TITLE: Exact sequence involving pushout
QUESTION [4 upvotes]: I think I have a bit of a misunderstanding of pushouts. Let $U, X, Y$ be objects of an abelian category $A$ with morphisms $f: U \rightarrow X$ and $g: U \rightarrow Y.$ Let $Z$ be the pushout of this diagram with natural morphisms $f': X \rightarrow Z$ and $g': Y \rightarrow Z.$ Then apparently we have a short exact sequence $$0 \rightarrow U \rightarrow X \oplus Y \rightarrow Z \rightarrow 0 $$ where the map $U \rightarrow X \oplus Y$ is defined as $u \mapsto (f(u), g(u))$ and the map $X \oplus Y \rightarrow Z$ is denoted by $(f', g').$ I cannot see how this is true. My book states the mapping of the short exact sequence in this way but I think the map $X \oplus Y \rightarrow Z$ should rather be $(-f', g')$ or $(f', -g')$ where $(-f', g')(x, y) = -f'(x) + g'(y).$ Am I right or am I misunderstanding something?

Here is a screenshot of the passage. The arrow notation is a bit different...

REPLY [4 votes]: $\require{AMScd}$In an abelian category, a commutative square
  \begin{CD}
U@>f>> X\\
@VgVV@VVg'V\\
Y@>>f'> Z
\end{CD}
  is a pushout if and only if the sequence
  $$U\xrightarrow{[f,g]}X\oplus Y\xrightarrow{\left[\genfrac{}{}{0pt}{}{g'}{-f'}\right]}Z\to 0$$
  is exact.

The exactness of the sequence is equivalent to the fact that $\left[\genfrac{}{}{0pt}{}{g'}{-f'}\right]$ is the cokernel of $[f,g]$.
The square is commutative if and only if $fg'=gf'$ (diagrammatic composition order) which is equivalent to
$$[f,g]\left[\genfrac{}{}{0pt}{}{g'}{-f'}\right]=fg'-gf'=0$$
On the other hand, if $[f,g]\left[\genfrac{}{}{0pt}{}{p}{q}\right]=fp+gq=0:U\to V$, then $fp=-gq$ hence there exists one and only one morphism $h:Z\to V$ such that $p=g'h$ and $-q=f'h$, hence
\begin{align}
\left[\genfrac{}{}{0pt}{}{p}{q}\right]
&=\left[\genfrac{}{}{0pt}{}{g'h}{-f'h}\right]\\
&=\left[\genfrac{}{}{0pt}{}{g'}{-f'}\right]h
\end{align}
thus proving that $\left[\genfrac{}{}{0pt}{}{g'}{-f'}\right]$ is the cokernel of $[f,g]$.


TITLE: Probability of "at least amount"
QUESTION [1 upvotes]: I have 4 six-sided dice.  One of them has values of [0,0,3,3,3,4], and the other three have values of [0,0,0,1,1,2].  The first dice has a 2/3 chance of rolling at least a 3.  If I roll the other three dice together, what are the odds of rolling at least a three?
I don't know what to call this, but that is the best way I know how to ask that question.  I have many dice like this in a game I play and I'd like to do some analysis.  Mostly six-sided dice with values ranging from zero to four, but there are some ten-sided dice with values ranging from zero to four, so a general formula would be great.
So as a secondary question, I'd also like to solve that if I have a ten-sided dice with values [0,0,0,0,0,0,0,4,4,4], and three of the aforementioned [0,0,0,1,1,2] dice, what are the odds of rolling at least a sum of 5?
Thanks!

REPLY [2 votes]: We solve the problem of the probability of getting at least $3$ using three dice that have faces $(0,0,0,1,1,2)$. Mainly it is to illustrate technique.
It is easier to find the probability $p$ of getting a sum less than $3$, that is, a sum of $0$, $1$, or $2$. Then the answer to your problem is $1-p$.
The following are all the ways we could end up with sum $\le 2$:
(i) all $0$;
(ii) two $0$ and a $1$;
(iii) two $0$ and a $2$;
(iv) one $0$ and two $1$;
We find the probability of each and add up.
(i) The probability is $\left(\frac{3}{6}\right)^2$.
(ii) The die that shows a $1$ can be chosen in $3$ ways. The probability it shows a $1$ is $\frac{2}{6}$. The probability the other two show $0$ is $\left(\frac{3}{6}\right)^2$. Thus we get probability $3\cdot \frac{2}{6}\cdot \left(\frac{3}{6}\right)^2$.
(iii) Essentially the same reasoning gives $3\cdot \frac{1}{6}\cdot \left(\frac{3}{6}\right)^2$.
(iv) The reasoning is the same as for (ii). We get $3\cdot \frac{3}{6}\cdot \left(\frac{2}{6}\right)^2$.
Add all the above probabilities to get $p$. Then as mentioned earlier the answer to your problem is $1-p$.
Remark: If problems get much more complicated, and there are quite a few, simulation may be the way to go.


TITLE: Approximating dense subspaces of Fréchet spaces
QUESTION [6 upvotes]: If $H$, $H_0$ are two separable Hilbert spaces and $H$ is continuously and densly embedded in $H_0$, it is possible to construct a sequence of linear operators
$$ P_n : H_0 \to H $$
such that for all $x \in H_0$ one has convergence $P_n x \to x$ in the $H_0$-norm.
The motivation is to generalize the idea of smoothing operators. For example, $H^2(\mathbb R)$ is densely embedded in $L^2(\mathbb R)$ and it is well known that one can approximate each $L^2$-function by an $H^2$-function. In this case $P_n$ could be a convolution with a smooth kernel with width $1/n$.
Can this be generalized to Fréchet spaces or even beyond? If $H$ and $H_0$ are Fréchet spaces rather than Hilbert spaces, is it always possible to construct a sequence of bounded operators $P_n$ as above?
In the Hilbert case one way to construct $P_n$ is by considering an unbounded, self-adjoint operator $A : D(A) \to H_0$, with $D(A)=H$, that represents the inner product via
$$ \langle v,w \rangle_H = \langle Av, Aw \rangle_{H_0}\,. $$
If $\{ P_\Omega\}$ is the spectral measure accociated to $A$, then we can use $P_{[-n,n]}$ as our sequence of operators. It is less clear to me, what to with Fréchet spaces.

