TITLE: Finiteness of conditional expectation if expectation is finite
QUESTION [2 upvotes]: I have $E(X) < \infty$. Under which conditions follows that $E(X|A)<\infty$ ?
(A is an event of the form {$Y=y$} if it should matter)
If I can use the formula $E(X|A)=\frac{E(X 1_A)}{P(A)}$ ($1_A$ the indicator function of the set A) then it would follow at least for $P(A)\neq0$, but I am not sure if this formula is applyable as I am working with the general definition (A random variable Y is called conditional expectation value and we write $Y=E(X|\mathcal{A})$ if Y is $\mathcal{A}$-measurable and for all $A \in \mathcal{A}$: $E(X1_A)=E(Y1_A)$.
How can I assert that $E(X|A) <\infty$? (Or if my way seems valid, how can I deduce the formula from the general definition?)

REPLY [5 votes]: (Since your $A$ and $\mathcal A$ are suspiciously close (and I wonder whether you do not use one for the other at least once in your question), let me use $G$ to denote the sigma-algebra.)
You recall the definition of $Y=E(X|G)$. An important point is that this random variable $Y$ is defined only if $X$ is integrable (that is, when $E(|X|)$ is finite), and that $Y$ is always integrable (otherwise $E(Y1_A)$ might not be defined for some $A$ in $G$).
Hence yes, $E(X|G)$ (which is a random variable) is (almost surely) finite, since it is integrable by definition.
For a reference, see the little blue book by David Williams, more commonly called Probability with martingales.


TITLE: Expressing the given integral in terms of summation
QUESTION [2 upvotes]: Can anyone help me in changing the integral into the given form: $$\lim_{n \to \infty}n^{2} \Biggl(\ \  \int\limits_{0}^{1} \sqrt[n]{1+x^{n}} \ \text{dx}-1 \Biggr) = \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{2}}$$
Once this is done, we know that the integral converges to $\frac{\pi^2}{12}$.
Added: One can generally see that $$(1+x)^{a} = \sum\limits_{k=0}^{\infty} { a \choose k} x^{k};  \qquad x \in [0,1], \ a \in (0,1)$$

REPLY [3 votes]: Express $(1+x)^{1/n}-1$ as power series (over $k$) by using $\binom{a}{k}=\frac{a(a-1)\dots (a-k+1)}{k!}$ for $a=1/n$. Integrate termwise.
Now think about why you can interchange limit and summation and take the limit for each summand.


TITLE: Evaluation of $\oint_C y(y-x^2)\, dx +x (1+y)^2 \, dy$ where $C$ is the closed curve $x^2+y^2 = 4$.
QUESTION [1 upvotes]: I want to evaluate

$$\oint_C y(y-x^2)\, dx +x (1+y)^2 \, dy ,$$ where $C$ is the closed curve $x^2+y^2 = 4$.

I want to know why the solution below using Green's Theorem is wrong:
$$\displaystyle I = \iint_{R} \frac{\partial}{\partial x} x(1+y^2)-\frac{\partial}{\partial y} y(y-x^2)\,dx\,dy = \iint_R 1+x^2+y^2 \,dx\,dy$$
But since $x^2+y^2 = 4$ this is simply $\displaystyle 5 \iint_R \,dx \, dy$ $ ~~~~~(\Leftarrow $ I suspect this might be faulty logic$)$.
$R$ is a projection of a circle of radius $2$, and the double integral a representation of its area, thus $$I =5 \pi (2)^2 = 20 \pi.$$
But the value of the integral is $12 \pi$, not $20 \pi$, so this must be wrong somewhere. But where?

REPLY [1 votes]: You're right about where the error occurs: It's true that $x^2 + y^2 = 4$ on $C$, but that's not true everywhere on $R$. The shape of $R$ suggests that a certain change of coordinates would be effective at the point of the error, though.


TITLE: Manifold with Atlases of different Dimension
QUESTION [0 upvotes]: Given a set M with a topology, will two different Atlases $A_1$ & $A_2$ have same dimension? Does the following argument hold : -
If at a point $x$ i have a chart $(U,\phi)$ in $A_1$ $(U,\psi)$ in $A_2$, then $\psi \circ \phi^{-1}$ is a homeomorphism (we cannot claim this is diffeomorphism. right ?). And we cannot have homeomorphism between eculidean space of different dimensions.
Since the last fact used is not easy to prove, is there any other proof for the original question ?

REPLY [1 votes]: If a chart maps to $\mathbb{R}^p$ and another maps to $\mathbb{R}^q$, their differential must be a map $\mathbb{R}^p\to \mathbb{R}^q$ with a two sided inverse. Since the differential is a linear map, having a two sided inverse equals being an isomorphism, which requires $p = q$.
The topological case is much harder, though, because you can't reduce your problem to linear algebra.


TITLE: If $\lim_{n\to\infty} a_n\cdot b_n=1$ and $\lim_{n\to\infty} b_n=\infty$ then $\lim_{n\to\infty} a_n=0$
QUESTION [0 upvotes]: It's been awhile since I last touched Calculus so I'm asking for help in a really basic question. I'm trying to prove the following theorem:

Let $a_n$ and $b_n$ be sequences so $\lim_{n\to\infty} a_n\cdot b_n=1$ and $\lim_{n\to\infty} b_n=\infty$. So we get $\lim_{n\to\infty} a_n=0$.

