: Section 2.6 numbers 3 and 13, and Section 2.10 numbers 1, 2, 4 and 8
15. Problem 5 in Section 2.9
16. Section 2.4: problems 1, 2 and 3
17. Consider the symbolic statement $(P\land Q)\Rightarrow (R\vee Q)$. Negate this statement. Then illustrate a truth table for the negation.
18. Section 2.4: problems 5, 6 and 7
19. Check Yourself problems 1-3 at the end of Section 2.5
20. Write down the two main steps for a proof by contraposition, then do the same for contradiction. (Hint: the two steps should start with "Assume" and "Show"!)
21. On Thursday, Christian presented the first half of the proof for part II number three on this worksheet. In your journal, write the second half of the proof!
22. In your journal, write all ten ways of expressing the statement: If f is a differentiable function, then f is a continuous function.
23. Problems 4(a)-(c), 8 and 9 in Section 2.4
24. Problem 4(d) and (e) in Section 2.4
25. Richard Hammack's book: Section 1.8 numbers 1, 4 and 5
26. Check Yourself problems 1 and 2 at the end of Section 3.2
27. Prove the Theorem: Let $A$, $B$ and $X$ be sets. If $X\subseteq A$ or $X\subseteq B$, then $X\subseteq A\cup B$.
28. Problems 15, 21 and 23 in Section 2.9


text:
## Saturday, October 25, 2014

### The logjam in infrastructure investment

 Quarter-on-quarter growth of the non-seasonally adjusted value of infrastructure projects that are classified as being Under Implementation' in the CMIE Capex database.

In the standard narrative, the first wave of infrastructure investment in India (2003-2010) choked owing to the ground-level difficulties of investment, including land acquisition, corruption scandals, contract disputes, etc. By this logic, if we solve these problems, we will have set the stage for the second wave of infrastructure investment. This is the sort of idea that motivated the Cabinet Committee on Investment' in the late days of the UPA.

There is, however, another logjam which needs to be solved, that of infrastructure financing. In today's Indian Express, I worry about the short-term measures which were taken during the first wave. Each of them felt expedient at the time, but with the benefit of hindsight, they were mistakes. Instead of looking for more short-term expedient solutions, we need to solve the deeper problems of the financial system in order to break this logjam.

The second wave of infrastructure financing, that we now require, needs work on both tracks: fixing the foundations of the financial system so that well-incentivised developers try to invest, and fixing the ground level difficulties so that this investment actually turns into smoothly working operating assets.

Acknowledgements : My thinking on these questions has been greatly aided by Josh Felman (IMF), Gopal Sarma (Bain Consulting) and Suyash Rai (NIPFP). The phrase `the long run has caught up with us' is by K. P. Krishnan.

LaTeX mathematics works. This means that if you want to say $10 you have to say \$10.


text:
## Weighted Composition Operators between different Bloch-type Spaces in Polydisk

### Zehua Zhou Department of Mathematics, Tianjin University Tianjin 300072, P. R. China E-mail: zehuazhou2003@yahoo.com.cn

$1$
Abstract
Let $\phi \left(z\right)=\left({\phi }_{1}\left(z\right),\dots ,{\phi }_{n}\left(z\right)\right)$  be a holomorphic self-map of ${U}^{n}$  and $\psi \left(z\right)$  a holomorphic function on ${U}^{n},$  where ${U}^{n}$  is the unit polydisk of ${\text{C}}^{n}.$  Let $p\ge 0,$  $q\ge 0$  , this paper gives some necessary and sufficient conditions for the weighted composition operator ${W}_{\psi ,\phi }$  induced by $\psi$  and $\phi$  to be bounded and compact between $p$  -Bloch space ${\mathcal{ℬ}}^{p}\left({U}^{n}\right)$  and $q$  -Bloch space ${\mathcal{ℬ}}^{q}\left({U}^{n}\right).$  Keywords Bloch-type space, Weighted composition operator, Composition operator.
2000 Mathematics Subject Classification 47B38, 47B33, 32A37, 32A30

1 Introduction

Let $\Omega$  be a bounded homogeneous domain in ${\text{C}}^{n}.$  The class of all holomorphic functions with domain $\Omega$  will be denoted by $H\left(\Omega \right).$  Let $\phi$  be a holomorphic self-map of $\Omega ,$  the composition operator ${C}_{\phi }$  induced by $\phi$  is defined by $\left({C}_{\phi }f\right)\left(z\right)=f\left(\phi \left(z\right)\right),$  for $z$  in $\Omega$  and $f\in H\left(\Omega \right)$  . If, in addition, $\psi$  is a holomorphic function defined on $\Omega ,$  the multiplication operator induced by $\psi$  is defined by ${M}_{\psi }f\left(z\right)=\psi \left(z\right)f\left(z\right),$  and weighted composition operators ${W}_{\psi ,\phi }$  induced by $\psi$  and $\phi$  is defined by $\left({W}_{\psi ,\phi }f\right)\left(z\right)=\psi \left(z\right)f\left(\phi \left(z\right)\right)$  for $z$  in $\Omega$  and $f\in H\left(\Omega \right).$  If let $\psi \equiv 1,$  then ${W}_{\psi ,\phi }={C}_{\phi },$  if let $\phi =Id$  , then ${W}_{\psi ,\phi }={M}_{\psi }.$  So we can regard weighted composition operator as a generalization of a multiplication operator and a composition operator. Let $K\left(z,z\right)$  be the Bergman kernel function of $\Omega$  , the Bergman metric ${H}_{z}\left(u,u\right)$  in $\Omega$  is defined by ${H}_{z}\left(u,u\right)=\frac{1}{2}{\sum }_{j,k=1}^{n}\frac{{\partial }^{2}logK\left(z,z\right)}{\partial {z}_{j}\partial {\overline{z}}_{k}}{u}_{j}{\overline{u}}_{k},$  where $z\in \Omega$  and $u=\left({u}_{1},\dots ,{u}_{n}\right)\in {\text{C}}^{n}.$  Following Timoney [1], we say that $f\in H\left(\Omega \right)$  is in the Bloch space $\mathcal{ℬ}\left(\Omega \right),$  if $\parallel f{\parallel }_{\mathcal{ℬ}\left(\Omega \right)}={sup}_{z\in \Omega }{Q}_{f}\left(z\right)<\infty ,$  where ${Q}_{f}\left(z\right)=sup\left\{\frac{|▽f\left(z\right)u|}{{H}_{z}^{\frac{1}{2}}\left(u,u\right)}:u\in {\text{C}}^{n}-\left\{0\right\}\right\},$  and $▽f\left(z\right)=\left(\frac{\partial f\left(z\right)}{\partial {z}_{1}},\dots ,\frac{\partial f\left(z\right)}{\partial {z}_{n}}\right),▽f\left(z\right)u={\sum }_{l=1}^{n}\frac{\partial f\left(z\right)}{\partial {z}_{l}}{u}_{l}.$  The little Bloch space ${\mathcal{ℬ}}_{0}\left(\Omega \right)$  is the closure in the Banach space $\mathcal{ℬ}\left(\Omega \right)$  of the polynomial functions.
Let $\partial \Omega$  denote the boundary of $\Omega$  . Following Timoney [2], for $\Omega ={B}_{n}$  the unit ball of ${\text{C}}^{n}$  , ${\mathcal{ℬ}}_{0}\left({B}_{n}\right)=\left\{f\in \mathcal{ℬ}\left({B}_{n}\right):{Q}_{f}\left(z\right)\to 0,\text{as}z\to \partial {B}_{n}\right\};$  for $\Omega =\mathcal{D}$  the bounded symmetric domain other than ${B}_{n}$  , $\left\{f\in \mathcal{ℬ}\left(\mathcal{D}\right):{Q}_{f}\left(z\right)\to 0,\text{as}z\to \partial \mathcal{D}\right\}$  isthe set of constant functions on $\mathcal{D}.$  So if $\mathcal{D}$  is a bounded symmetric domain other than the ball, we denote the ${\mathcal{ℬ}}_{0*}\left(\mathcal{D}\right)=\left\{f\in \mathcal{ℬ}\left(\mathcal{D}\right):{Q}_{f}\left(z\right)\to 0,\text{as}z\to {\partial }^{*}\mathcal{D}\right\}$  and also call it little Bloch space, here ${\partial }^{*}\mathcal{D}$  means the distinguished boundary of $\mathcal{D}$  .
The unit ball is the only bounded symmetric domain $\mathcal{D}$  with the property that ${\partial }^{*}\mathcal{D}=\partial \mathcal{D}.$  Let ${U}^{n}$  be the unit polydisk of ${\text{C}}^{n}.$  Timony [1] shows that $f\in \left({U}^{n}\right)$  if and only if $\parallel f{\parallel }_{=}|f\left(0\right)|+{sup}_{z\in {U}^{n}}{\sum }_{k=1}^{n}|\frac{\partial f}{\partial {z}_{k}}\left(z\right)|\left(1-|{z}_{k}{|}^{2}\right)<+\infty .$  This equality was the starting point for introducing the $p$  -Bloch spaces.
Let $p\ge 0,$  a function $f\in H\left({U}^{n}\right)$  is said to belong to the $p$  -Bloch space ${\mathcal{ℬ}}^{p}\left({U}^{n}\right)$  if $\parallel f{\parallel }_{p}=|f\left(0\right)|+{sup}_{z\in {U}^{n}}{\sum }_{k=1}^{n}|\frac{\partial f}{\partial {z}_{k}}\left(z\right)|{\left(1-|{z}_{k}{|}^{2}\right)}^{p}<+\infty .$  It is easy to show that ${\mathcal{ℬ}}^{p}\left({U}^{n}\right)$  is a Banach space with the norm $\parallel \cdot {\parallel }_{p}.$  It is easy to see that if ${lim}_{z\to \partial {U}^{n}}{\sum }_{k=1}^{n}|\frac{\partial f}{\partial {z}_{k}}\left(z\right)|\left(1-|{z}_{k}{|}^{2}{\right)}^{p}=0,$  then $f$  must be a constant. So, there is no sense to introduce the corresponding little $p$  -Bolch space in this way. We will say that the little $p$  -Bolch space ${\mathcal{ℬ}}_{0}^{p}\left({U}^{n}\right)$  is the closure of the polynomials in the $p$  -Bolch space.
In the recent years, there have been many papers focused on studying the composition operators in function spaces (say, for 1-dimensional case see [3-8], for n-dimensional case see [9-12]). More recently, S.Ohno, K. Stroethoff and R.H.Zhao in [8] discuss the weighted composition operators between-type spaces for 1-dimensional case.