REPLY [3 votes]: Thanks to Jochen, Matthew and Bill, this is a detailed proof for Fréchet spaces.
Proposition. Let $E$ be a separable Fréchet space with the bounded approximation property, $F$ a topological vector space, continuously and densely embedded in $E$. Then there exists a sequence of continuous linear maps $P_n : E \to F$, such that
$$
\forall x \in E\,:\, P_n x \to x \text{ in }E\,.
$$
Proof.
Let $(x_n)_{n \in \mathbb N}$ be a countable, dense sequence in $E$ and $(\|\cdot\|_n)_{n \in \mathbb N}$ an increasing fundamental system of seminorms. We assumed that $E$ has the bounded approximation property, hence there exists an equicontinuous sequence of linear maps $T_n : E \to E$ with finite rank that converge to $\operatorname{Id}_E$, uniformly on compact sets. By passing to a subsequence we can assume that
$$
\| T_n x_j - x_j \|_n \leq \frac 1n \text{ for }j \leq n\,.
$$
Due to equicontinuity there exists for each $m$, an $N_m \in \mathbb N$ and $C_m>0$ such that
$$
\forall n \in \mathbb N\,,\; \forall x \in E\,:\, \| T_n x \|_m \leq C_m \| x \|_{N_m}\,.
$$
For each $n$, the space $T_n(E)$ is finite dimensional. Let $n'=n'(n)$ be such that $\|\cdot\|_{n'}$ is a norm on $T_n(E)$. We can construct a map $S_n : T_n(E) \to F$ with
$$
\| S_ny - y \|_{n} \leq \frac 1n \| y \|_{n'}\,,
$$
for all $y \in T_n(E)$.
To see that this is possible choose a basis $y_1, \dots, y_m$ of $T_n(E)$ and note that it is sufficient to define $S_n(y_i) \in F$, such that $\| S_n(y_i) - y_i \|_{n}$ is small enough. This is possible, because $F$ is dense in $E$. Define $P_n = S_n T_n$.
We have to show convergence $P_n x \to x$. Fix $x \in E$ and a seminorm $\|\cdot\|_m$. For $n$ and $k$ satisfying $m ,\,N_m,\,N_{m'} \leq n$ and $k \leq n$ we have
\begin{align*}
\| P_n x &- x \|_m \leq \|S_n T_n(x-x_k) -T_n(x-x_k) \|_m + \| T_n(x-x_k) \|_m + \\
&\qquad\qquad\qquad
+ \|S_n T_n x_k - T_n x_k \|_m  +\| T_n x_k - x_k\|_m  + \| x_k - x\|_m \\
&\leq \frac 1n \| T_n(x-x_k) \|_{m'} + C_m \| x - x_k \|_{N_m} + \frac 1n \| T_n x_k \|_{m'} + \frac 1n + \|x_k - x \|_m \\
&\leq
\frac {C_{m'}}{n} \| x-x_k \|_{N_{m'}} +  C_m \| x - x_k \|_{N_m}
+ \frac {C_{m'}}n \left( \| x\|_{N_{m'}} + \| x-x_k \|_{N_{m'}} \right) + \\
&\qquad\qquad + \frac 1n + \|x_k - x \|_m \,.
\end{align*}
We see that by choosing $n$ large enough and $\|x - x_k\|_n$ small enough we can achieve convergence.


text:
\begin{document}

\title{Weak Quasicircles Have Lipschitz Dimension 1}

\author{David M. Freeman}
\address{University of Cincinnati Blue Ash College, Department of Math, Physics, and Computer Science, 9555 Plainfield Rd., Cincinnati, OH 45236, United States}
\email{david.freeman@uc.edu}

\subjclass[2020]{30L05 (Primary); 54F45, 30C65 (Secondary)}

\date{\today}

\commby{Nageswari Shanmugalingam}

\begin{abstract}
We prove that the Lipschitz dimension of any bounded turning Jordan circle or arc is equal to 1. Equivalently, the Lipschitz dimension of any weak quasicircle or arc is equal to 1.
\end{abstract}

\maketitle

\section{Introduction}

In \cite{CK13}, Cheeger and Kleiner introduced the concept of Lipschitz dimension and proved deep results about metric spaces of Lipschitz dimension at most 1. Subsequently, in \cite{David19}, David further developed various  properties of Lipschitz dimension. While studying the non-invariance of Lipschitz dimension under quasisymmetric mappings in a general metric space setting, David asks in Question 8.7 of \cite{David19} if every quasisymmetric image of the unit interval (that is, a \textit{quasiarc}) has Lipschitz dimension equal to 1. We answer this question in the affirmative, thus demonstrating that the Lipschitz dimension of the unit interval is invariant under quasisymmetric homeomorphisms. In fact, we prove something stronger: the Lipschitz dimension of the unit interval is invariant under weakly quasisymmetric homeomorphisms. This can be derived from the following theorem, which constitutes the main result of this paper (see \rf{S:defs} for definitions).

\begin{theorem}\label{T:main}
Bounded turning Jordan circles/arcs have Lipschitz dimension $1$.
\end{theorem}

Since bounded turning Jordan arcs need not be metrically doubling, not all bounded turning Jordan arcs are quasiarcs. On the other hand, every quasiarc is bounded turning. Analogous statements hold for Jordan circles. Therefore, in answer to \cite[Question 8.7]{David19}, we obtain the following corollary.

\begin{corollary}\label{C:main}
Quasicircles/arcs have Lipschitz dimension $1$.
\end{corollary}

We also point out results of Cheeger and Kleiner pertaining to spaces of Lipschitz dimension at most 1. In particular, via \cite[Theorem 1.7]{CK13}, we have the following corollary to \rf{T:main}.