I'm trying to prove this theorem by the limit definition but I can't seem to figure out how to find $\epsilon$. Is it possible to show how?

REPLY [0 votes]: Assume that $\lim a_n \neq 0$. That is, there is some $a^* \neq 0 \in \mathbb{R}$ s. th. for every $\varepsilon > 0$ there exists $N_{\varepsilon}$ s. th $|a^* - a_n| < \varepsilon$ for $ n > N_{\varepsilon}$. In particular $|a_n| > c > 0$ for some $c$.
Since $b_n \rightarrow \infty$ (i.e. for any $B \in \mathbb{N}$ there is $n_B$ s. th $b_n > B$ for $n > n_B$), consider the following argument for the product:
Let $B \in \mathbb{N}$ and $c > 0$ be arbitrary. Take $n_1$ s. th $b_n > N$ for all $n > n_1$. Take $n_2$ s. th. $|a_n| > c > 0$  for any $n > n_2$. Then for any $n > \max\{n_1, n_2\}$ we have:
$$ |a_n b_n| > c N $$
Since N can be chosen arbitrarily large, if follows that $\lim |a_n b_n| = \infty$. This is a contradiction, hence a cannot converge to any number other then $0$.
This still does not cover the case where $a_n$ doesn't converge at all. That is an exercise for you.


TITLE: Understanding a multivariable $\epsilon - \delta$ proof of the Sum Law for limits
QUESTION [0 upvotes]: Theorem
Let $f,g: A \to \mathbb R^m$ be well defined on $A \subset \mathbb R^k$.
If
$$f(x) \to b, g(x) \to c, \text{ for } x \to a, x \in A$ $$
then
$$f+g \to b +c$$
Proof
$\forall x \in A$ is
$$||(f+g)-(b+c)|| \le ||f-b||+||g-c||$$
[Q1: Where does this inequality come from? I thought it is the triangular inequality but I can not see how it is applied ]
For an arbitrary $\epsilon > 0$ we can choose
$\delta_1 > 0$ such that
$$||f-b|| < \epsilon /2$$ for all $x$ with $|x-a|| < \delta_1$
The same is written for $g,b \text{ and } \delta_2$
[Q2: Where does the above come from? Where did $...< \epsilon / 2$ come from?]
If we set $\delta = \min\{\delta_1,\delta_2\}$ then
$$||(f+g)-(b+c)|| < \epsilon \text{ for all } x\in A \text{ with } ||x-a||<\delta$$
I have literally used a half day on understanding this. Please clarify this for me.

REPLY [2 votes]: You want to prove that $f(x)+g(x)$ is near $b+c$ when $x$ is near $a$. This means that you have to look at the quantity
$$(f(x)+g(x))-(b+c)=(f(x)-b)+(g(x)-c))$$
when $x$ is near $a$. In the first place the triangle inequality gives
$$|(f(x)+g(x))-(b+c)|\leq|f(x)-b|+|g(x)-c|\ .\tag{1}$$
Since we want that the LHS is $<\epsilon$ we have to make sure that the RHS is $<\epsilon$ when $x$ is near $a$. When both summands on the RHS are $<{\epsilon\over2}$ this is guaranteed. Now $\lim_{x\to a}f(x)=b$ implies that there is a $\delta_1>0$ such that $0<|x-a|<\delta_1$ guarantees $|f(x)-b|<{\epsilon\over2}$, and similarly, there is a $\delta_2>0$ such that $0<|x-a|<\delta_2$ guarantees that $|g(x)-c|<{\epsilon\over2}$.
Since we want both estimates $<{\epsilon\over2}$ to be fulfilled simultaneously we  impose on $x$ the stronger condition $$0<|x-a|<\delta:=\min\{\delta_1,\delta_2\}>0\ .$$
When $x$ is so near to $a$ then  the required error bound
$$|(f(x)+g(x))-(b+c)|<\epsilon$$
is guaranteed.

REPLY [1 votes]: I'll go ahead and reproduce the proof with arguments to answer your questions.
Assuming your assumptions, we get that for $x \in A$
$$
||(f+g)(x)-(b+c)||=||f(x)+g(x)-(b+c)||=||f(x)-b+g(x)-c)||
$$
$$
\leq||f(x)-b||+||g(x)-c)|| \; \:\; \:\; \:\; \: (*)
$$
This should clear up your first question.
Now since $f$ and $g$ has a limit for $x\to a$ it means per definition that given some $\varepsilon >0$ we can find $\delta_f>0$, s.t. $\forall x \in B_{\delta_f}(a)$ it holds that
$$
||f(x)-b||<\varepsilon/2.\; \:\; \:\; \:\; \: (1)
$$
By the same argument there exists a $\delta_g>0$, s.t. $\forall x \in B_{\delta_g}(a)$ it holds that
$$
||g(x)-b||<\varepsilon/2. \; \:\; \:\; \:\; \: (2)
$$
The reason why we want the values to be less than $\varepsilon /2$ is due to the evaluation in equation $(*)$. Observe that if we choose $\delta = \text{min}\{\delta_f,\delta_g\}$ both equation $(1)$ and $(2)$ two hold for $x \in B_\delta(a)$.
Thus we see from $(*)$ that
$$
||(f+g)(x)-(b+c)||\leq||f(x)-b||+||g(x)-c)||< \varepsilon/2
+ \varepsilon/2 = \varepsilon.
$$
Which by definition means that $(f+g)(x) \to (b+c)$ for $x \to a$.


