text:
\begin{document}
\global\long\def\l{\lambda}
\global\long\def\ep{\epsilon}

\title{A Blow-Up Result for Dyadic Models of the Euler Equations}

\author{In-Jee Jeong and Dong Li}

\email{ijeong@math.princeton.edu, dli@math.ubc.ca}
\begin{abstract}
We partially answer a question raised by Kiselev and Zlatos in
\cite{MR2180809}; in the generalized dyadic model of the Euler
equation, a blow-up of $H^{1/3+\delta}$-norm occurs. We recover a
few previous blow-up results for various related dyadic models as
corollaries.
\end{abstract}
\maketitle

\section{Introduction}

In this paper, we consider the following infinite system of ordinary
differential equations (ODEs):
\begin{align}
\frac{da_{j}(t)}{dt}&
=\alpha\big(\lambda^{j}a_{j-1}^{2}(t)-\l^{j+1}a_{j}(t)a_{j+1}(t)\big)
\notag\\
& \qquad +\beta\big(\l^{j}a_{j-1}(t)a_{j}(t)-\l^{j+1}a_{j+1}^{2}(t)\big),\label{eq:KPO}
\end{align}
for $j\geq0$ and with the boundary condition $a_{-1}(t)\equiv0$.
The coefficients $\alpha$ and $\beta$ are usually taken to be nonnegative constants.
We will assume $\l=2$ throughout, but our results hold for arbitrary
$\l>1$ with proper adjustments of the parameters. The special case
$\alpha=1,\beta=0$ is often called the KP equations, which have appeared
in the literature almost simultaneously in two papers \cite{MR2038114,MR2095627}.
The opposite extreme $\alpha=0,\beta=1$  first appeared in Obukhov's
work \cite{Obukhov1971} and was suggested as an alternative to the
KP equations in \cite{MR2231615}. Hence (\ref{eq:KPO}) can be viewed
as a linear combination of these two models.

These types of infinite system of ODEs are called dyadic models of
the Euler equations. For a heuristic derivation of the KP equations
from the Euler equations, one can see \cite{MR2095627}
for an argument based on the wavelet expansion of a scalar function over dyadic
cubes. Alternatively, consider the Euler equations in $\mathbb{R}^{n}$ with periodic
boundary conditions and rewrite the equations in terms of the Fourier
coefficients of the velocity vector field. Then one obtains an infinite
system of ODEs for the evolution of Fourier coefficients which share
several structural similarities with (\ref{eq:KPO}). We will return
to this point after Lemma \ref{lem:KPO}, from which equations (\ref{eq:KPO})
appear naturally from some constitutive relations.

To state blow-up and regularity results for dyadic models, let us
first define analogues of the Sobolev norms in the space of sequences.
The $H^{s}$-norm of a solution $a=(a_{0},a_{1},...)$ at time $t$
is defined by the formula
\[
||a(t)||_{s}^{2}:=\sum_{j=0}^{\infty}2^{2sj}a_{j}^{2}(t).
\]
In particular, we define the energy $E(t)$ as the square of $H^{0}$-norm (or the usual $l^2$-norm):
\[
E(t):=\sum_{j=0}^{\infty}a_{j}^{2}(t).
\]
Regarding the KP equations, the following blow-up result have been
proved several times (we have listed the references in a more or
less chronological order):
\begin{thm*}[\cite{MR2038114,MR2095627,MR2231615,MR2180809,MR2746670}]
 For every nonzero initial data, the $H^{s}$-norm of any solution\footnote{The meaning of solution
 here requires some clarification. It is known (cf. Proposition 2.1 of \cite{MR2180809})
 that local wellposedness holds in $C_t^0 H^s$ with $s\ge 1$. Alternatively one can work with  finite energy
 Leray-Hopf type solutions (cf. \cite{MR3057168,MR2415066}) for which uniqueness is a subtle issue.}
becomes infinite in finite time for all $s>s_{cr}:=1/3$.
\end{thm*}
Now for the Obukhov equations, there is the following regularity result:
\begin{thm*}[\cite{MR2180809}]
 If the initial data have finite $H^{s}$-norm for some $s>1$, then
the corresponding solution exists globally and has finite $H^{s}$-norm for all $t\geq0$.
\end{thm*}
In \cite{MR2180809}, Kiselev and Zlatos raised the question of whether blow-up
can occur in the case $\alpha,\beta>0$ (or more generally $\operatorname{sgn}(\alpha)=
\operatorname{sgn}(\beta)$). Corollary \ref{cor:KPO} of this paper
answers this question affirmatively, at least in the case when $\beta$ is
small relative to $\alpha$. Our proof of blow-up is quite different
and seems to be simpler than the previous proofs for the KP equations
(\cite{MR2038114,MR2095627,MR2231615,MR2180809}). Roughly speaking, all the previous blow-up proofs
rest on the intuition that at least a fixed proportion
of the energy contained in the $j$th component must be transferred
to the higher components within a time scale of $\tau^{-j}$ for some
$\tau>0$. To achieve this, one has to make strong use of the ``positivity''
of the KP equations; that is, once we have $a_{j}(t_{0})\geq0$ for
some $j$ and $t_{0}$,  then $a_{j}(t)\geq0$ for all future $t>t_{0}$. This
positivity in turn implies that there is no ``backward'' transfer
of energy; to be more precise, if the initial data satisfy $a_{j_0+1}(0)\geq0$, then
$a_{j_0+1}(t)\ge 0$, and
\[
\frac{d}{dt}E_{j_{0}}(t):=\frac{d}{dt}\big(\sum_{j=0}^{j_{0}}a_{j}^{2}(t)\big)
=-2 \lambda^{j_0+1} a_{j_0}^2 a_{j_0+1}
\leq0
\]
for all  $t\geq0$. Unfortunately, this mechanism of forward energy transfer
 seems to break down once we have both $\alpha,\beta>0$. The proof in
\cite{MR2746670} still makes use of positivity but it appears to be different
from others; we will come back to their proof after Lemma \ref{lem:KPO}.

Next, let us consider the following system of equations, where there
is an extra ``dissipation'' term on the right hand side:

\begin{equation}
\frac{da_{j}}{dt}=\alpha\big(\lambda^{j}a_{j-1}^{2}-\l^{j+1}a_{j}a_{j+1}\big)+
\beta\big(\l^{j}a_{j-1}a_{j}-\l^{j+1}a_{j+1}^{2}\big)-\nu\l^{2\gamma j}a_{j},
\label{eq:NSE}
\end{equation}
again for $j\geq0$ and with $a_{-1}(t)\equiv0$. Here, $\alpha=1,\nu=1$
can be assumed with appropriate rescaling and $\gamma>0$ is a parameter
representing the intensity of the dissipation. In the special
case $\beta=0$, these equations are often called the dyadic Navier-Stokes
Equations (NSEs), and they are already studied quite extensively in
the literature. In particular, the following blow-up result has been
proved by Cheskidov in \cite{MR2415066}:
\begin{thm*}[\cite{MR2415066}]
 Consider equations (\ref{eq:NSE}) with $\alpha=1$, $\beta=0$,
$\nu=1$, and $\gamma<1/3$. For every $\delta>0$, there exists a
constant $M(\delta)$ such that if the initial data satisfy $a_{j}(0)\geq0$
for all $j\geq0$ and $||a(0)||_{\delta}>M(\delta)$, then $||a(t)||_{1/3+\delta}^{3}$
is not locally integrable on $[0,\infty)$.
\end{thm*}
In particular, any solution blows up in finite time in $H^{1/3+\delta}$-norm
for every $\delta>0$. Our proof of the main theorem recovers this
$H^{1/3+\delta}$-norm blow-up in Corollary \ref{cor:D}.

\section{Results and Conjectures}

To begin, we borrow a lemma from \cite{MR2180809} from which
the model (\ref{eq:KPO}) follows naturally. We omit the proof since it is immediate in view of
the (formal) energy conservation constraint.
\begin{lem}
\label{lem:KPO}Assume that real-valued functions $a_{j}(t)$ ($j\geq0$)
satisfy a system of ODEs of the form
\[
\frac{da_{j}(t)}{dt}=F_{j}(a(t))
\]
where
\begin{itemize}
\item for each $j\geq0$, the map $F_{j}$ is a quadratic function of $a(t)$;
\item $F_{j}$ can involve only $a_{j-1}(t)$, $a_{j}(t)$, and $a_{j+1}(t)$;
\item each term for $F_{j}$ has a factor of $\l^{j}$ times a constant
which is independent of $j$, i.e.
\begin{align*}
F_j= \sum_{\substack{\mu_1=\pm 1, 0\\ \mu_2 =\pm 1,0}} C_{\mu_1,\mu_2} \lambda^j a_{j+\mu_1}(t) a_{j+\mu_2}(t),
\end{align*}
where $C_{\mu_1,\mu_2}$ are constants independent of $j$;

\item and the energy $\sum a_{j}^{2}(t)$ is (formally) conserved.
\end{itemize}
Then the system is necessarily of the form (\ref{eq:KPO}).
\end{lem}

The Euler equations are, of course, energy conserving (for smooth
solutions) and have quadratic nonlinearity. The factor $\l^{j}$ was
inserted so that the dyadic model would share similar functional
estimates with the Euler equations. One can argue that to model 3D
Euler equations, the choice $\l=2^{5/2}$ is
appropriate\footnote{Roughly speaking, this is based on the estimate
that (here $P_{2^j}$ is the usual Littlewood-Paley projector adapted
to the frequency block $|\xi|\ \sim 2^{j}$ and one can think of $u$
as the velocity in Euler) $\| P_{2^j} u \cdot \nabla P_{2^j} u
\|_{L^{2}(\mathbb R^d)} \lesssim 2^{(\frac d2 +1)j} \| P_{2^j} u
\|_{L^2(\mathbb R^d)}^2.$ For $d=3$, the factor is $2^{5/2}$. }
 (see \cite{MR2038114}). Lastly, the fact that $F_{j}$
only consists of $a_{j-1}$ and $a_{j+1}$ certainly does not hold
in the case of the Euler equations, but certain ``locality of interactions''
assumptions are believed to hold in the theory of turbulence. For example
one can see \cite{MR1877599, 2005PhyD..207...91E}.