\begin{corollary}\label{C:embed}
If $\Gamma$ is a bounded turning Jordan circle or arc, then there exists a measure space $(X,\mu)$ such that $\Gamma$ admits a bi-Lipschitz embedding into $L_1(X,\mu)$.
\end{corollary}

Furthermore, via \cite[Theorem 1.11]{CK13}, we also have the following.

\begin{corollary}\label{C:inverse}
If $\Gamma$ is a bounded turning Jordan circle or arc, then $\Gamma$ is bi-Lipschitz homeomorphic to an inverse limit of an admissible inverse system of graphs.
\end{corollary}

Indeed, given a metric space $X$ and a Lipschitz light map $F:X\to\mathsf{R}$,  Cheeger and Kleiner explicitly construct an admissible inverse system of graphs whose inverse limit is bi-Lipschitz homeomorphic to $X$ (see \cite[Section 4]{CK13}).

\rf{T:main} is also relevant to the following result of David.

\begin{theorem}[Theorem 5.9 of \cite{David19}]\label{T:Carnot}
Let $\mathbb{G}$ denote a non-abelian Carnot group, and let $K\subset \mathbb{G}$ denote a compact subset of positive measure. Then the Lipschitz dimension of $K$ is equal to $\infty$.
\end{theorem}

Since the Lipschitz dimension of a compact space is bounded from below by its topological dimension (see \cite[Observation 1.4]{David19}), given an integer $n\geq1$, we note that the Lipschitz dimension of the product of $n$ bounded turning Jordan arcs is at least $n$. Furthermore, the Lipschitz dimension of the product of $n$ bounded turning Jordan arcs is bounded from above by $n$ (here we use \rf{T:main} and \cite[Proposition 3.1]{David19}). Therefore, the Lipschitz dimension of the product of $n$ bounded turning Jordan arcs is equal to $n$. Since Lipschitz dimension is invariant under bi-Lipschitz homeomorphisms, we arrive at the following corollary.

\begin{corollary}\label{C:nonembed}
Let $\mathbb{G}$ denote a non-abelian Carnot group. If $K\subset \mathbb{G}$ is a compact subset of positive measure, then $K$ does not admit a bi-Lipschitz embedding into a product of finitely many bounded turning Jordan arcs.
\end{corollary}

In connection with the title of this paper, we also point out the following theorem of Meyer. In particular, all the above results pertaining to bounded turning Jordan circles/arcs can be understood as results about weak quasicircles/arcs.

\begin{theorem}[Theorem 1.1 of \cite{Meyer11}]\label{T:Meyer}
A Jordan circle/arc $\Gamma$ is a weak quasicircle/arc if and only if $\Gamma$ is bounded turning.
\end{theorem}

Finally, we highlight the following result of Herron and Meyer, as it is foundational to our proof of \rf{T:main}.

\begin{theorem}[\cite{HM12}]
If $\Gamma$ is a bounded turning Jordan circle, then $\Gamma$ is bi-Lipschitz homeomorphic to some Jordan circle in $\mathcal{S}_1$.
\end{theorem}

Here $\mathcal{S}_1$ is the collection of all Jordan circles given by dyadic diameter functions constructed using the snowflake parameter $\sigma=1$ (see \cite{HM12} for definitions). This result allows us to distort any given bounded turning Jordan circle into a form more amenable to the construction of a Lipschitz light map into $\mathsf{R}$.

The organization of this paper is as follows. In \rf{S:defs} we provide a few key definitions. Then, in \rf{S:prelims}, we investigate various aspects of Jordan circles in $\mathcal{S}_1$. In \rf{S:lip}, we construct a $1$-Lipschitz mapping from any Jordan circle $\Gamma\in\mathcal{S}_1$ into the unit circle. Finally, in \rf{S:light} we prove that this mapping is Lipschitz light via a series of technical lemmas.

\begin{acknowledge}
The author thanks Guy C. David for enlightening dialogue regarding the results and theoretical context of both this paper and \cite{David19}.
\end{acknowledge}

\section{Basic Definitions}\label{S:defs}

We write $\mathsf{N}$ to denote the set $\{0,1,2,\dots\}$ consisting of non-negative integers, and $\mathsf{R}$ to denote the Euclidean line.

Given two metric spaces $X$ and $Y$ and a constant $L\geq 1$, we say that an embedding $f:X\to Y$ is \textit{$L$-Lipschitz} provided that, for all points $x,y\in X$, we have $d(f(x),f(y))\leq L\,d(x,y)$. Furthermore, an embedding is \textit{$L$-bi-Lipschitz} if it is also true that $d(x,y)\leq L\,d(f(x),f(y))$.

An embedding $f:X\to Y$ is \textit{quasisymmetric} provided there exists a homeomorphism $\eta:[0,\infty)\to[0,\infty)$ such that, for all points $x,y,z\in X$ and $t\in[0,\infty)$,
\[d(x,y)\leq t\,d(x,z)\quad\text{implies}\quad d(f(x),f(y))\leq \eta(t)d(f(x),f(z)).\]
An embedding $f:X\to Y$ is \textit{weakly quasisymmetric} provided there exists a constant $H\geq 1$ such that, for all $x,y,z\in X$,
\[d(x,y)\leq d(x,z)\quad \text{implies}\quad d(f(x),f(y))\leq H d(f(x),f(z)).\]
While all quasisymmetries are weak quasisymmetries (with $H:=\eta(1)$), in general, a weak quasisymmetry need not be a quasisymmetry. We refer to the weak quasisymmetric image of the unit circle as a \textit{weak quasicircle}, and such an image of the closed unit interval as a \textit{weak quasiarc}. Thus every quasicircle/arc is a weak quasicircle/arc. Conversely, by \cite[Theorem 4.9]{TV80}, every weak quasicircle/arc that is \textit{metrically doubling} is a quasicircle/arc. Here we say that a space $X$ is metrically doubling if there exists $D\geq 1$ such that that any open ball of radius $2r>0$ can be covered by $D$ open metric balls of radius $r$. For additional information about weak quasicircles and quasiarcs, we refer the reader to \cite{Meyer11} and references therein.