TITLE: quick approximation for largest fibonacci under a limit?
QUESTION [0 upvotes]: I asked in a previous post about finding a closed form for: $$\sum_{i=0}^{n}F_{3i}$$ which is the sum of the even fibs less than or equal to the nth even fib. the great answersI got showed me a very simple solution, and a fast way to compute it! In case you're wondering what the solution is, it is this: $$\sum_{i=0}^{n}F_{3i}=\frac{F_{3n+2}-1}{2}$$
So know, as a follow up question, I'm asking if there is a quick way to find the index of the largest Fibonacci number less than a specific limit? If the answer isn't O(1), then would it be faster to find that n then compute the fib(3n+2) with matrix exponentiation, or would it be faster to use a brute force approach?

REPLY [2 votes]: Well, for all $n\geq 0$, $F_n$  is the nearest integer to $\frac{1}{\sqrt 5}\left(\frac{1+\sqrt{5}}2\right)^n$. So if you want to find the largest $n$ such that $F_n<M$, you can compute:
$$n \approx \frac{\log (M\sqrt{5})}{\log \frac{1+\sqrt 5}2}$$
The size of $M$ and how close $M$ is to an $F_n$ affects how closely you need to approximate these various real numbers. This feels $O(1)$ for reasonable values of $M$.  For example, when $M=1000000$, you get $n=30$, and $F_{30}=832040$.
The matrix technique I showed in the prior problem gives you an approach that lets you find $n$ and also find $F_n$ fairly quickly.
Basically, given the $A$ from my answer before, compute:
$$A, A^2, A^4, \dots, A^{2^k},\dots$$
by repeated squaring until you "get too far." So $F_{2^{k-1}}<M\leq F_{2^{k}}$. Now use a binary search for $2^{k-1}\leq n < 2^k$. You can do that quickly because you already computed $A^{2^i}$ for $i=0\dots,k-1$.
I'm not sure about the complexity of this process, but it is certainly not much worse than $O(\log M)$. It's hard to do better than that since you are adding and multiplying numbers with up to $\log M$ bits. You can probably find $n$ faster if you don't also need to find $F_n$.
Now, if you want the largest $F_{3k+2}<M$, you might have to take the resulting $A^n$ and multiply it by $A^{-1}$ or $A^{-2}$ to get the appropriate maximum in that case.


TITLE: Partition of $\Bbb N$ into infinite number of infinite disjoint sets: please help me understand a specific proof
QUESTION [0 upvotes]: Partition of N into infinite number of infinite disjoint sets?
One answer for the above question about "Partition of N into infinite number of infinite disjoint sets" is the following (from Shai Covo):
Let
$$
A_0 = \lbrace 1,3,5,7,9,\ldots \rbrace
$$
and
$$
A_1 = \lbrace 2^n 1 : n \in \mathbb{N} \rbrace,
$$
$$
A_2 = \lbrace 2^n 3 : n \in \mathbb{N} \rbrace,
$$
$$
A_3 = \lbrace 2^n 5 : n \in \mathbb{N} \rbrace,
$$
$$
A_4 = \lbrace 2^n 7 : n \in \mathbb{N} \rbrace,
$$
$$
A_5 = \lbrace 2^n 9 : n \in \mathbb{N} \rbrace,
$$
$$
\cdots.
$$
Noting that for any two distinct elements $r_1$ and $r_2$ of $A_0$ it holds $2^{n_1}r_1 \neq 2^{n_2}r_2$ $\forall n_1,n_2 \in \mathbb{N}$, we have that the $A_i$ are disjoint. On the other hand, let $2k$, with $k \in \mathbb{N}$, be an arbitrary even natural number. Considering its prime factorization, it is necessarily of the form $2k = 2^n r$, where $n \in \mathbb{N}$ and $r \in A_0$. Hence $2k \in \cup _{i = 1}^\infty  A_i$, from which it follows that $\cup _{i = 1}^\infty  A_i = \lbrace 2,4,6,8,10,\ldots \rbrace$, and so $\mathbb{N} = \cup _{i = 0}^\infty  A_i$, with all the $A_i$ disjoint and countably infinite.
I am trying to figure out firstly in the above answer, how to formally prove that for any two distinct elements $r_1$ and $r_2$ of $A_0$ it holds $2^{n_1}r_1 \neq 2^{n_2}r_2$ $\forall n_1,n_2 \in \mathbb{N}$? And secondly, the above answer only proves that the set of even natural numbers is a subset of $\cup _{i = 1}^\infty A_{i}$ but it did not prove the other direction, may I ask how to prove that direction?
Thanks so much.

REPLY [1 votes]: 1) if $r_1$ and $r_2$ are odd and $2^{n_1}r_1=2^{n_2}r_2$, unique factorization proves that $n_1=n_2$ and $r_1=r_2$.
2) For every odd number $r$, all the numbers in $A_r$ are even, since they are of the form $2^nr$ with $n>0$. Thus, $\bigcup\limits_{r\text{ odd}}A_r$ is contained in the set of even numbers.


TITLE: If an equation has multiple roots of same value , What does it signify?
QUESTION [3 upvotes]: $(x-1)^{560}$ is said to have 560 roots. Now , all the roots must be 1(since it is the only factor tha can be formed ). And another example is $x^2 - 2x + 1 = 0 $ , Here $  x = 1$ ,only root as answer.
Q1 By saying $560$ roots , It does not give an difference to the answer( I mean that there is only root possible ) but does it give a difference  in the graph of the function ?.