In this paper, we discuss the boundedness and compactness of the weighted composition operators between $p$  -Bloch space and $q$  -Bloch space in polydisk, some new methods and techniques have been used because of the difference between topology boundary $\partial {U}^{n}$  and distinguished boundary ${\partial }^{*}{U}^{n}$  of ${U}^{n}$  , where $n>1$  , especially in the proof of Theorems 2 and 3 (need to be discussed according to the properties of the boundary). The results in this paper will extend corresponding results on the Bloch spaces (see [4-11]).
Our main results are the following:
Theorem 1 Let $\phi =\left({\phi }_{1},\dots ,{\phi }_{n}\right)$  be a holomorphic self-map of ${U}^{n}$  and $\psi \left(z\right)$  a holomorphic function of ${U}^{n},$  $p\ge 0,$  $q\ge 0.$  (i) If $p=1$  , then ${W}_{\psi ,\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)⟶{\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  is bounded if and only if
 $\begin{array}{c}{\sum }_{k,l=1}^{n}|\frac{\partial \psi }{\partial {z}_{k}}\left(z\right)|{\left(1-|{z}_{k}{|}^{2}\right)}^{q}ln\frac{4}{1-|{\phi }_{l}\left(z\right){|}^{2}}=O\left(1\right)\left(z\in {U}^{n}\right)\end{array}$ (1)
and
 $\begin{array}{c}|\psi \left(z\right)|{\sum }_{k,l=1}^{n}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{{\left(1-|{z}_{k}{|}^{2}\right)}^{q}}{1-|{\phi }_{l}\left(z\right){|}^{2}}=O\left(1\right)\left(z\in {U}^{n}\right).\end{array}$ (2)
(ii) If $0\le p<1$  , then ${W}_{\psi ,\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)⟶{\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  is bounded if and only if
 $\begin{array}{c}\psi \in {\mathcal{ℬ}}^{q}\left({U}^{n}\right)\end{array}$ (3)
and
 $\begin{array}{c}|\psi \left(z\right)|{\sum }_{k,l=1}^{n}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{{\left(1-|{z}_{k}{|}^{2}\right)}^{q}}{{\left(1-|{\phi }_{l}\left(z\right){|}^{2}\right)}^{p}}=O\left(1\right)\left(z\in {U}^{n}\right).\end{array}$ (4)
(iii) If $p>1$  , then ${W}_{\psi ,\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)⟶{\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  is bounded if and only if
 $\begin{array}{c}{\sum }_{k,l=1}^{n}|\frac{\partial \psi }{\partial {z}_{k}}\left(z\right)|\frac{{\left(1-|{z}_{k}{|}^{2}\right)}^{q}}{{\left(1-|{\phi }_{l}\left(z\right){|}^{2}\right)}^{p-1}}=O\left(1\right)\left(z\in {U}^{n}\right)\end{array}$ (5)
and
 $\begin{array}{c}|\psi \left(z\right)|{\sum }_{k,l=1}^{n}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{{\left(1-|{z}_{k}{|}^{2}\right)}^{q}}{{\left(1-|{\phi }_{l}\left(z\right){|}^{2}\right)}^{p}}=O\left(1\right)\left(z\in {U}^{n}\right).\end{array}$ (6)
Theorem 2 Let $\phi =\left({\phi }_{1},\dots ,{\phi }_{n}\right)$  be a holomorphic self-map of ${U}^{n}$  and $\psi \left(z\right)$  a holomorphic function of ${U}^{n}$  , $p\ge 0,q\ge 0.$  (i) If $p=1$  , then ${W}_{\psi ,\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)⟶{\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  is compact if and only if ${W}_{\psi ,\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)⟶{\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  is bounded and
 $\begin{array}{c}{\sum }_{k,l=1}^{n}|\frac{\partial \psi }{\partial {z}_{k}}\left(z\right)|{\left(1-|{z}_{k}{|}^{2}\right)}^{q}ln\frac{4}{1-|{\phi }_{l}\left(z\right){|}^{2}}=o\left(1\right)\left(\phi \left(z\right)\to \partial {U}^{n}\right),\end{array}$ (7)
and
 $\begin{array}{c}|\psi \left(z\right)|{\sum }_{k,l=1}^{n}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{{\left(1-|{z}_{k}{|}^{2}\right)}^{q}}{1-|{\phi }_{l}\left(z\right){|}^{2}}=o\left(1\right)\left(\phi \left(z\right)\to \partial {U}^{n}\right).\end{array}$ (8)
(ii) If $p>1$  , then ${W}_{\psi ,\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)⟶{\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  is compact if and only if ${W}_{\psi ,\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)⟶{\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  is bounded and
 $\begin{array}{c}{\sum }_{k,l=1}^{n}|\frac{\partial \psi }{\partial {z}_{k}}\left(z\right)|\frac{{\left(1-|{z}_{k}{|}^{2}\right)}^{q}}{{\left(1-|{\phi }_{l}\left(z\right){|}^{2}\right)}^{p-1}}=o\left(1\right)\left(\phi \left(z\right)\to \partial {U}^{n}\right),\end{array}$ (9)
and
 $\begin{array}{c}|\psi \left(z\right)|{\sum }_{k,l=1}^{n}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{{\left(1-|{z}_{k}{|}^{2}\right)}^{q}}{{\left(1-|{\phi }_{l}\left(z\right){|}^{2}\right)}^{p}}=o\left(1\right)\left(\phi \left(z\right)\to \partial {U}^{n}\right).\end{array}$ (10)
Remark 1 It is easy to show that if (1) or (5) or (7) or (9) holds, then $\psi \in {\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  and
 $\begin{array}{c}{\sum }_{k=1}^{n}|\frac{\partial \psi \left(z\right)}{\partial {z}_{k}}\left(z\right)|{\left(1-|{z}_{k}{|}^{2}\right)}^{q}=o\left(1\right)\left(\phi \left(z\right)\to \partial {U}^{n}\right).\end{array}$ (11)
In fact, (1) implies that ${\sum }_{k}^{n}|\frac{\partial \psi }{\partial {z}_{k}}\left(z\right)|{\left(1-|{z}_{k}{|}^{2}\right)}^{q}{\sum }_{l=1}^{n}ln\frac{4}{1-|{\phi }_{l}\left(z\right){|}^{2}}\le C$  for all $z\in {\text{C}}^{n}.$  The same reason for others.
Theorem 3 Let $\phi =\left({\phi }_{1},\dots ,{\phi }_{n}\right)$  be a holomorphic self-map of ${U}^{n}$  and $\psi \left(z\right)$  a holomorphic function of ${U}^{n}$  , $0\le p<1,$  $q\ge 0.$  Then ${W}_{\psi ,\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)⟶{\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  is compact if and only if ${W}_{\psi ,\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)⟶{\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  is bounded and
 $\begin{array}{c}{lim}_{|{\phi }_{l}\left(z\right)|\to 1}{\sum }_{k=1}^{n}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{\left(1-|{z}_{k}{|}^{2}{\right)}^{q}}{\left(1-|{\phi }_{l}\left(z\right){|}^{2}{\right)}^{p}}=0\end{array}$ (12)
for each $l\in \left\{1,...,n\right\}$  with $|{\phi }_{l}\left(z\right)|\to 1.$
Corollary 1 Let $\phi =\left({\phi }_{1},...,{\phi }_{n}\right)$  be a holomorphic self-map of ${U}^{n}$  , $p,q\ge 0.$  Then ${C}_{\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)\to {\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  is bounded if and only if there exists a constant $C$  such that ${\sum }_{k,l=1}^{n}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{\left(1-|{z}_{k}{|}^{2}{\right)}^{q}}{\left(1-|{\phi }_{l}\left(z\right){|}^{2}{\right)}^{p}}\le C,$  for all $z\in {U}^{n}.$
Proof Let $\psi \left(z\right)\equiv 1,z\in {U}^{n}$  , then $\frac{\partial \psi }{\partial {z}_{k}}\left(z\right)=0$  for all $k\in \left\{1,2,\cdots ,n\right\}.$  It is clear that condition ( 1 ) in Theorem 1, condition ( 7 ) in Theorem 2 and condition ( 9 ) in Theorem 3 hold, note that ${W}_{\psi ,\phi }={C}_{\phi }$  , the Corollary follows by combining the above Theorems. Similar to Corollary 1, the following Corollary follows.
Corollary 2 Let $\phi =\left({\phi }_{1},...,{\phi }_{n}\right)$  be a holomorphic self-map of ${U}^{n}.$  If $p\ge 1$  and $q\ge 0,$  then ${C}_{\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)\to {\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  is compact if and only if ${\sum }_{k,l=1}^{n}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{\left(1-|{z}_{k}{|}^{2}{\right)}^{q}}{\left(1-|{\phi }_{l}\left(z\right){|}^{2}{\right)}^{p}}\le C,$  for all $z\in {U}^{n},$  and ${\sum }_{k,l=1}^{n}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{\left(1-|{z}_{k}{|}^{2}{\right)}^{q}}{\left(1-|{\phi }_{l}\left(z\right){|}^{2}{\right)}^{p}}=o\left(1\right)\left(\phi \left(z\right)\to \partial {U}^{n}\right).$  If $0\le p<1\right)$  and $q\ge 0,$  Then ${C}_{\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)\to {\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  is compact if and only if ${\sum }_{k,l=1}^{n}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{\left(1-|{z}_{k}{|}^{2}{\right)}^{q}}{\left(1-|{\phi }_{l}\left(z\right){|}^{2}{\right)}^{p}}\le C,$  for all $z\in {U}^{n},$  and ${lim}_{|{\phi }_{l}\left(z\right)|\to 1}{\sum }_{k=1}^{n}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{\left(1-|{z}_{k}{|}^{2}{\right)}^{q}}{\left(1-|{\phi }_{l}\left(z\right){|}^{2}{\right)}^{p}}=0$  for each $l\in \left\{1,...,n\right\}$  with ${\phi }_{l}\left(z\right)|\to 1.$
Remark 2 If let $\phi =id:{U}^{n}⟶{U}^{n},$  then we can obtain the corresponding results about multiplication operator ${M}_{\psi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)\to {\mathcal{ℬ}}^{q}\left({U}^{n}\right).$  Throughout the remainder of this paper $C$  will denote a positive constant, the exact value of which will vary from one appearance to the next.