A \textit{Jordan circle} $\Gamma$ is a homeomorphic image of the unit circle. Given two points $x,y\in \Gamma$, we write $\Gamma(x,y)$ to denote a component of $\Gamma\setminus\{x,y\}$ of minimal diameter. We write $\Gamma[x,y]$ to denote the topological closure of $\Gamma(x,y)$; thus $\Gamma[x,y]=\Gamma(x,y)\cup\{x,y\}$. Analogously, a \textit{Jordan arc} $\Gamma$ is a homeomorphic image of the closed unit interval. In this setting, given two points $x,y\in\Gamma$, we write $\Gamma(x,y)$ to denote the connected component of $\Gamma\setminus\{x,y\}$. Again, $\Gamma[x,y]=\Gamma(x,y)\cup\{x,y\}$.

A Jordan circle or arc is said to be  \textit{bounded turning} provided that there exists a constant $C\geq 1$ such that, for all pairs of points $x,y\in\Gamma$, we have $\diam(\Gamma[x,y])\leq C\,d(x,y)$. In this case we say that $\Gamma$ is $C$-bounded turning. This property is at times referred to in the literature as \textit{linear connectivity}.

Given a metric space $(X,d)$ and $\delta>0$, a \textit{$\delta$-sequence} in $X$ is a finite sequence of points $\{x_i\}_{i=0}^n$ such that, for each $0\leq i\leq n-1$, we have $d(x_i,x_{i+1})\leq \delta$. A subset $U\subset X$ is said to be $\delta$-connected if every pair of points in $U$ is contained in a $\delta$-sequence in $U$. A \textit{$\delta$-component} of $X$ is a maximal $\delta$-connected subset of $X$.

We say that a map $F:X\to Y$ is \textit{Lipschitz light} provided there exists $C\geq 1$ such that $F$ is $C$-Lipschitz, and, for every $r>0$ and every subset $E\subset Y$ with $\diam(E)\leq r$, the $r$-components of $F^{-1}(E)$ have diameter bounded above by $C\,r$. Here we employ the definition of Lipschitz light used in \cite[Definition 1.2]{David19}. As shown by David in \cite[Section 1.4]{David19}, this definitions is equivalent to \cite[Definition 1.14]{CK13} for maps into $\mathsf{R}^n$ (for $n\geq1$).

A metric space $X$ has \textit{Lipschitz dimension} $n\in\mathsf{N}$ if $n$ is minimal such that there exists a Lipschitz light map $F:X\to\mathsf{R}^n$.

\section{Preliminary Results}\label{S:prelims}

Following \cite{HM12}, we view the unit circle $\mathsf{S}$ as $[0,1]/\{0,1\}$, the closed unit interval whose endpoints are identified. We equip $\mathsf{S}$ with the arc-length metric $\lambda$. That is, for two points $s,t\in\mathsf{S}$ such that $0\leq s\leq t\leq 1$, we have
\[\lambda(s,t):=\min\{t-s,1-(t-s)\}.\]
The space $\mathsf{S}$ inherits the usual left-to-right orientation on $[0,1]$.

Given $n\in\mathsf{N}$, we write $\mathcal{I}_n$ to denote the collection of $2^n$ closed dyadic intervals in $[0,1]$, each of length $2^{-n}$. We write $\hI_n$ to denote the collection $\cup_{m=0}^n\mathcal{I}_m$. Furthermore, we write $\hat{\mathcal{I}}$ to denote the collection $\cup_{n=0}^\infty\mathcal{I}_n$. Given an interval $I\in\hat{\mathcal{I}}$, we write $l(I)$ to denote the unique index $n\in\mathsf{N}$ such that $I\in \mathcal{I}_n$. For convenience, we use the language of a dyadic tree to describe intervals in $\hat{\mathcal{I}}$. In particular, given any $I\in\hat{\mathcal{I}}$, there are exactly two dyadic \textit{children} contained in $I$, and $I$ is contained in its unique dyadic \textit{parent} interval.

Similarly, we write $\mathcal{D}_n$ to denote the collection of $2^{n}$ dyadic endpoints of intervals in $\mathcal{I}_n$. For example, $\mathcal{D}_0=\{0\}=\{1\}$, $\mathcal{D}_1=\{0,1/2\}=\{1,1/2\}$, etc. Note that, for each $n\in\mathsf{N}$, we have $\mathcal{D}_n\subset\mathcal{D}_{n+1}$. We write $\mathcal{D}$ to denote $\cup_{n=0}^\infty\mathcal{D}_n$.

In order to utilize the catalogue $\mathcal{S}_1$, we rely upon notation and terminology from \cite{HM12}. Given a dyadic diameter function $\Delta$, the distance $d_\Delta$ on $\mathsf{S}$ is defined as
\[d_\Delta(x,y):=\inf\sum_{k=1}^N\Delta(J_k),\]
where the infimum is taken over all chains $J_1,\dots, J_N$ of intervals from $\hat{\mathcal{I}}$ joining $x$ to $y$. That is, $\{x,y\}$ is contained in the connected set  $J_1\cup\dots\cup J_N$. For each $n\in\mathsf{N}$, we define a distance $d_n$ on $\mathsf{S}$ using the \textit{truncated} diameter function $\Delta_n$, defined as follows. For $m\leq n$ and $I\in\mathcal{I}_m$, we define $\Delta_n(I):=\Delta(I)$. For every $m> n$ and $I\in \mathcal{I}_m$, we inductively define $\Delta_n(I)=\frac{1}{2}\Delta_n(\tilde{I})$, where $\tilde{I}\in\mathcal{I}_{m-1}$ denotes the dyadic parent of $I$. In analogy with $d_\Delta$, we then define
\[d_n(x,y):=\inf\sum_{k=1}^N\Delta_n(J_k),\]
where the infimum is taken over all chains $\{J_k\}_{k=1}^N$ joining $x$ to $y$. We write $\Gamma_n$ to denote the metric space $(\mathsf{S},d_n)$, and $\Gamma$ to denote $(\mathsf{S},d_\Delta)$. For $n\in\mathsf{N}$, we write $\diam_n(E)$ to denote the $d_n$-diameter of a set $E\subset\mathsf{S}$. Furthermore, we write $\diam_\Delta(E)$ to denote the $d_\Delta$-diameter of $E$.