REPLY [4 votes]: The roots of these functions
$$f(x)=(x-1)^{560} \\ g(x)=x^2-4x+4=(x-2)^2$$
are not equal.  This means that the graphs are also different.
I accept
$$f(x)=(x-1)^{560}, ~g(x)=(x-1)^2$$
Both functions get equal value at the point $x = 0,1,2.$ This means that the graphs pass through the same point, but it does not mean the graphs are the "same".
Just notice that:

If $x≠0,1,2$, then $$(x-1)^2≠(x-1)^{560}.$$

This means, graps can never be the "same".
When will the graph of two functions be the same?
Of course, when these functions are equivalent functions.
For example:
$$\color{blue}{f(x)=|x|} \qquad \color{red}{g(x)=\sqrt {x^2}}$$
Since the values of the functions are the same at all points, the graphs are also the same.


text:
\begin{document}

\title[An a priori bound of endomorphisms of $\bC\bP^k$]{An a priori bound of endomorphisms of $\bC\bP^k$ and a remark on the Makienko conjecture in dimension one}

\author[Y\^usuke Okuyama]{Y\^usuke Okuyama}
\address{
Division of Mathematics,
Kyoto Institute of Technology,
Sakyo-ku, Kyoto 606-8585 Japan}
\email{okuyama@kit.ac.jp}

\date{\today}

\begin{abstract}
Let $f$ be an endomorphism of $\bC\bP^k$ of degree $>1$, and assume that
for any cyclic Fatou component $W$ of $f$ having a period $p\in\bN$,
the equilibrium measure $\mu_f$ has a positive charge on the boundary of $W$
if and only if $f^{-p}(W)=W$. Then
we obtain a locally uniform a priori bound of the dynamics of $f$,
which in particular yields a Diophantine-type estimate of the dynamics of $f$
on its domaines singuliers. We also point out that
in the case of $k=1$, the statement of our assumption is related to
both the impossibility for the Julia set of $f$ to be the boundary of lakes of Wada
and the so called Makienko conjecture
on the non-emptiness of the residual Julia set of $f$.
\end{abstract}

\subjclass[2010]{32H50}
\keywords{a priori bound, domaine singulier,
pluripotential theory, Makienko conjecture, residual Julia set,
lakes of Wada}

\maketitle

\section{Introduction}\label{sec:intro}

Let $f$ be a holomorphic endomorphism of $\bP^k=\bC\bP^k$,
$k\in\bN$, of algebraic degree
$d>1$ (so of topological degree $d^k$).
By the seminal Forn{\ae}ss--Sibony \cite{FS94nato}
(see also the survey \cite{DSsurvey}),
the weak limit
\begin{gather*}
 \mu_f:=\lim_{n\to\infty}\frac{(f^*\omega_{\FS})^{\wedge k}}{d^{kn}}
\end{gather*}
exists on $\bP^k$, so that $\mu_{f^n}=\mu_f$ on $\bP^k$ for every $n\in\bN$,
where $\omega_{\FS}$ is the Fubini-Study K\"ahler form on $\bP^k$
normalized as $\omega_{\FS}^{\wedge k}(\bP^k)=1$.
Let us equip $\bP^k$ with the normalized
{\itshape chordal} metric $[x,y]_{\bP^k}(\le 1)$
 (see \eqref{eq:chordal} for the definition),
which is comparable to the metric
on $\bP^k$ induced by $\omega_{\FS}$.
The probability measure $\mu_f$,
which is called the {\itshape equilibrium measure} of $f$,
has no masses on pluripolar subsets in $\bP^k$ and
satisfies the $f$-balanced property $f^*\mu_f=d^k\cdot\mu_f$ on $\bP^k$.
Letting $\delta_x$ be the Dirac measure on $\bP^k$ at each $x\in\bP^k$,
for every $x\in\bP^k$ except for a pluripolar subset in $\bP^k$,
the weak convergence
\begin{gather}
 \lim_{n\to\infty}\frac{(f^n)^*\delta_x}{d^{kn}}=\mu_f
\label{eq:equidist}
\end{gather}
holds on $\bP^k$, so $\mu_f$ is mixing (so ergodic) under $f$
(see \cite{BD01,DS08,DS10} for more details).
Let us also recall a more classical Fatou-Julia strategy.
The (first) Julia set $J(f)$ of $f$ is by definition
all points in $\bP^k$ at each of which the family $\{f^n:n\in\bN\}$ is not normal,
and is a compact subset in $\bP^k$.
The (first) Fatou set $F(f)$ of $f$ is defined by $\bP^k\setminus J(f)$,
and each component of $F(f)$ is called a Fatou component of $f$.
Both $J(f)$ and $F(f)$ are totally invariant under $f$,
$f$ maps each Fatou component of $f$ properly to a Fatou component of $f$,
and the preimage of a Fatou component of $f$ under $f$ is the union of
(at most $d^k$) Fatou components of $f$. A Fatou component $W$ of $f$ is
said to be cyclic under $f$ if $f^p(W)=W$ for some $p\in\bN$; then
such $p$ is called {\itshape a} period of $W$ (under $f$),
and the minimal such $p$ is called the {\em exact period} of $W$ (under $f$).
In general, we have
$\supp\mu_f\subset J(f)$,
and we have $\supp\mu_f=J(f)$ if $k=1$ (see Section \ref{sec:lift} for more details).