2 Some Lemmas

Lemma 1 Let $f\in {\mathcal{ℬ}}^{p}\left({U}^{n}\right),$  (i) If $p=1,$  then $|f\left(z\right)|\le \left(\frac{1}{2}+\frac{1}{2nln2}\right)\left({\sum }_{l=1}^{n}ln\frac{4}{1-|{z}_{l}{|}^{2}}\right)\parallel f{\parallel }_{p};$  (ii) If $0\le p<1,$  then $|f\left(z\right)|\le \left(1+\frac{1}{1-p}\right)\parallel f{\parallel }_{p};$  (iii) If $p>1$  , then $|f\left(z\right)|\le \left(\frac{1}{n}+\frac{{2}^{p-1}}{p-1}\right)\left({\sum }_{l=1}^{n}\frac{1}{\left(1-|{z}_{l}{|}^{2}{\right)}^{p-1}}\right)\parallel f{\parallel }_{p}.$
Proof This Lemma can be proved by some integral estimates (if necessary, the proof can be omitted).
By the definition of $\parallel .{\parallel }_{p}$  , $|f\left(0\right)|\le \parallel f{\parallel }_{p},|\frac{\partial f\left(z\right)}{\partial {z}_{l}}|\le \frac{\parallel f{\parallel }_{p}}{\left(1-|{z}_{l}{|}^{2}{\right)}^{p}}\left(l\in \left\{1,2,\cdots ,n\right\}\right)$  and
 $\begin{array}{ccc}& & f\left(z\right)-f\left(0\right)={\int }_{0}^{1}\frac{df\left(tz\right)}{dt}dt={\sum }_{l=1}^{n}{\int }_{0}^{1}{z}_{l}\frac{\partial f}{\partial {\zeta }_{l}}\left(tz\right)dt,\end{array}$
So
 $\begin{array}{ccc}& & |f\left(z\right)|\le |f\left(0\right)|+{\sum }_{l=1}^{n}|{z}_{l}|{\int }_{0}^{1}\frac{\parallel f{\parallel }_{p}}{{\left(1-{t}^{2}|{z}_{l}{|}^{2}\right)}^{p}}dt\end{array}$
 $\begin{array}{ccc}& & \le \parallel f{\parallel }_{p}+\parallel f{\parallel }_{p}{\sum }_{l=1}^{n}{\int }_{0}^{|{z}_{l}|}\frac{1}{{\left(1-{t}^{2}\right)}^{p}}dt.\end{array}$ (13)
If $p=1,$
 $\begin{array}{c}{\int }_{0}^{|{z}_{l}|}\frac{1}{{\left(1-{t}^{2}\right)}^{p}}dt=\frac{1}{2}ln\frac{1+|{z}_{l}|}{1-|{z}_{l}|}\le \frac{1}{2}ln\frac{4}{1-|{z}_{l}{|}^{2}}.\end{array}$ (14)
It is clear that $ln\frac{4}{1-|{z}_{l}{|}^{2}}>ln4=2ln2,$  so
 $\begin{array}{c}1\le \frac{1}{2ln2}ln\frac{4}{1-|{z}_{l}{|}^{2}}\le \frac{1}{2nln2}{\sum }_{l=1}^{n}ln\frac{4}{1-|{z}_{l}{|}^{2}}.\end{array}$ (15)
Combining ( 13 ),( 14 ) and ( 15 ), we get $|f\left(z\right)|\le \left(\frac{1}{2}+\frac{1}{2nln2}\right)\left({\sum }_{l=1}^{n}ln\frac{4}{1-|{z}_{l}{|}^{2}}\right)\parallel f{\parallel }_{p}.$  If $p\ne 1,$
 $\begin{array}{ccc}& & {\int }_{0}^{|{z}_{l}|}\frac{1}{{\left(1-{t}^{2}\right)}^{p}}dt={\int }_{0}^{|{z}_{l}|}\frac{1}{\left(1-t{\right)}^{p}}\cdot \frac{1}{\left(1+t{\right)}^{p}}dt\end{array}$
 $\begin{array}{ccc}& & \le {\int }_{0}^{|{z}_{l}|}\frac{1}{\left(1-t{\right)}^{p}}dt=\frac{1-\left(1-|{z}_{l}|{\right)}^{-p+1}}{1-p}.\end{array}$ (16)
If $0\le p<1,$  ( 16 ) gives that ${\int }_{0}^{|{z}_{l}|}\frac{1}{{\left(1-{t}^{2}\right)}^{p}}dt\le \frac{1}{1-p},$  it follows from ( 13 ) that $|f\left(z\right)|\le \left(1+\frac{n}{1-p}\right)\parallel f{\parallel }_{p}.$  If $p>1,$  ( 16 ) gives that ${\int }_{0}^{|{z}_{l}|}\frac{1}{{\left(1-{t}^{2}\right)}^{p}}dt\le \frac{1-\left(1-|{z}_{l}{|}^{p-1}\right)}{\left(p-1\right)\left(1-|{z}_{l}|{\right)}^{p-1}}\le \frac{{2}^{p-1}}{\left(p-1\right)\left(1-|{z}_{l}^{2}|{\right)}^{p-1}},$  it follows from ( 13 ) that
 $\begin{array}{ccc}|f\left(z\right)|& \le & \parallel f{\parallel }_{p}+\frac{{2}^{p-1}}{p-1}\left({\sum }_{l=1}^{n}\frac{1}{\left(1-|{z}_{l}{|}^{2}{\right)}^{p-1}}\right)\parallel f{\parallel }_{p}\end{array}$
 $\begin{array}{ccc}& \le & \left(\frac{1}{n}+\frac{{2}^{p-1}}{p-1}\right)\left({\sum }_{l=1}^{n}\frac{1}{\left(1-|{z}_{l}{|}^{2}{\right)}^{p-1}}\right)\parallel f{\parallel }_{p}.\end{array}$
Now the Lemma is proved.
Lemma 2 Let $\phi$  be a holomorphic self-map of ${U}^{n}$  and $\psi \left(z\right)$  a holomorphic function on ${U}^{n},$  then ${W}_{\psi ,\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)\to {\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  is compact if and only if for any bounded sequence $\left\{{f}_{j}\right\}$  in ${\mathcal{ℬ}}^{p}\left({U}^{n}\right)$  which converges to zero uniformly on compact subsets of ${U}^{n}$  , we have $\parallel {W}_{\psi ,\phi }{f}_{j}{\parallel }_{q}\to 0,$  as $j\to \infty .$
Proof Assume that ${W}_{\psi ,\phi }$  is compact and suppose $\left\{{f}_{j}\right\}$  is a sequence in ${\mathcal{ℬ}}^{p}\left({U}^{n}\right)$  with ${sup}_{j\in \mathbf{N}}\parallel {f}_{j}{\parallel }_{p}<\infty$  and ${f}_{j}\to 0$  uniformly on compact subsets of ${U}^{n}.$  By the compactness of ${W}_{\psi ,\phi }$  we have that ${W}_{\psi ,\phi }\left({f}_{j}\right)=\psi {f}_{j}\circ \phi$  has a subsequence $\psi {f}_{{j}_{m}}\circ \phi$  which converges in ${\mathcal{ℬ}}^{q},$  say, to $g.$  By Lemma 1 we have that for any compact $K\subset {U}^{n}$  there is a positive constant ${C}_{K}$  independent of $f$  such that $|\psi \left(z\right){f}_{{j}_{m}}\left(\phi \left(z\right)\right)-g\left(z\right)|\le {C}_{K}\parallel \psi {f}_{{j}_{m}}\circ \phi -g{\parallel }_{q}$  for all $z\in K$  . This implies that $\psi \left(z\right){f}_{{j}_{m}}\left(\phi \left(z\right)\right)-g\left(z\right)\to 0$  uniformly on compacts of ${U}^{n}.$  Since $K$  is a compact subset of ${U}^{n},$  $|\psi \left(z\right)|\le C$  for all $z\in K$  and $\phi \left(K\right)$  isalso a compact subset of ${U}^{n},$  by the hypothesis, $|\psi \left(z\right){f}_{{j}_{m}}\left(\phi \left(z\right)\right)|\le C|{f}_{{j}_{m}}\left(\phi \left(z\right)\right)|$  converges to zero uniformly on $K$  . It follows from the arbitrary of $K$  that the limit function $g$  is equal to $0.$  Since it is true for arbitrary subsequence of $\left\{{f}_{j}\right\}$  we see that ${W}_{\psi ,\phi }{f}_{j}\to 0$  in ${\mathcal{ℬ}}^{q}.$  Conversely, $\left\{{g}_{j}\right\}$  be any sequence in the ball ${\mathcal{K}}_{M}={B}_{{\mathcal{ℬ}}^{p}}\left(0,M\right)$  of the space ${\mathcal{ℬ}}^{p}\left({U}^{n}\right).$  Since $\parallel {g}_{j}{\parallel }_{p}\le M<\infty ,$  by Lemma 1, $\left\{{g}_{j}\right\}$  is uniformly bounded on compact subsets of ${U}^{n}$  and hence normal by Montel's theorem. Hence we may extract a subsequence $\left\{{g}_{{j}_{m}}\right\}$  which converges uniformly on compact subsets of ${U}^{n}$  to some $g\in H\left({U}^{n}\right).$  It follows that $\frac{\partial {g}_{{j}_{m}}}{\partial {z}_{l}}\to \frac{\partial g}{\partial {z}_{l}}$  uniformly on compacts, for each $l\in \left\{1,...,n\right\},$  which implies $g\in {\mathcal{ℬ}}^{p}\left({U}^{n}\right)$  and $\parallel g{\parallel }_{p}\le M.$  Hence the sequence $\left\{{g}_{{j}_{m}}-g\right\}$  is such that $\parallel {g}_{{j}_{m}}-g{\parallel }_{p}\le 2M<\infty ,$  and converges to $0$  on compact subsets of ${U}^{n},$  by the hypothesis and Lemma 2, we have that $\psi {g}_{{j}_{m}}\circ \phi \to \psi g\circ \phi$  in ${\mathcal{ℬ}}^{q}.$  Thus the set ${W}_{\psi ,\phi }\left({\mathcal{K}}_{M}\right)$  is relatively compact, finishing the proof.
Lemma 3 Let $0\le p<1.$  If $f\in {\mathcal{ℬ}}^{p}\left({U}^{n}\right)$  , then $|f\left(z\right)-f\left(w\right)|\le \frac{2}{1-p}\parallel f{\parallel }_{p}{\sum }_{k=1}^{n}|{z}_{k}-{w}_{k}{|}^{1-p},z,w\in {U}^{n}.$
Proof For any $z=\left({z}_{1},{z}_{2},\cdots ,{z}_{n}\right),$  $w=\left({w}_{1},{w}_{2},\cdots ,{w}_{n}\right)\in {U}^{n},$  then $tz+\left(1-t\right)w\in {U}^{n}$  for $t\in \left[0,1\right].$  Denote $F\left(t\right)=f\left(tz+\left(1-t\right)w\right),$  then
 $\begin{array}{ccc}& & f\left(z\right)-f\left(w\right)=F\left(1\right)-F\left(0\right)={\int }_{0}^{1}{F}^{\prime }\left(t\right)dt={\int }_{0}^{1}\frac{d}{dt}\left(f\left(tz+\left(1-t\right)w\right)dt\end{array}$
 $\begin{array}{ccc}& & ={\sum }_{k=1}^{n}\left({z}_{k}-{w}_{k}\right){\int }_{0}^{1}\frac{\partial f}{\partial {\zeta }_{k}}\left(tz+\left(1-t\right)w\right)dt.\end{array}$ (17)
$f\in {\mathcal{ℬ}}^{p}\left({U}^{n}\right),$  so for $k\in \left\{1,2,\cdots ,n\right\}$  $\left(1-|{\zeta }_{k}{|}^{2}{\right)}^{p}|\frac{\partial f}{\partial {\zeta }_{k}}\left(\zeta \right)|\le \parallel f{\parallel }_{p}$  $|\frac{\partial f}{\partial {\zeta }_{k}}|\le \parallel f{\parallel }_{p}\left(1-|{\zeta }_{k}{|}^{2}{\right)}^{-p}\le \parallel f{\parallel }_{p}\left(1-|{\zeta }_{k}|{\right)}^{-p}.$  Let ${\zeta }_{k}=t{z}_{k}+\left(1-t\right){w}_{k}={w}_{k}+t\left({z}_{k}-{w}_{k}\right),$  note that if $|a|<1,|b|<1$  and $|a+b|<1$  then $|a|+|b|+|a+b|<2,$  so $1-|{\zeta }_{k}|=1-|{w}_{k}+t\left({z}_{k}-{w}_{k}\right)|\ge |1-|{w}_{k}|-t|{z}_{k}-{w}_{k}||,$
 $\begin{array}{c}|\frac{\partial f}{\partial {\zeta }_{k}}\left(tz+\left(1-t\right)w\right)|\le \parallel f{\parallel }_{p}\left(1-|{\zeta }_{k}{|}^{2}{\right)}^{-p}\le \parallel f{\parallel }_{p}{|1-|{w}_{k}|-t|{z}_{k}-{w}_{k}||}^{-p}.\end{array}$ (18)
It follows from ( 17 ) and ( 18 ) that
 $\begin{array}{ccc}& & |f\left(z\right)-f\left(w\right)|\le \parallel f{\parallel }_{p}{\sum }_{k=1}^{n}|{z}_{k}-{w}_{k}|{\int }_{0}^{1}{|1-|{w}_{k}|-t|{z}_{k}-{w}_{k}||}^{-p}dt\end{array}$
 $\begin{array}{ccc}& & =\parallel f{\parallel }_{p}{\sum }_{k=1}^{n}{\int }_{0}^{|{z}_{k}-{w}_{k}|}{|1-|{w}_{k}|-t|}^{-p}dt.\end{array}$ (19)
If $1-|{w}_{k}|\le |{z}_{k}-{w}_{k}|,$  then
 $\begin{array}{ccc}& & {\int }_{0}^{|{z}_{k}-{w}_{k}|}{|1-|{w}_{k}|-t|}^{-p}dt\end{array}$
 $\begin{array}{ccc}& & ={\int }_{0}^{1-|{w}_{k}|}\left(1-|{w}_{k}|-t{\right)}^{-p}dt+{\int }_{1-|{w}_{k}|}^{|{z}_{k}-{w}_{k}|}{\left(t-\left(1-|{w}_{k}|\right)\right)}^{-p}dt\end{array}$
 $\begin{array}{ccc}& & =\frac{1}{1-p}\left(\left(1-|{w}_{k}|{\right)}^{1-p}+\left(|{z}_{k}-{w}_{k}|-\left(1-|{w}_{k}|\right){\right)}^{1-p}\right)\end{array}$
 $\begin{array}{ccc}& & \le \frac{2}{1-p}\left(|{z}_{k}-{w}_{k}|{\right)}^{1-p}.\end{array}$ (20)
If $1-|{w}_{k}|>|{z}_{k}-{w}_{k}|,$  let $a=1-|{w}_{k}|,b=|{z}_{k}-{w}_{k}|,a>b$  and note that the fundamental inequality ${a}^{1-p}<\left(a-b{\right)}^{1-p}+{b}^{1-p},$  where $0\le p<1,$  so
 $\begin{array}{ccc}& & {\int }_{0}^{|{z}_{k}-{w}_{k}|}{|1-|{w}_{k}|-t|}^{-p}dt={\int }_{0}^{|{z}_{k}-{w}_{k}|}{\left(1-|{w}_{k}|-t\right)}^{-p}dt\end{array}$
 $\begin{array}{ccc}& & =\frac{1}{1-p}\left(\left(1-|{w}_{k}|{\right)}^{1-p}-\left(1-|{w}_{k}|-|{z}_{k}-{w}_{k}|{\right)}^{1-p}\right)\end{array}$
 $\begin{array}{ccc}& & <\frac{1}{1-p}|{z}_{k}-{w}_{k}{|}^{1-p}.\end{array}$ (21)
Combining  20  21  $\left(\right),\left(\right)$  and  19  $\left(\right)$  , this Lemma is proved.
Lemma 4 Let $0\le p<1.$  Every norm-bounded sequence in ${\mathcal{ℬ}}^{p}\left({U}^{n}\right)$  has a subsequence that converges uniformly on $\overline{{U}^{n}}$  .
Proof Let $\left\{{f}_{j}\right\}$  be a sequence in ${\mathcal{ℬ}}^{p}\left({U}^{n}\right),$  and $\parallel {f}_{j}{\parallel }_{p}\le C$  for $j=1,2,\cdots .$  It follows from Lemma 3 that $|{f}_{j}\left(z\right)-{f}_{j}\left(w\right)|\le C{\sum }_{k=1}^{n}|{z}_{k}-{w}_{k}{|}^{1-p}$  for $z,w\in {U}^{n}.$  Thus the family $\left\{{f}_{j}:n=1,2,\cdots \right\}$  is equicontinuous. By Lemma 1, $|{f}_{j}\left(z\right)|\le C\parallel {f}_{j}{\parallel }_{p}\le C,$  the family $\left\{{f}_{j}:n=1,2,\cdots \right\}$  is bounded uniformly on $\overline{{U}^{n}}$  . The statement of the lemma now follows from the Arzela-Ascoli Theorem.
Lemma 5 Let $p\in \left[0,1\right)$  and $\left\{{f}_{j}\right\}$  be a norm bounded sequence in ${\mathcal{ℬ}}^{p}\left({U}^{n}\right)$  which converges to zero uniformly on compacts of ${U}^{n}.$  Then it converges to zero uniformly on $\overline{{U}^{n}}.$
Proof Let $\varepsilon \in \left(0,1\right),$  $r\in \left(1-\varepsilon ,1\right),$  and $w\in r{U}^{n},$  then from Lemma 3 we have
 $\begin{array}{ccc}& & |{f}_{j}\left(z\right)|\le |{f}_{j}\left(w\right)|+C\parallel {f}_{j}{\parallel }_{p}{\sum }_{k=1}^{n}|{z}_{k}-{w}_{k}{|}^{1-p}\end{array}$
 $\begin{array}{ccc}& & \le |{f}_{j}\left(w\right)|+C{\sum }_{k=1}^{n}|{z}_{k}-{w}_{k}{|}^{1-p},z\in {U}^{n}.\end{array}$
Since ${sup}_{w\in r{U}^{n}}|{f}_{j}\left(w\right)|\to 0$  and since for each $z\in \overline{{U}^{n}}\r{U}^{n},$  $1-\varepsilon   there exists $\eta$  such that $1-\varepsilon <\eta   Choosing ${w}_{z}=\left(\eta {z}_{1},\eta {z}_{2},\cdots ,\eta {z}_{n}\right),$  then ${w}_{z}\in r{U}^{n}$  and $|{z}_{k}-\left({w}_{z}{\right)}_{k}{|}^{1-p}=|{z}_{k}-\eta {z}_{k}{|}^{1-p}=\left(1-\eta {\right)}^{1-p}|{z}_{k}|<{\varepsilon }^{1-p}\le \varepsilon .$  So ${lim}_{k\to \infty }{sup}_{z\in \overline{{U}^{n}}}|{f}_{j}\left(z\right)|\le C\varepsilon ,$  from which the result follows.
Combining Lemma 2 and Lemma 5, we can obtain the following Lemma at once.
Lemma 6 Let $\phi$  be a holomorphic self-map of ${U}^{n}$  and $\psi \left(z\right)$  a holomorphic function on ${U}^{n}.$  Let $0\le p<1$  and $q\ge 0,$  then ${W}_{\psi ,\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)\to {\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  is compact if and only if for any bounded sequence $\left\{{f}_{j}\right\}$  in ${\mathcal{ℬ}}^{p}\left({U}^{n}\right)$  which converges to zero uniformly on $\overline{{U}^{n}}$  we have $\parallel {W}_{\psi ,\phi }{f}_{j}{\parallel }_{q}\to 0,$  as $j\to \infty .$

