TITLE: Find all couples $(a,b) \in \mathbb{Z}^2$
QUESTION [3 upvotes]: I was asked to determine all possible couples $(a,b) \in \mathbb{Z}^2$ which satisfies $\text{gcd}(411,2016) = a411 + b2016$. Using the Euclidean division alogirthm, I have found that
\begin{align}
2016 &= 4 \cdot 411 + 372\\
 411 &= 1 \cdot 372 + 39\\
 372 &= 9 \cdot 39 + 21\\
  39 &= 1 \cdot 21 + 18\\
  21 &= 1 \cdot 18 + 3\\
  18 &= 3 \cdot 6
\end{align}
from which it folows that $\text{gcd}(411,2016) = 3$. Working back, I find that $3 = 21 \cdot 2016 - 103 \cdot 411$.
In order to determine the other solutions, I worked more generale: suppose we want to find $x,y \in \mathbb{Z}$ such that $ax + by = d$. Suppose that $(x,y), (x',y')$ are two solutions, then filling this in we find that $a(x-x') = b(y' - y)$. Supposing that $a,b$ are coprime (as is the case for my expansion of $3$, we must have that $a$ is a divisor of $y'-y$ and $b$ is a divisor of $x-x'$. Therefore, there are $k, k' \in \mathbb{Z}$ such that $ak = y'-y$ and $bk' = x-x'$. However, I have no idea how $k,k'$ are related. Any hints?

REPLY [2 votes]: Hint $ $ Suppose $\, a/b = a'/b'\,$ where $a'/b'$ is reduced (in lowest terms). Then we have
$$ \begin{array}{}\large \frac{y'-y}{x'-x}\, =\, \frac{a}b\, =\, \frac{\color{#0a0}{a'}}{\color{#c00}{b'}}\\[.3em]
\Rightarrow\,\ \begin{align}{y'-y} \:&=\: n\,\color{#0a0}{a'}\\ x'-x \:&=\: n\,\color{#c00}{b'}\end{align}
\end{array}$$
for some integer $n,\,$ by a basic principality property of reduced fractions (Unique Fractionization)


TITLE: Is $\frac{dx(t-T)}{d(t-T)}=\frac{dx(t-T)}{dt}$? Where T is a constant and t is an independent variable.
QUESTION [0 upvotes]: Given a function x(t) and a constant T, is the following relation true?:
$$\frac{dx(t-T)}{d(t-T)}=\frac{dx(t-T)}{dt}$$
If it is true, what is the proof?
Also, if this is true, does this mean that this statement is wrong?:
$$\frac{dx(t-T)}{d(t-T)}=\frac{dx(z)}{dz}=\frac{dx(t)}{dt}$$ where $z=t-T$.
Here is a snippet from Example 1.11 of Linear Systems and Signals by B. P. Lathi and Roger Green using this relation:

Applying input $x(t)$ to the system produces output $y(t) = \frac{d}{dt}x(t)$; delaying
this output by T produces $y(t − T) = \frac{d}{d(t−T)}x(t − T) = \frac{d}{dt}x(t − T)$. This is just the output of
the system to a delayed input $x(t−T)$. Since the T-delayed output of the system to input $x(t)$
equals the output of the system to the T-delayed input $x(t−T)$, the system is time invariant.

This part refers to proving the system $y(t)=\frac{d}{dt}x(t)$ to be time invariant.

REPLY [0 votes]: The proof is the chain rule, using $$\frac{d(t-T)}{dt} = 1\qquad\text{or its consequence}\qquad \frac{dt}{d(t-T)} = 1$$

Yes,
$$
\frac{dx(z)}{dz}=\frac{dx(t)}{dt}
$$
is wrong, where $z=t-T$.  Unless $x(t)$ is periodic with period $T$, then it is right.


TITLE: Numerical ODE solution with partial derivatives
QUESTION [0 upvotes]: We had a numerical method for the solution of an ODE of the form
$$y_{n+1}=y_n+h_n(f(t_n,y_n)+\frac{h_n}{2}f'(t_n,y_n))$$
Where $f(t_n,y_n)=3y-2t$
Now the solution treats the partial derivatives of $f'(t_n,y_n)$ as a sum.
Why do we use them as a sum $f_t+f_y$? Shouldn't it be a vector $(f_t,f_y)^T$

REPLY [1 votes]: As I wrote in several comments in previous questions, you want to use the second degree Taylor polynomial
$$
y(t+h)=y(t)+y'(t)h+\frac12y''(t)h^2+O(h^3)
$$
as approximation for the next point on the trajectory. The second derivative of $y$ is
$$
y''(t)=\frac{d}{dt}y'(t)
=\frac{d}{dt}f(t,y(t))=f_t(t,y(t))+f_y(t,y(t))y'(t) \\
=f_t(t,y(t))+f_y(t,y(t))f(t,y(t)).
$$
The only important point is that it is exactly this expression that will give a second order method, how you denote this is of secondary concern.
For your example you have correctly applied the formula, only it does not make sense to only half-way replace $f$ with its actual expression, you should continue to
$$
f'(t,y)=3(3y-2t)-2=9y-6t-2.
$$


TITLE: What constitues algebraic competency?
QUESTION [0 upvotes]: A bit of background to my question: last year I signed up for a refresher course in Maths at a university, because I seem to have forgotten most of what I learned in school about Maths but feel interest and need to get better at Maths though.
I take a course grouped into various segments. At the end of that (probably) multi-year journey stands the option to continue the course and study Maths at university level (to go for an associates degree or bachelor.)
In higher courses we will learn Calculus (which I never learned in school) and of course other advanced subjects. I read feedback from other students that being "algebraic competent" is very important to grok the more advanced topics, I also read on this website for example that "algebraic competency" is very important for Calculus and the lack thereof is often cause for struggling later on.
So, what would that "algebraic competency" be? Being able to manipulate even complex equations with ease? Or … something else? And how can I learn it?
Since this is my second try at the subject of Maths, and I also have more resources available to me than when I was young, I now have the option to already learn more about subjects that I can see will give me trouble later on.

REPLY [0 votes]: From my experience, Calculus teachers will expect you to know how to do some "basic algebra". They might give you a quick refresher at the start of the first semester but if at that point things are still fuzzy, you will feel that the course is heavy and hard to follow. So for example, being able to:

Isolate a variable ($2x+1=3x\Rightarrow x=1$)
Know your exponent and log laws
Know basic math symbols and notation such as $\mathbb{R}$, $\Rightarrow$, $\infty$, etc.
Knowing what a function is and their properties
Know your basic functions, their domains, ranges and graphs (not quite algebra but still useful)
Knowing vocabulary such as "constant", "variable", "factor", "polynomial", "degree", "term", etc.
Being able to use factorisation (of polynomials)

will all come in handy. You don't have to know everything from the start, but the more proficient you will be with this, the more smoothly Calculus will go.
Let me give you an example. In the first Calculus course, you will have to calculate limits like one:
\begin{equation}
\lim_{x\rightarrow\infty}\dfrac{x^2-x}{x-1}
\end{equation}
In order to do this, you have to see that $x^2-x=x(x-1)$ which will allow you to cancel out the $x-1$ on the numerator and denominator and get to the final answer (which is $\infty$). But if you are not used to do factorization, your eye might not see it and you might get stuck on the problem longer than normal (as in someone who is "algebraically competent").
The list I gave you already covers a lot but probably isn't complete. It's a good place to start.


TITLE: Electrostatic potential...in electrodynamic regime in the expression $P=U.I$
QUESTION [0 upvotes]: Let us consider a two terminal electrical component. It is typical to calculate the power it consumes by considering the voltage drop $U$ and the current passing through it $I$ and to say it absorbs a power $P=U.I$ (of course voltage and current should be oriented in the appropriate manner to say that it absorbs this power). I emphasize on the fact that this formula is also used when varying-in-time fields are considered (the instantaneous power reads $P(t)=U(t)I(t)$)
Now, from electromagnetism theory, we know (among other things) that:
$$\mathbf{E}=-\mathbf{\nabla} V - \partial_t \mathbf{A}$$
$$\mathbf{B}=-\mathbf{\nabla} \wedge \mathbf{A}$$
$$\mathbf{\nabla} \wedge \mathbf{B}=\mu_0 \mathbf{j} + \mu_0 \epsilon_0 \partial_t \mathbf{E}$$
In electrostatic regime, the relationship between voltage drop around two points $A$ and $B$, and electric field is simply:
$$U=\int_A^B \mathbf{E}.\mathbf{dl} $$
This is "consistant" with the general expression of the electric field I gave in term of potentials, assuming $\partial_t=0$
Question1:
What happens in the varying regime ? To what the $U$ in $P=U.I$ corresponds to ? Is it still the line integral of the electrical field ? If so, the voltage drop now depends on the precise path where the integration has been performed because we also integrate the quantity $\partial_t \mathbf{A}$. Then it is not really well defined because we don't need to specify "a path" when we talk about the voltage around a two terminal electrical component in varying regime.
Or maybe the answer is to take in general the "path independant" quantity:
$$U=\int_A^B (\mathbf{E}+\partial_t \mathbf{A}).\mathbf{dl} $$
Question2:
If the voltage drop is still defined as $U=\int_A^B \mathbf{E}.\mathbf{dl} $, how to make sense of the fact such definition depends on the exact path of integration ? Is there a way to make sense of it in the lumped-approximation regime, which means that we don't see spatial variation of electrical waves around the size of the electrical element we are considering ?

REPLY [0 votes]: Question1:
What happens in the varying regime ? To what the $U$ in $P=U.I$ corresponds to ? Is it still the line integral of the electrical field ?

Most usually in AC circuits, voltage is drop of electrostatic potential, the line integral of electrostatic part of the field (the Coulomb field that depends only on charge distribution). The power $P=UI$ is then rate of work of electrostatic forces only (due to electric charge present because of battery or other energy source). This work turns up as other form of energy (for example, as internal energy ("heat") in a resistor or as a magnetic energy in ideal inductor).
We can try to change definition of $U$ to integral of total electric field along the current path, but then (if there is significant induced electric field present in the current path), the power $P=UI$ will have different meaning and value: rate of work of total electric field, which turns up always as non-electromagnetic energy. For example, rate of work of total electric field in a real inductor with ohmic losses turns up as internal energy in the inductor. With good conductivity, this can be much lower than rate of work of electrostatic forces.
"Voltage" both in physics and engineering is unfortunately an overloaded term and confusion is ubiquituous. Usually it is best to stick with the meaning

voltage is drop of Coulomb potential (thus integral of the Coulomb field, a quantity that depends only on endpoints, not path).

However, sometimes when only total field is known or of interest (working of a voltmeter) term "voltage" is used to mean

voltage is integral of total electric field, which then is not function of endpoints only, but depends on the path choice; and the chosen path is often assumed to be the path of electric current.

These two definitions are at odds with each other in situations with induced electric field, so one should use only one definition at a time or find another more descriptive name for quantity 2), e.g. "total electric electromotive force".
In AC circuit theory, where currents change in time, the first definition is actually the standard one and is much less confusing.
For example, potential drop on an ideal inductor (no material core, no ohmic losses) in the pre-chosen positive direction is known to be
$$
p.d. = L\frac{dI}{dt}.
$$
This is a unique value, self-inductance $L$ does not depend on choice of integration path in space, only on geometry of the ideal inductor. However, if we used the second definition of "voltage", on ideal inductor it would always be zero (because ideal inductor is made of ideal conductor for which total electric field vanishes inside). That would not be very useful as it would be inconsistent with the standard exposition of AC circuits.

Question2:
If the voltage drop is still defined as $U=\int_A^B \mathbf{E}.\mathbf{dl} $, how to make sense of the fact such definition depends on the exact path of integration ?

"Voltage drop" is a very confusing term for this, it is better to say just "voltage". The sense is made by the fact that some preferred path is implicitly assumed (path of electric current) so this "voltage" has unique value. This is just the definition 2) described above.

Is there a way to make sense of it in the lumped-approximation regime, which means that we don't see spatial variation of electrical waves around the size of the electrical element we are considering ?

One could try to formulate everything in terms of definition 2) where only total field is used and the integration path is always the path the current takes. But it is not standard, see the problem with ideal inductor described above.
The solution is much simpler. In lumped element model of AC circuits, use the definition 1). The definition 2) is not needed.


TITLE: How to calculate the p-value
QUESTION [0 upvotes]: I'm following a statistics course for the first time and I'm wondering how can I find the difference beteween two indepentent samples.  I read the explanation but still couldn't understand how it works and I'm still stuck at finding the p-value. Here is the data:
$$\bar x_1=5.9$$
$$\bar x_2=4.1$$
$$\sigma_1=4.1$$
$$\sigma_2=3.7$$
$$n_1=42$$
$$n_2=47$$
And I want to test:
$$H_0:\mu_1-\mu_2>0$$
$$H_a:\mu_1-\mu_2<0$$
Lets say at a 0.05 level.  So I find SE:
$$SE^2=\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}$$
This gives SE = 0.83
So I calculate my t stat:
$$t=\frac{\bar x_1-\bar x_2}{SE}=2.16458$$
How can I calculate the p-value and what would it mean?

REPLY [1 votes]: The notation $\sigma_i$ rather than $s_i$ suggests you are treating the standard deviations as known rather than as being estimated based on the data. Furthermore, the way you find your standard error makes sense only if that is what you are doing. That is not realistic but is often done in early exercises. Therefore you should treat your test statistic $Z = (\overline x_1-\overline x_2)/\text{SE}$ as being normally distributed rather than as having a t-distribution.
Suppose your null hypothesis is that the means of the two populations are equal and the alternative hypothesis says only that they are not equal, without saying which direction the inequality goes. Then the p-value is the probability that $|Z|,$ the absolute value of $Z,$ exceeds the value actually observed, given that the null hypothesis is true. So, for example, suppose you have $|Z| = 2.3.$ then you need $\Pr(|Z|>2.3) = \Pr(Z>2.3) + \Pr(Z<-2.3) = 2\Pr(Z< -2.3).$ Normally you would find that probability by using either a table or a software package.


TITLE: Prove for a in $U_m$, $a^k \equiv 1 \pmod{m}$ for some k>1
QUESTION [0 upvotes]: Prove for $a$ in $U_m$, $a^k \equiv 1 \pmod{m}$ for some $k>1$
This is basically equivalent to proving that if gcd(a,m)=1, then the function $m|a^k-1$ has an integer solution.
Can't use theorems related to orders directly, can't use Euler's totient theorem

REPLY [0 votes]: $U_m$ is a finite group with identity $1$; denote its order by $\phi(m)$.  Then, for $a$ in $U_m$, $a^{\phi(m)}=1$.


TITLE: System differential equations operational method
QUESTION [1 upvotes]: I am trying  to solve this system of differential equation
$$x''-2y'+x=0$$
$$y''-2x'+y=e^{2t}$$
Then
$$D^2 x-2Dy+x=0$$
$$D^2 y-2Dx+y=e^{2t}$$
But I don't know what to do next
$$(D^2 +1)x-2Dy=0$$
$$(D^2 +1)y-2Dx=e^{2t}$$

REPLY [2 votes]: $$x(D^2 +1)-2Dy=0$$
$$y(D^2 +1)-2Dx=e^{2t}$$
Differentiate:
$$2D(D^2 +1)x-4D^2y=0$$
$$(D^2 +1)^2y-2D(D^2+1)x=(D^2+1)e^{2t}$$
Now sum both DE and solve.
$$(D^2 +1)^2y-4D^2y=(D^2+1)e^{2t}$$
More simply:
$$(D^2 -1)^2y=(D^2+1)e^{2t}$$
$$(D^2 -1)^2y=5e^{2t}$$


TITLE: Bijective functions questions.
QUESTION [1 upvotes]: I'm having some problems with these questions on functions being bijective or not and I would appreciate some help.