We say that a chain of dyadic intervals $\{I_i\}_{i=1}^N$ is \textit{minimal} provided that it consists of intervals with pairwise disjoint interiors and that no union of at least two distinct intervals from the chain forms an interval in $\hat{\mathcal{I}}$. In particular, if the union of intervals $\cup_{k=1}^MI_{i_k}$ from a minimal chain $\{I_i\}_{i=1}^N$ is equal to some interval $J\in\hat{\mathcal{I}}$, then $M=1$ and $J=I_{i_1}\in\{I_i\}_{i=1}^N$.

\begin{lemma}\label{L:minimal}
The definitions of $d_n$ and $d_\Delta$ are unchanged by the assumption that the chains of dyadic intervals utilized in these definitions are minimal.
\end{lemma}

\begin{proof}
We focus on the definition of $d_n$. The proof for $d_\Delta$ is the same. Suppose that $\{I_i\}_{i=1}^N$ is a chain of dyadic intervals joining $x$ and $y$ in $\mathsf{S}$. Suppose $I_j$ and $I_k$ have non-disjoint interiors. Since both $I_j$ and $I_k$ are dyadic, one must be a subset of the other. Without loss of generality, $I_j\subset I_k$. Therefore, the sum $\sum_{i=1}^N\Delta_n(I_i)$ can be decreased by eliminating the interval $I_j$ from $\{I_i\}_{i=1}^N$. It follows that $d_n$ can be defined only using chains consisting of intervals with pairwise disjoint interiors.

Next, suppose there exists $M\geq2$ and a subcollection $\{I_{i_k}\}_{k=1}^M\subset \{I_i\}_{i=1}^N$ such that $J=\cup_{k=1}^MI_{i_k}\in\hat{\mathcal{I}}$. Since $\Delta_n(J)\leq\sum_{k=1}^M\Delta_n(I_{i_k})$, the sum $\sum_{i=1}^N\Delta_n(I_i)$ can be decreased by replacing the intervals $\{I_{i_k}\}_{k=1}^M$ in $\{I_i\}_{i=1}^N$ with the single interval $J$. Since $N<+\infty$, such a replacement can happen at most finitely many times. It follows that $d_n$ can be defined only using minimal chains.
\end{proof}

For use below, we record the following technical lemma.

\begin{lemma}\label{L:minimalchain}
Assume $\{I_i\}_{i=1}^N$ is a minimal chain of dyadic intervals indexed such that, for $1\leq i\leq N-1$, the right endpoint of $I_i$ is the left endpoint of $I_{i+1}$. Under this assumption, there is either a unique interval or a unique pair of adjacent intervals from $\{I_i\}_{i=1}^N$ of maximal $d_0$-diameter. Write $i_*$ to denote the index of such a maximal interval. If $i_*>1$, then  $l(I_i)$ is strictly decreasing for $i=1,\dots,i_*-1$. If $i_*<N$, then $l(I_i)$ is strictly increasing for $i=i_*+1,\dots,N$.
\end{lemma}

\begin{proof}
We may assume that $\sigma:=\bigcup_{i=1}^NI_i\not=\mathsf{S}$, else $N=1$ and $I_1=\mathsf{S}$.

Suppose there are two distinct intervals $I_j$ and $I_k$ in $\{I_i\}_{i=1}^N$ of maximal $d_0$-diameter, where $j<k$ and $n:=l(I_j)=l(I_k)$. If these intervals are not adjacent, then the chain $\{I_{i}\}_{i=j+1}^{k-1}$ consisting of intervals from $\{I_i\}_{i=1}^N$ joins the right endpoint of $I_j$ to the left endpoint of $I_k$. Since $I_j$ and $I_k$ are not adjacent, the union $\cup_{i=j}^k I_i\subset \sigma$ contains at least three consecutive intervals from $\mathcal{I}_n$. Such a union must contain some interval $J$ from $\mathcal{I}_{n-1}$. It follows from the assumption that $\{I_i\}_{i=1}^N$ is minimal that the interval $J$ must be an element of $\{I_i\}_{i=1}^N$. However, this violates the assumption that $I_j$ and $I_k$ are of maximal $d_0$-diameter in $\mathcal{I}_n$. Therefore, the intervals $I_j$ and $I_k$ must be adjacent.

To verify the second part of the lemma, suppose $i_*>1$. If $i_*=2$ then the desired conclusion is trivial, so we may assume that $i_*\geq3$. Let $1\leq i\leq i_*-2$, and write $m:=l(I_i)$. Since $I_i$ lies to the left of $I_{i_*}$ in $\sigma$ and $\diam_0(I_{i_*})$ is strictly larger than $\diam_0(I_i)$, the interval $J$ of $\mathcal{I}_m$ immediately to the right of $I_i$ is contained in $\sigma$. By the minimality of $\{I_i\}_{i=1}^N$, no union of at least two intervals from $\{I_i\}_{i=1}^N$ is equal to $J$. Therefore, either $J=I_{i+1}$ or $J$ is strictly contained in $I_{i+1}$. If $J=I_{i+1}$, then (lest we violate the minimality of $\{I_i\}_{i=1}^N$), the midpoint of $I_i\cup I_{i+1}$ is an element of $\mathcal{D}_{m-1}$, and so $I_i\cup I_{i+1}\not\in\mathcal{I}_m$. In this case, since $I_{i+1}$ lies to the left of $I_{i_*}$ and $\diam_0(I_{i_*})>\diam_0(I_{i+1})$, the interior of the interval $K$ in $\mathcal{I}_m$ immediately to the right of $I_{i+1}$ is disjoint from the interior of $I_{i_*}$. It follows from the minimality of $\{I_i\}_{i=1}^N$ that $K$ cannot be strictly contained in any element of $\{I_i\}_{i=1}^N$ (since any dyadic interval strictly containing $K$ must also contain $I_{i+1}$), and so (again using minimality), we conclude that $K=I_{i+2}$. But this nevertheless violates the minimality of $\{I_i\}_{i=1}^N$, since $I_{i+1}\cup I_{i+2}\in\mathcal{I}_{m-1}$. Therefore, $J$ is strictly contained in $I_{i+1}$, and so $l(I_{i+1})<l(I_i)$.