3 The Proof of Theorem 1

First shows the sufficiency. If $f\in {\mathcal{ℬ}}^{p}\left({U}^{n}\right),$  a direct calculation gives
 $\begin{array}{ccc}& & {\sum }_{k=1}^{n}|\frac{\partial \left({W}_{\psi ,\phi }f\left(z\right)\right)}{\partial {z}_{k}}|\left(1-|{z}_{k}{|}^{2}{\right)}^{q}\end{array}$
 $\begin{array}{ccc}& & ={\sum }_{k=1}^{n}|\frac{\partial \psi \left(z\right)}{\partial {z}_{k}}f\left(\phi \left(z\right)\right)+\psi \left(z\right){\sum }_{l=1}^{n}\frac{\partial f}{\partial {w}_{l}}\left(\phi \left(z\right)\right)\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\left(1-|{z}_{k}{|}^{2}{\right)}^{q}\end{array}$ (22)
 $\begin{array}{ccc}& & \le {\sum }_{k=1}^{n}|\frac{\partial \psi \left(z\right)}{\partial {z}_{k}}||f\left(\phi \left(z\right)\right)|\left(1-|{z}_{k}{|}^{2}{\right)}^{q}\end{array}$
 $\begin{array}{ccc}& & +|\psi \left(z\right)|{\sum }_{k,l=1}^{n}|\frac{\partial f}{\partial {w}_{l}}\left(\phi \left(z\right)\right)\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\left(1-|{z}_{k}{|}^{2}{\right)}^{q}.\end{array}$ (23)
By Lemma 1, if $p=1$  , ( 22 ) gives that
 $\begin{array}{ccc}& & {\sum }_{k=1}^{n}|\frac{\partial \left({W}_{\psi ,\phi }f\left(z\right)\right)}{\partial {z}_{k}}|\left(1-|{z}_{k}{|}^{2}{\right)}^{q}\end{array}$
 $\begin{array}{ccc}& & \le C\left({\sum }_{k,l=1}^{n}|\frac{\partial \psi }{\partial {z}_{k}}\left(z\right)|{\left(1-|{z}_{k}{|}^{2}\right)}^{q}ln\frac{4}{1-|{\phi }_{l}\left(z\right){|}^{2}}\end{array}$
 $\begin{array}{ccc}& & +|\psi \left(z\right)|{\sum }_{k,l=1}^{n}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{{\left(1-|{z}_{k}{|}^{2}\right)}^{q}}{1-|{\phi }_{l}\left(z\right){|}^{2}}\right)\parallel f{\parallel }_{p}.\end{array}$ (24)
Combining ( 1 ), ( 2 ) and ( 24 ), we know ${W}_{\psi ,\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)⟶{\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  is bounded.
If $0\le p<1$  , by Lemma 1, ( 23 ) gives that
 $\begin{array}{ccc}& & {\sum }_{k=1}^{n}|\frac{\partial \left({W}_{\psi ,\phi }f\left(z\right)\right)}{\partial {z}_{k}}|\left(1-|{z}_{k}{|}^{2}{\right)}^{q}\end{array}$
 $\begin{array}{ccc}& & \le C\left({\sum }_{k,l=1}^{n}|\frac{\partial \psi }{\partial {z}_{k}}\left(z\right)|{\left(1-|{z}_{k}{|}^{2}\right)}^{q}\end{array}$
 $\begin{array}{ccc}& & +|\psi \left(z\right)|{\sum }_{k,l=1}^{n}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{{\left(1-|{z}_{k}{|}^{2}\right)}^{q}}{{\left(1-|{\phi }_{l}\left(z\right){|}^{2}\right)}^{p}}\right)\parallel f{\parallel }_{p}.\end{array}$ (25)
Combining ( 3 ), ( 4 ), Lemma 1 and ( 25 ), we know ${W}_{\psi ,\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)⟶{\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  is bounded.
If $p>1,$  by Lemma 1, ( 23 ) gives that
 $\begin{array}{ccc}& & {\sum }_{k=1}^{n}|\frac{\partial \left({W}_{\psi ,\phi }f\left(z\right)\right)}{\partial {z}_{k}}|\left(1-|{z}_{k}{|}^{2}{\right)}^{q}\end{array}$
 $\begin{array}{ccc}& & \le C\left({\sum }_{k,l=1}^{n}|\frac{\partial \psi }{\partial {z}_{k}}\left(z\right)|\frac{{\left(1-|{z}_{k}{|}^{2}\right)}^{q}}{{\left(1-|{\phi }_{l}\left(z\right){|}^{2}\right)}^{p-1}}\end{array}$
 $\begin{array}{ccc}& & +|\psi \left(z\right)|{\sum }_{k,l=1}^{n}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{{\left(1-|{z}_{k}{|}^{2}\right)}^{q}}{{\left(1-|{\phi }_{l}\left(z\right){|}^{2}\right)}^{p}}\right)\parallel f{\parallel }_{p}.\end{array}$ (26)
Combining ( 5 ), ( 6 ) and ( 26 ), we know ${W}_{\psi ,\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)⟶{\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  is bounded.
To show the necessity, assume that ${W}_{\psi ,\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)⟶{\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  is bounded, with
 $\begin{array}{c}\parallel {W}_{\psi ,\phi }f{\parallel }_{q}\le C\parallel f{\parallel }_{p}\end{array}$ (27)
for all $f\in {\mathcal{ℬ}}^{p}\left({U}^{n}\right).$  It is clear that $\psi \in {\mathcal{ℬ}}^{q}\left({U}^{n}\right),$  that is, ( 3 ) holds.
If $p=1$  , For fixed $l\left(1\le l\le n\right),$  we will make use of a family of test functions $\left\{{f}_{w}:w\in \text{C},|w|<1\right\}$  in ${\mathcal{ℬ}}^{p}\left({U}^{n}\right)$  defined as follows: let ${f}_{w}\left(z\right)=\frac{{z}_{l}}{1-\overline{w}{z}_{l}},$  then $\frac{\partial {f}_{w}\left(z\right)}{\partial {z}_{l}}=\frac{1}{\left(1-\overline{w}{z}_{l}{\right)}^{2}},\frac{\partial {f}_{w}\left(z\right)}{\partial {z}_{k}}=0\left(k\ne l\right)$  , $\parallel {f}_{w}{\parallel }_{p}\le \frac{4}{1-|w{|}^{2}}.$  It follows from ( 22 ) and ( 27 ) that ${\sum }_{k=1}^{n}|\frac{\partial \psi \left(z\right)}{\partial {z}_{k}}\frac{{\phi }_{l}\left(z\right)}{1-\overline{w}{\phi }_{l}\left(z\right)}+\psi \left(z\right)\frac{1}{\left(1-\overline{w}{\phi }_{l}\left(z\right){\right)}^{2}}\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\left(1-|{z}_{k}{|}^{2}{\right)}^{q}\le C\frac{1}{1-|w{|}^{2}},$  let $w={\phi }_{l}\left(z\right),$  then
 $\begin{array}{c}|\psi \left(z\right)|{\sum }_{k=1}^{n}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{{\left(1-|{z}_{k}{|}^{2}\right)}^{q}}{1-|{\phi }_{l}\left(z\right){|}^{2}}\le C+\parallel \psi {\parallel }_{q}\le C.\end{array}$ (28)
So for any $z\in {U}^{n}$  , $|\psi \left(z\right)|{\sum }_{k,l=1}^{n}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{{\left(1-|{z}_{k}{|}^{2}\right)}^{q}}{1-|{\phi }_{l}\left(z\right){|}^{2}}\le C,$  that is, ( 2 ) is valid.
If we set ${g}_{w}\left(z\right)=ln\frac{4}{1-\overline{w}{z}_{l}},$  then $\frac{\partial {g}_{w}\left(z\right)}{\partial {z}_{l}}=\frac{\overline{w}}{1-\overline{w}{z}_{l}},\frac{\partial {g}_{w}\left(z\right)}{\partial {z}_{k}}=0\left(k\ne l\right)$  , $\parallel {g}_{w}{\parallel }_{p}\le 2+ln4.$  It follows from ( 22 ) and ( 27 ) that ${\sum }_{k=1}^{n}|\frac{\partial \psi \left(z\right)}{\partial {z}_{k}}ln\frac{4}{1-\overline{w}{\phi }_{l}\left(z\right)}+\psi \left(z\right)\frac{\overline{w}}{1-\overline{w}{\phi }_{l}\left(z\right)}\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\left(1-|{z}_{k}{|}^{2}{\right)}^{q}\le C,$  let $w={\phi }_{l}\left(z\right)$  , ( 1 ) is valid. If $0\le p<1$  , $\psi ={W}_{\psi ,\phi }1\in {\mathcal{ℬ}}^{q}\left({U}^{n}\right),$  (3) is valid. For fixed $l\left(1\le l\le n\right),$  we will make use of a family of test functions $\left\{{f}_{w}:w\in \text{C}-\left\{0\right\},|w|<1\right\}$  in ${\mathcal{ℬ}}^{p}\left({U}^{n}$  defined as follows: set ${f}_{w}\left(z\right)={\int }_{0}^{{z}_{l}}{\left(1-\frac{{\overline{w}}^{2}}{|w{|}^{2}}{z}_{l}^{2}\right)}^{-p}d{z}_{l}$  with ${f}_{w}\left(0\right)=0.$  Since for $w\ne 0,$  $\frac{\partial {f}_{w}}{\partial {z}_{l}}={\left(1-\frac{{\overline{w}}^{2}}{|w{|}^{2}}{z}_{l}^{2}\right)}^{-p},\frac{\partial {f}_{w}}{\partial {z}_{i}}=0,\left(i\ne l\right),$  it is easy to show $\parallel {f}_{w}{\parallel }_{\alpha }=1.$  Lemma 1 gives $|{f}_{w}\left(z\right)|\le 1+\frac{n}{1-p},$  so by ( 3 ),
 $\begin{array}{c}{\sum }_{k=1}^{n}|\frac{\partial \psi }{\partial {z}_{k}}\left(z\right){f}_{w}\left(\phi \left(z\right)\right)\left(1-|{z}_{k}{|}^{2}{\right)}^{q}|\le C{\sum }_{k=1}^{n}|\frac{\partial \psi }{\partial {z}_{k}}\left(z\right)|{\left(1-|{z}_{k}{|}^{2}\right)}^{q}\le C.\end{array}$ (29)
For $z\in {U}^{n},$  it follows from ( 22 ) and ( 27 ) that ${\sum }_{k=1}^{n}|\frac{\partial \psi }{\partial {z}_{k}}\left(z\right){f}_{w}\left(\phi \left(z\right)\right)+\psi \left(z\right){\sum }_{l=1}^{n}\frac{\partial {f}_{w}\left(\phi \left(z\right)\right)}{\partial {\phi }_{l}}\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\left(1-|{z}_{k}{|}^{2}{\right)}^{q}\le C,$  combining ( 29 ), we get ${\sum }_{k=1}^{n}|\psi \left(z\right)\frac{\partial {f}_{w}\left(\phi \left(z\right)\right)}{\partial {\phi }_{l}}\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\left(1-|{z}_{k}{|}^{2}{\right)}^{q}\le C.$  For an arbitrary $z\in {U}^{n}$  with ${\phi }_{l}\left(z\right)\ne 0$  and set $w={\phi }_{l}\left(z\right)$  in the above inequality to obtain $|\psi \left(z\right)|{\sum }_{k=1}^{n}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{{\left(1-|{z}_{k}{|}^{2}\right)}^{q}}{{\left(1-|{\phi }_{l}\left(z\right){|}^{2}\right)}^{p}}\le C.$  If ${\phi }_{l}\left(z\right)=0,$  let $f\left(z\right)={z}_{l},$  it follows from ( 22 ) and ( 27 ) that $|\psi \left(z\right)||\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{{\left(1-|{z}_{k}{|}^{2}\right)}^{q}}{{\left(1-|{\phi }_{l}\left(z\right){|}^{2}\right)}^{p}}=|\frac{\partial \psi }{\partial {z}_{k}}\left(z\right){\phi }_{l}\left(z\right)+\psi \left(z\right)\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\left(1-|{z}_{k}{|}^{2}{\right)}^{q}\le C.$  So for all $z\in {U}^{n},$  $|\psi \left(z\right)|{\sum }_{k,l=1}^{n}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{{\left(1-|{z}_{k}{|}^{2}\right)}^{q}}{{\left(1-|{\phi }_{l}\left(z\right){|}^{2}\right)}^{p}}\le C,$  that is, ( 4 ) is valid.
If $p>1$  , similarly, the test functions would be ${f}_{w}\left(z\right)=\frac{1}{\overline{w}}\left(\frac{1-|w{|}^{2}}{{\left(1-\overline{w}{z}_{l}\right)}^{p}}-\frac{1}{{\left(1-\overline{w}{z}_{l}\right)}^{p-1}}\right),\left(|w|<1,w\ne 0,1\le l\le n\right)$  for proving ( 6 ), and ${g}_{w}\left(z\right)=\left(\frac{p}{{\left(1-\overline{w}{z}_{l}\right)}^{p-1}}-\frac{\left(p-1\right)\left(1-|w{|}^{2}\right)}{{\left(1-\overline{w}{z}_{l}\right)}^{p}}\right),\left(|w|<1,1\le l\le n\right)$  for proving ( 5 ), we omit the details. Now the proof of Theorem 1 is completed.