Our principal result is the following locally uniform {\em a priori} bound
of the dynamics of $f$.

\begin{mainth}[an {\itshape a priori} bound]\label{th:nonlinvanish}
Let $f$ be a holomorphic endomorphism of $\bP^k$ of algebraic degree $d>1$,
and suppose that {\em (*) for any cyclic Fatou component $W$ of $f$ having
a period $p\in\bN$, $\mu_f(\partial W)>0$ if and only if $f^{-p}(W)=W$.}
Then for every holomorphic endomorphism $g$ of $\bP^k$ of degree $>0$
and every non-empty open subset $D$ in $\bP^k$,
\begin{gather}
\lim_{n\to\infty}\frac{\sup_{y\in D}\log[f^n(y),g(y)]_{\bP^k}}{d^n+\deg g}=0.\label{eq:negative}
\end{gather}
\end{mainth}

We conclude this section with a few remarks on the conclusion \eqref{eq:negative}
and the assumption (*) in Theorem \ref{th:nonlinvanish}
since both are seemingly technical.
\begin{remark}\label{sec:rotation}
 Following Fatou \cite[Sec.\ 28]{Fatou1920deux},
 a cyclic Fatou component $W$ of $f$ having the exact period, say,
 $p\in\bN$ is called
 a {\itshape domaine singulier} (a singular domain, or a rotation domain)
 of $f$ if $f^p:W\to W$ is injective ({\itshape singulier} in French)
so biholomorphic.
The conclusion \eqref{eq:negative} in Theorem \ref{th:nonlinvanish} for $g=\Id_{\bP^k}$ is equivalent to that {\itshape for every {\em domaine singulier} $W$ of $f$ having the exact period $p\in\bN$ and every non-empty open subset $D\Subset W$,}
 \begin{gather}
 \lim_{n\to\infty}\frac{\log\sup_{y\in D}[f^{pn}(y),y]_{\bP^k}}{d^{pn}+1}=0.\label{eq:rotation}
 \end{gather}
 This {\itshape Diophantine-type} estimate \eqref{eq:rotation} of $f^p$ on
 a {\em domaine singulier} $W$
 has been known in \cite[Theorem 3]{OkuNonlinear}
 without assuming (*) but
 under the assumption (**) that $W$ is of maximal type
 in that, setting $q:=\min\{j\in p\bN:f^j|W\in G_0\}$,
 where $G_0$ is the component of the closed subgroup
 generated by $f^p|W$ in the biholomorphic automorphisms group $\Aut(W)$
 of $W$ containing $\Id_W$,
 there is a Lie groups isomorphism $G_0\to\mathbb{T}^k$
 that maps $f^q|W$ to $(e^{2i\pi\alpha_1},\ldots,e^{2i\pi\alpha_k})\in\mathbb{T}^k$
 for some $\alpha_1,\ldots,\alpha_k\in(\bR\setminus\bQ)/\bZ$; in general,
 $G_0$ at least contains a real $1$-dimensional torus $\bT^1$

 (Ueda \cite{Ueda98}).
\end{remark}

\begin{remark}\label{th:equivalent}
In the statement in the assumption (*), one implication is obvious;
if $f^{-p}(W)=W$, then $\supp\mu_f\subset\partial W$
by (the property of pluripolar subsets and) the above description of $\mu_f$,
so $\mu_f(\partial W)=1>0$.

In fact, it is always the case that
{\em for every cyclic Fatou component $W$ of $f$ having
a period $p\in\bN$, we have $\mu_f(\partial W)\in\{0,1\}$ and,
if $\mu_f(\partial W)=0$, then we have
$\mu_f(\partial U)=0$ for every component $U$ of
$\bigcup_{n\in\bN\cup\{0\}}f^{-n}(W)$};
for, by the ergodicity of $\mu_f$ under $f$ and $f^p(W)=W$,
we have not only
$\mu_f(\bigcup_{n\in\bN\cup\{0\}}f^{-pn}(\partial W))\in\{0,1\}$ but,
for every $n\in\bN$, also $f^{-p(n-1)}(\partial W)\subset f^{-pn}(\partial W)$.
Then by the $f$-invariance of $\mu_f$, we compute as
$\mu_f(\partial W)=\mu_f(f^{-pn}(\partial W))
=\mu_f(f^{-p(n-1)}(\partial W))
+\mu_f(f^{-pn}(\partial W)\setminus f^{-p(n-1)}(\partial W))
=\mu_f(\partial W)+\mu(f^{-pn}(\partial W)\setminus f^{-p(n-1)}(\partial W))$,
so $\mu_f(f^{-pn}(\partial W)\setminus f^{-p(n-1)}(\partial W))=0$,
and then $\mu_f(\partial W)

=\mu_f(\bigcup_{n\in\bN\cup\{0\}}f^{-pn}(\partial W))\in\{0,1\}$.