An analogous argument verifies the final assertion of the lemma.
\end{proof}

Since, for any $I\in\mathcal{I}$, we have $\Delta_n(I)\leq\Delta(I)$, it follows that, for any $x,y\in\mathsf{S}$,
\begin{equation}\label{E:leqn}
d_n(x,y)\leq d_\Delta(x,y).
\end{equation}
Therefore, for any set $E\subset \mathsf{S}$, we have $\diam_n(E)\leq \diam_\Delta(E)$. Furthermore, for any $n\in\mathsf{N}$ and $I\in\hI_n$ with endpoints $a,b$, via \cite[Lemma 3.1]{HM12}, we have
\begin{equation}\label{E:intn}
d_n(a,b)=\diam_{n}(I)=\Delta_n(I)=\Delta(I)=\diam_\Delta(I)=d_\Delta(a,b).
\end{equation}

\begin{lemma}\label{L:distconverge}
If $x,y\in \mathsf{S}$ and $n\in\mathsf{N}$, then
\[d_\Delta(x,y)\leq d_n(x,y)+2\max\{\Delta(I)\,|\,I\in\mathcal{I}_n\}.\]
In particular, $d_n(x,y)\to d_\Delta(x,y)$.
\end{lemma}
\begin{proof}

Let $M(n):=\max\{\Delta(I)\,|\,I\in\mathcal{I}_n\}$. Fix $x,y\in \mathsf{S}$, $n\in\mathsf{N}$, and let $0<\varepsilon<M(n)$ be given. Let $\{I_i\}_{i=1}^N$ be a minimal chain of dyadic intervals joining $x$ and $y$, indexed as in the assumptions of \rf{L:minimalchain}, such that $\sum_{i=1}^N\Delta_n(I_i)<d_n(x,y)+\varepsilon$. If $\{I_i\}_{i=1}^N\subset\hI_n$, then we are done, because $\Delta=\Delta_n$ on $\hI_n$. If not, then let $I_{i_*}$ denote an interval from $\{I_i\}_{i=1}^N$ of maximal $d_0$-diameter, and write $m:=l(I_{i_*})$. If $x$ and $y$ are contained in adjacent intervals $J$ and $K$ from $\mathcal{I}_{n}$, then
\[d_\Delta(x,y)\leq \Delta(J)+\Delta(K)\leq 2M(n)\leq d_n(x,y)+2M(n).\]
Therefore, we can assume that $x$ and $y$ are contained in non-adjacent intervals from $\mathcal{I}_n$. It follows from the minimality of $\{I_i\}_{i=1}^N$ that $m\leq n$. Therefore, either $l(I_1)\leq n$, or, by \rf{L:minimalchain}, there exists a maximal index $i$ such that $1\leq i_1<i_*$ and, if $1\leq i\leq i_1$, then $l(I_i)>n$. Similarly, either $l(I_N)\leq n$, or there exists a minimal index $i$ such that $i_*< i_2\leq N$ and, if $i_2\leq i\leq N$, then $l(I_i)>n$. Assume the existence of such $i_1$ and $i_2$ (else the following argument simplifies). Via \rf{L:minimalchain}, one can verify that the interval $\sigma_1:=\bigcup_{i=1}^{i_1}I_i$ is contained in some interval $J_1\in\mathcal{I}_n$ adjacent (on the left) to $I_{i_1+1}$. Similarly, $\sigma_2:=\bigcup_{i=i_2}^{N}I_i$ is contained in some interval $J_2\in\mathcal{I}_n$ adjacent (on the right) to $I_{i_2-1}$. Thus we have
\begin{align*}
d_\Delta(x,y)&\leq \Delta(J_1)+\sum_{i=i_1+1}^{i_2-1}\Delta(I_i)+\Delta(J_2)=\Delta(J_1)+\sum_{i=i_1+1}^{i_2-1}\Delta_n(I_i)+\Delta(J_2)\\
&\leq\sum_{i=1}^N\Delta_n(I_i)+2M(n)<d_n(x,y)+\varepsilon+2M(n).
\end{align*}
Since $\varepsilon>0$ was arbitrary, we are done.
\end{proof}

\section{Constructing a $1$-Lipschitz map $F_0:\Gamma\to\Gamma_0$}\label{S:lip}

Let $\Gamma$ denote a bounded turning Jordan circle or arc. Our first step towards the construction of a Lipschitz light map $F:\Gamma\to\mathsf{R}$ is to realize that it is sufficient to find a Lipschitz light map $F:\Gamma\to\Gamma_0$. This is because $\Gamma_0$ is easily seen to admit a Lipschitz light map into $\mathsf{R}$, and one can verify that the composition of a Lipschitz light map from $\Gamma$ to $\Gamma_0$ with a Lipschitz light map from $\Gamma_0$ into $\mathsf{R}$ is Lipschitz light (as noted in \cite[Section 5]{David19}). See also our comments at the outset of \rf{S:light}.

Next, we again recall the following result of Herron and Meyer.

\begin{theorem}[\cite{HM12}]
If $\Gamma$ is a bounded turning Jordan circle (or arc), then $\Gamma$ is bi-Lipschitz homeomorphic to some Jordan circle in $\mathcal{S}_1$ (or $\mathcal{S}_1'$).
\end{theorem}

Here $\mathcal{S}_1$ is defined as in \cite{HM12}. The collection $\mathcal{S}_1'$ can be analogously defined using dyadic diameter functions on the unit interval $[0,1]$. We remark that the validity of this extension of the main result of \cite{HM12} to Jordan arcs is pointed out by Herron and Meyer on page 605 of \cite{HM12}.