4 The Proof of Theorem 2

To show the sufficiency.
If $p=1$  . First assume conditions ( 7 ) and ( 8 ) hold, we need to prove ${W}_{\psi ,\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)⟶{\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  is compact. According to Lemma 2, assume that ${\parallel {f}_{j}\parallel }_{p}\le C,j=1,2,\dots ,$  and $\left\{{f}_{j}\right\}$  converges to zero uniformly on compact subsets of ${U}^{n},$  we need only prove that ${\parallel {W}_{\psi ,\phi }{f}_{j}\parallel }_{q}\to 0,$  as $j\to \infty .$  For every $\varepsilon >0,$  ( 7 ) and ( 8 ) imply that there exists a $r$  , $0  , such that
 $\begin{array}{c}{\sum }_{k,l=1}^{n}|\frac{\partial \psi }{\partial {z}_{k}}\left(z\right)|{\left(1-|{z}_{k}{|}^{2}\right)}^{q}ln\frac{4}{1-|{\phi }_{l}\left(z\right){|}^{2}}<\varepsilon \end{array}$ (30)
and
 $\begin{array}{c}{\sum }_{k,l=1}^{n}|\psi \left(z\right)\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{{\left(1-|{z}_{k}{|}^{2}\right)}^{q}}{1-|{\phi }_{l}\left(z\right){|}^{2}}<\varepsilon ,\end{array}$ (31)
whenever $dist\left(\phi \left(z\right),\partial {U}^{n}\right)  Since ${f}_{j}\left(\phi \left(0\right)\right)$  converges to zero, by ( 30 ) and ( 31 ), it follows from ( 24 ) that for large enough $j$
 $\begin{array}{c}{\sum }_{k=1}^{n}|\frac{\partial \left({W}_{\psi ,\phi }{f}_{j}\right)}{\partial {z}_{k}}\left(z\right)|\left(1-|{z}_{k}{|}^{2}{\right)}^{q}\le C\varepsilon ,\end{array}$ (32)
whenever $dist\left(\phi \left(z\right),\partial {U}^{n}\right)  If we set $\varepsilon =1,$  by ( 30 ) and ( 31 ),it is easy to show that ( 1 ) and ( 2 ) holds, from Remark 1, we know $\psi \in {\mathcal{ℬ}}^{q}\left({U}^{n}\right).$  On the other hand, if we write $E=\left\{w\in {U}^{n}:dist\left(w,\partial {U}^{n}\right)\ge r\right\},$  which is a closed subset of ${U}^{n}$  , then ${f}_{j}\left(w\right)$  and $\frac{\partial {f}_{j}}{\partial {w}_{l}}\left(w\right)\to 0$  uniformly on $E$  . So by ( 22 ) and the boundedness condition (2) in Theorem 1, we obtain for $dist\left(\phi \left(z\right),\partial {U}^{n}\right)\ge r,$
 $\begin{array}{ccc}& & {\sum }_{k=1}^{n}|\frac{\partial \left({W}_{\psi ,\phi }{f}_{j}\right)}{\partial {z}_{k}}\left(z\right)|\left(1-|{z}_{k}{|}^{2}{\right)}^{q}\le C\left(|{f}_{j}\left(\phi \left(z\right)\right)|+{\sum }_{l=1}^{n}|\frac{\partial {f}_{j}}{\partial {w}_{l}}\left(\phi \left(z\right)\right)|\right)\end{array}$
 $\begin{array}{ccc}& & =C\left(|{f}_{j}\left(w\right)|+{\sum }_{l=1}^{n}|\frac{\partial {f}_{j}}{\partial {w}_{l}}\left(w\right)|\right)\le C\varepsilon .\end{array}$ (33)
Since $\psi \left(0\right){f}_{j}\left(\phi \left(0\right)\right)\to 0$  for large enough $j,$  combining ( 32 ) and ( 33 ) we know ${\parallel {W}_{\psi ,\phi }{f}_{j}\parallel }_{q}\to 0,$  as $j\to \infty .$  If $p>1,$  note that ( 26 ), in a similar manner to the case $p=1$  , we can show the sufficiency, omit the details.
To show the necessity.
First we will prove the following Lemma:
Lemma 7 Let $p\ge 1$  and $q\ge 0$  . If ${W}_{\psi ,\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)⟶{\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  is compact, then
 $\begin{array}{c}|\psi \left(z\right)|{\sum }_{k,l=1}^{n}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{{\left(1-|{z}_{k}{|}^{2}\right)}^{q}}{{\left(1-|{\phi }_{l}\left(z\right){|}^{2}\right)}^{p}}=o\left(1\right)\left(\phi \left(z\right)\to \partial {U}^{n}\right).\end{array}$ (34)
Proof For any $\left\{{z}^{j}\right\}$  in ${U}^{n}$  with $\phi \left({z}^{j}\right)\to \partial {U}^{n}$  as $j\to \infty .$  Let ${A}_{j}=\phi \left({z}^{j}\right)=\left({a}_{1j},\dots ,{a}_{nj}\right),$  where ${a}_{kj}={\phi }_{k}\left({z}^{j}\right),1\le k\le n.$  Since $\phi \left({z}^{j}\right)\to \partial {U}^{n},$  as $j\to \infty$  , there exists some $s,\left(1\le s\le n\right),$  with $|{a}_{sj}|\to 1$  as $j\to \infty .$  Without loss of generality, we may assume that $s=1$  .
Case 1: If for some $l\left(1\le l\le n\right),$  $|{a}_{lj}|\to 1$  as $j\to \infty ,$  then set
 $\begin{array}{c}{f}_{j}\left(z\right)=\frac{{\left(1-|{a}_{lj}{|}^{2}\right)}^{2}}{{\left(1-{z}_{l}\overline{{a}_{lj}}\right)}^{p+1}}-\frac{1-|{a}_{lj}{|}^{2}}{{\left(1-{z}_{l}\overline{{a}_{lj}}\right)}^{p}},j=1,2,\cdots .\end{array}$ (35)
$\frac{\partial {f}_{j}}{\partial {z}_{k}}=0,\left(k\ne l\right),$  $\frac{\partial {f}_{j}}{\partial {z}_{l}}=\left(p+1\right)\overline{{a}_{lj}}\frac{{\left(1-|{a}_{lj}{|}^{2}\right)}^{2}}{{\left(1-{z}_{l}\overline{{a}_{lj}}\right)}^{p+2}}-p\overline{{a}_{lj}}\frac{1-|{a}_{lj}{|}^{2}}{{\left(1-{z}_{l}\overline{{a}_{lj}}\right)}^{p+1}},$  so $\parallel {f}_{j}{\parallel }_{p}\le C$  and $\left\{{f}_{j}\left(z\right)\right\}$  tends to zero uniformly on compact subsets of ${U}^{n}.$  So by Lemma 2, we know ${\parallel {W}_{\psi ,\phi }{f}_{j}\parallel }_{q}\to 0$  as $j\to \infty .$  It follows from ( 22 ) that
 $\begin{array}{ccc}& & {\sum }_{k=1}^{n}{\left(1-|{z}_{k}^{j}{|}^{2}\right)}^{q}|\frac{\partial \psi }{\partial {z}_{k}}\left({z}^{j}\right){f}_{j}\left(\phi \left({z}^{j}\right)\right)+\psi \left({z}^{j}\right)\frac{\partial {f}_{j}}{\partial {w}_{l}}\left(\phi \left({z}^{j}\right)\right)\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left({z}^{j}\right)|\end{array}$
 $\begin{array}{ccc}& & \le {\parallel {W}_{\psi ,\phi }{f}_{j}\parallel }_{q}\to 0.\end{array}$ (36)
Note that ${f}_{j}\left(\phi \left({z}^{j}\right)\right)=0$  and $\frac{\partial {f}_{j}}{\partial {z}_{l}}\left(\phi \left({z}^{j}\right)\right)=\overline{{a}_{lj}}\frac{1}{{\left(1-|{a}_{lj}{|}^{2}\right)}^{p}}$  , ( 36 ) gives that
 $\begin{array}{ccc}& & |\psi \left({z}^{j}\right)|{\sum }_{k=1}^{n}\frac{{\left(1-|{z}_{k}^{j}{|}^{2}\right)}^{q}}{{\left(1-|{a}_{lj}{|}^{2}\right)}^{p}}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left({z}^{j}\right)|\end{array}$
 $\begin{array}{ccc}& & \le C\frac{1}{|{a}_{lj}|}{\parallel {W}_{\psi ,\phi }{f}_{j}\parallel }_{q}\to 0\text{as}j\to \infty .\text{}\end{array}$ (37)
Case 2: If for some $l\left(2\le l\le n\right),$  $|{a}_{lj}|\to ̸1$  as $j\to \infty ,$  then we assume $|{a}_{lj}|\le \rho <1.$  If $p\ge 1$  , The compactness of ${W}_{\psi ,\phi }$  implies ${W}_{\psi ,\phi }$  is bounded, by Theorem 1 and Remark 1, ( 11 ) is valid. Set
 $\begin{array}{c}{f}_{j}\left(z\right)={z}_{l}{\left(\frac{1-|{a}_{1j}{|}^{2}}{1-{z}_{1}\overline{{a}_{1j}}}\right)}^{p},j=1,2,\cdots .\end{array}$ (38)
It is easy to show that $\parallel {f}_{j}{\parallel }_{p}\le C$  , and $\left\{{f}_{j}\right\}$  tends to zero uniformly on compact subsets of ${U}^{n}.$  It follows from ( 22 ) that
 $\begin{array}{ccc}& & |\psi \left({z}^{j}\right)|{\sum }_{k=1}^{n}{\left(1-|{z}_{k}^{j}{|}^{2}\right)}^{q}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left({z}^{j}\right)|\end{array}$
 $\begin{array}{ccc}& & \le {\parallel {W}_{\psi ,\phi }{f}_{j}\parallel }_{q}+C{\sum }_{k=1}^{n}{\left(1-|{z}_{k}^{j}{|}^{2}\right)}^{q}|\frac{\partial \psi }{\partial {z}_{k}}\left({z}^{j}\right)|\end{array}$
 $\begin{array}{ccc}& & +C|\psi \left({z}^{j}\right)|{\sum }_{k=1}^{n}\frac{{\left(1-|{z}_{k}^{j}{|}^{2}\right)}^{q}}{1-|{a}_{1j}{|}^{2}}|\frac{\partial {\phi }_{1}}{\partial {z}_{k}}\left({z}^{j}\right)|\end{array}$
 $\begin{array}{ccc}& & \le {\parallel {W}_{\psi ,\phi }{f}_{j}\parallel }_{q}+C{\sum }_{k=1}^{n}{\left(1-|{z}_{k}^{j}{|}^{2}\right)}^{q}|\frac{\partial \psi }{\partial {z}_{k}}\left({z}^{j}\right)|\end{array}$
 $\begin{array}{ccc}& & +C|\psi \left({z}^{j}\right)|{\sum }_{k=1}^{n}\frac{{\left(1-|{z}_{k}^{j}{|}^{2}\right)}^{q}}{\left(1-|{a}_{1j}{|}^{2}{\right)}^{p}}|\frac{\partial {\phi }_{1}}{\partial {z}_{k}}\left({z}^{j}\right)|\to 0.\end{array}$
So
 $\begin{array}{ccc}& & |\psi \left({z}^{j}\right)|{\sum }_{k=1}^{n}\frac{{\left(1-|{z}_{k}^{j}{|}^{2}\right)}^{q}}{\left(1-|{a}_{lj}{|}^{2}{\right)}^{p}}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left({z}^{j}\right)|\end{array}$
 $\begin{array}{ccc}& & \le \frac{1}{\left(1-{\rho }^{2}{\right)}^{p}}|\psi \left({z}^{j}\right)|{\sum }_{k=1}^{n}{\left(1-|{z}_{k}^{j}{|}^{2}\right)}^{q}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left({z}^{j}\right)|\to 0.\end{array}$ (39)
If $p=1$  , by Lemma 7, ( 8 ) is necessary, now we prove that ( 7 ) is also necessary.
In case 1, set
 $\begin{array}{c}{f}_{j}\left(z\right)={\left(ln\frac{4}{1-|{a}_{lj}{|}^{2}}\right)}^{-1}{\left(ln\frac{4}{1-{z}_{l}\overline{{a}_{lj}}}\right)}^{2},\end{array}$ (40)
It is easy to show that $\parallel {f}_{j}{\parallel }_{p}\le C$  , and $\left\{{f}_{j}\right\}$  tends to zero uniformly on compact subsets of ${U}^{n}.$  It follows from ( 22 ) that ${\sum }_{k=1}^{n}{\left(1-|{z}_{k}^{j}{|}^{2}\right)}^{q}|\frac{\partial \psi }{\partial {z}_{k}}\left({z}^{j}\right){f}_{j}\left(\phi \left({z}^{j}\right)\right)+\psi \left({z}^{j}\right)\frac{\partial {f}_{j}}{\partial {w}_{l}}\left(\phi \left({z}^{j}\right)\right)\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left({z}^{j}\right)|\le {\parallel {W}_{\psi ,\phi }{f}_{j}\parallel }_{q}\to 0.$  By a direct calculation and by ( 34 ), we obtain
 $\begin{array}{ccc}& & {\sum }_{k=1}^{n}|\frac{\partial \psi }{\partial {z}_{k}}\left(z\right)|{\left(1-|{z}_{k}{|}^{2}\right)}^{q}ln\frac{4}{1-|{\phi }_{l}\left(z\right){|}^{2}}\end{array}$
 $\begin{array}{ccc}& & \le {\parallel {W}_{\psi ,\phi }{f}_{j}\parallel }_{q}+2|\psi \left({z}^{j}\right)|{\sum }_{k=1}^{n}\frac{{\left(1-|{z}_{k}^{j}{|}^{2}\right)}^{q}}{{\left(1-|{a}_{lj}{|}^{2}\right)}^{p}}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left({z}^{j}\right)|\to 0.\end{array}$ (41)
In case 2, by remark 1, ${\sum }_{k=1}^{n}|\frac{\partial \psi }{\partial {z}_{k}}\left({z}^{j}\right)|{\left(1-|{z}_{k}^{j}{|}^{2}\right)}^{q}ln\frac{4}{1-|{\phi }_{l}\left({z}^{j}\right){|}^{2}}\le ln\frac{4}{1-{\rho }^{2}}{\sum }_{k=1}^{n}|\frac{\partial \psi }{\partial {z}_{k}}\left({z}^{j}\right)|{\left(1-|{z}_{k}^{j}{|}^{2}\right)}^{q}\to 0.$  Combining case 1 and case 2, we know ( 7 ) is necessary.
If $p>1.$  by Lemma 7, we know ( 10 ) is necessary, now we prove ( 9 ) is also necessary.
In case 1, set ${f}_{j}\left(z\right)=\left(p+1\right)\frac{1-|{a}_{lj}{|}^{2}}{{\left(1-{z}_{l}\overline{{a}_{lj}}\right)}^{p}}-p\frac{{\left(1-|{a}_{lj}{|}^{2}\right)}^{2}}{{\left(1-{z}_{l}\overline{{a}_{lj}}\right)}^{p+1}},j=1,2,\cdots .$  It is easy to show that $\parallel {f}_{j}{\parallel }_{p}\le C$  , and $\left\{{f}_{j}\right\}$  tends to zero uniformly on compact subsets of ${U}^{n}.$  Note that ${f}_{j}\left(\phi \left({z}^{j}\right)\right)=\frac{1}{{\left(1-|{a}_{lj}{|}^{2}\right)}^{p-1}}$  and $\frac{\partial {f}_{j}}{\partial {z}_{l}}\left(\phi \left({z}^{j}\right)\right)=0,$  it follows from ( 22 ) that
 $\begin{array}{ccc}& & {\sum }_{k=1}^{n}{\left(1-|{z}_{k}^{j}{|}^{2}\right)}^{q}|\frac{\partial \psi }{\partial {z}_{k}}\left({z}^{j}\right){f}_{j}\left(\phi \left({z}^{j}\right)\right)+\psi \left({z}^{j}\right)\frac{\partial {f}_{j}}{\partial {w}_{l}}\left(\phi \left({z}^{j}\right)\right)\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left({z}^{j}\right)|\end{array}$
 $\begin{array}{ccc}& & ={\sum }_{k=1}^{n}|\frac{\partial \psi }{\partial {z}_{k}}\left({z}^{j}\right)|\frac{{\left(1-|{z}_{k}^{j}{|}^{2}\right)}^{q}}{\left(1-|{a}_{lj}{|}^{2}{\right)}^{p-1}}\le {\parallel {W}_{\psi ,\phi }{f}_{j}\parallel }_{q}\to 0.\end{array}$ (42)
In case 2, by remark 1,
 $\begin{array}{ccc}& & {\sum }_{k=1}^{n}|\frac{\partial \psi }{\partial {z}_{k}}\left({z}^{j}\right)|\frac{{\left(1-|{z}_{k}^{j}{|}^{2}\right)}^{q}}{{\left(1-|{\phi }_{l}\left({z}^{j}\right){|}^{2}\right)}^{p-1}}\end{array}$
 $\begin{array}{ccc}& & \le \frac{1}{{\left(1-{\rho }^{2}\right)}^{p-1}}{\sum }_{k=1}^{n}|\frac{\partial \psi }{\partial {z}_{k}}\left({z}^{j}\right)|{\left(1-|{z}_{k}^{j}{|}^{2}\right)}^{q}\to 0.\end{array}$
Combining cas1 and case 2, we know ( 9 ) is necessary.
Now the proof of Theorem 2 is completed.