Finally, if in addition $\mu_f(\partial W)=0$, then
for every component $U$ of $\bigcup_{n\in\bN\cup\{0\}}f^{-n}(W)$,
by the $f$-invariance of $\mu_f$
and $f^p(W)=W$, we compute as
$0=\mu_f(\partial W)=\mu_f(\partial(f^{-(p-j)}(W)))\ge
\mu_f(\partial(f^j(W)))=\mu_f(\bigcup_{n\in\bN\cup\{0\}}f^{-pn}(\partial(f^j(W))))
\ge\mu_f(\partial U)\ge 0$ for some $j\in\{0,\ldots,p-1\}$,
so $\mu_f(\partial U)=0$.
\end{remark}

\begin{remark}
In the dimension $k=1$ case,
we have $\supp\mu_f=J(f)$, and by the Riemann-Hurwitz formula,
for any cyclic Fatou component $W$ of $f$ having a period $p\in\bN$,
if $f^{-p}(W)=W$, then $p\in\{1,2\}$.
Moreover, by Sullivan's no wandering domain theorem
\cite{Sullivan85}, for any Fatou component $U$ of $f$,
there is $N\in\bN\cup\{0\}$ such that $W:=f^N(U)$ is a cyclic Fatou
component of $f$; then $\mu_f(\partial W)\in\{0,1\}$
and, if in addition $\mu_f(\partial W)=0$, then
$\mu_f(\partial U)=0$ (seen in Remark \ref{th:equivalent}).
By the Riemann-Hurwitz formula again, there are at most two
Fatou components $U$ of $f$ such that $f^{-2}(U)=U$.

It seems open whether the statement in the assumption (*) is the case
for any $f$, even in the case $k=1$. We point out the following.
\begin{proposition}
 Let $f\in\bC(z)$ be of degree $d>1$. Then the following statements
 are equivalent$;$
 \begin{enumerate}
 \item for any Fatou component $U$ of $f$,
       $\mu_f(\partial U)>0$ if and only if $f^{-2}(U)=U$.
       \label{measure}
 \item for any Fatou component $U$ of $f$,
 $\partial U=J(f)$ if and only if $f^{-2}(U)=U$.
       \label{makienko}
 \end{enumerate}
Moreover, if the above statements are the cases, then
so are the followings$;$
\begin{enumerate}
\setcounter{enumi}{2}
 \item  the Julia set $J(f)$ of $f$ cannot be the boundary of lakes of Wada.
Here a compact subset $C$ in $\bP^1$ is called the {\em boundary of
lakes of Wada} if there are at least three distinct components of
$\bP^1\setminus C$, the boundary of each of which equals the whole $C$
$($introduced in \cite{Yoneyama17}$)$.
 \label{wada}
 \item $($the so called {\em Makienko conjecture} $(1990))$
the residual Julia set
 $J_0(f):=J(f)\setminus(\bigcup_{U:\text{a Fatou component of }f}\partial U)$
       of $f$
 is empty if and only if there is a Fatou component $U$ of $f$
       such that $f^{-2}(U)=U$.
       \label{residual}
\end{enumerate}
\end{proposition}

Indeed, the observations in the first paragraph yield
\eqref{measure}$\Leftrightarrow$\eqref{makienko},
\eqref{makienko}$\Rightarrow$\eqref{wada}, and
\eqref{measure}$\Rightarrow$\eqref{residual}.

For further studies, see e.g.\
 \cite{Qiao95,Morosawa97,SunYang03,CMMR09,HT18}.
\end{remark}

\section{Background}\label{sec:lift}
We say a function $H:\bC^{k+1}\to\bR\cup\{-\infty\}$
is log-homogeneous (of order $1$) if
for every $c\in\bC^*$, $H(cZ)=H(Z)+\log|c|$ on $\bC^{k+1}$.
Let $\|\cdot\|$ be the Euclidean norm on $\bC^{k+1}$.
The function $\log\|\cdot\|:\bC^{k+1}\to\bR\cup\{-\infty\}$ is
continuous, plurisubharmonic, and log-homogeneous
(of order $1$).
The complex Laplacian $\rd\rd^c$ is normalized as usual,
so in particular that
$\pi^*\omega_{\FS}=\rd\rd^c\log\|\cdot\|$ on $\bC^{k+1}\setminus\{0\}$,
where the origin $(0,\ldots,0)$ of the $\bC$-linear space
$\bC^{k+1}$ is denoted by $0$. More generally,
a function $H:\bC^{k+1}\to\bR\cup\{-\infty\}$
which is plurisubharmonic and log-homogeneous (of order $1$) is also called an
$\omega_{\FS}$-plurisubharmonic function {\itshape on} $\bP^k$;
indeed, the function $H-\log\|\cdot\|$ descends to $\bP^k$ so that
$\pi^*\bigl(\rd\rd^c(H-\log\|\cdot\|)+\omega_{\FS}\bigr)=\rd\rd^c H$ on $\bC^{k+1}\setminus\{0\}$, $\rd\rd^c(H-\log\|\cdot\|)+\omega_{\FS}\ge 0$ on $\bP^k$,
and $\bigl(\rd\rd^c(H-\log\|\cdot\|)+\omega_{\FS}\bigr)\wedge\omega_{\FS}^{k-1}$
is a probability measure on $\bP^k$.