Let us remark on the property of energy conservation. A formal calculation yields
that
\begin{align*}
\frac{d}{dt}E(t) &=2\sum_{j\geq0}a_{j}(\lambda^{j}a_{j-1}^{2}-\l^{j+1}a_{j}a_{j+1})\\
&=2\big(\sum_{j\geq1}\l^{j}a_{j-1}^{2}a_{j})-2(\sum_{j\geq0}\l^{j+1}a_{j}^{2}a_{j+1})=0.
\end{align*}
But in the above computation, an interchange of the order of
summation and differentiation must be justified, and it is
sufficient to require that $||a(t)||_{s}<\infty$ for $s>s_{cr}=1/3$.
However, for solutions with less regularity, this computation is no
longer valid and the dissipation of energy can indeed
occur\footnote{This is in some sense connected to the Onsager's
conjecture.}. In \cite{MR2746670} it was established that for every
initial condition with nonnegative components, the energy dissipates
to zero as $t\rightarrow\infty$. In particular, it implies
finite-time blow-up in every $H^{s}$ norm for $s>1/3$.

We are ready to state our main result.
\begin{thm}[The full model with diffusion]
\label{thm:KPOD}Consider the equations (\ref{eq:NSE}) with parameters
$\l=2$, $\alpha=1$, $\nu=1$, and $\beta\geq0$. For every $s>1/3$
and $\gamma<1/3$, there exists a value $\beta_{s,\gamma}>0$ such that
for $\beta\in[0,\beta_{s,\gamma})$, there exists a class of initial data for which the
corresponding solutions blow up in finite time in $H^{s}$-norm.
More precisely, for each solution $a(t)$,
 $\|a(t)\|_s^2$ is not locally integrable on $[0,\infty)$.
\end{thm}
\begin{rem}
As will be clear from our proof, the initial data $a(0)=(a_j(0))_{j=0}^{\infty}$ can even taken
to be compactly supported, in the sense that for some integer $j_0>0$, $a_j(0)=0$ for all $j\ge j_0$.
\end{rem}

\begin{rem}
Instead of considering only nearest neighborhood interactions (i.e $a_{j}$, $a_{j-1}$, $a_{j+1}$),
one can generalize the full model \eqref{eq:NSE} to arbitrarily finitely many (or even infinitely many with sufficiently fast
decay of interaction) neighborhood interactions. It is expected that our method of proof also
carriers over to this case.
\end{rem}

\begin{rem}
Although Theorem \ref{thm:KPOD} settles the blow-up of \eqref{eq:NSE} more or less satisfactorily,
the proof itself (albeit simple) gives little information
on the transfer of energy mechanism in the model. On the other hand, the previous proofs on the blow-up of KP
model \emph{do} respect
the details of the dynamics and give some insight of the cascade mechanism. In light of this,
it is still desirable to give a
more "dynamic" proof in this flavor. After all, one of the main reasons for
studying the dyadic models is to understand  energy cascade and even turbulence transport.
\end{rem}

Before the proof of Theorem \ref{thm:KPOD}, we state
two direct corollaries which simply correspond to cases $\nu=0$ and
$\beta=0$, respectively.

\begin{cor}[The full model with no diffusion]
\label{cor:KPO}Consider the equations (\ref{eq:KPO}) with parameters
$\l=2$, $\alpha=1$, and $\beta\geq0$, and fix $s>1/3$ together
with $(2^{s}-2^{1-2s})/(1+2^{1-3s})>\beta$. Then for every nonnegative
initial data (that is, $a_{j}(0)\geq0$ for all $j$), there is finite-time
blow up in $H^{s}$-norm.
\end{cor}

\begin{rem}
Here and below (in Corollary \ref{cor:D}), the blow-up of $H^s$-norm is again
understood as that $\| a(t) \|_s^2$ is not locally integrable on $[0,\infty)$.

\end{rem}

For example, when $\beta<6/5$, every nonnegative initial data blow-up
in the $H^{1}$-norm.
\begin{cor}[KP with diffusion]
\label{cor:D} Consider the equations (\ref{eq:NSE}) with $\l=2$,
$\alpha=1$, $\beta=0$, $\nu=1$, and $\gamma<1/3$. For every $s>1/3$,
let $\theta=\theta({s,\gamma})$ be a constant such that
\begin{align*}
&-\frac 43 <\theta<2(s-1), \\
&-\frac 43 < \theta <-4\gamma.
\end{align*}
There exists a constant $C=C(s,\gamma,\theta)>0$, such that once
the initial data $(a_j(0))_{j=0}^{\infty} \in H^s$ satisfy
\[
\sum_{j=0}^{\infty}2^{ j(\theta+1)}a_{j}(0)>C,
\]
there is finite-time blow up in $H^{s}$-norm. \end{cor}
\begin{proof}[Proof of  Theorem \ref{thm:KPOD}]
We fix some $s>1/3$ and assume towards contradiction that
$||a(t)||_{s}^2$ is locally integrable\footnote{This in turn would
imply that one can freely interchange summation and differentiation
in the argument below. Alternatively, one can recast the equations
into integral (in time) formulation and justify passing the limit
under the integral.} on $[0,\infty)$.
 By setting $b_{j}(t):=\l^{j}a_{j}(t)$, we simplify
the equation as follows:
\begin{equation}
\frac{db_{j}}{dt}=\big(\lambda^{2}b_{j-1}^{2}-b_{j}b_{j+1}\big)+\beta\big(\l b_{j-1}b_{j}-\l^{-1}b_{j+1}^{2}\big)-\l^{2\gamma j}b_{j}.\label{eq:NSE-1}
\end{equation}
Then we consider the sum
\[
A(t):=\sum_{j=0}^{\infty}b_{j}^{2}w^{-j}
\]
where $w>1$ is a constant to be optimized later. We observe that
if $w^{-1}\leq\l^{-2}\cdot2^{2s}$, then $A(t)$ is also integrable since $A(t)\leq||a(t)||_{s}^{2}$
for all $t \geq 0$. Then we consider the quantities
\[
\frac{d}{dt}(b_{j}w^{-j})=(\l^{2}b_{j-1}^{2}w^{-j}-b_{j}b_{j+1}w^{-j})+
\beta(\l b_{j-1}b_{j}w^{-j}-\l^{-1}b_{j+1}^{2}w^{-j})-\l^{2\gamma j}b_{j}w^{-j}
\]
and sum them over all $j\geq0$. By the Cauchy-Schwartz inequality, the infinite sum appearing on the right hand side
is bounded in absolute value by $\mathrm{const}\cdot (A(t)+ \sqrt{A(t)}) $, and therefore the sum can be rearranged
whenever $A(t)$ is finite.