5 Proof of Theorem 3

Proof of Theorem 3. First assume ${W}_{\psi ,\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)⟶{\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  is bounded and ( 12 ) holds, we prove that ${W}_{\psi ,\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)\to {\mathcal{ℬ}}^{q}\left({U}^{n}\right)$  is compact. By Lemma 6, assume that ${\parallel {f}_{j}\parallel }_{p}\le C,j=1,2,...,$  and $\left\{{f}_{j}\right\}$  converges to zero uniformly on $\overline{{U}^{n}},$  we need to prove that ${\parallel {W}_{\psi ,\phi }{f}_{j}\parallel }_{q}\to 0,$  as $j\to \infty .$  Note that the boundedness of ${W}_{\psi ,\phi }$  and Theorem 1, ( 1 ) and ( 2 ) hold. So
 $\begin{array}{c}|\psi \left(z\right)|\le C\parallel \psi {\parallel }_{q}\le C,|\psi \left(z\right)|{\sum }_{k=1}^{n}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{\left(1-|{z}_{k}{|}^{2}{\right)}^{q}}{\left(1-|{\phi }_{l}\left(z\right){|}^{2}{\right)}^{p}}\le C\end{array}$ (43)
for $l\in \left\{1,2,\cdots ,n\right\}.$  By the assumption as $j\to \infty$
 $\begin{array}{c}{sup}_{z\in \overline{{U}^{n}}}|{f}_{j}\left(z\right)|\to 0\end{array}$ (44)
 $\begin{array}{ccc}& & {\sum }_{k=1}^{n}|\frac{\partial \left({W}_{\psi ,\phi }{f}_{j}\left(z\right)\right)}{\partial {z}_{k}}|\left(1-|{z}_{k}{|}^{2}{\right)}^{q}\end{array}$
 $\begin{array}{ccc}& & ={\sum }_{k=1}^{n}|\frac{\partial \psi \left(z\right)}{\partial {z}_{k}}{f}_{j}\left(\phi \left(z\right)\right)+\psi \left(z\right){\sum }_{l=1}^{n}\frac{\partial {f}_{j}}{\partial {w}_{l}}\left(\phi \left(z\right)\right)\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\left(1-|{z}_{k}{|}^{2}{\right)}^{q}\end{array}$
 $\begin{array}{ccc}& & \le |{f}_{j}\left(\phi \left(z\right)\right)|{\sum }_{k=1}^{n}|\frac{\partial \psi }{\partial {z}_{k}}\left(z\right)|{\left(1-|{z}_{k}{|}^{2}\right)}^{q}\end{array}$
 $\begin{array}{ccc}& & +{\sum }_{k=1}^{n}{\sum }_{l=1}^{n}|\frac{\partial {f}_{j}}{\partial {w}_{l}}\left(\phi \left(z\right)\right)||\psi \left(z\right)||\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{\left(1-|{z}_{k}{|}^{2}{\right)}^{q}}{\left(1-|{\phi }_{l}\left(z\right){|}^{2}{\right)}^{p}}\end{array}$
 $\begin{array}{ccc}& & \le C|{f}_{j}\left(\phi \left(z\right)\right)|+{\sum }_{k=1}^{n}{\sum }_{k=1}^{n}|\frac{\partial {f}_{j}}{\partial {w}_{l}}\left(\phi \left(z\right)\right)||\psi \left(z\right)||\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{\left(1-|{z}_{k}{|}^{2}{\right)}^{q}}{\left(1-|{\phi }_{l}\left(z\right){|}^{2}{\right)}^{p}}\end{array}$
 $\begin{array}{ccc}& & \le C\varepsilon +{\sum }_{k=1}^{n}{\sum }_{l=1}^{n}|\frac{\partial {f}_{j}}{\partial {w}_{l}}\left(\phi \left(z\right)\right)||\psi \left(z\right)||\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{\left(1-|{z}_{k}{|}^{2}{\right)}^{q}}{\left(1-|{\phi }_{l}\left(z\right){|}^{2}{\right)}^{p}}.\end{array}$ (45)
For every $\varepsilon >0$  , $l\in \left\{1,...,n\right\},$  ( 12 ) implies that there exists an $r$  , $0  , such that
 $\begin{array}{c}|\psi \left(z\right)|{\sum }_{k=1}^{n}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{\left(1-|{z}_{k}{|}^{2}{\right)}^{q}}{\left(1-|{\phi }_{l}\left(z\right){|}^{2}{\right)}^{p}}<\varepsilon ,\end{array}$ (46)
whenever $|{\phi }_{l}\left(z\right)|>r.$  Note that ${U}^{n}=\left\{z\in {U}^{n}:|{\phi }_{l}\left(z\right)|>r\right\}\bigcup \left\{z\in {U}^{n}:|{\phi }_{l}\left(z\right)|\le r\right\}.$  For each $l\in \left\{1,2,\cdots ,n\right\}.$  If $|{\phi }_{l}\left(z\right)|>r,$  from ( 46 ) and $\parallel {f}_{j}{\parallel }_{p}\le C$  , we have
 $\begin{array}{c}|\frac{\partial {f}_{j}}{\partial {w}_{l}}\left(\phi \left(z\right)\right)||\psi \left(z\right)||\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{\left(1-|{z}_{k}{|}^{2}{\right)}^{q}}{\left(1-|{\phi }_{l}\left(z\right){|}^{2}{\right)}^{p}} (47)
If $|{\phi }_{l}\left(z\right)|\le r,$  by the Cauchy's estimate applied to the function $g\left({w}_{l}\right)=f\left({w}_{1},...,{w}_{l-1},{w}_{l},{w}_{l+1}...,{w}_{n}\right),$  of one variable, if $|{w}_{l}|\le r,$  we have $|\frac{\partial {f}_{j}}{\partial {w}_{l}}\left(w\right)|\le C{sup}_{|{w}_{l}|\le \frac{1+r}{2}}|{f}_{j}\left(w\right)|\le C{sup}_{z\in \overline{{U}^{n}}}|{f}_{j}\left(z\right)|,$  for some $C>0$  independent of $f.$  From this, ( 43 ) and ( 44 ) we have
 $\begin{array}{ccc}& & |\frac{\partial {f}_{j}}{\partial {w}_{l}}\left(\phi \left(z\right)\right)||\psi \left(z\right)||\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left(z\right)|\frac{\left(1-|{z}_{k}{|}^{2}{\right)}^{q}}{\left(1-|{\phi }_{l}\left(z\right){|}^{2}{\right)}^{p}}\end{array}$
 $\begin{array}{ccc}& & \le C{sup}_{z\in \overline{{U}^{n}}}|{f}_{j}\left(z\right)|\to 0\end{array}$ (48)
as $j\to \infty .$  So From ( 45 ),( 47 ) and ( 48 ), and since ${lim}_{j\to \infty }{f}_{j}\left(\phi \left(0\right)\right)=0,$  we obtain ${\parallel {C}_{\phi }{f}_{j}\parallel }_{q}\to 0,$  as $j\to \infty ,$  from which the result follows.
Now suppose that ${W}_{\psi ,\phi }:{\mathcal{ℬ}}^{p}\left({U}^{n}\right)\to {\mathcal{ℬ}}^{q}\left({\text{C}}^{n}\right)$  is compact. Then ${W}_{\psi ,\phi }$  is bounded, and by Theorem 1 we know that ( 4 ) holds. Now we prove that condition ( 12 ) holds. Let $\left\{{z}^{j}\right\}$  be a sequence in ${U}^{n}$  and ${w}^{j}=\phi \left({z}^{j}\right)=\left({w}_{1}^{j},...,{w}_{n}^{j}\right).$  For each $l\in \left\{1,2,\cdots ,n\right\}$  with ${lim}_{j\to \infty }|{w}_{l}^{j}|\to 1.$  Let ${f}_{j}\left(z\right)=\frac{1-|{w}_{l}^{j}{|}^{2}}{\left(1-{z}_{l}\overline{{w}_{l}^{j}}{\right)}^{p}},j=1,2,...,$  then $\frac{\partial {f}_{j}}{\partial {z}_{l}}\left(z\right)=p\overline{{w}_{l}^{j}}\frac{1-|{w}_{l}^{j}{|}^{2}}{\left(1-{z}_{l}\overline{{w}_{l}^{j}{|}^{2}}{\right)}^{p+1}}.$  As in the proof of Theorem 2 we can prove that $\parallel {f}_{j}{\parallel }_{p}\le 1+{2}^{p+1}p,$  for all $j\in \mathbf{N},$  i.e. $\left\{{f}_{j}\right\}$  is bounded on ${\mathcal{ℬ}}^{p}\left({U}^{n}\right).$  It is easy to see that ${f}_{j}$  tends to zero uniformly on compact subsets of ${\text{C}}^{n}.$  So by Lemma 2, we know ${\parallel {W}_{\phi }{f}_{j}\parallel }_{q}\to 0$  as $j\to \infty$  and notice that ${f}_{j}\left(\phi \left({z}^{j}\right)\right)=\left(1-|{w}_{l}^{j}{|}^{2}{\right)}^{1-p},\frac{\partial {f}_{j}}{\partial {w}_{l}}\left(\phi \left({z}^{j}\right)\right)=p\overline{{w}_{l}^{j}}\frac{1}{\left(1-|{w}_{l}^{j}{|}^{2}{\right)}^{p}},$  consequently, by ( 22 )
 $\begin{array}{ccc}& & {\sum }_{k=1}^{n}{\left(1-|{z}_{k}^{j}{|}^{2}\right)}^{q}|\frac{\partial \psi }{\partial {z}_{k}}\left({z}^{j}\right){f}_{j}\left(\phi \left({z}^{j}\right)\right)+\psi \left({z}^{j}\right)\frac{\partial {f}_{j}}{\partial {w}_{l}}\left(\phi \left({z}^{j}\right)\right)\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left({z}^{j}\right)|\end{array}$
 $\begin{array}{ccc}& & \le {\parallel {W}_{\psi ,\phi }{f}_{j}\parallel }_{q}.\end{array}$ (49)
So ${\sum }_{k=1}^{n}p|{w}_{l}^{j}||\psi \left({z}^{j}\right)|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left({z}^{j}\right)|\frac{{\left(1-|{z}_{k}^{j}{|}^{2}\right)}^{q}}{\left(1-|{w}_{l}^{j}{|}^{2}{\right)}^{p}}\le \parallel \psi {\parallel }_{q}{\left(1-|{w}_{l}^{j}{|}^{2}\right)}^{1-p}+{\parallel {W}_{\psi ,\phi }{f}_{j}\parallel }_{q}.$  Hence $|\psi \left({z}^{j}\right)|{\sum }_{k=1}^{n}\frac{\left(1-|{z}_{k}^{j}{|}^{2}{\right)}^{q}}{\left(1-|{w}_{l}^{j}{|}^{2}{\right)}^{p}}|\frac{\partial {\phi }_{l}}{\partial {z}_{k}}\left({z}^{j}\right)|\le \frac{1}{p|{w}_{l}^{j}|}\left(C\left(1-|{w}_{l}^{j}{|}^{2}{\right)}^{1-p}+{\parallel {W}_{\psi ,\phi }{f}_{j}\parallel }_{q}\right)\to 0$  as $j\to \infty .$  Now the proof of Theorem 3 is finished.
References

1. R.Timoney, Bloch Function in Several Complex Variables, I, Bull.London Math.Soc. 12(1980),241-267.
2. R.Timoney, Bloch Function in Several Complex Variables, II, J. Reine Angew Math., 319(1980):1-22.
3. C. C. Cowen and B. D. MacCluer, Composition operators on spaces of analytic functions, CRC Press, Boca Raton, FL,1995.
4. K. Madigan and A. Matheson, Compact composition operators on the Bloch space, Trans. Amer. Math. Soc. 347(1995): 2679-2687.
5. K. Madigan, Compact composition operators on analytic Lipschitz space, Proc. Amer. Math. Soc. 119(1993): 465-473.
6. M. D. Contreras and A. G. Hernandez-Diaz, Weighted composition operators in weighted Banach spaces of ananytic functions, J. Austral. Math. Soc. (Series A), 69(2000):41-46.
7. Alfonso Montes-Rodriguez, Weighted composition operators on weighted banach spaces of analytic functions, J. London Math. Soc. (2)61(2000):872-884.
8. S.Ohno, K. Stroethoff and R.H.Zhao, Weighted composition operators between-type spaces, Rocky Mountain Journal of Mathematics,Vol33, No.1(2003):191-215.
9. J. H. Shi and L. Luo, Composition operators on the Bloch space of several complex variables, Acta Math. Sinica, English Series, 16(2000):85-98.
10. Z. H. Zhou and J. H. Shi, Compact composition operators on the Bloch space in polydisk, Science in China (Series A), 44(2001): 286-291.
11. Z. H. Zhou, Composition operators on the Lipschitz spaces in polydisk. Science in China (Series A), 46(2003): 33-38.
12. Z. H. Zhou and J. H. Shi, Compactness of composition operators on the Bloch space in classical bounded symmetric domains, The Michigan Mathematical Journal, 50(2002): 381-405.
13. J. P. D'Angelo, Several complex variables and geometry of real hypersurfaces, CRC Press, 1993.