Since Lipschitz dimension is invariant under bi-Lipschitz homeomorphisms, we may work exclusively with Jordan circles in $\mathcal{S}_1$ (or arcs in $\mathcal{S}_1'$). We will only present the details for weak quasicircles; the details for weak quasiarcs are analogous. Thus, given a curve $\Gamma=(\mathsf{S},d_\Delta)\in\mathcal{S}_1$, we construct a Lipschitz light map $F:\Gamma\to\Gamma_0$.

We will need the following map $f$ in order to achieve this goal, which we will refer to as a \textit{folding map}. Given an interval $I\subset \mathsf{S}$, divide $I$ into two dyadic subintervals of equal length with disjoint interiors, and denote these two subintervals by $I^0$ and $I^1$, respectively. We also divide $I$ into four consecutive dyadic subintervals of equal length with disjoint interiors, and denote these four subintervals by $I^{00}$, $I^{01}$, $I^{10}$, and $I^{11}$, respectively. Thus $I^0=I^{00}\cup I^{01}$ and $I^1=I^{10}\cup I^{11}$. Assume these intervals are indexed (in binary) such that adjacent intervals proceed consecutively along the positive orientation in $\mathsf{S}$. Finally, divide each of $I^{01}$ and $I^{10}$ into two dyadic subintervals of equal length with disjoint interiors, and denote these subintervals by $I^{010}$, $I^{011}$, $I^{100}$, and $I^{101}$, respectively. Thus $I^{01}=I^{010}\cup I^{011}$ and $I^{10}=I^{100}\cup I^{101}$. Again we index these intervals such that their order reflects the positive orientation of $\mathsf{S}$. The map $f:I\to I$ is defined by its action on these subintervals. It maps
\begin{itemize}
	\item{$I^{00}$ linearly onto $I^0$ in an orientation preserving manner,}
	\item{$I^{010}$ linearly onto $I^{10}$ in an orientation preserving manner,}
	\item{$I^{011}$ linearly onto $I^{10}$ in an orientation reversing manner,}
	\item{$I^{100}$ linearly onto $I^{01}$ in an orientation reversing manner,}
	\item{$I^{101}$ linearly onto $I^{01}$ in an orientation preserving manner, and}
	\item{$I^{11}$ linearly onto $I^1$ in an orientation preserving manner.}
\end{itemize}

We note that this definition of a folding map can be scaled linearly and applied to any interval. Thus, for any $n\in\mathsf{N}$, let $I\in\mathcal{I}_n$. If the two dyadic children $I'$ and $I''$ of $I$ satisfy $\Delta(I')=\Delta(I'')=\frac{1}{2}\Delta(I)$, then we define the map $f_n:I\to I$ to be the identity map. Thus $f_n$ is an isometry from $(I,d_{n+1})\to(I,d_n)$. If $\Delta(I')=\Delta(I'')=\Delta(I)$, then we define the map $f_n:I\to I$ to be a folding map. The map $f_n:\Gamma_{n+1}\to\Gamma_n$ is defined in this manner on each interval $I\in\mathcal{I}_n$.

\begin{lemma}\label{L:fold}
For each $n\in\mathsf{N}$, the map $f_n:\Gamma_{n+1}\to\Gamma_n$ is $1$-Lipschitz.
\end{lemma}

\begin{proof}
We examine the image of an interval $I\in\hat{\mathcal{I}}$ under the map $f_n$. If $I\in \hI_n$, then $f_n(I)=I\in\hat{\mathcal{I}}$. If $I\in\hat{\mathcal{I}}\setminus\hI_{n+1}$, then $f_n(I)\in\hat{\mathcal{I}}$ and $\diam_n(f_n(I))\leq \diam_{n+1}(I)$. If $I\in\mathcal{I}_{n+1}$ (the only remaining possibility), then $f_n$ fixes the endpoints of $I$, and $I\subset f_n(I)$. In particular, if $f_n$ is the identity on $\tilde{I}$ (the dyadic parent of $I$), then $f_n(I)=I$. If $f_n$ is a folding map on $\tilde{I}$, then $f_n(I)$ is the union of $I$ with an interval $J\in\mathcal{I}_{n+2}$ that is adjacent to $I$. Moreover, it is straightforward to verify that
\begin{align*}
\diam_n(f_n(I))&=\diam_n(I\cup J)=\diam_n(I)+\diam_n(J)\\
	&=\frac{3}{4}\diam_n(\tilde{I})=\frac{3}{4}\diam_{n+1}(I)<\diam_{n+1}(I).
\end{align*}
Therefore, given a chain $\{I_i\}_{i=1}^N$ of dyadic intervals, $\{f_n(I_i)\}_{i=1}^N$ can be written as a chain of dyadic intervals $\{I_j'\}_{j=1}^{N'}$ such that $\cup_{i=1}^Nf_n(I_i)=\cup_{j=1}^{N'}I_j'$, and
\[\sum_{i=1}^N\diam_n(f_n(I_i))=\sum_{j=1}^{N'}\diam_n(I_j')=\sum_{j=1}^{N'}\Delta_n(I_j').\]
Thus, if $\{I_i\}_{i=1}^N$ joins $x$ and $y$, then $\{I_j'\}_{j=1}^{N'}$ joins $f_n(x)$ and $f_n(y)$, and
\[d_n(f_n(x),f_n(y))\leq\sum_{j=1}^{N'}\Delta_n(I_j')=\sum_{i=1}^N\diam_n(f_n(I_i))\leq\sum_{i=1}^N\Delta_{n+1}(I_i).\]
It follows from the definition of $d_{n+1}(x,y)$ that $d_n(f_n(x),f_n(y))\leq d_{n+1}(x,y)$.
\end{proof}