 We therefore obtain:
\begin{eqnarray}
\frac{d}{dt}(\sum_{j=0}^{\infty}b_{j}w^{-j}) & = & \l^{2}\sum_{j=1}^{\infty}b_{j-1}^{2}w^{-j}-\sum_{j=0}^{\infty}b_{j}b_{j+1}w^{-j}\label{eq:blowup}\\
 & + & \beta\l\sum_{j=1}^{\infty}b_{j-1}b_{j}w^{-j}-\beta\l^{-1}\sum_{j=0}^{\infty}b_{j+1}^{2}w^{-j}-\sum_{j=0}^{\infty}\l^{2\gamma j}b_{j}w^{-j}.\nonumber
\end{eqnarray}
We note in advance that again by the Cauchy-Schwartz inequality,
\begin{equation}
(\sum_{j=0}^{\infty}b_{j}w^{-j})^{2}\leq(\sum_{j=0}^{\infty}w^{-j})(\sum_{j=0}^{\infty}b_{j}^{2}w^{-j})=\frac{A(t)}{1-w^{-1}}\label{eq:A}
\end{equation}
holds. Then first four terms on the right hand side of (\ref{eq:blowup})
can be estimated as follows:
\begin{eqnarray*}
 &  & \l^{2}w^{-1}\sum_{j=1}^{\infty}b_{j-1}^2 w^{-{j-1}}
 -w^{1/2}\sum_{j=0}^{\infty}(b_{j}w^{-\frac{j}{2}})(b_{j+1}w^{-\frac{j+1}{2}})\\
 &  & +\beta\l w^{-1/2}\sum_{j=1}^{\infty}(b_{j-1}w^{-\frac{j-1}{2}})(b_{j}w^{-j})
 -\beta\l^{-1}w\sum_{j=0}^{\infty}b_{j+1}^2 w^{-{(j+1)}}\\
 & \geq &  (\lambda^2 w^{-1} -w^{1/2}-\beta\l w^{-1/2}-\beta \lambda^{-1} w
 ) \cdot A(t).
\end{eqnarray*}
Regarding the last term, we have
\begin{eqnarray*}
-\sum_{j=0}^{\infty}\l^{2\gamma j}b_{j}w^{-j} & = & -\sum_{j=0}^{\infty}(b_{j}w^{-\frac{j}{2}})(\l^{2\gamma j}w^{-\frac{j}{2}})\geq-(\sum_{j=0}^{\infty}b_{j}^{2}w^{-j})^{\frac{1}{2}}(\sum_{j=0}^{\infty}\l^{4\gamma j}w^{-j})^{\frac{1}{2}}\\
 & = & -\frac 1 {1-\lambda^{4\gamma} w^{-1}} \cdot(\sum_{j=0}^{\infty}b_{j}^{2}w^{-j})^{\frac{1}{2}}\\
 & \geq & - \frac 1 {1-\lambda^{4\gamma} w^{-1}} (\eta A(t)+\frac{1}{4\eta}),
\end{eqnarray*}
where $\eta>0$ is a constant. Here we have assumed that $\l^{4\gamma}w^{-1}<1$.
Adding above two estimates together with (\ref{eq:A}), we conclude
for some $C_{2}>0,$
\[
\frac{d}{dt}(\sum_{j=0}^{\infty}b_{j}w^{-j})\geq C_{1}\cdot(\sum_{j=0}^{\infty}b_{j}w^{-j})^{2}-C_{2}.
\]
Therefore, if we have $C_{1}>0$ and $\sum b_{j}(0)w^{-j}>\sqrt{C_{2}/C_{1}}$, then
$\sum b_{j}w^{-j}$ will not be locally integrable on $[0,\infty)$, which is a contradiction
to the fact that $A(t)$ is locally integrable. To this end, we need
\[
C_{1}=(1-w^{-1}) \cdot\Bigl(\l^{2}w^{-1}-w^{1/2}-\beta\l w^{-1/2}-\beta\l^{-1}w-\frac {\eta} {1-\lambda^{4\gamma} w^{-1}}
\Bigr)>0.
\]
But at the expense of choosing $\eta$ sufficiently small, it is enough
to have
\begin{align*}
\l^{2}w^{-1}-w^{1/2}-\beta\l w^{-1/2}-\beta\l^{-1}w>0.
\end{align*}
Therefore, to conclude the proof, we need:
\begin{align}
w^{-1} & \le \l^{-2}2^{2s} \notag \\
w^{-1} & <  \l^{-4\gamma} \notag \\
\l^{2}w^{-1} - w^{1/2}&>\beta(\l w^{-1/2}+\l^{-1}w), \label{w_ineq}
\end{align}
where $\l=2$. Assuming for the moment that $\beta=0$, we have $\l>w^{3/4}$
from the last inequality which gives restrictions $s>1/3$, $\gamma<1/3$.
On the other hand, it is clear now that once we have $s>1/3,\gamma<1/3$,
we can choose $w$ in a way that for small $\beta>0$, all the above three
inequalities are satisfied.
\end{proof}

\begin{proof}[Proof of Corollary \ref{cor:KPO}]
Since $\nu=0$, the second inequality of \eqref{w_ineq} is not needed.
One can just choose $w^{-1} =2^{2(s-1)}$ and this gives
\begin{align*}
\frac {2^{s}-2^{1-2s}}{1+2^{1-3s}}>\beta.
\end{align*}
\end{proof}
\begin{proof}[Proof of Corollary \ref{cor:D}]
Since $\beta=0$, the conditions on $w^{-1}$ in \eqref{w_ineq} take the form
\begin{align*}
&2^{-\frac 43} <w^{-1} \le 2^{2(s-1)}, \\
&2^{-\frac 43}<w^{-1}<2^{-4\gamma}.
\end{align*}
Denoting $w^{-1}=2^{\theta}$ then yields the result.
\end{proof}

Let us close by presenting a few conjectures which would complement
or generalize regularity and blow-up results currently known. We first
explain the result of \cite{MR2746670}: recall that we have already
mentioned their dissipation of energy result. But they also proved
the existence of so-called ``self-similar solutions'', which are
natural analogues of the fixed point in the forced case. To be specific,
consider the forced KP equations:
\begin{align*}
\frac{d} {dt} a_{j}(t) & =  \lambda^{j}a_{j-1}^{2}(t)-\l^{j+1}a_{j}(t)a_{j+1}(t),\,\,\,(j\geq1)\\
\frac{d}{dt}a_{0}(t)  &= -\l a_{0}(t)a_{1}(t)+f_{0},
\end{align*}
where $f_{0}>0$ is a constant. Then it is immediate that there exists
a unique fixed point which have finite energy. This fixed point satisfies
$\bar{a}_{j}=\mathrm{const}\cdot\l^{-j/3}$ so it has finite $H^{s}$-norm
precisely for $s<1/3$. In \cite{MR2337019,MR2600714}, it was established
that this fixed point is the unique global attractor of the dynamics.

When there is no forcing, there does not exist nontrivial fixed points.
However, self-similar solutions are the correct analogues; we define
a solution self-similar if for every $j\geq0$, $a_{j+1}(t)/a_{j}(t)$
is constant in time. From this requirement, it is straightforward
to check that the solution must have the form
\[
a_{j}(t)=\frac{c_{j}}{t-t_{0}}
\]
for some constants $c_{j}$ and $t_{0}>0$ which satisfy the recurrence (for $\lambda=2$)
\[
c_{j}c_{j+1}=2^{-j}c_{j}+c_{j-1}^{2}/2
\]
for all $j\geq0$ with $c_{-1}=0$. The choice of $c_{0}>0$ uniquely
determines the whole sequence and the self-similar solution, modulo
the choice of $t_{0}>0$ which is independent. The hard part is to
show that there exists a value of $c_{0}>0$ (which turns out to be
unique) such that the self-similar solution $a(t)$ has finite energy.
Then it is not hard to see that the self-similar solution satisfies
$c_{j}\sim\mathrm{const}\cdot\l^{-j/3}$. Note this  power-law decay in
$j$ can already be noticed from our proof; the scale $a_{j}(t)\sim\l^{-j/3}$
roughly corresponds to the case where we have equalities in the Cauchy-Schwartz
inequalities used in the proof. Now it is very desirable to show that
the self-similar solutions are the global attractors of the unforced
dynamics. If we believe in the convergence towards self-similar ones,
it is natural to conjecture that in the KP equations, the $H^{s}$-norms
remain finite for all $s<1/3$. This finiteness of smaller Sobolev
norms are partially obtained in the works \cite{MR3057168,MR2844828}.
Also, one can revert all inequalities in our proof and try to get some a
priori estimates on the solution, which look similar to some regularity
results proved in \cite{MR3057168,MR2844828}. Finally, it is tempting
to believe that such self-similar solutions also exist for our equation,
at least when $\beta$ is small. One can write down the recurrence
relation as above but this relation is now more complicated.

\section*{Acknowledgements}
I. Jeong would like to thank his advisor Prof. Ya.G. Sinai for many
helpful discussions. I. Jeong was supported in part by a Samsung Fellowship. D. Li was supported in part by an Nserc
discovery grant.

\bibliographystyle{plain}
\bibliography{dyadic}

\end{document}


TITLE: Prove that $A^c = (A^c∩B)∪(A^c∩B^c)$
QUESTION [0 upvotes]: Could someone help me with this problem?
I have no idea how to attempt to solve it nor do I know where I can ask for help besides her.

REPLY [1 votes]: Set unions and intersections distribute over each other, so we can rewrite
$$
(\overline{A}\cap B) \cup (\overline{A} \cap \overline{B})
$$
Instead as:
$$
\overline{A} \cap (B\cup\overline{B})
$$
The union of a set and its complement is just the universal set $U$ in which $A$ and $B$ reside, and the intersection of any set with $U$ in turn is just that set:
$$
\overline{A} \cap (B\cup\overline{B}) = \overline{A}\cap U = \overline{A}
$$

You can also prove it directly, like so.

Let $X:=(\overline{A}\cap B) \cup (\overline{A} \cap \overline{B})$ be your set in question.
Take any arbitrary $x\in \overline{A}$. By definition, $x\in U = B\cup\overline{B}$, so either $x\in B$, or $x\in \overline{B}$. In the former case, $x\in(\overline{A}\cap B) \Rightarrow x\in X$, while in the latter case, $x\in(\overline{A}\cap \overline{B}) \Rightarrow x\in X$. So we have proved $x\in\overline{A}\Rightarrow x\in X$.
In the other direction, take any $x\in X$. Because $X$ is a union of sets, either $x\in (\overline{A}\cap B)$, or $x\in(\overline{A}\cap \overline{B})$, but both of these statements imply $x\in \overline{A}$. So we have also proved $x\in X \Rightarrow x\in\overline{A}$.
Combine both implicational directions, and universally instantiate $x$, and we have in fact proved $\forall x :x\in X \Leftrightarrow x\in\overline{A}$. This is the definition of equality of sets, therefore $X = \overline{A}$.