$1$  Supported in part by the National Natural Science Foundation of China (Grand No.10371091), and LiuHui Center for Applied Mathematics, Nankai University and Tianjin University


text:
79 views
O(n-1)+O(n)=O(n)
O(n/2)+O(n)=O(n)
but O(1)+O(2)+O(3)+...+O(n)=O(n(n+1)/2)=O(n^2)
why?
| 79 views

+1 vote

When you add constant time complexities N times, you end up getting N.O(1) = O(N).

• O(n-1)+O(n)=O(n) ; we are not adding n times .

by Boss (35.7k points)
0

So when we are solving computations where O(1) or O(constant) is involved, then how do we know whether to involve that in the calculation or to ignore it. For e.g if O(1)+O(2)+O(3)+...+O(logn)+O(n) is there, then do we ignore the constant parts of the beginning?

f(a)<=O(x): It means for function f(a) upper value cant go beyond x, so max value this function can attain  is x.(not quoting from any textbooks just in simple terms i briefed it)

And as far as we are concerned we need to give closest upper bound as we can write many upper bound for above given expression.

exp1+exp2+......+exp N=exp   always take max (exp1,exp2+...exp N) in calculating upper bound if like this is the given qsn and write every exp as max of the given, if for example exp1 is max than exp1+exp1+exp1+.....exp1(ntimes) this way you will give nearest and closest to upper bound for given expression.

Now coming to your qsn why ,

O(n-1)+O(n)=O(n) , we need to give upperbound for this expression ,as we can see max value it can attain is n, let see how

for O(n-1) we can write its upper bound as O(n) and for exp2 its already said max it can has is O(n).

so, you can write it as O(n)+O(n)=O(2n)=O(n) as we know O(c.n)=O(n)

Similarly, O(n/2)=O(n) and whole expression can be written as O(n)+O(n)=O(n)

now for your better understanding, we can rewrite above given exprsn as O(n)+O(n-1)+O(n-2)+O(n-3)+O(n-4)+...........+O(1) to find its upper bound you can again rewrite as O(n)+O(n)+O(n)+O(n)......+O(n)(n times)=O(n*n)=O(n^2)

by Active (3.9k points)
0

while trying to solve this, intuitively i got two approaches, one is as you mentioned, the other one is as follows:

O(1)+O(2)+O(3)+...+O(n-2)+O(n-1)+O(n)

=>O(1)+O(2)+O(3)+...+O(n-2)+O(n) [since O(n-1)+O(n)=O(n)]

=>O(1)+O(2)+O(3)+...+O(n-2)+O(n)

=>O(1)+O(2)+O(3)+...+O(n)  [since O(n-2)+O(n)=O(n)]

after n steps

=>O(1) +O(n)=O(n)

So where exactly am in going wrong in this approach?

+1 vote


text:
# ML Wiki

## Chi-Squared Ranking

The $\chi^2$ statistics measures the lack of independence between $f$ and $y$ and can be compared to $\chi^2$ with one degree of freedom

it's zero when $f$ and $y$ are independent

$\chi^2$ measures how the results of an observation differs from the result expected according to the null hypothesis

• lower values indicate less dependency
• so for $\chi^2$ we want to take biggest values


text:
# Two-dimensional disk distribution functions¶

galpy contains various disk distribution functions, both in two and three dimensions. This section introduces the two-dimensional distribution functions, useful for studying the dynamics of stars that stay relatively close to the mid-plane of a galaxy. The vertical motions of these stars may be approximated as being entirely decoupled from the motion in the plane.

## Types of disk distribution functions¶

galpy contains the following distribution functions for razor-thin disks: galpy.df.dehnendf, galpy.df.shudf, and galpy.df.schwarzschilddf. These are the distribution functions of Dehnen (1999AJ….118.1201D), Shu (1969ApJ…158..505S), and Schwarzschild (the Shu DF in the epicycle approximation, see Binney & Tremaine 2008). Everything shown below for dehnendf can also be done for shudf and schwarzschilddf. The Schwarzschild DF is primarily included as an educational tool; it is not a true steady-state DF, because it uses the approximate energy from the epicycle approximation rather than the true energy, and is fully superseded by the Shu DF, which is a good steady-state DF.

These disk distribution functions are functions of the energy and the angular momentum alone. They can be evaluated for orbits, or for a given energy and angular momentum. At this point, only power-law rotation curves are supported. A dehnendf instance is initialized as follows

>>> from galpy.df import dehnendf
>>> dfc= dehnendf(beta=0.)

This initializes a dehnendf instance based on an exponential surface-mass profile with scale-length 1/3 and an exponential radial-velocity-dispersion profile with scale-length 1 and a value of 0.2 at R=1. Different parameters for these profiles can be provided as an initialization keyword. For example,

>>> dfc= dehnendf(beta=0.,profileParams=(1./4.,1.,0.2))

initializes the distribution function with a radial scale length of 1/4 instead.

We can show that these distribution functions have an asymmetric drift built-in by evaluating the DF at R=1. We first create a set of orbit-instances and then evaluate the DF at them

>>> from galpy.orbit import Orbit
>>> os= [Orbit([1.,0.,1.+-0.9+1.8/1000*ii]) for ii in range(1001)]
>>> dfro= [dfc(o) for o in os]
>>> plot([1.+-0.9+1.8/1000*ii for ii in range(1001)],dfro)

Or we can plot the two-dimensional density at R=1.

>>> dfro= [[dfc(Orbit([1.,-0.7+1.4/200*jj,1.-0.6+1.2/200*ii])) for jj in range(201)]for ii in range(201)]
>>> dfro= numpy.array(dfro)
>>> from galpy.util.bovy_plot import bovy_dens2d
>>> bovy_dens2d(dfro,origin='lower',cmap='gist_yarg',contours=True,xrange=[-0.7,0.7],yrange=[0.4,1.6],xlabel=r'$v_R$',ylabel=r'$v_T$')

## Evaluating moments of the DF¶

galpy can evaluate various moments of the disk distribution functions. For example, we can calculate the mean velocities (for the DF with a scale length of 1/3 above)

>>> dfc.meanvT(1.)
# 0.91715276979447324
>>> dfc.meanvR(1.)
# 0.0

and the velocity dispersions

>>> numpy.sqrt(dfc.sigmaR2(1.))
# 0.19321086259083936
>>> numpy.sqrt(dfc.sigmaT2(1.))
# 0.15084122011271159

and their ratio

>>> dfc.sigmaR2(1.)/dfc.sigmaT2(1.)
# 1.6406766813028968

In the limit of zero velocity dispersion (the epicycle approximation) this ratio should be equal to 2, which we can check as follows

>>> dfccold= dehnendf(beta=0.,profileParams=(1./3.,1.,0.02))
>>> dfccold.sigmaR2(1.)/dfccold.sigmaT2(1.)
# 1.9947493895454664

We can also calculate higher order moments

>>> dfc.skewvT(1.)
# -0.48617143862047763
>>> dfc.kurtosisvT(1.)
# 0.13338978593181494
>>> dfc.kurtosisvR(1.)
# -0.12159407676394096

We already saw above that the velocity dispersion at R=1 is not exactly equal to the input velocity dispersion (0.19321086259083936 vs. 0.2). Similarly, the whole surface-density and velocity-dispersion profiles are not quite equal to the exponential input profiles. We can calculate the resulting surface-mass density profile using surfacemass, sigmaR2, and sigma2surfacemass. The latter calculates the product of the velocity dispersion squared and the surface-mass density. E.g.,

>>> dfc.surfacemass(1.)
# 0.050820867101511534

We can plot the surface-mass density as follows

>>> Rs= numpy.linspace(0.01,5.,151)
>>> out= [dfc.surfacemass(r) for r in Rs]
>>> plot(Rs, out)

or

>>> plot(Rs,numpy.log(out))

which shows the exponential behavior expected for an exponential disk. We can compare this to the input surface-mass density

>>> input_out= [dfc.targetSurfacemass(r) for r in Rs]
>>> plot(Rs,numpy.log(input_out)-numpy.log(out))

which shows that there are significant differences between the desired surface-mass density and the actual surface-mass density. We can do the same for the velocity-dispersion profile

>>> out= [dfc.sigmaR2(r) for r in Rs]
>>> input_out= [dfc.targetSigma2(r) for r in Rs]
>>> plot(Rs,numpy.log(input_out)-numpy.log(out))

That the input surface-density and velocity-dispersion profiles are not the same as the output profiles, means that estimates of DF properties based on these profiles will not be quite correct. Obviously this is the case for the surface-density and velocity-dispersion profiles themselves, which have to be explicitly calculated by integration over the DF rather than by evaluating the input profiles. This also means that estimates of the asymmetric drift based on the input profiles will be wrong. We can calculate the asymmetric drift at R=1 using the asymmetric drift equation derived from the Jeans equation (eq. 4.228 in Binney & Tremaine 2008), using the input surface-density and velocity dispersion profiles

>>> dfc.asymmetricdrift(1.)
# 0.090000000000000024

which should be equal to the circular velocity minus the mean rotational velocity

>>> 1.-dfc.meanvT(1.)
# 0.082847230205526756

These are not the same in part because of the difference between the input and output surface-density and velocity-dispersion profiles (and because the asymmetricdrift method assumes that the ratio of the velocity dispersions squared is two using the epicycle approximation; see above).

## Using corrected disk distribution functions¶

As shown above, for a given surface-mass density and velocity dispersion profile, the two-dimensional disk distribution functions only do a poor job of reproducing the desired profiles. We can correct this by calculating a set of corrections to the input profiles such that the output profiles more closely resemble the desired profiles (see 1999AJ….118.1201D). galpy supports the calculation of these corrections, and comes with some pre-calculated corrections (these can be found here). For example, the following initializes a dehnendf with corrections up to 20th order (the default)

>>> dfc= dehnendf(beta=0.,correct=True)

The following figure shows the difference between the actual surface-mass density profile and the desired profile for 1, 2, 3, 4, 5, 10, 15, and 20 iterations

and the same for the velocity-dispersion profile

galpy will automatically save any new corrections that you calculate.

All of the methods for an uncorrected disk DF can be used for the corrected DFs as well. For example, the velocity dispersion is now

>>> numpy.sqrt(dfc.sigmaR2(1.))
# 0.19999985069451526

and the mean rotation velocity is

>>> dfc.meanvT(1.)
# 0.90355161181498711

and (correct) asymmetric drift

>>> 1.-dfc.meanvT(1.)
# 0.09644838818501289

That this still does not agree with the simple dfc.asymmetricdrift estimate is because of the latter’s using the epicycle approximation for the ratio of the velocity dispersions.

## Oort constants and functions¶

galpy also contains methods to calculate the Oort functions for two-dimensional disk distribution functions. These are known as the Oort constants when measured in the solar neighborhood. They are combinations of the mean velocities and derivatives thereof. galpy calculates these by direct integration over the DF and derivatives of the DF. Thus, we can calculate

>>> dfc= dehnendf(beta=0.)
>>> dfc.oortA(1.)
# 0.43190780889218749
>>> dfc.oortB(1.)
# -0.48524496090228575

The K and C Oort constants are zero for axisymmetric DFs

>>> dfc.oortC(1.)
# 0.0
>>> dfc.oortK(1.)
# 0.0

In the epicycle approximation, for a flat rotation curve A =- B = 0.5. The explicit calculates of A and B for warm DFs quantify how good (or bad) this approximation is

>>> dfc.oortA(1.)+dfc.oortB(1.)
# -0.053337152010098254

For the cold DF from above the approximation is much better

>>> dfccold= dehnendf(beta=0.,profileParams=(1./3.,1.,0.02))
>>> dfccold.oortA(1.), dfccold.oortB(1.)
# (0.49917556666144003, -0.49992824742490816)

## Sampling data from the DF¶

We can sample from the disk distribution functions using sample. sample can return either an energy–angular-momentum pair, or a full orbit initialization. We can sample 4000 orbits for example as (could take two minutes)

>>> o= dfc.sample(n=4000,returnOrbit=True,nphi=1)

We can then plot the histogram of the sampled radii and compare it to the input surface-mass density profile

>>> Rs= [e.R() for e in o]
>>> hists, bins, edges= hist(Rs,range=[0,2],normed=True,bins=30)
>>> xs= numpy.array([(bins[ii+1]+bins[ii])/2. for ii in range(len(bins)-1)])
>>> plot(xs, xs*exp(-xs*3.)*9.,'r-')

E.g.,

We can also plot the spatial distribution of the sampled disk

>>> xs= [e.x() for e in o]
>>> ys= [e.y() for e in o]
>>> figure()
>>> plot(xs,ys,',')

E.g.,

We can also sample points in a specific radial range (might take a few minutes)

>>> o= dfc.sample(n=1000,returnOrbit=True,nphi=1,rrange=[0.8,1.2])

and we can plot the distribution of tangential velocities

>>> vTs= [e.vxvv[2] for e in o]
>>> hists, bins, edges= hist(vTs,range=[.5,1.5],normed=True,bins=30)
>>> xs= numpy.array([(bins[ii+1]+bins[ii])/2. for ii in range(len(bins)-1)])
>>> dfro= [dfc(Orbit([1.,0.,x]))/9./numpy.exp(-3.) for x in xs]
>>> plot(xs,dfro,'r-')

The agreement between the sampled distribution and the theoretical curve is not as good because the sampled distribution has a finite radial range. If we sample 10,000 points in rrange=[0.95,1.05] the agreement is better (this takes a long time):

We can also directly sample velocities at a given radius rather than in a range of radii. Doing this for a correct DF gives

>>> dfc= dehnendf(beta=0.,correct=True)
>>> vrvt= dfc.sampleVRVT(1.,n=10000)
>>> hists, bins, edges= hist(vrvt[:,1],range=[.5,1.5],normed=True,bins=101)
>>> xs= numpy.array([(bins[ii+1]+bins[ii])/2. for ii in range(len(bins)-1)])
>>> dfro= [dfc(Orbit([1.,0.,x])) for x in xs]
>>> plot(xs,dfro/numpy.sum(dfro)/(xs[1]-xs[0]),'r-')

galpy further has support for sampling along a given line of sight in the disk, which is useful for interpreting surveys consisting of a finite number of pointings. For example, we can sampled distances along a given line of sight

>>> ds= dfc.sampledSurfacemassLOS(30./180.*numpy.pi,n=10000)

which is very fast. We can histogram these

>>> hists, bins, edges= hist(ds,range=[0.,3.5],normed=True,bins=101)

and compare it to the predicted distribution, which we can calculate as

>>> xs= numpy.array([(bins[ii+1]+bins[ii])/2. for ii in range(len(bins)-1)])
>>> fd= numpy.array([dfc.surfacemassLOS(d,30.) for d in xs])
>>> plot(xs,fd/numpy.sum(fd)/(xs[1]-xs[0]),'r-')

which shows very good agreement with the sampled distances

galpy can further sample full 4D phase–space coordinates along a given line of sight through dfc.sampleLOS.