Are the following functions bijective?
a) $f\colon \Bbb N \to \Bbb N$ defined as: $f(n)=n^2$.
b) $f\colon [-1,1]\to[0,1]$ defined as: $f(n) = n^2$.
c) $f\colon \Bbb R^2 \to \Bbb R^2$ defined as: $f(x,y)=(x+y,x-y)$.
d) $f\colon \Bbb Z^2 \to \Bbb Z^2$ defined as:
$$f(x,y)=\begin{cases}(x,y+1), &\text{if $x$ and $y$ are both even or odd}\\(x+1,y),&\text{otherwise}\end{cases}$$

I have two questions from these sets of questions.
For (b) I said the $0$ from the $[0, 1]$ set does not have a pre-image because no number from the $[-1, 1]$ set squared equals $0$. Therefore it's not surjective so it's not bijective. I am not sure if I am going about this question the right way, or if it's even right.
And also for (c), from what I understand, the input is a set of pairs of rational numbers and so is the output, however, I don't know how to prove it through disproving injection or surjection. Any help would be appreciated.
UPDATE: I attempted to do D and this is what I got. I first tried to prove surjectivity. So, if I have an element [20, 10] for example, you can't get to there because the function only outputs pairs of numbers that are 1 apart. So this means its not surjective so it cannot be bijective. Is this correct?

REPLY [1 votes]: I’ll do (c) as a model and leave (d) for you to try.
To show that $f$ is bijective, you must show that it is both injective and surjective.

$f$ is injective. Suppose that $\langle x_0,y_0\rangle,\langle x_1,y_1\rangle\in\Bbb R^2$ and $f(x_0,y_0)=f(x_1,y_1)$; then $\langle x_0+y_0,x_0-y_0\rangle=\langle x_1+y_1,x_1-y_1\rangle$, so $x_0+y_0=x_1+y_1$ and $x_0-y_0=x_1-y_1$. Adding these two equations, we find that $2x_0=2x_1$ and hence that $x_0=x_1$; subtracting this from $x_0+y_0=x_1+y_1$, we find that $y_0=y_1$. Thus, $\langle x_0,y_0\rangle=\langle x_1,y_1\rangle$, and $f$ is therefore injective. (This is the standard approach to proving injectivity: assume that the function takes two things in the domain to the same thing in the range, and show that the things in the domain were actually the same thing.)
$f$ is surjective. Let $\langle u,v\rangle\in\Bbb R^2$ be arbitrary; we must find $\langle x,y\rangle\in\Bbb R^2$ such that $f(x,y)=\langle u,v\rangle$, i.e., such that $\langle x+y,x-y\rangle=\langle u,v\rangle$, or, finally, $x+y=u$ and $x-y=v$. Adding these two equations, we find that $2x=u+v$, so we must set $x=\frac12(u+v)$. Subtracting the second from the first gives us the equation $2y=u-v$, from which we see that we must set $y=\frac12(u-v)$. You can now check that with these values for $x$ and $y$ we do indeed find that $f(x,y)=\langle u,v\rangle$, and it follows that $f$ is surjective.


TITLE: Different proof for the inequality $b \gcd(a,c)^2+ a \gcd(b,c)^2-c \gcd(a,b)^2 \le abc$?
QUESTION [4 upvotes]: In Encyclopedia of Distances, there is given a distance on natural numbers without zero:
$$d(a,b) = \log\left( \frac{ab}{\gcd(a,b)^2}\right)\,.$$
Taking (again for a reference see the Encyclopedia of Distances), the Schoenberg transform of this metric we get the metric:
$$D(a,b) = 1-\exp\big(-d(a,b)\big) = 1-\frac{\gcd(a,b)^2}{ab}\,.$$
Since this metric satisfies the triangle inequality, we get:
$$D(a,b) \le D(a,c) + D(c,b)\,,$$
and the inequality, after some algebra,
$$b \gcd(a,c)^2+ a \gcd(b,c)^2-c \gcd(a,b)^2 \le abc$$
follows for three natural numbers $a,b,c$.
My question is, if there is a more direct proof for this?

REPLY [1 votes]: Write
$$abc-b\gcd(a,c)^2-a\gcd(b,c)^2=c\left(a-\frac{\gcd(a,c)^2}{c}\right)\left(b-\frac{\gcd(b,c)^2}{c}\right)-\frac{\gcd(a,c)^2\gcd(b,c)^2}{c}.$$
Since $ac\ge\gcd(a,c)^2$ and $bc\ge\gcd(b,c)^2$, we see that
$$abc-b\gcd(a,c)^2-a\gcd(b,c)^2\ge-\frac{\gcd(a,c)^2\gcd(b,c)^2}{c}.$$
We only need to show that
$$\frac{\gcd(a,c)^2\gcd(b,c)^2}{c}\leq c\gcd(a,b)^2.$$
But this is equivalent to
$$\gcd(a,c)\gcd(b,c)\leq c\gcd(a,b).$$
However, this follows from
$$\gcd(xy,mn)\geq \gcd(x,m)\gcd(y,n)$$
by taking $x=a$, $y=c$, $m=c$, and $n=b$.  From this proof, we can also see that the inequality $$b\gcd(a,c)^2+a\gcd(b,c)^2-c\gcd(a,b)^2\leq abc$$ is an equality iff $a=c$ or $b=c$.


TITLE: The classical definition for the $sign$ function does not interact well with complex numbers
QUESTION [0 upvotes]: The classic definition for the sign function is [1]
$$sign(x)=
\begin{cases}
\frac{x}{|x|}, & \text{if } x \neq 0 \\
0, & \text{if } x = 0
\end{cases}$$
But when the $sign$ function is applied to the complex numbers you get a continuous function instead of the nice step like function that exists for the sign in the reals. Wouldn't it make more sense to use
$$f(a+bi)=
\begin{cases}
\frac{a}{|a|}+\frac{b}{|b|}i, & \text{if } a+bi \neq 0 \\
0, & \text {if } a+bi = 0\end{cases}$$?
[1] I didn't use the inequality definition of the sign function is because "inequalities are not well defined in the complex plane"/"the complex numbers are not well ordered".

REPLY [1 votes]: Actually, the original definition works perfectly for complex numbers. Except at $z=0$, it gives a unit direction vector for $z$, as is exactly the case for the reals. (We regard positive reals as "pointing to the right" and negative reals as "pointing to the left"; $0$ doesn't point anywhere.)
It coincides with the real version for real values of $z$.
It is NOT continuous at $z=0$, as is also the case with the reals. It is continuous everywhere else, as is also the case with the reals.
Seems pretty good to me.


TITLE: How prove this inequality $\left[\frac{n}{\sqrt{3}}\right]+1>\frac{n^2}{\sqrt{3n^2-5}}$
QUESTION [2 upvotes]: let $k$ is postive integer,and for any postive integer $n\ge 2$,
show that:
  $$\left[\dfrac{n}{\sqrt{3}}\right]+1>\dfrac{n^2}{\sqrt{3n^2-5}}>\dfrac{n}{\sqrt{3}}$$
where $[x]$ is  the largest integer not greater than $x$

REPLY [3 votes]: Let $q_n = \left\lfloor \frac{n}{\sqrt{3}}\right\rfloor + 1$. When $n \ge 11\sqrt{3}$, we have
$$\frac{n}{\sqrt{3}q_n} \ge \frac{\frac{n}{\sqrt{3}}}{\frac{n}{\sqrt{3}}+1} = 1 - \frac{\sqrt{3}}{n+\sqrt{3}} \ge \frac{11}{12}
\quad\implies\quad \frac{n^2}{3q_n^2} \ge \left(\frac{11}{12}\right)^2 > \frac{5}{6}
$$
Be definition, $\left\lfloor \frac{n}{\sqrt{3}}\right\rfloor$ is the largest integer
less than or equal to $\frac{n}{\sqrt{3}}$. This implies
$$\frac{n}{\sqrt{3}} < q_n \quad\implies\quad 3q_n^2 - n^2 > 0 \quad\implies\quad 3 q_n^2 - n^2 \ge 2$$
The last inequality is true because $3q_n^2 - n^2$ is an integer and there is no integer solution for the equation $3 q^2 - n^2  = 1$.
Combine these, we find for any $n \ge 20 > 11\sqrt{3}$, we  have
$$3 n^2 - \frac{n^4}{q_n^2} = 3\left(\frac{n^2}{3q_n^2}\right)(3q_n^2 - n^2) > 3 \left( \frac{5}{6}\right) 2 = 5$$
This leads to
$$3n^2 - 5 > \frac{n^4}{q_n^2} \quad\iff\quad q_n > \frac{n^2}{\sqrt{3n^2 -5}} \quad\text{ for } n \ge 20 \tag{*1}$$
By brute force, one can verify RHS$(*1)$ also work for $2 \le n \le 19$.
As pointed out by Macavity in comment, the largest admissible $k$ for $n = 5$ is $5$.
This means the maximum value of $k$ which works for all $n$ is indeed $5$.


text:
{\bf Problem.} Joe will randomly select two letters from the word CAMP, four letters from the word HERBS, and three letters from the word GLOW.  What is the probability that he will have all of the letters from the word PROBLEM?  Express your answer as a common fraction.
{\bf Level.} Level 5
{\bf Type.} Counting & Probability
{\bf Solution.} Each letter in the word PROBLEM appears exactly once among the words CAMP, HERBS, and GLOW.  Therefore, in order to have all of the letters to spell PROBLEM, Joe must select both M and P when choosing two letters from CAMP.  The probability of this is $1/\binom{4}{2}=1/6$.  Also, he must select the letters E, R, and B when choosing four letters from the word HERBS.  Among the $\binom{5}{4}=5$ ways of choosing these letters, 2 of them contain all of the letters E, R, and B.  Therefore, the probability that he will select E, R, and B from HERBS is 2/5.  Finally, Joe must select L and O among the 3 letters he chooses from GLOW.  Among the $\binom{4}{3}=4$ ways of choosing these letters, 2 of them contain both L and O.  Therefore, the probability that he will select L and O from GLOW is $2/4=1/2$.  Altogether, the probability that he will select all the letters from the word PROBLEM is $\left(\frac{1}{6}\right)\left(\frac{2}{5}\right)\left(\frac{1}{2}\right)=\boxed{\frac{1}{30}}$.

set_name: MATH


text:
\begin{document}
\title{A rigidity theorem for H\'{e}non maps}
\keywords{Polynomial automorphisms, H\'{e}non maps, rigidity, non-escaping sets,  \short{2}, Reinhardt domains}

\subjclass{Primary: 32H02  ; Secondary : 32H50}
\author{Sayani Bera, Ratna Pal, Kaushal Verma}

\address{SB: School of Mathematics, Ramakrishna Mission Vivekananda Educational and Research Institute, PO Belur Math, Dist. Howrah,
West Bengal 711202, India}
\email{sayanibera2016@gmail.com}

\address{RP: Department of Mathematics, Indian Institute of Science Education and Research, Pune, Maharashtra-411008, India}
\email{ratna.math@gmail.com}

\address{KV: Department of Mathematics, Indian Institute of Science, Bangalore 560 012, India}
\email{kverma@iisc.ac.in}

\begin{abstract}
The purpose of this note is two fold. First, we study the relation between a pair of H\'{e}non maps that share the same forward and backward non-escaping sets. Second, it is shown that there exists a continuum of \short{2}'s that are biholomorphically inequivalent and finally, we provide examples of \short{2}'s that are neither Reinhardt nor biholomorphic to Reinhardt domains.
\end{abstract}

\maketitle

\section{Introduction}

\no Let $P, Q$ be a pair of holomorphic polynomials in the plane and $J_P, J_Q$ the corresponding Julia sets. If $P$ and $Q$ commute, it is known that
$J_P = J_Q$. Conversely, if $J_P = J_Q$, a basic question that has been studied rather extensively is to find relations between $P$ and $Q$. For example, a result of Beardon \cite{Be1}, that builds on the work of Baker--Er\"{e}menko \cite{BE}, shows that
\[
P \circ Q = \sigma \circ Q \circ P
\]
where $\sigma(z) = az + b$, $\vert a \vert = 1$ and $\sigma(J_P) = J_P$. In other words, two polynomials with the same Julia set necessarily commute up to a rigid motion, of the kind described above, that preserves the shared Julia set. Other related work on this question for polynomials can be found for example in \cite{Be2} and \cite{SS}, while the case of rational functions is dealt with in \cite{LP} and \cite{D} and the references therein.

\medskip

The purpose of this note is to prove a suitable analogue of this result in
$\mathbb{C}^2$. This will be a consequence of a more general rigidity theorem. To explain this, let $\mathcal{H}$ be the family of polynomial automorphisms of
$\mathbb{C}^2$ of the form
\begin{equation}
H = H_m \circ H_{m-1} \circ \cdots \circ H_1
\end{equation}
where
\begin{equation}
H_j(x, y) = (b_jy, p_j(y) - \delta_j x)
\end{equation}
with $p_j$ a polynomial of degree $d_j \ge 2$ and $b_j\delta_j \not= 0$. The degree of $H$ is $d = d_1d_2 \ldots d_m$. In what follows, the phrase {\it H\'{e}non map} will refer to a map as in (1.1). The only difference between these expressions and the more commonly accepted normal form of a H\'{e}non map lies in the non-zero scalars $b_j$. The reasons for this apparent deviation from standard nomenclature will be explained later -- they certainly do not include any implication of generality of any sort.

\medskip

Let us recall in brief the basic dynamical objects associated with such maps, keeping in mind that the extra constants $b_j$ do not really influence the calculations in a significant manner -- see \cite{BS} for details. For $R > 0$, let
\begin{align*}
V^+_R &= \{ (x,y) \in \mathbb C^2: \vert x \vert < \vert y \vert, \vert y \vert > R \},\\
V^-_R &= \{ (x,y) \in \mathbb C^2: \vert y \vert < \vert x \vert, \vert x \vert > R \},\\
V_R &= \{ (x, y) \in \mathbb C^2: \vert x \vert, \vert y \vert \le R \}.
\end{align*}
For a given $H \in \mathcal{H}$, or for that matter a compact family of H\'{e}non maps, there exists $R > 0$ such that
\[
H(V^+_R) \subset V^+_R, \; H(V^+_R \cup V_R) \subset V^+_R \cup V_R
\]
and
\[
H^{-1}(V^-_R) \subset V^-_R, \; H^{-1}(V^-_R \cup V_R) \subset V^-_R \cup V_R.
\]
Let
\[
K^{\pm}_H = \{(x, y) \in \mathbb C^2 : \;\text{the sequence}\; \left(H^{\pm n}(x, y) \right) \; \text{is bounded} \}
\]
be the set of non-escaping points. Then $K^{\pm} \subset V_R \cup V^{\mp}_R$ and
\begin{equation}\label{escape}
\mathbb C^2 \sm K^{\pm}_H = \bigcup_{n=0}^{\infty} (H^{\mp n})(V^{\pm}_R)
\end{equation}
In fact, $\mathbb C^2 \sm K^{\pm}$ is the set of points for which
$\Vert H^{\pm n}(x, y) \Vert \rightarrow \infty$ as $n \rightarrow \infty$. Each $H \in \mathcal{H}$ extends meromorphically to $\mathbb{P}^2$ with an isolated indeterminacy point at $I^+ = [1:0:0]$ and the same holds for $H^{-1}$ which has $I^{-} = [0:1:0]$ as its lone indeterminacy point. The Green's functions
\[
G^{\pm}_H(x, y) = \lim_{n \rightarrow \infty} \frac{1}{d^n} \log^+ \Vert H^{\pm n}(x, y) \Vert
\]
are plurisubharmonic on $\mbb C^2$ and non-negative everywhere, pluriharmonic on $\mbb C^2 \sm K^{\pm}_H$ and vanish precisely on $K^{\pm}_H$. By construction,
\[
G^{\pm}_H \circ H = d G^{\pm}_H.
\]
Both $G^{\pm}_H$ have logarithmic growth near infinity, i.e., there exist $R>0$ and $C>0$ such that
\begin{equation}\label{L1}
\log^+ \lvert y \rvert-C\leq G_H^+ (x,y) \leq \log^+ \lvert y \rvert+C
\end{equation}
for $(x,y)\in \overline{V_R^+ \cup V_R}$, and
\begin{equation}\label{L2}
\log^+ \lvert x \rvert-C\leq G_H^+ (x,y) \leq \log^+ \lvert x \rvert+C.
\end{equation}
for $(x,y)\in \overline{V_R^- \cup V_R}$. Hence
\begin{equation}\label{L3}
G_H^\pm (x,y)\leq \max \{\log^+ \lvert x\rvert, \log^+ \lvert y \rvert \}+ C
\end{equation}
for all $(x,y)\in \mathbb{C}^2$. Let $J^{\pm}_H = \partial K^{\pm}_H$. It turns out that $G_H^\pm$ are the pluricomplex Green's functions for $K^{\pm}_H$ respectively. The supports of the positive closed $(1,1)$ currents
\[
\mu^{\pm}_H = dd^c G^{\pm}_H
\]
are $J^{\pm}_H$ and $\mu_H = \mu^+_H \wedge \mu^-_H$ is an invariant measure for $H$. Since we will be dealing with more than one H\'{e}non map, these objects carry a subscript to distinguish them from their counterparts associated with other maps. Here is the main result:

\begin{thm} \label{rigidity}
Let $H$ be a H\'{e}non map. Suppose that $F \in {\rm Aut}(\mbb C^2)$ is such that $F(K^{\pm}_H) = K^{\pm}_H$.  Then $F$ is a polynomial automorphism.  Furthermore,
\begin{itemize}
\item[(i)] if $\deg(F) = 1$, then $F$ is of the form
\[
(x,y)\mapsto (cx+d, ey+f)
\]
and
\item[(ii)] if $\deg(F)\geq 2$, then either $F$ or $F^{-1}$ is a H\'{e}non map and accordingly there exists a linear map $C(x, y) = (\delta_- x, \delta_+ y)$ with $\vert \delta_\pm \vert = 1$ such that
\[
F^{\pm 1} \circ H = C \circ H \circ F^{\pm 1}.
\]
In addition, there exist positive integers $m_0, n_0$ and an affine automorphism $\sigma$ such that $F^{\pm m_0}=\sigma \circ H^{n_0}$ depending on whether $F$ or $F^{-1}$ is a H\'{e}non map.
\item[(iii)] There exists a H\'{e}non map $R$ such that any H\'{e}non map $F$ satisfying $F(K_H^\pm)=K_H^\pm$ is of the form $F=\sigma_F R^{r_F}$ for some affine automorphism $\sigma_F$ and for some integer $r_F \geq 1$.
\end{itemize}