TITLE: Definition of subobject
QUESTION [3 upvotes]: In the definition of subobject there is an equivalence relation defined on monomorphisms into a fixed codomain.
My question is that how do we know that the collection of monomorphisms into a fixed codomain is a set?
Is it obvious? or Is it because we can define equivalence relation not only on a set but a class? I don't know.
Please help. Thank you.

REPLY [3 votes]: It's not satisfied in general. This leads to the notion of a well-powered category.
An example of a category which is not well-powered is the partial order $V \cup \{\infty\}$, where $V$ is some Grothendieck universe and $\infty$ is some new element with $x \leq \infty$ for all $x \in V$.
See MO/93853 for abelian categories which are not well-powered.


text:
\begin{document}
	\title{Chaos in convolution operators on the space of entire functions of infinitely many complex variables}
	\author{Blas M. Caraballo \thanks{The first named author is supported by CAPES and CNPq } and Vin\'icius V. F\'avaro\thanks{The second named author is supported by FAPEMIG Grant APQ-03181-16; and CNPq Grant 310500/2017-6.}}

	\date{}
	\maketitle

	\vspace{-0.2 cm}

	\begin{center}
		\textit{Dedicated to the memory of Professor Jorge Mujica (1946-2017)}
	\end{center}

	\vspace{0.2 cm}

	\begin{abstract}
		A classical result of Godefroy and Shapiro states that every nontrivial convolution operator on the space $\mathcal{H}(\mathbb{C}^n)$ of entire functions of several complex variables is hypercyclic. In sharp contrast with this result F\'avaro and Mujica show that no translation operator on the space $\mathcal{H}(\mathbb{C}^\mathbb{N})$ of entire functions of infinitely many complex variables is hypercyclic. In this work we study the linear dynamics of convolution operators on $\mathcal{H}(\mathbb{C}^\mathbb{N})$. First we show that no convolution operator on $\mathcal{H}(\mathbb{C}^\mathbb{N})$ is neither cyclic nor $n$-supercyclic for any positive integer $n$. After we study the notion of Li--Yorke chaos in non-metrizable topological vector spaces and we show that  every nontrivial convolution operator on $\mathcal{H}(\mathbb{C}^\mathbb{N})$ is Li--Yorke chaotic.

		{\small \bigskip\noindent\textbf{MSC2010:}  47A16, 47B38, 32A15}

		{\small \medskip\noindent\textbf{Keywords:}   $n$-supercyclicity, cyclicity, Li-Yorke chaos, convolution operators, holomorphic functions of infinitely many complex variables.}
	\end{abstract}

	\section{Introduction}
	Let $V$ be a subset of a Hausdorff topological complex vector space $E$ and let $T\colon E\to E$ be a continuous linear operator (from now on we just write operator). The \emph{orbit of $V$ under $T$}, denoted by orb$_T(V)$, is the subset of $E$ given by
	\[\textnormal{orb}_T(V)=\bigcup_{k=0}^\infty T^k(V).\]
	If $V=\{x\}$ is a singleton and orb$_T(V)=\{T^kx : k\in\N_0\}$ is dense in $E$, where $\N_0=\{0,1,2,3,\ldots\}$, then $T$
	is said to be \emph{hypercyclic} and $x$ a \emph{hypercyclic vector} for $T$. If the linear space generated by orb$_T(V)$ is dense in $E$, then $T$
	is said to be \emph{cyclic}	and $x$ a \emph{cyclic vector} for $T$. If $V=\textnormal{span}\{x\}$ and orb$_T(V)=\C\cdot\{T^kx : k\in\N_0\}$ is dense in $E$, then $T$ is said to be \emph{supercyclic} and $x$ a \emph{supercyclic vector} for $T$. Finally, if $V$ is a vector subspace of dimension $n$ and orb$_T(V)$ is dense in $E$, then $T$
	is said to be \emph{$n$-supercyclic} and $V$ a \emph{supercyclic subspace} for $T$.

	Hypercyclicity is the most important concept in linear dynamics and it has received considerable attention in the last 25 years.  References \cite{bay, grosse2011linear} provide deep and detailed surveys of the theory. The notion of chaos in linear dynamics was introduced by Godefroy and Shapiro \cite{godefroy} in 1991. They adopted the Devaney's definition of chaos. Recall that an operator on a Fréchet space is \emph{chaotic} if it is hypercyclic and it has a dense set of periodic points.

	There are several important notions of chaos and some authors have started to study this notions in the context of linear dynamics. In addition to the notions defined above we mention the first mathematical definition of chaos given in 1975 by Li and Yorke in  \cite{li1975period}, which is currently known as Li--Yorke chaos. This classical notion of Li-Yorke chaos was introduced for maps defined on metric spaces as follow: Given a metric space $(M,d)$ and a continuous map $f\colon M\rightarrow M$, we recall that a pair $(x,y)\in M\times M$ is called a \emph{Li-Yorke pair} for $f$ if
	$$
	\liminf_{n\rightarrow\infty} d(f^n(x),f^n(y))=0 \textrm{ and } \limsup_{n\rightarrow\infty}  d(f^n(x),f^n(y))>0.$$
	A \emph{scrambled set} for $f$ is a subset $S$ of $M$ such that $(x, y)$ is a Li–Yorke pair for $f$ whenever $x$ and $y$ are distinct points in $S$. The map $f$ is said to be \emph{Li–Yorke chaotic} if there exists an uncountable scrambled set for $f$. By \cite[Theorem 9]{Li-Yorke} hypercyclicity implies chaos in the sense of Li--Yorke.

	In this paper we are mainly interested in the linear dynamics of convolution operators on spaces of entire functions of infinitely many complex variables. We remark that several results on linear dynamics of operators on spaces of entire functions of infinitely many complex variables have appeared in the last few decades. See for instance \cite{aronbes, bay2, BBFJ, bes2012, CDSjmaa, vinicius2016hypercyclic, FMNach, GS, goswinBAMS, MPS, peterssonjmaa}.

	A classical result due to Godefroy and Shapiro \cite{godefroy}  states that every nontrivial convolution o\-pe\-ra\-tor on $\mathcal{H}(\C^n)$ is hypercyclic. Moreover,  A. Bonilla and K.-G. Grosse-Erdmann \cite{bonilla2006theorem} showed that these convolution operators are even frequently hypercyclic, which is a stronger notion than hypercyclicity. In sharp contrast with these results, Fávaro and Mujica \cite{vinicius2016hypercyclic} proved that no convolution operator on $\mathcal{H}(\C^\N)$  can be hypercyclic. At first sight this result may look surprising, since it is well known that every $f\in\mathcal{H}(\mathbb{C}^\mathbb{N})$ depends only of  finitely many variables (see \cite[p. 162]{dineen2012complex}). Based on these facts, the following question arises:
	\begin{center}
		Do the convolution operators on $\mathcal{H}(\mathbb{C}^\mathbb{N})$ satisfy some notion of the linear dynamics weaker than hypercyclicity?

	\end{center}

	Note that the notions of $1$-supercyclicity and supercyclicity are equivalent and that the following diagram holds:
	\[\xymatrix@=7mm{\textnormal{hypercyclicity} \ar@2{->}[r] & \textnormal{supercyclicity}\ar@2{->}[r]\ar@2{->}[d]&\textnormal{cyclicity}\\
		&n\textnormal{-supercyclicity}\ar@2{}[ru]}\]

	An $n$-supercyclic operator need not be cyclic, for $n=2,3,\ldots$ (for an example in infinite dimension see \cite{bourdon2004}). Hilden and Wallen \cite{hilden1974some} proved that no operator on $\C^n$ can be supercyclic $(n=2,3\ldots)$. So, $n$-supercyclicity does not imply supercyclicity, in general. For properties and results about supercyclicity and $n$-supercyclicity we refer to \cite{bourdon2004,feldman2002n,herrero1991limits,hilden1974some}.

	In sharp contrast with the aforementioned result of Godefroy and Shapiro we will show that no convolution operator on $\mathcal{H}(\C^\N)$  can be neither cyclic nor $n$-supercyclic for any positive integer $n$ (Theorem \ref{nsuper}).
	So we may rewrite the last question in the following way:

	\begin{center}
		Are the convolution operators on $\mathcal{H}(\mathbb{C}^\mathbb{N})$ at least Li--Yorke chaotic?
	\end{center}

	Since $\mathcal{H}(\C^\N)$ is a non-metrizable complete locally convex space, the classical notion of Li--Yorke chaos does not make sense in this context. Recently T. Arai \cite{arai2017devaney} introduced the notion of Li-Yorke chaos for an action of a group on an uniform space. Since every topological vector space is an uniform space, we will adopt the Arai's definition of Li-Yorke chaos. Using this definition we will prove that the last question has positive answer, i.e., every convolution operator on $\mathcal{H}(\mathbb{C}^\mathbb{N})$  is Li--Yorke chaotic (Theorem \ref{TLY}). It is worth to mention that the criteria that appear in the literature to prove that an operator satisfies (or not) some kind of linear chaos are in general for operators defined on $F$-spaces. Since $\mathcal{H}(\C^\N)$ is not a metric space we will use no criterion to prove the first result (Theorem \ref{nsuper}). However, to show the second result (Theorem \ref{TLY}) we will adapt a criterion obtained by Bernardes \textit{et al} \cite{Li-Yorke} for operators on Fréchet spaces to operators on Hausdorff topological vector spaces. This criterion is the key of the proof.