Set $\ell(k):=\binom{k+1}{2}\in\bN$ so that
$\bC^{k+1}\wedge\bC^{k+1}\cong\bC^{\ell(k)}$ (cf.\ \cite[\S 8.1]{Kobayashi98})
and, for simplicity,
also denote by $\|\cdot\|$ the Euclidean norm on $\bC^{\ell(k)}$.
Let $\pi:\bC^{k+1}\setminus\{0\}\to\bP^k$
be the canonical projection.
The chordal metric on $\bP^k$ is
\begin{gather}
 [x,y]_{\bP^k}:=\|Z\wedge W\|(\|Z\|\cdot\|W\|)\le 1,\quad x,y\in\bP^k,\label{eq:chordal}
\end{gather}
where $Z\in\pi^{-1}(x),W\in\pi^{-1}(y)$.

Let $f$ be a holomorphic endomorphism of $\bP^k=\bC\bP^k$
of algebraic degree $d>1$. A lift of $f$ is
an ordered $(k+1)$-tuple $F=(F_0,F_1,\ldots,F_k)\in(\bC[z_0,z_1,\ldots,z_k]_d)^{k+1}$
of homogeneous polynomials of degrees $d$
in the indeterminants $z_0,\ldots,z_{k+1}$,
which is unique up to multiplication in $\bC^*$, such that
$\pi\circ F=f\circ\pi$
on $\bC^{k+1}\setminus\{0\}$
(and that $F^{-1}(0)=\{0\}$).
Fix a lift $F$ of $f$. Then the uniform limit
\begin{gather}
 G^F:=\lim_{n\to\infty}\frac{\log\|F^n\|}{d^n}:\bC^{k+1}\setminus\{0\}\to\bR\label{eq:escape}
\end{gather}
exists (\cite{HP94}) and, setting $G^F(0):=-\infty$,
the function $G^F:\bC^{k+1}\to\bR\cup\{-\infty\}$ is
continuous, plurisubharmonic,
and log-homogeneous (of order $1$).
By Ueda \cite[Theorem 2.2]{Ueda94},
the region of pluriharmonicity of $G^F$

coincides not only with $\pi^{-1}(F(f))$
but also with $\pi^{-1}(\tilde{F}(f))$, where $\tilde{F}(f)$
is the set of all points in $\bP^k$ at each of which
the family $\{f^{n_j}:\bP^k\to(\bP^k,[x,y]_{\bP^k}):j\in\bN\}$ is normal
for {\em some} sequence $(n_j)$ in $\bN$ tending to $\infty$ as $j\to\infty$.

\section{Proof of Theorem \ref{th:nonlinvanish}}
\label{sec:vanish}
Let $f$ be a holomorphic endomorphism of $\bP^k=\bC\bP^k$
of algebraic degree $d>1$, and assume (*) that
for any cyclic Fatou component $W$ of $f$ having a period $p\in\bN$,
$\mu_f(\partial W)>0$ if and only if $f^{-p}(W)=W$.

Suppose to the contrary that
there are a holomorphic endomorphism $g$ of $\bP^k$ of degree $>0$
and a domain $D$ in $\bP^k$
such that \eqref{eq:negative} does not hold, so that
there is a sequence $(n_j)$ in $\bN$ tending to $\infty$ as $j\to\infty$ such that
\begin{gather}
\lim_{j\to\infty}\frac{\sup_{y\in D}\log[f^{n_j}(y),g(y)]_{\bP^k}}{d^{n_j}+\deg g}<0.\label{eq:quick}
\end{gather}
Then $D\subset F(f)$, and let $U$ be the Fatou component of $f$ containing $D$.
Since $\deg g>0$,
we have $\lim_{j\to\infty}f^{n_{j+1}-n_j}=\Id_{g(D)}$ locally uniformly
on $g(D)$, and then there is $N\in\bN\cup\{0\}$ such that $V:=f^{n_N}(U)(=g(U))$
is a cyclic Fatou component of $f$ having the exact period, say, $p\in\bN$
and satisfies $\deg(f^p:V\to V)=1$. Then $f^{-p}(V)\neq V$ since $($the topological
degree of $f)=d^k>1$, so that under the assumption (*), we have
\begin{gather}
 \mu_f(\partial U)=0\label{eq:preimage}
\end{gather}
(seen in Remark \ref{th:equivalent}).

Also fix a lift $G$ of $g$. By \eqref{eq:chordal} and \eqref{eq:escape},
the family $\bigl\{(\log|F^{n_j}\wedge G|)/d^{n_j}:j\in\bN\bigr\}$
is locally uniformly bounded from above on $\bC^{k+1}$.
Moreover, by \eqref{eq:chordal}, \eqref{eq:escape}, and $J(f)\neq\emptyset$,
we have $\limsup_{j\to\infty}\sup_{\{G^F=0\}}
(\log|F^{n_j}\wedge G|)/d^{n_j}\ge 0>-\infty$.
Hence by \eqref{eq:chordal}, \eqref{eq:escape}, and a version of
Hartogs lemma for a sequence of plurisubharmonic functions
(see \cite[Theorem 4.1.9(a)]{Hormander83} or \cite[Theorem 1.1.1]{Azarin80}),
taking a subsequence of $(n_j)$ if necessary,
there is a plurisubharmonic function $\phi$ on $\bC^{k+1}$ such that
\begin{gather*}
 \phi:=\lim_{j\to\infty}\frac{\log\|F^{n_j}\wedge G\|}{d^{n_j}+\deg g}(\le G^F)
\quad\text{in }L^1_{\loc}(\bC^{k+1},m_{2k+2}),
\end{gather*}
where $m_{2k+2}$ denotes the $(2k+2)$-dimensional
Lebesgue measure on $\bC^{k+1}\cong\bR^{2k+2}$.
The function $\phi$ is also log-homogeneous
(of order $1$).