\no Conversely, if $F$ and $H$ be two H\'{e}non maps satisfying
\[
F\circ H=C\circ H \circ F
\]
where $C:(x,y)\mapsto (\delta_- x, \delta_+ y)$ with $\lvert \delta_\pm \rvert =1$ and $C(K_H^\pm)=K_H^\pm$, then $F(K_H^\pm)=K_H^\pm$.
\end{thm}

\begin{cor}
Let $F, H$ be H\'{e}non maps such that $K^{\pm}_H = K^{\pm}_F$. Then there exists a linear map $C(x, y) = (\delta_-x, \delta_+y)$ with $\vert \delta_{\pm} \vert = 1$ such that $F \circ H = C \circ H \circ F$. In particular, if each of $F, H$ are finite compositions of maps of the form
\[
(x, y) \mapsto (y, p(y) - x)
\]
with $p$ a monic polynomial (such maps are volume preserving), and $K^{\pm}_H = K^{\pm}_F$, then $F \circ H = H \circ F$.
\end{cor}

In the volume preserving case, it follows from \cite{Bi} that if $d_F$ and $d_H$ are the degrees of $F, H$ respectively, then there are positive integers $k, l$ such that $d_F^k = d_H^l$ and there exists an affine automorphism $\sigma$ such that
$F^k =\sigma \circ H^l $.

\medskip

The proof of Theorem 1.1 requires several steps and begins by using an idea of Buzzard--Fornaess \cite{BF}. The hypotheses imply that $G^{\pm}_H \circ F$ vanishes precisely on $K_H^{\pm}$ and is pluriharmonic outside it. Its restriction to vertical slices of the form $x=$ constant is harmonic away from a compact set and therefore admits a harmonic conjugate modulo a period that apriori depends on the slice under consideration. However, the pluriharmonicity of $G^{\pm} \circ F$ away from $K^{\pm}_H$ shows that the period is independent of the vertical slice and this shows that $G^{\pm+}_H \circ F$ and $G^{\pm}_H$ agree upto a multiplicative constant. The logarithmic growth of $G^{\pm}_H$ then shows that $F$  must be a polynomial. Next, by Jung's theorem, $F$ can be written as a composition of affine and elementary maps and four cases arise depending on the nature of the first and last terms in this composition. Using the mapping properties of H\'{e}non maps with respect to the filtration and the hypothesis that $F$ preserves $K^{\pm}_H$ allows us to eliminate two cases. The remaining possibilities correspond precisely to the conclusion that either $F$ or $F^{-1}$ is a H\'{e}non map. Again, $F(K^{\pm}_H) = K^{\pm}_H$ implies that $K^{\pm}_H \subset K^{\pm}_F$ and the rigidity theorem of Dinh--Sibony \cite{DS} now shows that $J^{\pm}_H = J^{\pm}_F$ and hence $G^{\pm}_H = G^{\pm}_F$. Working with the associated B\"{o}ttcher functions then shows that $F$(or $F^{-1}$) and $H$ must commute up to a linear map $C$.

\medskip

The connection to \short{2}'s comes via the observation, due to Fornaess \cite{Fo}, that the sub-level sets of the Green's functions of H\'{e}non maps are examples of \short{2}'s. Recall that a \short{2} $\subset \mbb C^2$ is a domain satisfying the following properties: It can be exhausted by an increasing sequence of biholomorphic images of the ball $\mbb B^2 \subset \mbb C^2$, the Kobayashi metric on it vanishes identically, but it admits a non-constant plurisubharmonic function that is bounded above. We show that \short{2}'s arising as sub-level sets of the Green's functions of H\'{e}non maps are not Reinhardt. Furthermore, if the H\'{e}non map is chosen as a small perturbation of a hyperbolic polynomial in one variable, then the sub-level sets of the associated Green's functions cannot even be biholomorphic to a Reinhardt domain.

\medskip

Finally, a word about the choice of the normal form of the maps in $\mathcal H$: every map in $\mathcal{H}$ is, by Friedland--Milnor \cite{FM}, conjugate to a finite composition of maps of the form $(x, y) \mapsto (y, p(y) - ax)$ for some $a \not= 0$. The natural starting point then would be to consider the family of maps in which each $b_j = 1$ -- these are the so called generalized H\'{e}non maps. If we work with such maps, the proof of the main theorem shows that every automorphism that preserves the associated non-escaping sets must be of the form (1.1) and there is no apriori reason for concluding that $b_j = 1$. The point, therefore, is that all calculations are done in a fixed coordinate system dictated by the choice of the map we begin with. Hence, the choice of the family
$\mathcal{H}$ as in (1.1) is not an artificial one but is naturally induced by the situation; it also serves to state the main theorem more succinctly.

\medskip

Finally, the one variable analogue of the main theorem is given in the Appendix. It supplements the existing known results on maps that share the same Julia set.

\section{B\"{o}ttcher functions and their relation with the Green's functions}
\no
Let $H=H_m \circ H_{m-1}\circ \cdots \circ H_1$, where $H_j(x,y)=(b_jy,p_j(y)-\delta_j x)$ with $p_j$ a polynomial of degree $d_j$ with highest degree coefficient $c_j$ and $b_j \delta_j \not= 0$. Then $H(x,y)=(H_1(x,y),H_2(x,y))$ where the degree of $H_1$ is strictly less than that of $H_2$ when regarded as a polynomial in $y$ and in fact
\begin{equation}\label{H1}
H_2(x,y)=c_H y^d+q(x,y)
\end{equation}
where
\[
c_H = \prod_{j=1}^m {c_j}^{d_{j+1}\cdots d_m}
\]
with the convention that $d_{j+1}\cdots d_m=1$ when $j=m$, $d=d_j \cdots  d_1$ and $q$ is a polynomial in $x,y$ of degree strictly less than $d$.

\medskip

\no
Similarly, $H^{-1}(x,y)=(H_1'(x,y),H_2'(x,y))$ where the degree of $H_2'$ is strictly less than that of $H_1'$ when regarded as a polynomial in $x$ and
\begin{equation}\label{H2}
H_1'(x,y)=c_H' x^d+q'(x,y)
\end{equation}
where
\[
c_H'=\prod_{j=1}^m {\left({c_j}\delta_j^{-1}\right)}^{d_{j-1}\cdots d_1} b_j^{-d_j d_{j-1} \cdots d_1}
\]
with the convention that $d_{j-1}\cdots d_1=1$ when $j=1$ and $q'$ is a polynomial in $x,y$ of degree strictly less than $d$.

\medskip
\no Keeping track of the various steps in the proof of Proposition 5.2 in \cite{HO} shows that:

\begin{prop}\label{Bot}
For a given H\'{e}non map $H$, there exist non-vanishing holomorphic functions
$\phi_H^\pm: V_R^\pm \rightarrow \mathbb{C}$  such that
\[
\phi_H^+\circ H(x,y)=c_H{(\phi_H(x,y))}^d
\]
in $V_R^+$ and
\[
\phi_H^-\circ H^{-1}(x,y)=c_H'{(\phi_H^-(x,y))}^d
\]
in $V_R^-$. Further,
\[
\phi_H^+(x,y)\sim y \text{ as } \lVert(x,y)\rVert\rightarrow \infty \text{ in } V_R^+
\]
and
\[
\phi_H^-(x,y)\sim x \text{ as } \lVert(x,y)\rVert\rightarrow \infty \text{ in } V_R^-.
\]

\end{prop}
\begin{proof}
Let $H^n(x,y)=({(H^{n})}_1(x,y),{(H^{n})}_2(x,y))$. Note that $y_n={(H^{n})}_2(x,y)$ is a polynomial in $x$ and $y$ of degree $d^{n}$.
Consider the telescoping product
\begin{equation}\label{tele}
y.\frac{{y_1}^{\frac{1}{d}}}{y}.\cdots.\frac{y_{n+1}^{\frac{1}{d^{n+1}}}}{y_{n}^{\frac{1}{d^{n}}}}\cdots
\end{equation}
which will be shown to converge. From (\ref{H1}),

\begin{eqnarray}\label{bott}
\frac{y_{n+1}^{\frac{1}{d^{n+1}}}}{y_{n}^{\frac{1}{d^{n}}}}&=&\frac{{\left(c_H y_n^d+q(x_n,y_n)\right)}^{\frac{1}{d^{n+1}}}}{y_n^{\frac{1}{d^n}}}\nonumber\\
&=&{\left(c_H+\frac{q(x_n,y_n)}{y_n^d}\right)}^{\frac{1}{d^{n+1}}}\nonumber\\
&=& {c_H}^{\frac{1}{d^{n+1}}}{\left(1+\frac{q(x_n,y_n)}{c_H y_n^d}\right)}^{\frac{1}{d^{n+1}}}.
\end{eqnarray}

Choose $R>0$ sufficiently large so that

\[
\left\lvert \frac{q(x_n,y_n)}{c_H y_n^d}\right\rvert <1
\]
for all $\lvert y\rvert >R$ and for all $n\geq 1$.
Let ${\left(1+\frac{q(x_n,y_n)}{c_H y_n^d}\right)}^{\frac{1}{d^{n+1}}}$ be the principal branch of the $d^{n+1}$-th root of
$\left(1+\frac{q(x_n,y_n)}{c_H y_n^d}\right)$.

\medskip
\no
Now note that the convergence of the product
\[
y.\frac{{y_1}^{\frac{1}{d}}}{y}.\cdots.\frac{y_{n+1}^{\frac{1}{d^{n+1}}}}{y_{n}^{\frac{1}{d^{n}}}}\cdots.
\]
is equivalent to the convergence of the series
\begin{equation}\label{Log}
{\rm{Log}}\ y+ {\rm{Log}} \left(\frac{{y_1}^{\frac{1}{d}}}{y}\right)+\cdots + {\rm{Log}}\left(\frac{y_{n+1}^{\frac{1}{d^{n+1}}}}{y_{n}^{\frac{1}{d^{n}}}}\right)+\cdots.
\end{equation}
There exists $M>0$ (depending on the coefficients of the polynomial $q$) such that
\begin{equation}\label{zero}
\left\lvert\frac{q(x_n,y_n)}{y_n^d}\right\rvert\leq \frac{M}{\lvert y_n \rvert}\leq \frac{\tilde{M}}{{\lvert y\rvert}^{d^n}}
\end{equation}
for $R>0$ sufficiently large. Thus the series (\ref{Log}) converges and consequently the series (\ref{tele}) also converges.

\medskip
\no
The function
\[
\phi_H^+(x,y)={c_H}^{-\frac{1}{d-1}}\lim_{n\rightarrow \infty}y_{n}^{\frac{1}{d^{n}}}={c_H}^{-\frac{1}{d-1}}\left(y.\frac{{y_1}^{\frac{1}{d}}}{y}.\cdots.\frac{y_{n+1}^{\frac{1}{d^{n+1}}}}{y_{n}^{\frac{1}{d^{n}}}}\cdots\right)
\]
is clearly well-defined. Further, it follows from (\ref{bott}) and (\ref{zero}) that
\[
\phi_H^+(x,y)\sim y
\]
as $\lVert (x,y)\rVert\rightarrow \infty$ in $V_R^+$.

\medskip
\no
Since
\begin{eqnarray*}
H^{n+1}(x,y)&=&H({(H^n)}_1(x,y), {(H^n)}_2(x,y))\\
&=&({(H^{n+1})}_1(x,y), c_H{(H^n)}_2^d(x,y)+q({(H^n)}_1(x,y),{(H^n)}_2(x,y))),
\end{eqnarray*}
it follows that
\begin{eqnarray*}
{(H^n)}_2(H(x,y))&=&

c_H {(H^n)}_2^d(x,y)(1+L(x,y))
\end{eqnarray*}
where $L(x,y)\rightarrow 0$ as $\lVert (x,y)\rVert\rightarrow \infty$ in $V_R^+$.

\medskip
\no
Therefore,
\begin{eqnarray*}
\phi_H^+(H(x,y))&=&{c_H}^{-\frac{1}{d-1}}\lim_{n\rightarrow \infty} (H^n)_2^{\frac{1}{d^n}}(H(x,y))\\
&=&c_H\lim_{n\rightarrow \infty} c_H^{\frac{1}{d^n}} {\left({c_H}^{-\frac{1}{d-1}}{(H^n)}_2^{\frac{1}{d^n}}(x,y){(1+L(x,y))}^{\frac{1}{d^{n+1}}}\right)}^d\\
&=&c_H {\left(\phi_H^+(x,y)\right)}^d.
\end{eqnarray*}

\medskip
\no
Consider $H^{-n}(x,y)=({(H^{-n})}_1(x,y),{(H^{-n})}_2(x,y))$ and note that $x_n={(H^{-n})}_1(x,y)$ is a polynomial in $x$ and $y$ of degree $d^{n}$.
As before, consider the telescoping product
\begin{equation*}
x.\frac{{x_1}^{\frac{1}{d}}}{x}.\cdots.\frac{x_{n+1}^{\frac{1}{d^{n+1}}}}{x_{n}^{\frac{1}{d^{n}}}}\cdots
\end{equation*}
which can be shown to be convergent. Thus, we can
define
\[
\phi_H^-(x,y)={c_H'}^{-\frac{1}{d-1}}\lim_{n\rightarrow \infty}x_{n}^{\frac{1}{d^{n}}}.
\]
The properties of $\phi_H^-$ can be established as in the case of $\phi_H^+$.