	For our purpose it is enough to present the definition of Li--Yorke chaos for an operator $T$ on a Hausdorff topological vector space $E$ as follow: A pair $(x,y)\in E\times E$ is said to be \emph{asymptotic} for $T$ if for any neighborhood of zero $U$, there exists $k\in\N$ such that $T^n(x-y)\in U$ for every $n\geq k$, that is, if $T^n(x-y)\to0.$
	A pair $(x,y)\in E\times E$ is said to be \emph{proximal} for $T$ if for any neighborhood of zero  $U$, there exists $n\in\N$ such that $T^n(x-y)\in U$, that is, if the sequence $\{T^n(x-y)\}$ has a subsequence converging to zero.

	A pair $(x,y)\in E\times E$ is said to be a \emph{Li--Yorke pair} for $T$ if it is proximal, but it is not asymptotic. In other words, $(x,y)$ is a Li--Yorke pair for $T$ if and only if the sequence $\{T^n(x-y)\}$ does not converge to zero, but it has a subsequence converging to zero.

	A \emph{scrambled set} for $T$ is a subset $S$ of $E$ such that $(x,y)$ is a Li--Yorke pair for $T$ whenever $x$ and $y$ are distinct points in $S$. Finally, we say that  $T$ is \emph{Li--Yorke chaotic} if there exists an uncountable scrambled set for $T$.

	It is easy to check that if $E$ is metrizable and we consider a translation-invariant metric (this metric exists by definition of metrizability), then both definitions of Li--Yorke chaos coincide.

	\section{Preliminaries}

	Given the topological product $\C^{\N}=\prod_{n=1}^{\infty}\C$, we consider the complex vector space of all entire functions $f\colon\C^{\N}\to\C$, which is denoted by $\mathcal{H}(\C^\N)$. It is well known that there are only two usual locally convex topologies on $\mathcal{H}(\C^\N)$: the compact open topology $\tau_0$ and its bornological associated topology $\tau_\delta$ (see \cite{Barroso1971,dineen2012complex}). It is also known that, with both topologies, $\mathcal{H}(\C^\N)$ is separable. For details and properties of these topologies we refer to \cite{ansemil1979topological,Barroso1971,barroso1979some}.

	For each $n\in\N$ we consider the canonical inclusion $J_n\colon\C^n\to\C^\N$, the canonical projection $\pi_n\colon\C^\N\to\C^n$ and the corresponding linear applications
	\[J_n^*\colon f\in\mathcal{H}(\C^\N)\to f\circ J_n\in\mathcal{H}(\C^n), \ \hspace{1cm} \ \pi_n^*\colon f_n\in\mathcal{H}(\C^n)\to f_n\circ \pi_n\in\mathcal{H}(\C^\N).\]
	Since $\pi_n\circ J_n=Id_{\C^n}$ it follows that
	\begin{equation}\label{complemented}
	J_n^*\circ\pi_n^*=Id_{\mathcal{H}(\C^n)}, \textrm{ for each } n\in\N.
	\end{equation}
	So $\mathcal{H}(\C^n)$ can be seen as the vector subspace of $\mathcal{H}(\C^\N)$ of all  entire functions on $\C^\N$ that depend only of the $n$ first variables, through the injective application $\pi_n^*$, for each $n\in\N$. It is easy to check that
	\begin{equation}
	\pi_1^*(\mathcal{H}(\C))\subset\pi_2^*(\mathcal{H}(\C^2))\subset\cdots\subset\pi_n^*(\mathcal{H}(\C^n))\subset\cdots\subset\mathcal{H}(\C^\N).\label{inclusao}
	\end{equation}

	By \cite[p. 162]{dineen2012complex} or \cite[Corolário 38]{Barroso1971}
	\begin{equation}\label{barro}
	\mathcal{H}(\C^\N)=\bigcup_{n=1}^\infty\{f_n\circ\pi_n\colon f_n\in\mathcal{H}(\C^n)\}=\bigcup_{n=1}^\infty\pi_n^*(\mathcal{H}(\C^n)).
	\end{equation}
	Also, by \cite[Proposition 1.3]{ansemil1979topological} the topology $\tau_\delta$, which was independently introduced by Nachbin \cite{nachbin1970espaces} and Couré \cite{coeure1970fonctions} coincides with the inductive limit topology of the Fréchet spaces $\mathcal{H}(\C^n)$, $n\in\N$, that is,
	\[(\mathcal{H}(\C^\N),\tau_\delta)=\textnormal{ind}_{n\in\N}\mathcal{H}(\C^n),\]
	where $\mathcal{H}(\C^n)$ is endowed with its usual topology, the compact open topology. More precisely, $\tau_\delta$ is the strongest locally convex topology on $\mathcal{H}(\C^\N)$, which becomes the applications $\pi_n^*$ continuous. If $\tau$ represents any of the topologies $\tau_0$, $\tau_\delta$ on $\mathcal{H}(\C^\N)$ then the linear operators
	\[J_n^*\colon f\in(\mathcal{H}(\C^\N),\tau)\to f\circ J_n\in\mathcal{H}(\C^n) \ \hspace{1cm} \ \pi_n^*\colon f_n\in\mathcal{H}(\C^n)\to f_n\circ \pi_n\in(\mathcal{H}(\C^\N),\tau)\]
	are continuous and it follows from \eqref{complemented} that $\mathcal{H}(\C^n)$  is topologically isomorphic to a complemented subspace of $(\mathcal{H}(\C^\N),\tau)$. In particular, $\pi_n^*(\mathcal{H}(\C^n))$ is a closed proper subspace of $(\mathcal{H}(\C^\N),\tau)$. For background information on these topologies we refer the reader to the book of Dineen  \cite{dineen2012complex}.

	Finally we recall that the \emph{translation operator by $\xi \in \C^\N$}, $$\tau_{\xi}\colon \mathcal{H}(\C^\N) \rightarrow \mathcal{H}(\C^\N) $$ is given by $(\tau_{\xi}f)(x) = f(x-\xi)$ for every $x \in \C^\N$. Analogously we define translation operators on $\mathcal{H}(\C^n)$ for each $n\in\N.$

	\begin{remark}\label{inclu}\rm
		It is interesting to note that, if $\xi\in\C^\N$ is such that $\pi_n(\xi)=0$, then the translation operator $\tau_\xi$ on $\mathcal{H}(\C^\N)$ coincides with the identity operator on $\pi_n^*(\mathcal{H}(\C^n))$, that is, $\tau_\xi|_{\pi_n^*(\mathcal{H}(\C^n))}=Id$. Hence ${\pi_n^*(\mathcal{H}(\C^n))}$ is a closed proper subspace of $\mathcal{H}(\C^\N)$ and $\tau_\xi$-invariant.
	\end{remark}

	\subsection{Convolution operators on $\mathcal{H}(\C^\N)$}

	In this section we prove some technical results about convolution operators that we need to show the main results of this work.

	\begin{definition}\rm
		A \textit{convolution operator} on $\mathcal{H}(\C^\N)$
		is a continuous linear mapping
		$$ L\colon \mathcal{H}(\C^\N) \rightarrow \mathcal{H}(\C^\N) $$
		such that $L(\tau_{\xi}f) = \tau_{\xi}(Lf)$ for every $f \in \mathcal{H}(\C^\N)$ and $\xi \in \C^\N$. Analogously we define convolution operators on $\mathcal{H}(\C^n)$ for each $n\in\N.$
	\end{definition}

	\begin{lemma}\label{Lm1}
		Let $f\in\mathcal{H}(\C^\N)$, $k\in\N$ and $T$ be a linear operator on $\mathcal{H}(\C^\N)$. Then there is $N\in\N$ such that  $T^if=\pi_{N}^*((T^if)\circ J_N)$ for every $i=0,\ldots,k$. Moreover, if $f=f_n\circ\pi_n$ with $f_n\in\mathcal{H}(\C^n)$ for some $n\in\N$, and $L$ is a linear operator on $\mathcal{H}(\C^\N)$ that commutes with all the translation operators (in particular, if $L$ is a convolution operator), then
		\[L^kf=\pi_{n}^*((L^kf)\circ J_n)=\pi_n^*\circ J_{n}^*(L^kf)\]
		for every $k\in\N_0$.
	\end{lemma}
	\begin{proof}
		By \eqref{barro} we may write

		$$f=f_{n_0}\circ\pi_{n_0}, Tf=f_{n_1}\circ\pi_{n_1},\ldots, T^kf=f_{n_k}\circ\pi_{n_k},$$ with $f_{n_i}\in\mathcal{H}(\C^{n_i})$ for every $i=0,\cdots,k$. Let $N=\max\{n_i: i=0,\ldots, k\}$ and $\xi\in\C^\N$ be such that $\pi_N(\xi)=0$. Then $\pi_{n_i}(\xi)=0$ for every $i=0,\cdots,k$ and
		\[\tau_\xi(T^{i}f)(x)=(T^{i}f)(x-\xi)=f_{n_i}\circ\pi_{n_i}(x-\xi)=f_{n_i}(\pi_{n_i}(x)-\pi_{n_i}(\xi))=f_{n_i}(\pi_{n_i}(x))=(T^{i}f)(x)\]
		for every $x\in\C^\N$ and every $i=0,\cdots,k$. On the other hand, given any $x=(x_j)\in\C^\N$, if we take $\xi=(0,\ldots,0,x_{N+1},x_{N+2},\ldots)\in\C^\N$, then
		\begin{align*}
		(T^if)(x)&=\tau_\xi(T^if)(x)=(T^if)(x-\xi)=(T^if)(x_1,\ldots,x_N,0,0,\ldots)\\
		&=(T^if)(J_N\circ\pi_N(x))=(T^if)\circ J_N\circ\pi_N(x).
		\end{align*}
		Thus $T^if=\pi_N^*((T^if)\circ J_N)$ for every $i=0,\ldots,k$.