By the log-homogeneity of both $G^F$ and $\phi$,
$\phi-G^F$ on $\bC^{k+1}\setminus\{0\}$ descends to
a function $\bP^k\to\bR_{\le 0}\cup\{-\infty\}$,
which is not only upper semicontinuous on $\bP^k$
but also plurisubharmonic on $U$, and
by \eqref{eq:chordal}, \eqref{eq:escape}, and \eqref{eq:quick},
the open subset $\{\phi-G^F<0\}$ in $\bP^k$ contains $D(\subset U)$
except for an at most $m_{2k+2}$-null subset in $D$. Hence
$U\cap\{\phi-G^F<0\}\neq\emptyset$.
By \eqref{eq:chordal}, \eqref{eq:escape},
the upper semicontinuity of $\phi-G^F$,
and a version of
Hartogs lemma for a sequence of plurisubharmonic functions
(see \cite[Theorem 4.1.9(b)]{Hormander83}),
we also have $\phi-G^F\equiv 0$ on $J(f)$,
so in particular $\phi=G^F$ on $\pi^{-1}(\partial U)$.

Let us define the {\itshape locally bounded} function
\begin{gather*}
 \psi:=\begin{cases}
	\max\{\phi,G^F-1\} & \text{on }\pi^{-1}(U)\\
	G^F & \text{on }\bC^{k+1}\setminus \pi^{-1}(U),
       \end{cases}
\end{gather*}

which is still plurisubharmonic on $\bC^{k+1}$;
for, it is upper semicontinuous on $\bC^{k+1}$
(since so is $\phi$, $\phi=G^F$ on $\pi^{-1}(\partial U)$, and $\phi\le G^F$)
and plurisubharmonic on $\bC^{k+1}\setminus(\pi^{-1}(\partial U)\cup\{0\})$,
and satisfies the mean value inequality at each point
$Z_0\in\pi^{-1}(\partial U)\cup\{0\}$
on each complex line passing through $Z_0$
(since so does $\phi$, $\phi=G^F$ on $\pi^{-1}(\partial U)$, and $\phi\le G^F$).
The function $\psi$ is also log-homogeneous (of order $1$).
By the log-homogeneity of both $G^F$ and $\psi$,
the function $\psi-G^F$ also descends to a function on $\bP^k$.

Following the manner in \cite[\S 2.1]{BermanBoucksom10},

let us also denote by $\rd\rd^c\psi$ (resp.\ $\rd\rd^c G^F$)
the current {\em on} $\bP^k$ whose pullback under
$\pi:\bC^{k+1}\setminus\{0\}\to\bP^k$
coincides (the genuine) $\rd\rd^c\psi$ (resp.\ the genuine $\rd\rd^c G^F$)
{\em on} $\bC^{k+1}\setminus\{0\}$.
Then both the $k$-th {\itshape Bedford--Taylor wedge products}
$(\rd\rd^c\psi)^{\wedge k}$ and $(\rd\rd^c G^F)^{\wedge k}$
{\itshape on} $\bP^k$ of $\rd\rd^c\psi$ and $\rd\rd^c G^F$,
respectively, exist and
are {\itshape probability} measures on $\bP^k$. From the definition of $G^F$,
the latter probability measure $(\rd\rd^c G^F)^{\wedge k}$ on $\bP^k$
is nothing but the equilibrium measure
$\mu_f$ of $f$. We claim that
$(\rd\rd^c\psi)^{\wedge k}=\mu_f$ on $\bP^k$;
for, by the definition of $\psi$, we have
$(\rd\rd^c\psi)^{\wedge k}=(\rd\rd^c G^F)^{\wedge k}=\mu_f$
on $\bP^k\setminus\overline{U}$, and
by $\supp\mu_f\subset J(f)$ and the vanishing $\mu_f(\partial U)=0$
in \eqref{eq:preimage}, we also have
$(\rd\rd^c\psi)^{\wedge k}(\overline{U})
=1-(\rd\rd^c\psi)^{\wedge k}(\bP^k\setminus\overline{U})
=1-\mu_f(\bP^k\setminus\overline{U})
=\mu_f(\overline{U})

=0$.
Hence the claim holds.

Once this claim is at our disposal, we have
$\psi-G^F\ge 0$ (indeed $=0$)
on $(\rd\rd^c\psi)^{\wedge k}$($=\mu_f$)-almost everywhere $\bP^k$,
and then by a version of the Bedford--Taylor domination principle
(\cite[Corollary 2.5]{BEGZ10}; for a summary on the properties of
plurisubharmonic {\itshape weights} on {\itshape big} line bundles
over complex compact manifolds, which applies to
our $\omega_{\FS}$-plurisubharmonic functions {\itshape on} $\bP^k$,
see \cite[\S 2]{BermanBoucksom10}),
we have $\psi\ge G^F$ on $\bC^k\setminus\{0\}$. In particular
$\phi=\psi\ge G^F$ on $\pi^{-1}(U)$, which contradicts
$U\cap\{\phi-G^F<0\}\neq\emptyset$. \qed

\begin{acknowledgement}
 This research was partially supported by JSPS Grant-in-Aid
 for Scientific Research (C), 15K04924.
\end{acknowledgement}

\def\cprime{$'$}

config: arxiv