\end{proof}
\no
Since
\[
G_H^+(x,y)=\lim_{n\rightarrow \infty}\frac{1}{d^n}\log\left \lVert{(H^n)}_2(x,y)\right\rVert
\]
in $V_R^+$, it follows that
\begin{equation}\label{Green1}
G_H^+=\log \lvert\phi_H^+\rvert+\frac{1}{d-1}\log \lvert c_H\rvert
\end{equation}
in $V_R^+$. Similarly
\begin{equation}\label{Green2}
G_H^-=\log \lvert\phi_H^-\rvert+\frac{1}{d-1}\log \left\lvert {c_H'}\right\rvert
\end{equation}
in $V_R^-$.

\begin{rem}
The uniqueness of $\phi_H^\pm$ follows immediately once we establish (\ref{Green1}) and (\ref{Green2}) along with their properties recorded in Proposition \ref{Bot}.
\end{rem}

\section{Proof of Theorem 1.1}

\no The first step is to show that
\begin{center}
{\it $F$ is a polynomial automorphism:}
\end{center}

\medskip

\no Since $F(K_H^\pm)=K_H^\pm$, it follows that $G_H^\pm\circ F=0$ on $K_H^\pm$. Fix $x \in \mathbb{C}$ and consider
\[
g_{x}(y)=G_H^+\circ F (x,y)
\]
which is harmonic outside the compact set $K_H^+ \cap (\{x\}\times \mathbb{C})$. Since it is harmonic outside a large disk of radius $R>0$, $g_{x}$ has a harmonic conjugate $h_{x}$ in $\{\vert y \vert > R\}$ with period $c_{x}$. Therefore
\[
\psi_{x}(y)=g_{x}(y)-c_{x}\log \lvert y\rvert+ i h_{x}(y)
\]
is holomorphic in $\{\lvert y\rvert >R\}$. Further, note that
\[
\left\lvert\exp(-\psi_{x}(y))\right\rvert \leq {\lvert y\rvert}^{c_{x}}
\]
which means that $\exp(-\psi_{x}(y))$ has at most a pole at infinity and consequently,
\[
\exp(-\psi_{x}(y))=y^k \exp f(y)
\]
where $f$ is a holomorphic function in $\{\lvert x\rvert >R\}$ having a removable singularity at infinity. Here $k = k_x$ is a positive integer. Taking absolute values and then log, we get
\[
g_{x}(y)-c_{x}\log \lvert y \rvert=-k \log \lvert y \rvert-{\rm{Re}}(f(y))
\]
in $\{\lvert y\rvert >R\}$. Therefore,
\[
g_{x}(y)=b_{x}\log \lvert y\rvert +O(1)
\]
in $\{\lvert y\rvert >R\}$ and since $g \ge 0$ everywhere,
\[
g_{x}(y)=b_{x}\log^+ \lvert y\rvert +O(1)
\]
in $\mathbb{C}$. Since the $O(1)$ term is bounded near infinity, $b_x$ is in fact
the period of $g_x(y)$.

\medskip

It turns out that $b_x$ is independent of $x$. To prove this, let us work in a small neighbourhood of a fixed $x_0$ and $R$ is large enough as before. Let $p, q$ be two distinct points near $x_0$ and let $I$ be the straight line segment joining them. Then
\[
\Sigma = \{(x, y) : x \in I, \vert y \vert = R \}
\]
is a smooth real 2-surface with two boundary components namely,
\[
\{(p, y) : \vert y \vert = R \} \cup \{ (q, y): \vert y \vert = R \}.
\]
Thanks to Stokes' theorem,
\[
b_p - b_q = \int_{\partial \Sigma} d^c (G^+_H \circ F) = \int_{\Sigma} dd^c(G^+_H \circ F) = 0
\]
where the last equality holds due to the pluriharmonicity of $G^+_H \circ F$ on $V^+_R$. Hence $b_x$ is locally constant and therefore constant everywhere. Let us write $b_x = b^+$ for all $x \in \mathbb{C}$.

\medskip

\no
Now $g_{x_0}(y)=b^+ \log^+ \lvert y \rvert+ O(1)$ and on the other hand $G_H^+(x,y)=\log^+ \lvert y\rvert +O(1)$ in $V_R^+$ (See (3.4) in \cite{BS}). The difference $g_{x_0}(y)-b^+ G_H^+(y)$ is therefore harmonic at each $y$ for which $(x_0,y)\in \mathbb{C}^2\setminus K_H^+$ with a removable singularity at infinity and vanishes for $(x_0,y)\in K_H^+$. Thus $g_{x_0}(y)=b^+G_H^+(x_0,y)$ for each $y\in \mathbb{C}$.

\medskip
\no
The same argument shows that $G_H^+\circ F-b^+G_H^+ \equiv 0$ in $\Delta(x_0;r_0)\times \mathbb{C}$. The difference is pluriharmonic in $\mathbb{C}^2\setminus K_H^+$ which is connected and it vanishes in $K_H^+$. Therefore, we have $G_H^+\circ F=b^+G_H^+$ in $\mathbb{C}^2$. The same proof applies to $G_H^- \circ F$.
Thus
\begin{equation}
G_H^\pm \circ F=b^\pm G_H^\pm
\end{equation}
everywhere in $\mathbb{C}^2$.

\medskip
\no
It remains to show that $F$ has polynomial growth in $\mathbb{C}^2$. Let $(x,y)\in \mathbb{C}^2$ and suppose that $F(x,y)\in \overline{V_R\cup V_R^+}$. Then it follows from (\ref{L1}), (\ref{L2}) and (\ref{L3}) that
\begin{equation*}
\log \lvert F_2 \rvert-C \leq G_H^+(F(x,y))=b^+ G_H^+(x,y)\leq b^+ (\max \{\log^+\lvert x \rvert, \log^+\lvert y \rvert\}+C).
\end{equation*}
Since $\lvert F_1\rvert\leq \lvert F_2\rvert+R$,
\begin{equation}\label{Pol1}
\lvert F(x,y)\rvert \leq L \max \{{(\lvert x \rvert+1)}^{b^+}, {(\lvert y \rvert+1)}^{b^+}\}
\end{equation}
for some $L>1$.

\medskip
\no
Similarly, if $F(x,y)\in V_R^-$, then
\begin{equation*}
\log \lvert F_1 \rvert-C \leq G_H^-(F(x,y))=b^- G_H^-(x,y)\leq b^- (\max \{\log^+\lvert x \rvert, \log^+\lvert y \rvert\}+A).
\end{equation*}
Therefore, using $\lvert F_1\rvert\leq \lvert F_2\rvert+R$, we obtain
\begin{equation}\label{Pol2}
\lvert F(x,y)\rvert \leq L \max \{{(\lvert x \rvert+1)}^{b^+}, {(\lvert y \rvert+1)}^{b^+}\}
\end{equation}
for some $L>1$.

\medskip
\no
Combining (\ref{Pol1}) and (\ref{Pol2}), it follows that $F$ has polynomial growth throughout $\mathbb{C}^2$ and hence $F$ is itself a polynomial automorphism.

\medskip

\no {\it Remark:} Starting with the assumption that $F$ preserves $K^{\pm}_H$ (which are exactly the zero sets of $G^{\pm}_H)$, (3.1) shows that
\[
F : \{G^{\pm}_H < \alpha\} \rightarrow \{G^{\pm}_H < b^{\pm} \alpha \}
\]
is biholomorphic for every $\alpha > 0$. As mentioned earlier, sub-level sets of $G^{\pm}$ are examples of {\it Short} $\mathbb{C}^2$'s and hence every $F$ that preserves $K^{\pm}_H$ gives rise to a biholomorphism between a pair of {\it Short} $\mathbb{C}^2$'s. This biholomorphism is in fact a polynomial map.

\medskip

There are two possibilities depending on whether $\deg(F) = 1$ or $\deg(F) \ge 2$. We begin with the latter case.

\subsection*{Case 1:} Suppose that $\deg(F) \ge 2$.

\medskip
\no
Since $F$ is a polynomial automorphism, Jung's theorem (see \cite{J}) shows that $F$ can be written as a composition of affine and elementary maps in $\mathbb{C}^2$. Recall that an elementary map is of the form
\[
e(x, y) = (\alpha x + p(y), \beta y + \gamma)
\]
where $\alpha \beta \not= 0$ and $p(y)$ is a polynomial in $y$. Thus four cases arise:

\medskip
\no
{\it{Case (i)}}: Let
\[
F=a_1\circ e_1 \circ a_2 \circ e_2\circ \cdots \circ a_k \circ e_k
\]
for some $k\geq 1$ where the $a_i$'s are non-elementary affine maps and the
$e_i$'s are non-affine elementary maps.
For simplicity, assume that
\[
F=a_1\circ e_1 \circ a_2 \circ e_2.
\]
Let
\[
a_1(x,y)=(\alpha_1 x+ \beta_1 y+ \delta_1,\alpha_2 x+ \beta_2 y+ \delta_2 ).
\]
for $\alpha_2\neq 0$ and consider the maps
\[
a_1^2(x,y)=(\alpha_2 x+ \beta_2 y+ \delta_2, s_2y+r_2)
\]
and
\begin{equation}\label{aff}
a_1^1(x,y)=(bx+cy,y)
\end{equation}
where $b \not= 0, c=\alpha_1/\alpha_2$, $r_2=(\delta_1-c\delta_2)/b$ and $s_2=(\beta_1-c\beta_2)/b$. For any non-zero $b$, it turns out that $a_1= a_1^1\circ \tau\circ a_1^2$ where $\tau(x,y)=(y,x)$. Writing $a_2$ in a similar fashion, we get
\[
F=a_1^1\tau a_1^2 e_1 a_2^1\tau a_2^2 e_2.
\]
Note that both $a_1^2 e_1 a_2^1$ and $a_2^2 e_2$ are elementary maps and we denote them by $E_1$ and $E_2$ respectively. Thus
\[
F=a_1^1 \tau E_1 \tau E_2
\]
where $E_i(x,y)=(m_i x+p_i(y), k_i y+ r_i)$ and the $p_i$'s are polynomials in
$y$ of degree at least $2$ for $i=1,2$.

\medskip
\no
Assume $c \neq 0$ in (\ref{aff}). Choose an unbounded sequence ${(x_n,y_n)}_{n\geq 1}\subseteq K_H^-\cap V_R^+$. Since both $\tau E_1$ and $\tau E_2$ are H\'{e}non maps, it follows that $(x_n',y_n')=\tau E_1 \tau E_2(x_n, y_n)\in V_R^+$ for all $n\geq 1$. Here and in what follows, $R$ may have to be increased finitely many times so that the filtration works for all H\'{e}non maps that will be encountered. Further, the sequence ${(x_n',y_n')}_{n\geq 1}$ is   unbounded. Take $\lvert b\rvert$ sufficiently small so that $\lvert c\rvert > \lvert b\rvert >0$ and thus
\[
\lvert b x_n'+cy_n'\rvert \geq (\lvert c\rvert-\lvert b\rvert)\lvert y_n'\rvert,
\]
which implies
\begin{equation}\label{a}
\lvert x_n''\rvert \geq (\lvert c\rvert-\lvert b\rvert)\lvert y_n''\rvert
\end{equation}
where $(x_n'',y_n'')=a_1^1(x_n',y_n')=(b x_n'+ c y_n', y_n')$. Since $(x_n,y_n)\in K_H^-\cap V_R^+$ and
\[
F(K_H^-)=K_H^-\subseteq V_R \cup V_R^+
\]
it follows that $\Vert (x_n'',y_n'') \Vert \rightarrow \infty$ as $n\rightarrow \infty$ in $V_R^+$. Since $\overline{K_H^-}=K_H^-\cup I^-$ in $\mathbb{P}^2$ (see \cite{DS}), it follows that $(x_n'', y_n'')$ approaches the indeterminacy point $I^-$ as $n\rightarrow \infty$. Consequently,
\begin{equation}\label{b}
\frac{\lvert x_n''\rvert}{\lvert y_n''\rvert}\rightarrow 0
\end{equation}
as $n\rightarrow \infty$. Comparing (3.5) and (\ref{b}), $c = 0$. Hence $F$ is a H\'{e}non map.

\medskip
\no
{\it{Case (ii):}} Let
\[
F=a_1\circ e_1\circ a_2 \circ\cdots\circ e_{k-1} \circ a_k
\]
for some $k\geq 2$.
For simplicity, assume that $F=a_1\circ e_1 \circ a_2$. As in the previous case we can write
\[
F=a_1^1 \tau a_1^2 e_1 a_2^1 \tau a_2^2
\]
where $a_1^1(x,y)=(bx+cy,y)$ and $ \tau a_2^2(x,y)=(s_2 y+r_2, \alpha_2 x+ \beta_2 y+ \delta_2)$.

\medskip
\no
{\it{Claim}:} There exists a sequence ${(x_n,y_n)}_{n \geq 1}\subseteq K_H^+ \cap V_R^-$ with $\lvert x_n \rvert \geq \lvert y_n\rvert\geq n$ for all $n\geq 1$

\medskip
\no
Suppose no such sequence exists. Choose $(x_n,y_n)\in K_H^+\cap V_R^-$ such that  $\lvert y_n \rvert$ is bounded by a fixed constant $M>1$ for all $n\geq 1$  and $\lvert x_n \rvert \rightarrow \infty$ as $n\rightarrow \infty$; of course, $M$ depends on the sequence $(x_n, y_n)$. Without loss of generality, let the H\'{e}non map $H:(x,y)\mapsto (y, p(y)-\delta x)$ with $\delta\neq 0$. Then there exists a subsequence $\{(x_{n_k}, y_{n_k})\}\subset  K_H^+ \cap V_R^-$  such that $\{H(x_{n_k}, y_{n_k})\}\subset  K_H^+ \cap V_R^-$ and consequently,
\[
\lvert y_{n_k} \rvert \geq \lvert p(y_{n_k})-\delta x_{n_k}\rvert  \geq \lvert \delta\rvert\lvert x_{n_k} \rvert-\lvert p(y_{n_k})\rvert.
\]
Since $\lvert y_{n_k}\rvert \leq M$ for all $k\geq 1$, the sequence $\{x_{n_k}\}$ also turns out to be bounded which is a contradiction.

\medskip
\no
Since $\overline{K_H^+}=K_H^+ \cup I^+$ in $\mathbb{P}^2$,
\[
\frac{\vert y_n \vert}{\vert x_n \vert}\rightarrow 0 \text{ as } n\rightarrow \infty
\]
and thus
\begin{equation}\label{eps}
\lvert y_n \rvert \leq \epsilon_n\lvert x_n\rvert
\end{equation}
for all $n\geq 1$ with $\epsilon_n \rightarrow 0$.

\medskip
\no
Now $\tau a_2^2(x_n,y_n)=(s_2 y_n+r_2, \alpha_2 x_n + \beta_2 y_n + \delta_2)$ and
\begin{eqnarray}
\lvert \alpha_2 x_n + \beta_2 y_n +\delta_2\rvert &\geq& \lvert \alpha_2\rvert \lvert x_n\rvert-\lvert\beta_2\rvert \lvert y_n\rvert-\lvert \delta_2\rvert \nonumber\\
&\geq& (\lvert \alpha_2 \rvert - \epsilon_n \lvert \beta_2\rvert)\lvert x_n\rvert-\lvert\delta_2\rvert \nonumber\\
&\geq& \frac{1}{2}\lvert \alpha_2 \rvert  \lvert x_n \rvert-\lvert \delta_2 \rvert  \nonumber\\
&\geq& \frac{1}{2} \lvert \alpha_2\rvert \lvert y_n \rvert-\lvert \delta_2 \rvert \geq \lvert s_2\rvert \lvert y_n\rvert + \lvert r_2\rvert \lvert y_n\rvert-\lvert \delta_2\rvert  \nonumber
\end{eqnarray}
for all  $n\geq n_0$. The last inequality follows since for $\lvert s_2\rvert$ and $\lvert r_2\rvert$ sufficiently small, one can choose ${\lvert\alpha_2\rvert}/{2}\geq\lvert s_2\rvert+ \lvert r_2\rvert$

\medskip
\no
Since $\lvert y_n\rvert \rightarrow \infty$ as $n\rightarrow \infty$, we get
\[
\lvert \alpha_2 x_n + \beta_2 y_n +\delta_2\rvert \geq \lvert s_2 \rvert \lvert y_n\rvert + \lvert r_2\rvert \geq \lvert s_2 y_n + r_2\rvert
\]
and
\[
\lvert \alpha_2 x_n + \beta_2 y_n +\delta_2\rvert \geq R
\]
for sufficiently large $n$.