		Now, suppose that $f=f_n\circ\pi_n$ and $L$ is a linear operator on $\mathcal{H}(\C^\N)$ that commutes with all the translation operators. Choosing $\xi\in\C^\N$ such that $\pi_n(\xi)=0$ we get
		\begin{align*}
		\tau_\xi(f)=f, \hspace{0.3cm} \tau_\xi(Lf)=L(\tau_\xi f)=Lf, \dots,
		\tau_\xi(L^kf)=L(\tau_\xi(L^{k-1}f))=L^kf,
		\end{align*}
		for every $k\in\N_0$. Following the same lines of the first part of the proof we obtain
		\[L^kf=\pi_{n}^*((T^kf)\circ J_n),\]
		for every $k\in\N_0$.
	\end{proof}

	This lemma tells us that the operator $\pi_n^*\circ J_{n}^*$ acts as the identity on $\textnormal{Orb}_L(f)$, whenever $f=f_n\circ\pi_n\in\mathcal{H}(\C^\N)$ and $L$ is a convolution operator.

	If $f=f_n\circ\pi_n\in\mathcal{H}(\C^\N)$ with $f_n\in\mathcal{H}(\C^n)$, and $\xi\in\C^\N$ is not difficult to verify that $\tau_\xi f=(\tau_{\pi_n(\xi)}f_n)\circ\pi_n$. In this sense the following question is quite natural:

	\medskip

	\textit{
		Does every convolution operator $L$ on $(\mathcal{H}(\C^\N),\tau)$ satisfy $Lf=L_nf_n\circ\pi_n$, where $L_n$ is a convolution operator on $\mathcal{H}(\C^n)$?
	}

	\medskip

	The following lemma gives a positive answer to this question.

	\begin{lemma}\label{GJD}
		Let $L$ be a convolution operator on $(\mathcal{H}(\C^\N),\tau)$.
		\begin{enumerate}
			\item [(a)] The operator $L_n:=J_n^*\circ
			L\circ\pi_n^*\colon\mathcal{H}(\C^n)\to \mathcal{H}(\C^n)$, $n\in\N$, is a convolution operator. We say that $L_n$ is \emph{the convolution operator on $\mathcal{H}(\C^n)$ associated to $L$}.
			\item[(b)] \[L(f_n\circ\pi_n)=(L_nf_n)\circ\pi_n, \ \ \textnormal{for every} \ \ f_n\in\mathcal{H}(\C^n) \ \textnormal{and} \ n\in\N.\]
			\item[(c)] $L$ is a scalar multiple of the identity on $\mathcal{H}(\C^\N)$ if and only if $L_n$ is a scalar multiple of the identity on $\mathcal{H}(\C^n)$, for every $n\in\N$.
		\end{enumerate}
	\end{lemma}
	\begin{proof}
		Let $n\in\N$, $f_n\in\mathcal{H}(\C^n)$ and $ f:=f_n\circ\pi_n\in\mathcal{H}(\C^\N)$.
		\begin{enumerate}
			\item [(a)]  Note that
			\begin{equation}\label{GJ}
			L_nf_n=J_n^*\circ L\circ\pi_n^*(f_n)=J_n^*\circ L(f)=(Lf)\circ J_n.
			\end{equation}
			Let $a\in\C^n$. We want to show that $\tau_a\circ L_n=L_n\circ\tau_a$. Applying \eqref{GJ} we have
			\begin{align*}
			[\tau_a\circ L_n](f_n)(z)&=\tau_a(L_nf_n)(z)=(L_nf_n)(z-a)=(Lf)\circ J_n(z-a)=(Lf)(J_n(z)-J_n(a))\\
			&=[\tau_{J_n(a)}(Lf)](J_n(z)),
			\end{align*}
			for every $z\in\C^n$ and so $[\tau_a\circ L_n](f_n)=[\tau_{J_n(a)}(Lf)]\circ J_n$. Using the fact that $L$ is a convolution operator we get
			\begin{align*}
			[\tau_a\circ L_n](f_n)&=[L(\tau_{J_n(a)}f)]\circ J_n=[L((\tau_af_n)\circ \pi_n)]\circ J_n=[L\circ \pi_n^*(\tau_af_n)]\circ J_n\\&=J_n^*\circ L\circ \pi_n^*(\tau_af_n)
			=[L_n\circ\tau_a](f_n).
			\end{align*}

			\item [(b)] Applying Lemma \ref{Lm1} to the entire function $f_n\circ\pi_n$ and using \eqref{GJ} we get
			\begin{equation}\label{GJchave}
			L(f_n\circ\pi_n)=\pi_n^*((L(f_n\circ\pi_n)) \circ J_n)=\pi_n^*(L_nf_n)=(L_nf_n)\circ\pi_n.
			\end{equation}
			\item[(c)] Let $\lambda\in\C$ be such that $Lg=\lambda g$ for every $g\in\mathcal{H}(\C^\N)$. Then
			\[(L_nf_n)\circ\pi_n=Lf=(\lambda f_n)\circ\pi_n.\]
			Since $\pi_n^*$ is injective, it follows that $L_nf_n=\lambda f_n$. Therefore $L_n$ is a scalar multiple of the identity.

			Conversely, suppose that for each $n\in\N$ there exists $\lambda_n\in\C$ such that $L_nf_n=\lambda_nf_n$ for every $f_n\in\mathcal{H}(\C^n)$. It is not difficult to verify that, if $g\in\pi_n^*(\mathcal{H}(\C^n))$, then $Lg=\lambda_ng$. Note that to prove the assertion it suffices to show that $\lambda_n=\lambda_m$ for any $n,m\in\N$. So, let $n,m\in\N$ with $n\leq m$. By (\ref{inclusao}) we may choose $g\in\mathcal{H}(\C^\N)$ such that $g\neq0$ and $g\in\pi_n^*(\mathcal{H}(\C^n))\subset\pi_m^*(\mathcal{H}(\C^m))$. Thus $\lambda_ng=Lg=\lambda_mg$ and since $g\neq0$ it follows that $\lambda_n=\lambda_m$.		\end{enumerate}
	\end{proof}

	Below we list some remarks about the previous lemma.

	\begin{remark}
		\begin{enumerate}
			\item[$(1)$] For $\xi\in\C^\N$, the convolution operator  $(\tau_\xi)_n=\tau_{\pi_n(\xi)}$ is a concrete example of convolution operator which is associated to the translation $\tau_{\xi}$.
			\item [$(2)$] The fact that $L$ commutes with all the translation operators is very important to show Lemma \ref{GJD}(b). In fact, if $L$ is a linear operator on $\mathcal{H}(\C^\N)$ that does not commute  with all the translation operators and $f=f_n\circ\pi_n\in\mathcal{H}(\C^\N)$ for some $n\in\N$, then it follows from Lemma \ref{Lm1} that there exists $N\in\N$, $N\geq n$ (not necessarily equal), such that $f=f_N\circ\pi_N$ and $Lf=\pi_N^*((Lf)\circ J_N)$. Now, using \eqref{GJ} we have $Lf=(L_Nf_N)\circ\pi_N$. Hence, we may not ensure that $L$ can be factored in the form $Lf=(L_nf_n)\circ\pi_n$.
			\item [$(3)$] Lemma \ref{GJD}(b) allows us to write the orbit  Orb$_L(f)$ in terms of convolution operators on $\mathcal{H}(\C^n)$, that is, if $f=f_n\circ\pi_n\in\mathcal{H}(\C^\N)$ and $L$ is a convolution operator on $(\mathcal{H}(\C^\N),\tau)$, then
			\[Lf=(L_nf_n)\circ\pi_n, \hspace{0,3cm}L^2f=L((L_nf_n)\circ\pi_n)=L_n(L_nf_n)\circ\pi_n=(L_n^2f_n)\circ\pi_n,\]
			and proceeding by induction it follows that
			\[L^kf=(L_n^kf_n)\circ\pi_n,\]
			for every $k\in\N_0$.
		\end{enumerate}
	\end{remark}