## Non-axisymmetric, time-dependent disk distribution functions¶

galpy also supports the evaluation of non-axisymmetric, time-dependent two-dimensional DFs. These specific DFs are constructed by assuming an initial axisymmetric steady state, described by a DF of the family discussed above, that is then acted upon by a non-axisymmetric, time-dependent perturbation. The DF at a given time and phase-space position is evaluated by integrating the orbit backwards in time in the non-axisymmetric potential until the time of the initial DF is reached. From Liouville’s theorem, which states that phase-space volume is conserved along the orbit, we then know that we can evaluate the non-axisymmetric DF today as the initial DF at the initial point on the orbit. This procedure was first used by Dehnen (2000).

This is implemented in galpy as galpy.df.evolveddiskdf. Such a DF is setup by specifying the initial DF, the non-axisymmetric potential, and the time of the initial state. For example, we can look at the effect of an elliptical perturbation to the potential like that described by Kuijken & Tremaine. To do this, we set up an elliptical perturbation to a logarithmic potential that is grown slowly to minimize non-adiabatic effects

>>> from galpy.potential import LogarithmicHaloPotential, EllipticalDiskPotential
>>> lp= LogarithmicHaloPotential(normalize=1.)

This perturbation starts to be grown at tform=-150 over a time period of tsteady=125 time units. We will consider the effect of this perturbation on a very cold disk (velocity dispersion $$\sigma_R = 0.0125\,v_c$$) and a warm disk ($$\sigma_R = 0.15\,v_c$$). We set up these two initial DFs

>>> idfcold= dehnendf(beta=0.,profileParams=(1./3.,1.,0.0125))
>>> idfwarm= dehnendf(beta=0.,profileParams=(1./3.,1.,0.15))

and then set up the evolveddiskdf

>>> from galpy.df import evolveddiskdf
>>> edfcold= evolveddiskdf(idfcold,[lp,ep],to=-150.)
>>> edfwarm= evolveddiskdf(idfwarm,[lp,ep],to=-150.)

where we specify that the initial state is at to=-150.

We can now use these evolveddiskdf instances in much the same way as diskdf instances. One difference is that there is much more support for evaluating the DF on a grid (to help speed up the rather slow computations involved). Thus, we can evaluate the mean radial velocity at R=0.9, phi=22.5 degree, and t=0 by using a grid

>>> mvrcold, gridcold= edfcold.meanvR(0.9,phi=22.5,deg=True,t=0.,grid=True,returnGrid=True,gridpoints=51,nsigma=6.)
>>> mvrwarm, gridwarm= edfwarm.meanvR(0.9,phi=22.5,deg=True,t=0.,grid=True,returnGrid=True,gridpoints=51)
>>> print(mvrcold, mvrwarm)
# -0.0358753028951 -0.0294763627935

The cold response agrees well with the analytical calculation, which predicts that this is $$-0.05/\sqrt{2}$$:

>>> print(mvrcold+0.05/sqrt(2.))
# -0.000519963835811

The warm response is slightly smaller in amplitude

>>> print(mvrwarm/mvrcold)
# 0.821633837619

although the numerical uncertainty in mvrwarm is large, because the grid is not sufficiently fine.

We can then re-use this grid in calculations of other moments of the DF, e.g.,

>>> print(edfcold.meanvT(0.9,phi=22.5,deg=True,t=0.,grid=gridcold))
# 0.965058551359
>>> print(edfwarm.meanvT(0.9,phi=22.5,deg=True,t=0.,grid=gridwarm))
# 0.915397094614

which returns the mean rotational velocity, and

>>> print(edfcold.vertexdev(0.9,phi=22.5,deg=True,t=0.,grid=gridcold))
# 0.0560531474616
>>> print(edfwarm.vertexdev(0.9,phi=22.5,deg=True,t=0.,grid=gridwarm))
# 0.0739164830253

which gives the vertex deviation in rad. The reason we have to calculate the grid out to 6nsigma for the cold response is that the response is much bigger than the velocity dispersion of the population. This velocity dispersion is used to automatically to set the grid edges, but sometimes has to be adjusted to contain the full DF.

evolveddiskdf can also calculate the Oort functions, by directly calculating the spatial derivatives of the DF. These can also be calculated on a grid, such that we can do

>>> oortacold, gridcold, gridrcold, gridphicold= edfcold.oortA(0.9,phi=22.5,deg=True,t=0.,returnGrids=True,gridpoints=51,derivGridpoints=51,grid=True,derivphiGrid=True,derivRGrid=True,nsigma=6.)
>>> oortawarm, gridwarm, gridrwarm, gridphiwarm= edfwarm.oortA(0.9,phi=22.5,deg=True,t=0.,returnGrids=True,gridpoints=51,derivGridpoints=51,grid=True,derivphiGrid=True,derivRGrid=True)
>>> print(oortacold, oortawarm)
# 0.575494559999 0.526389833249

It is clear that these are quite different. The cold calculation is again close to the analytical prediction, which says that $$A = A_{\mathrm{axi}}+0.05/(2\sqrt{2})$$ where $$A_{\mathrm{axi}} = 1/(2\times0.9)$$ in this case:

>>> print(oortacold-(0.5/0.9+0.05/2./sqrt(2.)))
# 0.0022613349141670236

These grids can then be re-used for the other Oort functions, for example,

>>> print(edfcold.oortB(0.9,phi=22.5,deg=True,t=0.,grid=gridcold,derivphiGrid=gridphicold,derivRGrid=gridrcold))
# -0.574674310521
>>> print(edfwarm.oortB(0.9,phi=22.5,deg=True,t=0.,grid=gridwarm,derivphiGrid=gridphiwarm,derivRGrid=gridrwarm))
# -0.555546911144

and similar for oortC and oortK. These warm results should again be considered for illustration only, as the grid is not sufficiently fine to have a small numerical error.

The grids that have been calculated can also be plotted to show the full velocity DF. For example,

>>> gridcold.plot()

gives

which demonstrates that the DF is basically the initial DF that has been displaced (by a significant amount compared to the velocity dispersion). The warm velocityd distribution is given by

>>> gridwarm.plot()

which returns

The shift of the smooth DF here is much smaller than the velocity dispersion.

## Example: The Hercules stream in the Solar neighborhood as a result of the Galactic bar¶

We can combine the orbit integration capabilities of galpy with the provided distribution functions and see the effect of the Galactic bar on stellar velocities. By backward integrating orbits starting at the Solar position in a potential that includes the Galactic bar we can evaluate what the velocity distribution is that we should see today if the Galactic bar stirred up a steady-state disk. For this we initialize a flat rotation curve potential and Dehnen’s bar potential

>>> from galpy.potential import LogarithmicHaloPotential, DehnenBarPotential
>>> lp= LogarithmicHaloPotential(normalize=1.)
>>> dp= DehnenBarPotential()

The Dehnen bar potential is initialized to start bar formation four bar periods before the present day and to have completely formed the bar two bar periods ago. We can integrate back to the time before bar-formation:

>>> ts= numpy.linspace(0,dp.tform(),1000)

where dp.tform() is the time of bar-formation (in the usual time-coordinates).

We initialize orbits on a grid in velocity space and integrate them

>>> ins= Orbit(numpy.array([[[1.,-0.7+1.4/100*jj,1.-0.6+1.2/100*ii,0.] for jj in range(101)] for ii in range(101)]))
>>> ins.integrate(ts,[lp,dp])

We can then evaluate the weight of these orbits by assuming that the disk was in a steady-state before bar-formation with a Dehnen distribution function. We evaluate the Dehnen distribution function at dp.tform() for each of the orbits (evaluating the distribution function only works for an Orbit with a single object, so we need to unpack the Orbit instance that contains all orbits)

>>> dfc= dehnendf(beta=0.,correct=True)
>>> out= [[dfc(o(dp.tform())) for o in j] for j in ins]
>>> out= numpy.array(out)

This gives

>>> from galpy.util.bovy_plot import bovy_dens2d
>>> bovy_dens2d(out,origin='lower',cmap='gist_yarg',contours=True,xrange=[-0.7,0.7],yrange=[0.4,1.6],xlabel=r'$v_R$',ylabel=r'$v_T$')

Now that galpy contains the evolveddiskdf described above, this whole calculation is encapsulated in this module and can be done much more easily as

>>> edf= evolveddiskdf(dfc,[lp,dp],to=dp.tform())
>>> mvr, grid= edf.meanvR(1.,grid=True,gridpoints=101,returnGrid=True)

The gridded DF can be accessed as grid.df, which we can plot as before

>>> bovy_dens2d(grid.df.T,origin='lower',cmap='gist_yarg',contours=True,xrange=[grid.vRgrid[0],grid.vRgrid[-1]],yrange=[grid.vTgrid[0],grid.vTgrid[-1]],xlabel=r'$v_R$',ylabel=r'$v_T$')

For more information see 2000AJ….119..800D and 2010ApJ…725.1676B. Note that the x-axis in the Figure above is defined as minus the x-axis in these papers.


text:
# A magnetic material has a magnetization of 2350 A/m and produces a flux density of 3.142 mWb/m2. Then, the relative permeability of the material is:

This question was previously asked in
Junior Executive (ATC) Official Paper 3: Held on Nov 2018 - Shift 3
View all JE ATC Papers >
1. 11.67
2. 17.47
3. 12.47
4. 16.67

Option 4 : 16.67
Free
Junior Executive (ATC) Official Paper 1: Held on Nov 2018 - Shift 1
20626
120 Questions 120 Marks 120 Mins

## Detailed Solution

Concept:

Magnetic field strength or field intensity (H) is the amount of magnetizing force.

Magnetic flux density (B) is the amount of magnetic force induced on the given body due to the magnetizing force H.

The relation between B and H is,

B = μH

Where,

μ = μ0 μr =  Permeability of the material

μ0 = Absolute permeability = 4π x 10-7 H/m

μr = relative permeability

Calculation:

Given-

I = 2350 A/m , B = 3.142 mWb/m2

B = μ0 (I + H)

$$(I+H)=\frac{3.14\times10^{-3}}{4 \times 3.14\times10^{-7}}$$

H = 150

We know that,

B = μH

$$μ=\frac{3.14\times10^{-3}}{150}$$

μ = 20 x 10-6

The relative permeability of the material is

∵ μ = μ0μr

$$μ_r=\frac{20\times10^{-6}}{4\times3.14\times10^{-7}}$$

μr ≈ 16.67


text:
# Math Help - Prove by contradiction

1. ## Prove by contradiction

Prove by contradiction that if r is irrational, then r^(1/2) is irrational

2. Originally Posted by hebby
Prove by contradiction that if r is irrational, then r^(1/2) is irrational
Where are you stuck? This is straight fwd
Hint: If x is rational, what can you say about x^2?

3. well then r^2 is rational as well, eg)4^2 is rational..then what?...thats it?

4. Originally Posted by hebby
well then r^2 is rational as well, eg)4^2 is rational..then what?...thats it?
yep !!

so if sqrt(r) is rational => r is rational (Contradiction !!)

5. Thanks, but we r going the other way around....ie if r is rational then sqrt r is rational.

6. Originally Posted by hebby
Thanks, but we r going the other way around....ie if r is rational then sqrt r is rational.
Absolutely not.
We are saying "if sqrt r is rational then r is rational"
Plz note the difference carefully

7. but the questions says if r is irrational then the sqrt r is irrational so we have to work this way right?

8. Originally Posted by hebby
but the questions says if r is irrational then the sqrt r is irrational so we have to work this way right?
dnt confuse yourself. both the statements are equivalent. one implies the other and vice-versa.

if sqrt r is rational then r is rational (we proved it)
but r is irrational (as per the question)
that mean sqrt r can't be rational because otherwise r will be rational (a contradiction)
hence sqrt r must be irrational !!


text:
##### Which statement allows us to conclude

 Algebra Tutor: None Selected Time limit: 1 Day

Jun 16th, 2015

----------------------------------------------------------------------------------

The statement that allows us to conclude that line m and line n are parallel is

B.   $\angle2\cong\angle7$

-------------------------------------------------------------------------------

The reason for this is as follows.

$\angle6\cong\angle7$ (because vertically opposite angles are equal)

Now, if the two lines are parallel then

$\angle6\cong\angle2$ (because corresponding angles are equal)

So, from the above two expressions, we can can conclude that

$\angle2\cong\angle7$ (because if a = b and b = c then a = c)

----------------------------------------------------------------------------------

Jun 16th, 2015

...
Jun 16th, 2015
...
Jun 16th, 2015
Oct 24th, 2016
check_circle


text:
Four Integrals in One Inequality

Solution

By Chebyshev's inequality (this is where we need function $f(x)\,$ to be increasing)

$\displaystyle \int_a^bxf(x)dx\ge\frac{1}{b-a}\left(\int_a^bxdx\right)\left(\int_a^bf(x)dx\right)=\frac{a+b}{2}\int_a^bf(x)dx.$

Similarly,

$\displaystyle \int_a^bx^3f(x)dx\ge\frac{a^3+a^2b+ab^2+b^3}{4}\int_a^bf(x)dx.$

By the Cauchy-Schwarz inequality,

$\displaystyle \int_a^bf^2(x)dx\ge\frac{1}{b-a}\left(\int_a^bf(x)dx\right)^2.$

By the AM-GM inequality, $\displaystyle \frac{a+b}{2}\cdot\frac{a^3+a^2b+ab^2+b^3}{4}\ge a^2b^2.\,$ Multiplying the three inequality yields the required one.

Acknowledgment

Dan Sitaru has kindly posted the problem of his from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page. Leo Giugiuc has commented with his solution.