\medskip
\no
Thus for a sequence ${(x_n,y_n)}_{n\geq 1}\subseteq K_H^+\cap V_R^-$ with $\lvert x_n \rvert \geq \lvert y_n \rvert \geq n$, it turns out that $\tau a_2^2 (x_n,y_n)\in V_R^+$ for sufficiently large $n$. Then
\[
(x_n',y_n')= \tau a_1^2 e_1 a_2^1 \tau a_2^2 (x_n,y_n)\in V_R^+
\]
and consequently
\[
\lvert b x_n'+c y_n'\rvert\leq (\lvert b\rvert+\lvert c\rvert)\lvert y_n'\rvert.
\]
Hence
\begin{equation}\label{contra}
\lvert y_n''\rvert \geq \frac{1}{(\lvert b \rvert+ \lvert c \rvert)} \lvert x_n''\rvert
\end{equation}
where $(x_n'',y_n'')=a_1^1(x_n',y_n')$.

\medskip
\no
Now since $F(K_H^+)=K_H^+$,
\[
(x_n'',y_n'')=F(x_n,y_n)\in K_H^+\cap V_R^-
\]
for sufficiently large $n\geq 1$ and $\Vert (x_n'',y_n'')\Vert \rightarrow \infty$ as $n\rightarrow \infty$. By (\ref{eps}), it follows that
\[
\lvert y_n''\rvert \leq \epsilon_n \lvert x_n''\rvert
\]
where $\epsilon_n\rightarrow 0$ as $n\rightarrow \infty$ which clearly contradicts (\ref{contra}). Thus $F$ cannot be of this form.

\medskip
\no
{\it{Case (iii):}} Let
\[
F=e_1\circ a_1 \circ e_2 \circ a_2 \circ \cdots \circ e_k \circ a_k
\]
for some $k\geq 1$. Note that $F^{-1}$ has a form as in Case 1. Since $F^{-1}$ also keeps $K_H^{\pm}$ invariant, it follows that $F^{-1}$ is a H\'{e}non map.

\medskip
\no
{\it{Case (iv):}} Let
\[
F=e_1\circ a_1 \circ e_2 \circ a_2 \circ \cdots\circ a_{k-1}\circ e_{k-1} \circ e_k
\]
for some $k\geq 1$. For simplicity, let us work with
\[
F=e_1\circ a_1 \circ e_2.
\]
As in the previous cases, we can write
\[
F=e_1 a_1^1\tau a_1^2 e_2
\]
and consequently,
\[
\tau F=\tau e_1 a_1^1\tau a_1^2 e_2.
\]
Note that both $\tau e_1 a_1^1$ and $\tau a_1^2 e_2$ are H\'{e}non maps.

\medskip
\no
Let $(x,y)\in K_H^- \cap V_R^+$, then $\tau F(x,y)\in K_H^- \cap V_R^+$. Thus $F(x,y)\notin K_H^-$ contradicting the hypothesis. Therefore, $F$ cannot be of this form.

\medskip
\no
Finally, the conclusion is that if $F(K_H^{\pm})=K_H^\pm$ for some automorphism $F$ in $\mathbb{C}^2$, then either $F$ or $F^{-1}$ must be a  H\'{e}non map. For the sake of definiteness, we assume that $F$ is a H\'{e}non map.

\medskip

\begin{center}
{\it The Green's functions of $H$ and $F$ coincide:}
\end{center}
\medskip
\no
Now $F(K_H^\pm)=K_H^\pm$ and consequently, $F^n(K_H^+)=K_H^+ \subseteq V_R \cup V_R^-$ for all $n\geq 1$.  Since the forward orbit of any $z\in \mathbb{C}^2\setminus K_F^-$ (See (\ref{escape})) eventually enters in $V_R^+$ and converges to the point at infinity $I^-$, we see that
\[
K_H^+\subseteq K_F^+.
\]
A rigidity theorem of Dinh-Sibony (\cite{DS}) asserts that the $dd^c$-closed currents $\mu_H^+$ and $\mu_F^+$ which are both supported on $K^+_F$ must agree up to a multiplicative constant. Therefore,
\[
J_H^+ = \text{supp} \; \mu_H^+=\text{supp} \; \mu_F^+ = J_F^+.
\]
Hence, the pluricomplex Green functions of these sets (namely, $J_H^+$ and $J_F^+$) are the same, i.e.,
\[
G_H^+=G_F^+
\]
and in the same vein, it can be shown that
\[
G_H^-=G_F^-
\]
in $\mathbb{C}^2$.

\medskip
\no
Now by (\ref{Green1})
\begin{equation*}
G_H^+=\log \lvert \phi_H^+\rvert+\frac{1}{d_H-1}\log \lvert c_H \rvert \text{ and } G_F^+=\log \lvert \phi_F^+\rvert +\frac{1}{d_F-1}\log \lvert c_F\rvert.
\end{equation*}
on $V_R^+$.

\medskip
\no
Since $\phi_H^+$ and $\phi_F^+$ are both asymptotic to $y$ as $\lVert(x,y)\rVert\rightarrow \infty$ in $V_R^+$, it follows that
\begin{equation}\label{const}
\frac{1}{d_H-1}\log \lvert c_H \rvert=\frac{1}{d_F-1}\log \lvert c_F \rvert
\end{equation}
and consequently
\begin{equation*}\label{Phi+}
\phi_H^+ \equiv \phi_F^+
\end{equation*}
in $V_R^+$. Similarly,
\begin{equation*}\label{Phi-}
\phi_H^- \equiv \phi_F^-
\end{equation*}
in $V_R^-$. From now on we shall write $\phi^\pm$ for $\phi_H^\pm \equiv \phi_F^\pm$.

\medskip
\no
By (\ref{const}),
\[
c_H^{d_F}c_F=c_F^{d_H}c_H\delta_+
\]
for some $\delta_+$ with $\lvert \delta_+\rvert=1$.

\medskip
\no
By Proposition \ref{Bot},
\begin{equation*}
\phi^+ \circ F \circ H(x,y)=c_F{(\phi^+ \circ H (x,y))}^{d_F}=c_F c_H^{d_F}{(\phi^+(x,y))}^{d_H d_F}
\end{equation*}
and similarly,
\begin{equation*}
\phi^+ \circ H \circ F(x,y)=c_H{(\phi^+ \circ F (x,y))}^{d_H}=c_H c_F^{d_H}{(\phi^+(x,y))}^{d_H d_F}.
\end{equation*}
Therefore,
\begin{equation*}\label{equality}
\phi^+ ( F \circ H) =\delta_+ \phi^+ ( H \circ F)
\end{equation*}
on $V_R^+$.
Since
\begin{equation*}
\phi^+ \circ F \circ H(x,y)\sim {(F\circ H)}_2(x,y)
\end{equation*}
and
\begin{equation*}
\phi^+ \circ H \circ F(x,y)\sim {(H\circ F)}_2(x,y)
\end{equation*}
as $\lVert(x,y)\rVert\rightarrow \infty$ in $V_R^+$, it follows that for a fix $x_0 \in \mathbb{C}$,
\[
{(F\circ H)}_2(x_0,y)- \delta_+{(H\circ F)}_2(x_0,y)\sim 0 \text{ as } \lvert y \rvert \rightarrow \infty.
\]
The expression on the left is a polynomial in $y$ and hence
\[
{(F\circ H)}_2(x_0,y)=\delta_+{(H\circ F)}_2(x_0,y)
\]
for all $y\in \mathbb{C}$. Therefore,
\begin{equation}\label{rel1}
{(F\circ H)}_2 \equiv \delta_+{(H\circ F)}_2
\end{equation}
in $\mathbb{C}^2$. To study the first components of these maps, we again appeal to Proposition \ref{Bot} which shows that
\begin{equation*}
(\phi^-\circ H(x,y))={\left(\frac{1}{c_H'}\right)}^{\frac{1}{d_H}}{\left(\phi^-(x,y)\right)}^{\frac{1}{d_H}}
\end{equation*}
for $(x,y)\in H^{-1}(V_R^-)$
and
\begin{equation*}
(\phi^-\circ F(x,y))={\left(\frac{1}{c_F'}\right)}^{\frac{1}{d_F}}{\left(\phi^-(x,y)\right)}^{\frac{1}{d_F}}
\end{equation*}
for $(x,y)\in F^{-1}(V_R^-)$, where principal roots are being taken. Thus
\begin{equation*}
\phi^-\circ H \circ F(x,y)={\left( \frac{1}{c_H'}\right)}^{\frac{1}{d_H}}{\left( \frac{1}{c_F'}\right)}^{\frac{1}{d_H d_F}}{\left( \phi^-(x,y)\right)}^{\frac{1}{d_H d_F}}
\end{equation*}
and
\begin{equation*}
\phi^-\circ F \circ H(x,y)={\left( \frac{1}{c_F'}\right)}^{\frac{1}{d_F}}{\left( \frac{1}{c_H'}\right)}^{\frac{1}{d_H d_F}}{\left( \phi^-(x,y)\right)}^{\frac{1}{d_H d_F}}
\end{equation*}
for all $(x,y)\in {(H\circ F)}^{-1}(V_R^-)\cap {(F\circ H)}^{-1}(V_R^-)=U $, say. Note that $U$ is an non-empty open neighborhood of $I^+=[1:0:0]$ in $\mathbb{P}^2$.

\medskip
\no
Thus
\[
\phi^-\circ (H\circ F)=\delta_- (F\circ H)
\]
for all $z\in U$ where
\begin{equation*}
{\left(\frac{1}{c_H'}\right)}^{\frac{1}{d_H}}{\left(\frac{1}{c_F'}\right)}^{\frac{1}{d_H d_F}}=\delta_- {\left(\frac{1}{c_F'}\right)}^{\frac{1}{d_F}}{\left(\frac{1}{c_H'}\right)}^{\frac{1}{d_H d_F}}.
\end{equation*}
Therefore,
\begin{equation*}\label{delta1}
{(c_H')}^{d_F}(c_F'){(\delta_-)}^{d_H d_F} = {(c_F')}^{d_H}(c_H').
\end{equation*}
By (\ref{Green2}), it follows that
\begin{equation*}
\lvert \delta_-\rvert =1
\end{equation*}
as in the previous case.

\medskip
\no
Now for a fixed $c\neq 0$, there exists $\epsilon>0$ sufficiently small such that
\[
U_{\epsilon,c}=\{[{1}/{y}:c:1]:0\neq \lvert y\rvert <\epsilon\}
\]
is contained inside $U\subset V_R^-$ with $I^+$ as a boundary point. Therefore, we can choose a sequence $[x_n:c:1]\in U_{\epsilon,c}$ with $x_n\rightarrow \infty$. Because $(F\circ H)(x_n,c), (H\circ F)(x_n,c)\in V_R^-$ as a consequence of  $(x_n,c)$ being in $U$, we have
$$
{(F\circ H)}_1(x_n,c),{(H\circ F)}_1(x_n,c)\rightarrow \infty
$$
as $n\rightarrow \infty$.

\medskip
\no
Now since $\phi^-(x,y)\sim x$ as $\lVert(x,y)\rVert\rightarrow \infty$,
\begin{equation*}
{(H\circ F)}_1(x_n,c)-\delta_-{(F\circ H)}_1(x_n,c)\rightarrow 0
\end{equation*}
as $n\rightarrow \infty$. The expression on the left is a polynomial in $x$ for each fixed $c$ and thus
\begin{equation*}
{(H\circ F)}_1(x,c)=\delta_-{(F\circ H)}_1(x,c)
\end{equation*}
for all $x\in \mathbb{C}$. Thus using the same argument as in the previous case, we get
\begin{equation}\label{rel2}
{(F\circ H)}_1 \equiv \delta_-{(H\circ F)}_1
\end{equation}
in $\mathbb{C}^2$.

\medskip
\no
Hence using (\ref{rel1}) and (\ref{rel2}), we get
\[
F\circ H= C\circ H\circ F
\]
where $C(x, y) = (\delta_- x,\delta_+ y)$ with $\lvert \delta_\pm\rvert=1$.

\medskip
\no
Similarly, if $F^{-1}$ is a H\'{e}non map, then
\[
F^{-1}\circ H= C\circ H\circ F^{-1}
\]
with $C(x, y) = (\delta_- x,\delta_+ y)$, $\lvert \delta_{\pm} \vert = 1$ as above.

\medskip
\begin{center}
{\it Some iterates of $F$ and $H$ agree upto an affine automorphism:}
\end{center}
 \medskip
 \no
Let $d_F, d_H$ be the degrees of $F, H$ respectively. We use the following fact (See Theorem 1.5, \cite{Bi}): {\it If for $m, n \in \mathbb{N}$, the condition that $d_F^m\leq d_H^n$ implies that $d_F^m$ divides $d_H^n$, then there
exist $n_0,m_0 \in \mathbb{N}$ such that $d_F^{m_0}=d_H^{n_0}$.}

\medskip
\noindent

Suppose that $d_F^m \le d_H^n$ for some $m, n \in \mathbb{N}$. Then
\[
L=H^n \circ F^{-m}
\]
satisfies $L(K_H^\pm)=K_H^\pm$. By the aforementioned arguments, either $L$ or $L^{-1}$ is a H\'{e}non map. We claim that $L^{-1}$ cannot be a H\'{e}non map for if it were, then
\begin{equation}\label{LH}
G^{\pm}_{L^{-1}}=G_H^\pm
\end{equation}
and
\begin{equation*}
G^+_{L^{-1}} \circ L^{-1}= d_{L^{-1}} G^+_{L^{-1}}
\end{equation*}
where $d_{L^{-1}}$ is the degree of $L^{-1}$. Using (\ref{LH}) and the fact that $G^{\pm}_H = G^{\pm}_F$,
\begin{equation*}
G^+_{L^{-1}} \circ L^{-1} = G_H^+ \circ F^m \circ H^{-n}=\frac{d_F^m}{d_H^n}G_H^+
\end{equation*}
in $\mathbb{C}^2$. Thus
\[
d_{L^{-1}} = \frac{d_F^m}{d_H^n} \le
 1
\]
which is a contradiction. Hence $L$ is a H\'{e}non map. Let $d_L$ be the degree of $L$. The same reasoning applied to $G^{\pm}_L$ now shows that
\[
d_L = \frac{d_H^n}{d_F^m}
\]
and hence $d_F^m$ divides $d_H^n$. Thus there exist $m_0, n_0 \in \mathbb{N}$ such that $d_F^{m_0}=d_H^{n_0}$.

\medskip

Let $L_{m_0, n_0} = F^{m_0} \circ H^{-n_0}$. Now two cases arise: the degree of $L_{m_0, n_0}$ is either $1$ or at least $2$. In the former case, there is nothing to prove. In the latter case, either $L_{m_0, n_0}$ or its inverse is a H\'{e}non map. If $L_{m_0, n_0}$ is H\'{e}non, then
\[
\deg(F^{m_0}) = \deg(L_{m_0, n_0}) \cdot \deg(H^{n_0})
\]
shows that the degree of $L_{m_0, n_0}$ is $1$. If the inverse of $L_{m_0, n_0}$ is H\'{e}non, then running the same argument with
\[
H^{n_0} = L^{-1}_{m_0, n_0} \circ F^{m_0}
\]
shows that the degree of $L_{m_0, n_0}$ is $1$ again. In all cases then, $F^{m_0} = \sigma \circ H^{n_0}$ for some affine automorphism $\sigma$.
\medskip

\no \subsection*{Case 2:} Suppose that $\deg(F) = 1$.

\medskip
\no
Let $F(x,y)=(cx+ay+d, bx+ey+f)$. Now $F(K_H^\pm)=K_H^\pm$, thus $(F\circ H)(K_H^\pm)=K_H^\pm$.  Since $\deg(F\circ H)\geq 2$, either $(F\circ H)$ or $H^{-1}\circ F^{-1}$ is H\'enon. A degree count shows that  $H^{-1}\circ F^{-1}$ is never a H\'{e}non map. Thus, $(F\circ H)$ is a H\'{e}non map and consequently $a=0$. Now $F$ extends holomorphically to $\mathbb{P}^2$ and $F([1:0:0])=[1:0:0]$. Thus $b=0$ and consequently, $F(x,y)=(cx+d, ey+f)$.