	\section{Linear dynamics of convolution operators on $\mathcal{H}(\C^\N)$}
	In this section we will study the linear dynamics of convolution operators on $\mathcal{H}(\C^\N)$. We start by proving that convolution operators on $\mathcal{H}(\mathbb{C}^{\mathbb{N}})$ are neither cyclic nor $n$-supercyclic for any $n\in\N$. This result improves a result of Fávaro and Mujica \cite{vinicius2016hypercyclic}, which states that no convolution operator on $\mathcal{H}(\C^\N)$ is hypercyclic.
	\begin{theorem}\label{nsuper}
		\begin{enumerate}
			\item [(a)] No convolution operator on $\mathcal{H}(\C^\N)$ is cyclic.
			\item [(b)] No convolution operator on $\mathcal{H}(\C^\N)$ is $n$-supercyclic, for any  $n\in\N$.
		\end{enumerate}
	\end{theorem}

	\begin{proof}
		Let $L$ be a convolution operator on $\mathcal{H}(\C^\N)$.
		\begin{enumerate}
			\item [(a)] Let $f\in\mathcal{H}(\C^\N)$. Then we may write  $f=f_n\circ\pi_n$ with $f_n\in\mathcal{H}(\C^n)$. By Lemma \ref{Lm1} the orbit of $f$ under $L$ is
			\[\textnormal{Orb}_L(f)=\{L^kf : k\in\N_0\}=\{\pi_{n}^*((L^kf)\circ J_n) : k\in\N_0\}\subset\pi_n^*(\mathcal{H}(\C^n)).\]
			Since $\pi_n^*(\mathcal{H}(\C^n))$ is a closed  proper subspace of $(\mathcal{H}(\C^\N),\tau)$, we have
			\[\overline{\textrm{span } \textnormal{Orb}_L(f)}^\tau\subset\pi_n^*(\mathcal{H}(\C^n)).\]
			Therefore $\textrm{span } \textnormal{Orb}_L(f)$  cannot be a dense subset of $(\mathcal{H}(\C^\N),\tau)$. So there is not a cyclic entire function for any convolution operator on $(\mathcal{H}(\C^\N),\tau)$. Hence $L$ is not cyclic.

			\item [(b)] Let $n\in\N, n>1$,  and let $V$ be an $n$-dimensional vector subspace of $\mathcal{H}(\C^\N)$ with generators $f_1,\ldots,f_n$. Then $L^k(V)$ is a vector subspace of $\mathcal{H}(\C^\N),$ generated by  $L^kf_1,\cdots,L^kf_n$, and with dimension less than or equal to $n$, for every $k\in\N_0$. If we write $f_1=f_{m_1}\circ\pi_{m_1}, \cdots,f_n=f_{m_n}\circ\pi_{m_n}$, with $f_{m_i}\in\mathcal{H}(\C^{m_i})$, for every $i=1,\cdots,n$, then it follows from Lemma \ref{Lm1} and (\ref{inclusao}) that
			\[L^k(V)\subset\pi_{m_1}^*(\mathcal{H}(\C^{m_1}))+\cdots+\pi_{m_n}^*(\mathcal{H}(\C^{m_n}))\subset \pi_{m}^*(\mathcal{H}(\C^{m})),\]
			for every  $k\in\N_0$, where $m:=\max\{m_i:  i=1,\dots, n\}$. Therefore
			\[\textnormal{orb}_L(V)=\bigcup_{k=0}^\infty L^k(V)\subset \pi_{m}^*(\mathcal{H}(\C^{m}))\]
			and so $\textnormal{orb}_L(V)$ cannot be dense in$(\mathcal{H}(\C^\N),\tau)$. Thus, no finite-dimensional subspace of $\mathcal{H}(\C^\N)$ is supercyclic for $L$. Hence $L$ is not $n$-supercyclic.
		\end{enumerate}

	\end{proof}

	\subsection{Li--Yorke chaos for convolution operators on $\mathcal{H}(\C^\N)$}

	The main result of this section is the following:

	\begin{theorem}\label{TLY}
		Every nontrivial convolution operator on $(\mathcal{H}(\C^\N),\tau)$ is Li--Yorke chaotic.
	\end{theorem}

	To show this result we will prove a characterization of Li--Yorke chaos that involves the existence of semi-irregular vector. This characterization was obtained by Bernardes Jr \textit{et al} \cite{Li-Yorke} for operators on Fréchet spaces. We will generalize this fact for Hausdorff topological vector spaces.	The definition below is known for Fréchet spaces.

	Let $E$ be a Hausdorff topological vector space and let $T$ be an operator on $E$. A vector $x\in E$ is said to be a \emph{semi-irregular vector} for $T$ if the sequence $(T^nx)$ does not converge to zero, but it has a subsequence converging to zero. It is easy to see that  $(x,y)\in E\times E$ is a Li--Yorke pair for $T$ if and only if $x-y$
	is a semi-irregular vector for $T$.

	As it was mentioned in \cite{Li-Yorke}, the notion of semi-irregularity makes sense only for infinite-dimensional spaces. An easy application of the Jordan form implies that there are no semi-irregular vectors for operators on finite-dimensional spaces.

	\begin{theorem}\label{TsI}
		Let $E$ be a Hausdorff topological vector space, and let $T$ be an operator on $E$. The following assertions are equivalent:
		\begin{enumerate}
			\item [(i)] $T$ is Li--Yorke chaotic.
			\item [(ii)] $T$ admits a Li--Yorke pair.
			\item [(iii)] $T$ admits a semi-irregular vector.
		\end{enumerate}
	\end{theorem}
	\begin{proof}Since the implications
		(i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) are immediate we just need to show that
		(iii) $\Rightarrow$ (i).  Let $x$ be a semi-irregular vector for $T$. Then for every $\alpha, \lambda\in\C$, with $\alpha\neq\lambda$, the sequence $\{T^n(\alpha x-\lambda x)\}$ does not converge to zero, but it has a subsequence converging to zero. Hence span$\{x\}$ is an uncountable scrambled set for $T$ and thus $T$ is Li--Yorke chaotic.
	\end{proof}

	Now we are able to prove the main result of this section.

	\begin{proof}[Proof of Theorem \ref{TLY}] Let $L$ be a nontrivial convolution operator on $(\mathcal{H}(\C^\N),\tau)$. We will show that $L$ has a semi-irregular entire function. By Lemma \ref{GJD}(c) there exists a nontrivial convolution operator $L_n$ associated to $L$. Since $L_n$ is a nontrivial convolution operator on $\mathcal{H}(\C^n)$, it follows from the classical result of Godefroy and Shapiro that $L_n$ is hypercyclic. In particular there exists a semi-irregular function $f_n\in\mathcal{H}(\C^n)$ for $L_n$. If we set $f=f_n\circ\pi_n\in\mathcal{H}(\C^\N)$, then $f$ is a semi-irregular function for $L$. In fact, since
		\begin{equation}\label{w}
		L^kf=(L_n^kf_n)\circ\pi_n=\pi_n^*\left(L_n^kf_n\right), \ \ \textnormal{for every} \ k\in\N_0,
		\end{equation}
		it follows immediately from \eqref{w} that the sequence $(L^kf)_{k=0}^\infty$ has a subsequence converging to zero. On the other hand, if $L^kf\to0$ in the topology of $(\mathcal{H}(\C^\N),\tau)$, then
		\[L_n^kf_n=J_n^*\circ\pi_n^*\left(L_n^kf_n\right)=J_n^*(L^kf)\to0 \ \textnormal{in} \ \mathcal{H}(\C^n)\]
		when $k\to\infty$, but this contradicts the fact  that $f_n$ is a semi-irregular function for $L_n$. Therefore $(L^kf)_{k=0}^\infty$ does not  converge to zero, and so $f$ is a semi-irregular function for $L$. Applying Theorem \ref{TsI} we obtain the desired.
	\end{proof}

config: arxiv


TITLE: Volume of sphere $x^2+y^2+z^2 =a^2$ and over lemniscate $r^2=a^2\cos2\theta$
QUESTION [0 upvotes]: Volume of sphere $x^2+y^2+z^2 =a^2$ and over lemniscate $r^2=a^2\cos2\theta$
Now I am having trouble finding limits of integration.
I know limits for $z$ are $ \pm (a^2 - (x^2 + y^2))$ and limits for $r$ are $\pm (a \sqrt{\cos2\theta})$.
How do I find limits of $\theta$ ?
Thanks

REPLY [1 votes]: Hint:
In order to find the limits for $\theta$ notice that $\theta\in(-\pi/4,\pi/4)$ gives a curve in the right half plane. Also $\theta \in(\pi/4, 3\pi/4)$ gives a curve in the left half plane. Maybe drawing the graph can help you.


TITLE: Proof of every measurable cardinal carries a normal measure
QUESTION [1 upvotes]: I'm reading the proof of Theorem 10.20 in Set Theory by Jech and I don't understand the last argument.
The theorem says every measurable cardinal carries a normal measure. The proof goes: Let $U$ be a nonprincipal $\kappa$-complete ultrafilter on $\kappa$. For $f$ and $g$ in $\kappa^\kappa$, define $f<g$ if and only if $\{\alpha<\kappa:f(\alpha)<g(\alpha)\}\in U$. Now let $f:\kappa\rightarrow\kappa$ be the least function with the property that for all $\gamma<\kappa$, $\{\alpha:f(\alpha)>\gamma\}\in U$. Let $D=\{X\subset\kappa: f^{-1}(X)\in U\}$.
Then we claim $D$ is a normal measure: Let $h$ be a regressiv efunction on a set $X\in D$. Let $g$ be the function defined by $g(\alpha)=h(f(\alpha))$. As $g(\alpha)<f(\alpha)$ for all $\alpha\in f^{-1}(X)$, we have $g<f$. Here is the part I don't understand: It follows by minimality of $f$ that $g$ is constant on some $Y\in U$. Hence $h$ is constant on $f(Y)$ and $f(Y)\in D$.
Why does it follow from the minimality of $f$ that $g$ is constant on some $Y\in U$? And $h$ is constant on $f(Y)$ and $f(Y)\in D$?