\medskip

\begin{center}
{\it{There exists a H\'{e}non map $R$ which generates every $F$ that preserves $K^{\pm}_H$}:}
\end{center}
\medskip
\no
Suppose $F(K_H^\pm)=K_H^\pm$ and $F$ is not affine. Without loss of generality, assume $F$ to be a H\'{e}non map. Let $R$ be a H\'{e}non map such that $R(K_H^\pm)=K_H^\pm$ and $\deg(R)\leq \deg(F)$ for all $F$ (non-affine) which preserve $K_H^\pm$. Now since $$
J_H^\pm=J_F^\pm=J_R^\pm,
$$
applying a same arguments used previously, it follows that  there exist $m_0$ and $n_0$ such that
\[
F^{m_0}=\gamma_F R^{n_0}
\]
for some affine automorphism $\gamma_F$. Hence  ${\deg(F)}^{m_0}={\deg(R)}^{n_0}$  and since $\deg(F)\geq \deg(R)$, it follows that $n_0\geq m_0$.  Thus $\deg(R)$ devides $\deg(F)$. Let $\deg(F)=r_F\deg(R) $. consider the automorphism
\[
\sigma_F=F\circ R^{-r_F }.
\]
Clearly, $\sigma_F (K_H^\pm)=K_H^\pm$. Therefore $\sigma_F$ turns out to be affine using the same set of arguments as in the previous case. Thus
\[
F=\sigma_F \circ R^{r_F}.
\]

\medskip
\no
For the converse, first note that $C\circ H$ is also a H\'{e}non map. Since
\[
C\circ H(K_H^+)=K_H^+,
\]
it follows that $K_H^+ \subset K_{(C\circ H)}^+$.  Also,
\[
H(K_{(C\circ H)}^+)=C^{-1}(K_{(C\circ H)}^+)
\]
which in turn gives $K_{(C\circ H)}^+ \subset K_H^+$. Thus
\begin{equation*}
K_{(C\circ H)}^+=K_H^+.
\end{equation*}
\medskip
\no
Since $F\circ H=C\circ H \circ F$, it follows that
\[
(C\circ H)(F(K_H^+))=F(K_H^+).
\]
Thus
\begin{equation}\label{1}
F(K_H^+)\subset K_{(C\circ H)}^+=K_H^+ .
\end{equation}
Indeed, if not, then there exists $F(p)\in F(K_H^+)$ such that $F(p)\notin K_{(C\circ H)}^+$. Therefore,
\[
\Vert {(C\circ H)}^n (F(p)) \Vert \rightarrow \infty
\]
in $V_R^+$. Also, ${(C\circ H)}^n (F(p))=F(p_n)$ with $p_n\in K_H^+$ for all $n\geq 1$ and $\Vert p_n \Vert \rightarrow \infty$ in $V_R^-$. Thus, on the one hand, $F(p_n) \in K_H^+ \cap V_R^+$ for all $n \ge 1$ and on the other, $\Vert F(p_n) \Vert \rightarrow \infty$. This is not possible since $K^+_H \subset V_R \cup V_R^-$.

\medskip
\no
Further, since $F\circ H=C\circ H \circ F$, it follows that
\[
H(F^{-1}(K_H^+))=F^{-1}(K_H^+)
\]
and therefore,
$F^{-1}(K_H^+)\subset K_H^+$, i.e.,
\begin{equation}\label{2}
K_H^+\subset F(K_H^+).
\end{equation}
Combining (\ref{1})and (\ref{2}), we get that $F(K_H^+)=K_H^+$. Similarly, it follows that $F(K_H^-)=K_H^-$. Thus
\[
F(K_H^\pm)=K_H^\pm
\]
and this completes the proof of Theorem 1.1

\section{Two remarks on \short{2}'s}

\no In this section, we will (i) show the existence of a continuum of biholomorphically non--equivalent \short{2}'s and (ii) demonstrate the existence of \short{2}'s that are not Reinhardt and not even biholomorphic to Reinhardt domains.

\medskip

\no We begin with the first theme. Let $H_1$ and $H_2$ be hyperbolic H\'{e}non maps with only one attracting periodic point at the origin. Then ${\rm int}(K_{H_1}^+)$ and ${\rm int}(K_{H_2}^+)$ are Fatou--Bieberbach domains containing the origin. For $c > 0$, the sub-level sets
\[
\Om_1^c=\{z \in \mbb C^2: G_{H_1}^+(z)< c\}
\]
and
\[
\Om_2^c=\{z \in \mbb C^2: G_{H_2}^+(z)<c\}
\]
are \short{2}'s and it is evident that ${\rm int}(K_{H_i}^+) \subset \Om_i^c$ for $i = 1, 2$.

\begin{prop}\label{biholomorphism}
Fix $c > 0$ and suppose that $\phi : \Om_1^c \rightarrow \Om_2^c$ is a biholomorphism. Then $\phi(K_{H_1}^+)=K_{H_2}^+.$
\end{prop}

\begin{proof}
Let $\Om_1= {\rm int}(K_{H_1}^+)$ and $\Om_2= {\rm int}(K_{H_2}^+)$ be the Fatou--Bieberbach domains containing the origin corresponding to the maps $H_1$ and $H_2$. Also,
\begin{align}\label{Julia}
J_1^+=\partial \Om_1=\partial K_{H_1}^+ \text{ and }J_2^+= \partial \Om_2=\partial K_{H_2}^+.
\end{align}

\no Note that $\phi(\Om_1)$ is Fatou--Bieberbach domain in $\Om_2^c$. Since $G_{H_2}^+$ is bounded above (by $c$) on it, it follows that $G_{H_2}^+$ must be constant there. Suppose that $G_{H_2}^+ \equiv \alpha$ on $\phi(\Om_1)$. Now suppose there exists $z \in \phi(\Om_1) \setminus K_{H_2^+}$ such that $G_{H_2}^+(z)=\alpha >0$. Since $K_{H_2}^+$ is closed, there exists a ball contained in
$\phi(\Om_1) \setminus K_{H_2^+}$ on which $G_{H_2}^+ \equiv \alpha$. This is not possible since $G_{H_2}^+$ is a non--constant pluriharmonic function on $\mbb C^2 \setminus K_{H_2}^+$. Hence $\phi(K_{H_1}^+)  \subset K_{H_2}^+.$ A similar argument applied to $\phi^{-1}$ shows that $\phi^{-1}(K_{H_2}^+) \subset K_{H_1}^+$ and hence $\phi(K_{H_1}^+)=K_{H_2}^+$. From \ref{Julia}, we also see that
$\phi(\Om_1)=\Om_2.$
\end{proof}

\begin{thm}\label{continuum}
There exists a continuum of biholomorphically non--equivalent \short{2}'s.
\end{thm}
\begin{proof}
Corollary 6.8 of \cite{Wolf} (see \cite{BS3} also) shows that for $s$ in the open interval $(3,4)$, there exist $a = a(s), c = c(s) \in C_H$ ($C_H$ is the main cardioid of Mandelbrot set) such that the H\'{e}non map
\[
H_{a,c}(z,w)=(aw+z^2+c,z)
\]
is hyperbolic, has only one attracting fixed point, say $p_{(a,c)}$ in $\mbb C^2$ and if $\Om_{a, c}$ is the associated basin of attraction, then $J^+_{a,c}=\partial\Om_{a,c}$ and the Hausdorff dimension $J_{a,c}^+$ is $s$.

\medskip\no
Let $G^+_{a,c}$ denote the Green's function corresponding to the H\'{e}non maps $H_{a,c}.$ Fix $C >0$ and consider the class of \short{2}'s given by
\[ \Om_{a,c}^{C}=\{ z \in \mbb C^2: G_{a,c}^+< C\}.\]
where $a=a(s)$ and $c=c(s)$ for $s \in (3,4).$

\medskip\no
Let $s_0 \not= s_1 \in (3,4)$ and set $a_0=a(s_0),c_0=c(s_0)$ and $a_1=a(s_1),c_1=c(s_1)$ as above.

\medskip\no
\textit{Claim:} $\Om_{a_0,c_0}^{C}$ is not biholomorphically equivalent to $\Om_{a_1,c_1}^{C}.$

\medskip\no
Any biholomorphism $\phi: \Om_{a_0,c_0}^{C} \to \Om_{a_1,c_1}^{C}$ must satisfy $\phi(\Om_{a_0,c_0})=\Om_{a_1,c_1}$ by Proposition \ref{biholomorphism} and therefore
\[
\phi(J_{a_0,c_0}^+)=\phi(\partial\Om_{a_0,c_0})=\partial\Om_{a_1,c_1}=J_{a_1,c_1}^+
\]
since $\phi$ is biholomorphic near $J_{a_0,c_0}^+$. Then it must  be true that
\[
s_1=\text{dim}_H J_{a_1,c_1}^+=\text{dim}_H \phi(J_{a_0,c_0}^+)=\text{dim}_H J_{a_0,c_0}^+=s_0,
\]
which is a contradiction.

\end{proof}
\no
Let $H$ be a H\'{e}non map as above and $G_H^\pm$ be the Green's functions of
$H$. The sub-level sets of $G_H^+$
\[
\Omega_c=\{z\in \mathbb{C}^2: G_H^+ < c\}.
\]
are known to be \short{2}'s by \cite{Fo}.

\begin{prop}
$\Omega_c$ is not Reinhardt for any $c>0$.
\end{prop}

\begin{proof}
Let us assume that $\Omega_c$ is a Reinhardt domain for some $c>0$. Then by Proposition \ref{biholomorphism}, any torus action preserving $\Omega_c$ also preserves $K_H^+$.  Let $(a,b)\in K_H^+$. Then since $K_H^+$ is also Reinhardt, the circle of radius $\lvert a \rvert$  lying in the horizontal line $y=b$ is in $K_H^+$.  Consider the disc $\mathcal{D}_b=\{(x,y):\lvert x\rvert \leq \lvert a \rvert, y=b\}$. Since $G_H^+$ restricted on the horizontal line $y=b$ is a subharmonic function  and $G_H^+$ vanishes on the boundary of $\mathcal{D}_b$, the maximum principle shows that $G_H^+$ vanishes on all of $\mathcal{D}_b$. Thus $\mathcal{D}_b \subset K_H^+$. Now if we can choose $b$ large enough, in particular $\lvert b\rvert >R$, then $\mathcal{D}_b$ and thus $K_H^+$ intersects $V_R^+$ which is clearly a contradiction. That points $(a, b) \in K_H^+$ exist with $\vert b \vert$ large enough is explained in Case $2$ of the subsection in which $F$ is shown to be a H\'{e}non map.
\end{proof}

\no The question that remains is whether $\Om_c$ can be biholomorphic to a Reinhardt domain. We do not know if this is possible, but it turns out that the \short{2}'s arising as sub-level sets of $H_{a, c}$ (as defined above) cannot be biholomorphic to a Reinhardt domain.

\medskip

\no To show this, recall that pseudoconvex Reinhardt domains are logarithmically convex. In fact, more is true -- they are also locally convexifiable near almost every point on their boundary. Here, a boundary point $p$ of a pseudoconvex Reinhardt domain, say $D \subset \mathbb{C}^k$, is said to be locally convexifiable if there is a neighbourhood $U$ of $p$ in $\mbb C^k$ and a biholomorphism $\phi$ from $U$ onto its image such that $\phi(U \cap D) \subset \mbb C^k$ is convex. Note that no assumptions are being made about the smoothness of $\pa D$ near $p$. We include a proof of this for the sake of completeness.

\begin{prop}\label{Green_Reinhardt}
Let $D \subset \mbb C^k$, $k \ge 2$, be a pseudoconvex Reinhardt domain. Then $D$ is locally convexifiable near every boundary point except possibly for a set of measure zero.
\end{prop}
\begin{proof}
For $1 \le j \le k$, let $Z_j=\{(z_1,\ldots,z_k) \in \mbb C^k: z_j=0\}$. Let
\[
Z=\bigcup_{j=1}^k Z_j
\]
be the union of these hyperplanes.

\medskip

\no \textit{Claim:} $D$ is locally convexifiable near every $z_0=(z_1^0,\hdots,z_k^0) \in \partial D \setminus Z$.

\medskip\no
For every $1 \le j \le k$, choose $w_j \in \mbb C$ such that $e^{w_j}=z_j^0$. There exists a sufficiently small $r_j>0$ and a branch of the multi-valued logarithm, say ${\rm Log}_j$ defined on the disc
$\Delta(w_j, r_j) \subset \mbb C$ such that
\begin{align}\label{log}
{\rm Log}_j\Big({\rm exp}\big(\Delta(w_j, r_j)\big)\Big)=\Delta(w_j,r_j)\text{ and } {\rm Log}_j\big({\rm exp}(z)\big)=z
\end{align}
for every $z \in \Delta(w_j, r_j)$. Let $R=(r_1,\hdots,r_k)$ and $\Delta^k(w;R)$ the polydisc at the point $w=(w_1,\hdots,w_k)$. Let
\[
\psi(z_1,\hdots,z_k)=(\exp z_1,\hdots,\exp z_k)
\]
and note that $V=\psi(\Delta^k(w;R))$ is a neighbourhood of $z_0$ due to the local injectivity of $\psi$. Let $W=V \cap D$ and define $\phi$ on $V$ as
\[
\phi(z_1,\hdots,z_k)=\big({\rm Log}_1(z_1),\hdots,{\rm Log}_k(z_k)\big).
\]
From (\ref{log}), $\phi$ is injective on $V$. To prove the claim, it is enough to show that $\phi(W)$ is a convex open set in $\mbb C^k.$

\medskip\no
Let $p,q \in \phi(W)$ where $p=(p_1,\hdots,p_k)$ and $q=(q_1,\hdots,q_k)$. Then there exist $p',q' \in W$ such that $p=\phi(p'),q=\phi(q')$. Let
\[
s_1=(\log|p'_1|,\hdots,\log|p'_k|) \text{ and }s_2=(\log|q'_1|,\hdots,\log|q'_k|).
\]
But $|p_j'|=\exp(\Re p_j)$ and $|q_j'|=\exp(\Re q_j)$ for every $1 \le j \le k$ and hence
\[
s_1=(\Re p_1,\hdots, \Re p_k) \text{ and }s_2=(\Re q_1,\hdots, \Re q_k).
\]
For $\la \in [0,1]$ let
\[
s_\la=\la s_1+(1-\la)s_2 \text { and }p_\la=\la p+(1-\la)q.
\]
Then $\Re p_\la=s_\la$ and $s_\la \in \log D$ where
\[
\log D =\{(\log |z_1|,\hdots,\log|z_k|):(z_1,\hdots,z_k) \in D\}.
\]
This means there exists $z_\la=(z_1^\la,\hdots,z_k^\la) \in \Om$ such that
$s^\la_j=\log|z_j^\la|$ for $1 \le j \le k$ where
$s_\la=(s_1^\la,\hdots,s_k^\la)$. Since $p,q \in \phi(W) \subset \Delta^k(w;R)$ and $\Delta^k(w;R)$ is convex, $p_\la \in \Delta^k(w;R)$ for all $\lambda \in [0,1]$. Let $\psi(p_\la) =(\tilde{p}_1^\la,\hdots\tilde{p}_k^\la)\in V$. Now
\begin{align}\label{la point}
\log|\tilde{p}_j^\la|=\log \exp (\Re p_j^\la)=\Re p_j^\la=s_j^\la.
\end{align}
As $D$ is Reinhardt, $s_\la \in \log D$ implies that
\[
\{(z_1,\hdots,z_k) \in \mbb C^k: \log|z_j|=s_j^\la \text{ for every } 1 \le j \le k\} \subset D.
\]
By (\ref{la point}), $\psi(p_\la) \in D$ and hence $p_\la \in \phi(W)$ for all $\la \in [0,1]$. This proves the claim.
\end{proof}

\begin{thm}
There exist \short{2}'s that are not biholomorphically equivalent to a Reinhardt domain.
\end{thm}
\begin{proof}
Let $\Om_{a(s),c(s)}^{C_0}$ be the \short{2}'s considered in the proof of Theorem \ref{continuum}. Recall that ${\rm dim}_H(J^+_{a(s),c(s)})=s$, $s \in (3,4).$

\medskip\no
For a fixed $s \in (3,4)$ and $C>0$, suppose that there exists a biholomorphism
\[
\phi : \Om_{a(s),c(s)}^{C} \ra G
\]
where $G \subset \mbb C^2$ is a Reinhardt domain. Let
\[
D=\phi\big({\rm int}(K^+_{a(s),c(s)})\big) \subset G.
\]
Now $\phi$ induces a $\mbb T^2$-action on $\Om_{a(s),c(s)}^{C}$ which must preserve ${\rm int}(K^+_{a(s),c(s)})$ by Proposition \ref{biholomorphism}. It follows that $D$ must itself be a Reinhardt domain. $D$ is also pseudoconvex being a Fatou-Bieberbach domain. From Proposition \ref{Green_Reinhardt}, $D$ is  locally convexifiable near almost every boundary point. This implies that the Hausdorff dimension of $\pa D$ is $3$ almost everywhere. But since
\[
J^+_{a(s),c(s)}=\partial K^+_{a(s),c(s)}=\phi^{-1}(\partial D)
\]
and $\phi^{-1}$ is biholomorphic near $\pa D$, the same is true for
$J^+_{a(s),c(s)}$. This cannot be true since its Hausdorff dimension is $s > 3$ at every point.
\end{proof}

\section{Appendix}
\no
Let $P$ be a polynomial of degree $d\geq 2$ in the complex plane and let $K_P$ be the filled Julia set of $P$, i.e.,
\[
K_P=\{z\in \mathbb{C}:  \text{the sequence} \; \{P^n(z)\}\text{ is bounded}\}.
\]
\begin{thm}
Let $Q$ be an entire function satisfying $Q(K_P) = K_P$. Then $Q$ is a polynomial whose Julia set coincides with that of $P$.
\end{thm}

To prove this, observe that a similar set of arguments as in the first part of the proof of Theorem {\ref{rigidity}} shows that if $g$ is the Green's function of $\mbb C \setminus K_P$ with pole at infinity, then $g \circ Q$ is a constant multiple of $g$. Combining this with the fact that $g$ behaves like
$\log \vert z \vert$ near $z = \infty$, it follows that $Q$ has polynomial growth and hence must be a polynomial.

\medskip

Thus, it suffices to consider the following situation: $P$ and $Q$ are polynomials such that
\[
Q(K_P)=K_P.
\]

\medskip
\no
Let $z\in \partial K_P=J_P$, then  there exist sequences $\{z_n\}\in {\rm{int}}(K_P)$ and $\{w_n\}\in I_P$ such that $z_n, w_n\rightarrow z$ and thus $Q(z_n), Q(w_n)\rightarrow Q(z)$ as $n\rightarrow \infty$. Now note that $Q(I_P)=I_P$ where $I_P$ is the unbounded component of the Fatou set $F_P$ of $P$. Further, by open mapping theorem $Q({\rm{int}}(K_P))\subset {\rm{int}}(K_P)$. Thus $Q(z)\in J_P$ and we get that
\[
Q(J_P)\subset J_P.
\]
Let $z\in J_P$, then there exists $y\in K_P$ such that $Q(y)=z$. Again by open mapping theorem since $Q({\rm{int}}(K_P))\subset {\rm{int}}(K_P)$, we have
\[
J_P\subset Q(J_P).
\]
Therefore,
\[
Q(J_P)=J_P
\]
and since $J_Q$ is the smallest closed invariant (under $Q$) subset in $\mathbb{C}$, we get
\begin{equation}\label{J}
J_Q\subset J_P.
\end{equation}
Now we  have started with the assumption that $Q(K_P)=K_P$. Thus $K_P\subset K_Q$. Let there exists some $z \in K_Q$ such that $z \notin K_P$, then by (\ref{J}), there exists a small ball $B_z$ with center at $z$ such that $B_z\subset  \rm{int}(K_Q)$ and $B_z\cap K_P=\emptyset$. This contradicts (\ref{J}). Thus
$K_P=K_Q$ and consequently, $J_P=J_Q$. Conversely, if $J_P=J_Q$, then $K_P= K_Q$.

\medskip
\no
To conclude, the assumption $Q(K_P)=K_P$ implies that $J_P=J_Q$. In fact, these are equivalent if both $P$ and $Q$ are polynomials.  As indicated in the introduction, $J_P$ and $J_Q$ coincide precisely when
\[
P\circ Q=\sigma \circ Q \circ P
\]
where $\sigma:z\mapsto az+b$ with $\lvert a\rvert=1$ and $\sigma (J_P)=J_P$.

config: arxiv


TITLE: Difficulty proving formula containing the adjugate and determinant of a matrix
QUESTION [0 upvotes]: This is what I need to prove:
You have an invertible matrix $A \in M_3(\Bbb R^3)$.
Prove that $\operatorname{adj}(\operatorname{adj}(A))=\det{(A)}^{n-2}A$
The proof goes as follows:
$\operatorname{adj}(\operatorname{adj}(A))=\operatorname{adj}(\det{(A)}A^{-1})$
$= \det{(\det{(A)}A^{-1})}(\det{(A)}A^{-1})^{-1}$
$= \det{(A)}^n\det{(A^{-1})}\det{(A)}^{-1}A$
I'm having trouble going from the second to the last step (The proof isn't complete, but I understand everything except that part).
I have already proven that:
$\det{(A^{-1})} = \frac{1}{\det{A}}$
and that
$A^{-1}=(\det{A})^{-1}\operatorname{adj}(A)$
So I think I need to use one of those but I can't seem to find anything useful.
Thanks in advance!

REPLY [1 votes]: Everything boils down to realize that for any scalar $\;c\in\Bbb R\;$ ,we have $\;\det(cA)=c^n\det A\;$ , where $\;n=\;$ the number of rows in the square matrix $\;A\;$ , and thus
$$\det\left(\det(A)\cdot A^{-1}\right)\left(\det(A)\cdot A^{-1}\right)^{-1}=\frac{(\det A)^n\det\left(A^{-1}\right)}{\det(A)}A$$
and now reach the end.


text:
\begin{document}

\vskip 1cm

\title[Generalized maximum principle]{Generalized maximum principle in optimal control}
\author{E.~R.~Avakov, G.~G.~Magaril-Il'yaev}

\address{Institute of Control Sciences of the Russian Academy of Sciences}
\address{Moscow State University}

\maketitle

For an optimal control problem, the concept of a~strong local infimum is introduce, for
which necessary conditions consisting of some family of ``maximum principles'' are
formulated. If a~function delivers a~strong local minimum in this problem (and therefore,
a~strong local infimum), then this family contains the classical Pontryagin maximum
principle  (see \cite{P},~\cite{IT}). As a~corollary, we derive generalized necessary
conditions for a~strong local minimum for a~problem of the calculus of variations.
Examples  are given to show that the necessary conditions obtained in the present paper
generalize and strengthen classical results.

It is worth noting that some ideas for necessary conditions of this kind are contained in
the book by R.~V.~Gamkrelidze \cite{Gam}, where the Pontryagin maximum principle is
derived as a~corollary to the maximum principle for a~more general problem stated in
terms of generalized controls. We also employ this idea, but from somewhat different
positions and in a~less general setting, when the generalized controls are  sliding
regime control. On the one hand, this constraint is quite sufficient for our purposes,
and on the other hand, it leads to a~simpler set of tools.

\vskip15pt

Let $[t_0,t_1]$ be a line interval, $U$~be a~nonempty subset of~$\mathbb R^r$,
$\varphi\colon \mathbb R\times\mathbb R^n\times \mathbb R^r\to \mathbb R^n$~be a~mapping of
variables $t\in \mathbb R$, $x\in\mathbb R^n$ and $u\in U$, and $f\colon\mathbb
R^n\times\mathbb R^n\to \mathbb R^{m_1}$, $g\colon\mathbb R^n\times\mathbb R^n\to
\mathbb R^{m_2}$ be mappings of variables $\zeta_i\in\mathbb R^n$, $i=1,2$.

Consider the following optimal control problem
\begin{multline}\label{poc}
f_0(x(t_0),x(t_1))\to\min,\quad \dot x =\varphi(t,x,u(t)),\quad u(t)\in U,\\
f(x(t_0),x(t_1))\le0, \quad g(x(t_0),x(t_1))=0,
\end{multline}
where $x\cd\in AC([t_0,t_1],\mathbb R^n)$ (is an absolutely continuous vector function on
$[t_0,t_1]$) and $u\cd\in L_\infty([t_0,t_1],\mathbb R^r)$.

In what follows we assume that the {\it mapping~$\varphi$ is continuous together with its
derivative with respect to~$x$ on~$\mathbb R\times\mathbb R^n\times \mathbb R^r$, and the mappings $f$
and~$g$ are continuously differentiable on  $\mathbb R^n\times\mathbb R^n$}.

A~function $x\cd\in AC([t_0,t_1],\mathbb R^n)$ is called {\it admissible} in problem \eqref{poc} if  $f(x(t_0),x(t_1))\le0$, $g(x(t_0),x(t_1))=0$
and there exists $u\cd\in
L_\infty([t_0,t_1],\mathbb R^r)$ such that $\dot x(t)=\varphi(t,x(t),u(t))$ and $u(t)\in U$ for almost all $t\in[t_0,t_1]$,

\begin{definition} \rm
We say that a function $\wx\cd\in C([t_0,t_1],\mathbb R^n)$ delivers
a~{\it strong local infimum} in problem~\eqref{poc} if  $f(\wx(t_0),\wx(t_1))\le0$,
$g(\wx(t_0),\wx(t_1))=0$, there exist a~neighbourhood~$V$ of the point~$\wx\cd$ and a~sequence
$\{x_N\cd\}$, $N\in\mathbb N$, of admissible functions in~\eqref{poc} such that
$f_0(x(t_0),x(t_1))\ge f_0(\wx(t_0),\wx(t_1))$ for any admissible function $x\cd\in V$ and
$x_N\cd$ converges uniformly to $\wx\cd$ as $N\to\infty$.
\end{definition}

Clearly, if  a~pair $(\wx\cd,\wu\cd)$ delivers a~{\it strong minimum} in problem \eqref{poc}, then $\wx\cd$~is
a~strong local infimum in this problem. On the other hand, if
a~function $\wx\cd$ delivers a~strong local infimum in~\eqref{poc}, $\wx\cd$ is admissible, and
$\wu\cd$ is the corresponding control, then the pair $(\wx\cd,\wu\cd)$ delivers a~strong minimum
in this problem.

Given arbitrary $k\in\mathbb N$ and tuples $\ov u\cd=(u_1\cd,\ldots,u_k\cd)\in
(L_\infty([t_0,t_1]),\mathbb R^r))^k$ and $\ov\alpha\cd=(\alpha_1\cd,\ldots,\alpha_k\cd)\in
(L_\infty([t_0,t_1]))^k$, where $\alpha_i(t)\ge0$, $\alpha_i(t)\ne0$, $i=1,\ldots,k$, and
$\sum_{i=1}^k\alpha_i(t)=1$ for almost all $t\in[t_0,t_1]$,
we associate with the control system specifying the constraints in problem \eqref{poc}
the following {\it extended} (relaxation) control system
\begin{multline}\label{rpoc}
\dot x =\sum_{i=1}^k\alpha_i(t)\varphi(t,x,u_i(t)),\quad u_i(t)\in U,\quad i=1,\ldots,k,\\
f(x(t_0),x(t_1))\le0, \quad g(x(t_0),x(t_1))=0.
\end{multline}

A triple $(x\cd,\ov u\cd,\ov \alpha\cd)$ ($x\cd\in AC([t_0,t_1],\mathbb R^n))$ is called {\it
admissible} for system~\eqref{rpoc} if  it satisfies all constraints in~\eqref{rpoc}.

Let us introduce some notation. We let $\la
\lambda,x\ra=\sum_{i=i}^n\lambda_ix_i$
denote a~linear functional $\lambda=(\lambda_1,\ldots,\lambda_n)\in(\mathbb R^n)^*$
evaluated at a~point
$x=(x_1,\ldots,x_n)^T\in\mathbb R^n$ ($T$~is the transpose). By $(\mathbb R^n)^*_+$ we denote the set
of positive functionals on  $\mathbb R^n$. The adjoint operator to a~linear operator $\Lambda\colon \mathbb R^n\to\mathbb
R^m$ is denoted by~$\Lambda^*$.

Given a~fixed function $\wx\cd$, the partial derivatives of mappings  $f$ and $g$
with respect to $\zeta_1$ and~$\zeta_2$ at a~point $(\wx(t_0),\wx(t_1))$ will be
briefly denoted by $\widehat f_{\zeta_i}$ and $\widehat g_{\zeta_i}$, $i=1,2$, respectively.

\begin{theorem}[generalized maximum principle]\label{T1}
If a function $\wx\cd\in AC([t_0,t_1],\mathbb R^n)$ delivers a~strong local infimum in problem \eqref{poc}, then
for any $k\in\mathbb N$, $\wou\cd=(\wu_1\cd,\ldots,\wu_k\cd)$ and
$\woa\cd=(\widehat\alpha_1\cd,\ldots,\widehat\alpha_k\cd)$ such that the triple
$(\wx\cd,\wou\cd,\woa\cd)$ is admissible for the control system \eqref{rpoc}, there exist
a~nonzero tuple  $(\lambda_0,\lambda_f,\lambda_g)\in \mathbb R_+\times(\mathbb
R^{m_1})^*_+\times(\mathbb R^{m_2})^*$ and a~vector function $p\cd\in AC([t_0,t_1],(\mathbb
R^n)^*)$ such that the following conditions hold:
\begin{itemize}
\item[$1)$] the stationarity condition  with respect to $x\cd$
$$
\dot p(t) =-p(t)\sum_{i=1}^k\widehat\alpha_i(t)\varphi_x(t,\wx(t),\wu_i(t)),
$$
\item[2)] the transversality condition
$$
p(t_0)=\lambda_0{\widehat {f}}_{0\zeta_1}+{\widehat {f}_{\zeta_1}}^*\lambda_f+{\widehat
{g}_{\zeta_1}}^*\lambda_g,\quad p(t_1)=-\lambda_0{\widehat {f}}_{0\zeta_2}-{\widehat
{f}_{\zeta_2}}^*\lambda_f-{\widehat {g}_{\zeta_2}}^*\lambda_g,
$$
\item[3)] the complementary slackness condition
$$
\la \lambda_f, f(\wx(t_0),\wx(t_1))\ra=0,
$$
\item[4)] the maximum condition for almost all $t\in[t_0,t_1]$
\begin{multline*}
\max_{u\in U}\widehat\alpha_i(t)\la p(t),\varphi(t,\wx(t),u)\ra=
\widehat\alpha_i(t)\la p(t),\varphi(t,\wx(t),\wu_i(t))\ra,\\
i=1,\ldots,k,
\end{multline*}
\begin{equation*}
\max_{u\in U}\la p(t),\varphi(t,\wx(t),u)\ra=\la p(t),\dot {\wx}(t)\ra.
\end{equation*}
\end{itemize}

Moreover, if for some $k\in\mathbb N$ and a~triple $(\wx\cd, \ov u\cd,\ov\alpha\cd)$
admissible for the control system~\eqref{rpoc}, conditions {\rm 1)--4)} hold  only when
$\lambda_0\ne0$, then there exists a~sequence of functions $x_N\cd$, $N\in\mathbb N$
admissible in problem~\eqref{poc} such that $x_N\cd\to\wx\cd$ as $N\to\infty$ uniformly
on $[t_0,t_1]$.
\end{theorem}

The first assertion of the theorem constitutes a~family of relations
(parameterized by all possible finite tuples $(\wou\cd,\woa\cd)$ such that
the triple $(\wx\cd,\wou\cd,\woa\cd)$ is admissible for the control system \eqref{rpoc}), of which each
has the form of a~maximum principle. Furthermore, if  $(\wx\cd,\wu\cd)$ is a~strong
minimum in problem \eqref{poc}, then this family contains (with $k=1$, $\wu_1\cd=\wu\cd$ and
$\widehat\alpha_1\cd=1$) the classical Pontryagin maximum principle.