REPLY [1 votes]: By minimality of $f$, we have $$\{\alpha:g(\alpha)\le\gamma\}\in U $$ for some $\gamma,$ so by $\kappa$-completeness, there is a $\beta \le \gamma$ such that $\{\alpha:g(\alpha)=\beta\}\in U.$
(Your other questions can be answered by straightforward definition chasing. For instance that $h$ is constant on $f(Y)$ is a trivial consequence of $g$ being constant on $Y.$)


TITLE: maximizing the area of a triangle with given angle and opposite side
QUESTION [2 upvotes]: Let $\Delta ABC$ have given angle $\theta$ at vertex A and opposite side BC of given length $a$.
I would like to find the maximum area of such a triangle in terms of $\theta$ and $a$.
I think the triangle of maximum area is isosceles, but how can I show this
(if this is correct)?

(I realize that this can also be worked using calculus, but I am looking for a more elementary solution.)

REPLY [0 votes]: Because you don't really know where point A should be relative to the segment BC, let me suggest the following: Take the perpendicular bisector of BC and at C construct half of angle A on one side of BC and the complement of half of angle A at C on the other side of BC. Suppose the new sides of those two angles meet the perpendicular bisector of BC at D and E respectively.  Let O be the midpoint of segment DE.  Then any point on the arc BEC of the circle with center O could be the point A, because an angle at a point on a circle measures half of the arc that it "subtends", so for any point A on that arc, angle BAC measures the same as angle BEC, which is the desired angle A.  (To see this, recall that segment BC is perpendicular to DE -- say they meet at F -- and because we built angle FCE to be the complement of half of angle A, angle FEC is half of angle A, so angle BEC is equal to angle A.)  The choice of point A on the arc BEC that gives the greatest area is the one that is the greatest perpendicular distance away from segment BC, and that is the point E, halfway along that arc.  So, yes, the greatest area comes from an isosceles triangle: the lengths EB and EC are congruent.


text:
\section{Introduction}

The general scenario for information reconciliation is one in which two parties have two sets of correlated data with some discrepancies between them. The situation is equivalent to transmit the data from one party to the other through a noisy channel, akin in the satellite scenario described by Maurer~\cite{Maurer_93}.

In a Quantum Key Distribution (QKD) protocol, errors are generated in the communications channel either by the interaction of the quantum information carrier with the environment, by imperfections in the QKD device or by an eavesdropper. The two parties participating in the communication, Alice and Bob, thus have two sets of correlated data from which a common set must be extracted. This problem has been previously subject to consideration \cite{Brassard_94,Watanabe_07,Watanabe_08,Leverrier_08,Elkouss_09}. It is a process known as key distillation, that requires a discussion carried over an authenticated classical channel. It is interactive in the sense that it needs communications through the channel. Since it can also be listened by an eavesdropper, it is important to minimize the amount of information that have to be transmitted in the reconciliation process. Any extra information limits the performance of the QKD implementation. In theory one could minimize the information leakage using a highly interactive protocol, but in practical applications this would lead to a prohibitively large communication overhead through the network, limiting also the effective keyrate.

It is in this scenario where modern Forward Error Correction (FEC) is an interesting solution. The idea is to make use of FEC's inherent advantage of requiring a single channel use to reconcile the two sets.
In~\cite{Elkouss_09} it was analyzed the use of a discrete number of Low-Density Parity-Check (LDPC) codes optimized for the binary symmetric channel. As a consequence the efficiency exhibited an staircase-like behaviour: each code was used within a range of error rates and the reconciliation efficiency was maximized only in the region close to the code's threshold.

In this work, we develop the idea of using LDPC codes optimized for the binary symmetric channel. We take these codes as an starting point and develop a rate compatible information reconciliation protocol with an efficiency close to optimal. In particular, the proposed protocol builds codes that minimize the exchanged information for error probabilities between $1\%$ and $10\%$\footnote{The maximum error thresholds for extracting an absolute secret key in a QKD protocol is $11\%$~\cite{Gisin_02}.}, the expected values in real implementations of QKD systems.

This solution addresses the rate adaptation problem (open problem 2) from the recent review paper of Matsumoto~\cite{Matsumoto_09} in which he lists the problems that an LDPC solution should overcome in order to compare advantageously to current interactive reconciliation solutions.

The paper is organised as follows:
In Section~\ref{sec-ratecompir} the main ideas are discussed. A new Information Reconciliation Protocol able to adapt to different channel parameters is presented and its asymptotic behavior discussed.
In Section~\ref{sec-results} the results of a practical implementation of the protocol are shown. In particular we have analyzed the rate compared to the optimal value and the reconciliation efficiency.


TITLE: How do I show that a complex number $z$ is a root of $z^6-1$?
QUESTION [3 upvotes]: This problem would be less confusing to me if one was not subtracted from $z^6$.
Here is the problem:
If $z=cos(\frac{\pi}{3})+jsin(\frac{\pi}{3})$, show that $z$ is a root of $z^6-1$.
My approach to this problem involved using de Moivre's theorem to convert $z$ to it's complex polar form. I think converting it to complex polar form first makes it easy to identify the root:

$$z=cos(\frac{\pi}{3})+jsin(\frac{\pi}{3})$$
$$a=cos(\frac{\pi}{3}), b=sin(\frac{\pi}{3})$$
$$r=\sqrt{cos(\frac{\pi}{3})^2+sin(\frac{\pi}{3})^2}$$
$$r=1$$
$$\theta=Tan^1(\frac{b}{a})$$
$$\theta=Tan^-1(\frac{sin(\frac{\pi}{3})}{cos(\frac{\pi}{3})})$$
$$\theta=\frac{\pi}{6}$$
$$\therefore z=1\angle{\frac{\pi}{6}}$$

This part is the confusing bit. Because $1$ is subtracted from $z^6$, I decided to expand the whole thing and then show that $z$ is a root of $z^6-1$.
$$(cos(\frac{\pi}{3})+jsin(\frac{\pi}{3}))^6-1$$
$$(\frac{1}{2}+\frac{\sqrt{3}}{2}j)^6-1$$

Noting that $j^6=-1$...
$$(\frac{1}{2}-\frac{\sqrt{3}}{2})^6-1$$

The equation becomes...
$$(-\frac{13}{32})-1$$

Which equates to...
$$-\frac{45}{32}$$

Convert to complex polar form...
$$z^6=-\frac{45}{32}\angle{0rad}$$
$$z^6=(-\frac{45}{32}\angle{0rad})^6$$
$$z^6=(-\frac{45}{32})^6\angle{(0)}^6$$
$$z^6=(-\frac{45}{32})^6\angle{6(0+2\pi n)}$$
$$z^6=(-\frac{45}{32})^6\angle{(12\pi-11\pi n)}$$
$$z^6=(-\frac{45}{32})^6\angle{(\pi n)}$$
Noting that $z$ is raised to the first power, make $n=1$
$$z^6=(-\frac{45}{32})^6\angle{\pi (1)}$$

I get...
$$z^6=(-\frac{45}{32})^6\angle{\pi}$$

Clearly the complex polar form I got is not equal to that of $z$. Can someone lead me to the right path?

REPLY [2 votes]: Not really sure what our OP is trying to do here; what the intended method might be.  But I'm pretty sure that
$\theta = \text{Tan}^{-1} \left ( \dfrac{\sin(\pi/3)}{\cos (\pi/3)} \right ) = \text{Tan}^{-1} ( \tan (\pi/3)) = \dfrac{\pi}{3} \ne \dfrac{\pi}{6}, \tag 1$
so it seems like the first wrong turn was made at this point.  And I can't see the logical connection 'twixt
$\left ( \dfrac{1}{2} + j \dfrac{\sqrt 3}{2} \right)^6 \tag 2$
and
$\left (\dfrac{1}{2} - \dfrac{\sqrt3}{2} \right )^6; \tag 3$
so the OP's remarks from this point on are pretty much difficult for me to decipher.
Nevertheless, we do have de Moivre's formula,
$(\cos \theta + j\sin \theta)^n = \cos (n\theta) + j \sin(n\theta); \tag 4$
and if we set
$z = \cos \dfrac{\pi}{3} + j\sin \dfrac{\pi}{3}, \tag 5$
then
$z^6 = \left ( \cos \dfrac{\pi}{3} + j\sin \dfrac{\pi}{3} \right)^6 = \cos \left (6 \dfrac{\pi} 3 \right ) + j \sin \left (6 \dfrac{\pi}{3} \right) = \cos 2 \pi + j \sin 2\pi = \cos 2 \pi = 1,  \tag 6$
whence
$z^6 - 1 = 0. \tag 7$