\vskip10pt

As a corollary to Theorem \ref{T1} we obtain generalized conditions for strong local minimum
in the simplest problem of the classical calculus of variations.

Let a~function $L\colon \mathbb R\times\mathbb R^n\times\mathbb R^n\to\mathbb R$
of variables  $t\in\mathbb R$, $x\in\mathbb R^n$ and $\dot x\in\mathbb R^n$ be continuous together
with its partial derivatives with respect to~$x$, $\dot x$, and $x_i\in\mathbb R^n$, $i=0,1$.
Consider the problem
\begin{equation}\label{var1}
\int_{t_0}^{t_1}L(t,x(t),\dot x(t))\,dt\to\min,\quad x(t_0)=x_0,\quad x(t_1)=x_1.
\end{equation}

Given a fixed $\wx\cd$, we write for brevity $\widehat L(t)=L(t,\wx(t),\dot\wx(t))$,
and similarly for the derivatives of~$L$ with respect to $x$ and~$\dot x$.

We let  $\mathcal A^k$ denote the set of tuples
$\ov\alpha\cd=(\alpha_1\cd,\ldots,\alpha_k\cd)$ introduced before the definition of the
control system \eqref{rpoc}.

\begin{theorem}\label{T2}
If a~function $\wx\cd\in AC([t_0,t_1],\mathbb R^n)$ delivers a~strong local minimum in
problem \eqref{var1}, then $\widehat L_{\dot x}\cd\in AC([t_0,t_1],\mathbb R^n)$ and for
any $k\in\mathbb N$, $(\wu_1\cd,\ldots,\wu_k\cd)\in (L_\infty([t_0,t_1],\mathbb R^n))^k$
and $(\widehat\alpha_1\cd,\ldots,\widehat\alpha_k\cd)\in\mathcal A^k$ such that, for
almost all $t\in[t_0,t_1]$
\begin{equation}\label{usl}
\dot\wx(t)=\sum_{i=1}^k\widehat\alpha_i(t)\wu_i(t)\,\,\,\,\text{and}\,\,\,\,\, \widehat
L(t)=\sum_{i=1}^k \widehat\alpha_i(t)L(t,\wx(t),\wu_i(t)),
\end{equation}
the following conditions are satisfied:
\begin{itemize}
\item[$1)$] the generalized Euler equation
$$
-\frac{d}{dt}\widehat L_{\dot x}(t)+\sum_{i=1}^k
\widehat\alpha_i(t)L_x(t,\wx(t),\wu_i(t))=0,
$$
\item[2)] the generalized Weierstrass condition
\begin{multline*}
L(t,\wx(t),u)-L(t,\wx(t),\wu_i(t))-\la L_{\dot
x}(t,\wx(t),\wu_i(t)),u-\wu_i(t)\ra\ge0,\\i=1,\ldots,k,
\end{multline*}
for all $u\in\mathbb R^n$.
\item[3)] If $L$ is twice differentiable with respect to~$\dot x$, then the
generalized Legendre condition holds
$$
\widehat L_{\dot x\dot x}(t,\wx(t),\wu_i(t))\ge0,\quad i=1,\ldots,k.
$$
\end{itemize}
\end{theorem}

It is clear that for  $k=1$, $\wu_1\cd=\dot\wx\cd$ and $\widehat\alpha_1\cd=1$ conditions
\eqref{usl} hold trivially and conditions 1)--3) pass into the classical strong minimum
conditions in problem~\eqref{var1}.

\vskip10pt

The first example illustrates how a~strong local infimum can be found with the help of Theorem~\ref{T1}.

\subsection*{Example 1}

Let $f\colon [0,1]\to\mathbb R$, $g\colon \mathbb R\to\mathbb R$, $m$  be an even number.
Consider the optimal control problem
\begin{multline}\label{exam1}
J(x\cd,u\cd)=\int_0^1((x(t)-f(t))^m+g(u(t)))\,dt\to\min,\quad \dot x=u,\\ |u(t)|
\ge1,\quad x(0)=0,\quad x(1)=f(1).
\end{multline}

We shall assume that the function $f$ is absolutely continuous, $f(0)=0$, and $|\dot f(t)|\le1$,
$|\dot f(t)|\ne1$ for almost all $t\in[0,1]$.
We also assume that~$g$ is continuous on $\mathbb R$, $g(-1)=g(1)$, and $g(u)>g(1)$ for  $|u|>1$.

Our aim is to evaluate the infimum of the functional $J(x\cd,u\cd)$ and find a~sequence
of admissible pairs $(x_N\cd,u_N\cd)$, $N\in\mathbb N$ in problem~\eqref{exam1} on which
the sequence $J(x_N\cd,u_N\cd)$ converges to this infimum.

We transform problem \eqref{exam1} to the equivalent Mayer problem
\begin{multline}\label{exam2}
x_2(1)-x_2(0)\to\min,\quad \dot x_1=u,\quad \dot x_2=(x_1-f(t))^m+g(u),\\
|u(t)|\ge1, \quad x_1(0)=0,\quad x_1(1)=f(1).
\end{multline}

Using Theorem~\ref{T1}, we shall try to find a~function which delivers a~strong local
infimum in this problem. If such a~function $\widehat{\ov x}\cd=(\wx_1\cd,\wx_2\cd)$
is found, then by definition $x_2(1)-x_2(0)\ge\wx_2(1)-\wx_2(0)$ for all admissible
functions $\ov x\cd$ from some neighbourhood of~$\widehat{\ov x}\cd$ and there exists
a~sequence of admissible for~\eqref{exam2} functions  $\ov x_N\cd=(x_{1N}\cd,
x_{2N}\cd)$ that converges uniformly to $\widehat{\ov x}\cd$ as $N\to\infty$. It follows that
$J(x\cd,u\cd)\ge \wx_2(1)-\wx_2(0)$ for all admissible pairs in problem \eqref{exam1} in which $x\cd$ lies
in some neighbourhood of $\wx_1\cd$. Setting
$x_{1N}\cd=x_N\cd$, we find a~sequence of
pairs $(x_N\cd,u_N\cd)$ admissible in problem \eqref{exam1},
where $u_N\cd=\dot x_N\cd$, such that $J(x_N\cd,u_N\cd)\to \wx_2(1)-\wx_2(0)$ as $N\to\infty$.

We apply Theorem~\ref{T1} with $k=2$. By this theorem if  tuples $(\widehat
\alpha_1\cd,\widehat \alpha_2\cd)$ and $(\wu_1\cd,\wu_2\cd)$ are such that
\begin{equation}\label{exam3}
\begin{aligned}
\dot{\wx}_1(t)&=\widehat \alpha_1(t)\wu_1(t)+\widehat\alpha_2(t)\wu_2(t),\\
\dot{\wx}_2(t)&=\widehat \alpha_1(t)g(\wu_1(t))+\widehat\alpha_2(t)g(\wu_2(t))+
(\wx_1(t)-f(t))^m
\end{aligned}
\end{equation}
and $\wx_1(0)=0$, $\wx_1(1)=f(1)$, then there exist a~nonzero set of Lagrange multipliers
$(\lambda_0,\lambda_1,\lambda_2)$, where $\lambda_0\ge0$, and an absolutely continuous function
$p\cd$, such that
\begin{multline}\label{exam4}
\dot p_1=- p_2m(\wx(t)-f(t))^{m-1},\quad \dot p_2=0,\quad p_1(0)=\lambda_1,\quad p_1(1)=-\lambda_2,\\
p_2(0)=p_2(1)=-\lambda_0
\end{multline}
and
\begin{equation}\label{exam5}
\max_{u\in
U}(p_1(t)u+p_2(t)((\wx(t)-f(t))^{m}+g(u)))\!=p_1(t)\dot\wx_1(t)+p_2(t)\dot\wx_2(t)
\end{equation}
for almost all $t\in[t_0,t_1]$.

By examining relations \eqref{exam3}, \eqref{exam4} and \eqref{exam5} one can show that they define uniquely,
up to an additive constant, the function
$(\wx_1\cd,\wx_2\cd)$  (where $\wx_1(t)=f(t)$, $\wx_2(t)=g(1)t+c$ for any
$c\in\mathbb R$ and all $t\in[t_0,t_1]$) and (assuming  $\lambda_0=1$) the Lagrange multipliers
$\lambda=(1,0,0)$ and $p=(0,1)$. Here it suffices to put $\wu_1(t)\equiv 1$,
$\wu_2(t)\equiv-1$, which gives $\widehat\alpha_1(t)=(1+\dot f(t))/2$,
$\widehat\alpha_2(t)=(1-\dot f(t))/2$ for almost all $t\in[t_0,t_1]$.

So, $t\mapsto (f(t),\ g(1)t+c)$ is the only trajectory  suspected for a~strong local infimum in problem~\eqref{exam2} for each
$c\in\mathbb R$. Note that this trajectory is not admissible for this problem. We claim that it delivers a~strong local infimum.

Indeed, $\lambda_0\ne0$, for otherwise relations \eqref{exam4} and \eqref{exam5}
would hold only for $\lambda_1=\lambda_2=0$. Hence, by the second assertion
of the theorem there exists a~sequence of functions
$(x_{1N}\cd, \,x_{2N}\cd)$ which are admissible for problem \eqref{exam2}  and uniformly
converge to $(\wx_1\cd,\, \wx_2\cd)$ as $N\to\infty$.

Additionally, it is clear that  $x_2(1)-x_1(0)\ge g(1)=\wx_2(1)-\wx_2(0)$  for any admissible function $x\cd=(x_{1}\cd, \,x_{2}\cd)$,
and hence by definition $\wx\cd$~delivers the global infimum in problem~\eqref{exam2}.
Hence, the infimum of the functional
$J(x\cd,u\cd)$ is $g(1)$, and by the above there exists a~sequence
of admissible pairs in problem \eqref{exam1} on which this functional converges to~$g(1)$.

Let us construct directly a~sequence of admissible pairs $(x_N\cd\, u_N\cd)$ in problem \eqref{exam1} such that
$J(x_N\cd,u_N\cd)\to g(1)$ as $N\to\infty$. Let $N\in\mathbb
N$. We split the interval $[0,1]$ into  $N$~intervals: $[s/N,\,(s+1)/N]$, $s=0,\ldots,N-1$.
Define $b_N(s)=f(s/N)-(s/N)$ and $c_N(s)=f((s+1)/N)+(s+1)/N$, $s=0,\ldots,N-1$. It is
easily checked that $((c_N(s)-b_N(s))/2)\in [s/N,(s+1)/N]$, $s=0,\ldots,N-1$.

Consider the sequence $x_{N}\cd$ defined by
\begin{equation*}
x_{N}(t)=
\begin{cases} t+b_N(s),\qquad t\in[s/N, \,(c_N(s)-b_N(s))/2],\\[10pt]
-t+c_N(s),\quad \ t\in[(c_N(s)-b_N(s))/2, \,(s+1)/N],
\end{cases}
\end{equation*}
$s=0,\ldots,N-1$.
Each $x_{N}(t)$ is a broken line (with slopes $\pm1$  of the segments and which interpolates $f\cd$ at the points
$s/N$, $s=0,\ldots,N$) and $x_{N}(t)$ uniformly converges to~$f\cd$. The sequence of pairs
$(x_{N}\cd,\,u_N\cd)$, where $u_N\cd=\dot x_{N}\cd$, is admissible in problem \eqref{exam1}, because
$|u_{N}(t)|=1$ for almost all $t\in[t_0,t_1]$, and since  $J(x_{N}\cd, u_{N}\cd)\to g(1)$
as $N\to\infty$, which is clear.

\vskip10pt

The following example shows that even in the classical setting the above assertions are
capable of delivering additional information about the strong minimum compared to with
known necessary conditions.

\subsection*{Example 2}

Let $L\colon \mathbb R\times\mathbb R\to\mathbb R$. Consider the following
variational calculus problem
\begin{equation}\label{z1}
\int_{0}^{1}L(x,\dot x)\,dt\to\min,\quad x(0)=x(1)=0.
\end{equation}
Assume that the function $L$ is continuously differentiable, $L_x(0,0)=L_{\dot x}(0,0)=0$ and
$L(0,\dot x)=0$ for any~$\dot x$ (a~typical situation when $L(x,\dot x)=xf(\dot x)$,
where the function~$f$ is continuously differentiable and $f(0)=0$).

It is an elementary matter to verify that the function $\wx\cd=0$ satisfies the Pontryagin maximum principle
($\dot x=u$, $u\in U=\mathbb R$). The next result is proved using Theorem~\ref{T2}.

\begin{proposition}
If a function $\wx\cd=0$ delivers a~strong local minimum  in problem \eqref{z1}, then the function $\dot x\to L_x(0,\dot x)$ is linear.
\end{proposition}

\begin{proof}
We apply the theorem with $k=2$. It is clear that conditions \eqref{usl} are satisfied for any $u_1<0$, $u_2>0$ and
$\alpha_1=u_2/(u_2-u_1)$, $\alpha_2=-u_1/(u_2-u_1)$. Hence, the Euler equation holds, which in this case reads as
\begin{equation}\label{eqe}
u_2L_x(0,u_1)=u_1L_x(0,u_2).
\end{equation}
Setting here $u_1=-1$, $u_2=1$, we find
\begin{equation}\label{z6}
L_{x}(0,-1)=-L_{x}(0,1).
\end{equation}

Let $u\in\mathbb R$ and $u\ne0$. If $u<0$, then from \eqref{eqe} for $u_1=u$ and $u_2=1$ we find that
$$
L_x(0,u)=L_{x}(0,1)u.
$$
If $u>0$, then again from \eqref{eqe} with $u_2=u$ and $u_1=-1$ and taking into account \eqref{z6}, we have
$$
L_x(0,u)=-L_{x}(0,-1)u=L_{x}(0,1)u.
$$
If $u=0$, then by the condition $L_x(0,0)=0$, and so $L_x(0,u)=L_{x}(0,1)u$ for any $u\in\mathbb R$.
\end{proof}

In fact a more general fact holds. Assume that in problem \eqref{z1}\enskip
$\dot x=u$ and $u(t)\in U$ for almost all $t\in [t_0,t_1]$, where $U$~is an arbitrary set,
but $0\in{\rm int}\,U$. No differentiability of~$L$ with respect to~$u$ is required. If  $\wx\cd=0$
delivers a~strong local infimum in this problem, then using Theorem~\ref{T1} and
arguing as in the proposition, we find that the function $u\mapsto L_x(0,u)$ is linear on
some interval with centre at the origin.

Thus, Theorem \ref{T1} is a~strengthening of the Pontryagin maximum principle.

\smallskip

\textbf{Acknowledgement.} The authors are sincerely grateful to Revaz Valer'yanovich
Gamkrelidze for helpful discussions.


TITLE: Spectral density of Liouville CFTs
QUESTION [2 upvotes]: I find the spectral density of Liouville CFT in two papers, but they are quite different. It's confusing for me.
In more details, the eq. (1.4) in https://arxiv.org/abs/1702.00423 says that the spectral density of  Liouville CFT is given by polynomial dependence, however, the eq. (104) in https://arxiv.org/abs/1412.5205 suggests that it is given by exponential dependence (Cardy formula).
What does cause this difference? normalization? including non-physical states?

REPLY [2 votes]: The polynomial density of Yin et al is correct. It is the density of primary states.
Cardy's formula is supposed to give the density of all states, not just primary states, and it is exponential. Of course there are many more descendent states than primary states.
What I do not understand in your second reference is why they find Cardy's formula when computing the density of primary states, and why they identify the density of primary states in Liouville theory with the Plancherel measure on their quantum group. These look like two mistakes.