Given any $m\leq n\in\mathsf{N}$, we define $F_{m,n}:=f_m\circ f_{m+1}\circ \dots \circ f_n:\Gamma_{n+1}\to \Gamma_m$. If $m=n$, then we understand $F_{m,m}=F_{n,n}$ to denote $f_n$. As a composition of $1$-Lipschitz maps (\rf{L:fold}), each map $F_{m,n}:\Gamma_{n+1}\to\Gamma_m$ is $1$-Lipschitz. Furthermore, this sequence of maps induces a $1$-Lipschitz map $F_m:(\mathcal{D},d_\Delta)\to(\Gamma_m,d_m)$. To see this, let $x\in \mathcal{D}$, and let $k\in\mathsf{N}$ be the smallest integer such that $x\in \mathcal{D}_k$. For any $n\geq k$, the map $f_n$ fixes the set $\mathcal{D}_k$. Therefore, if $n\geq k$, then $f_n(x)=x$. If $n\geq k>m$, we then observe that $\lim_{n\to+\infty}F_{m,n}(x)=F_{m,k}(x)$. In this case, define $F_m(x):=F_{m,k}(x)$. If $n>m\geq k$, then define $F_m(x):=x$.

To see that $F_m$ thus defined is $1$-Lipschitz on $\mathcal{D}$, let $x,y$ denote any two points in $\mathcal{D}$. Choose $k\in\mathsf{N}$ such that $x,y\in \mathcal{D}_k$ and  $k>m$. Via \rf{E:leqn} and the fact that, for all $n\geq m$ the map $F_{m,n}$ is $1$-Lipschitz, we have
\[d_m(F_m(x),F_m(y))=d_m(F_{m,k}(x),F_{m,k}(y))\leq d_{k+1}(x,y)\leq d_\Delta(x,y).\]
Since $\Gamma$ is complete, $\mathcal{D}$ is dense in $\Gamma$ (cf.\,\cite[Section 3.2]{HM12}), and $F_m$ is Lipschitz on $\mathcal{D}$, it is then straightforward to extend $F_m$ such that $F_m:\Gamma\to\Gamma_m$ is $1$-Lipschitz.

We note that we may also view the maps $F_{m,n}$ as acting on $\Gamma$. Moreover, it follows from \rf{E:leqn} that $F_{m,n}:\Gamma\to\Gamma_m$ is $1$-Lipschitz. With this in mind, we prove the following lemma.

\begin{lemma}\label{L:uniform}
For each $m\in\mathsf{N}$, the maps $F_{m,n}:\Gamma\to\Gamma_m$  uniformly converge to $F_m:\Gamma\to\Gamma_m$ as $n\to+\infty$.
\end{lemma}

\begin{proof}
Fix $m\in\mathsf{N}$ and $\varepsilon>0$. Choose $M\in\mathsf{N}$ such that
\[n\geq M \quad\text{implies} \quad\max\{\Delta(I)\,|\,I\in\mathcal{I}_n\}<\varepsilon/2.\] For any $x\in\Gamma$, there exists a nested sequence of dyadic intervals $I_n\in\mathcal{I}_n$ such that, for every $n\in\mathsf{N}$, we have $I_n\subset I_{n-1}$ and $x\in I_n$. Furthermore, there exists $x_n\in\mathcal{D}_n$ such that $x_n\in I_n$, and so $d_\Delta(x_n,x)\to0$. Assuming $n>\max\{m,M\}$, we have $F_m(x_{n+j})=F_{m,n+j}(x_{n+j})$. If $j=0$, write $w_{n,0}:=x_n$. If $j\geq1$, write $w_{n,j}:=f_{n+1}\circ\dots \circ f_{n+j}(x_{n+j})$. In either case, we note that $F_m(x_{n+j})=F_{m,n+j}(x_{n+j})=F_{m,n}(w_{n,j})$. Since $f_{n+j}(I_n)=I_n$, we have $w_{n,j}\in I_n$. Thus,
\[F_m(x)=\lim_{j\to+\infty} F_m(x_{n+j})=\lim_{j\to+\infty}F_{m,n}(w_{n,j})\in F_{m,n}(I_n).\]
Combining these observations, we find that, for $n\geq \max\{m,M\}$, we have
\begin{align*}
d_m(F_{m,n}(x),F_m(x))&\leq d_m(F_{m,n}(x),F_{m,n}(x_{n}))+d_m(F_{m,n}(x_{n}),F_m(x))\\
&\leq d_\Delta(x,x_{n})+\diam_m(F_{m,n}(I_n))\\
&\leq 2\diam_\Delta(I_n)\leq 2\max\{\Delta(I)\,|\,I\in\mathcal{I}_n\}<\varepsilon.
\end{align*}
It follows that $F_{m,n}:\Gamma\to\Gamma_m$ is uniformly convergent to $F_m:\Gamma\to\Gamma_m$.
\end{proof}

\section{Proving that $F_0:\Gamma\to\Gamma_0$ is Lipschitz light}\label{S:light}

Our goal in this section is to verify the existence of a constant $C\geq 1$ such that, for any subset $E\subset \Gamma_0$, the $\diam_0(E)$-components of $F_0^{-1}(E)$ have $d_\Delta$-diameter bounded above by $C\diam_0(E)$. Via the following lemma, this will be sufficient to prove that $F_0:\Gamma\to\Gamma_0$ is Lipschitz light, and thus (via the comments at the outset of \rf{S:lip}) that $\Gamma$ has Lipschitz dimension equal to $1$.

\begin{lemma}\label{L:def}
Suppose there exists a constant $C\geq 1$ such that $F:\Gamma\to\Gamma_0$  is $C$-Lipschitz, and, for any subset $E\subset \Gamma_0$ such that $\diam_0(E)>0$, the $\diam_0(E)$-components of $F^{-1}(E)$ have $d_\Delta$-diameter bounded by $C\,\diam_0(E)$. This implies that, for any $r>0$ and any subset $E\subset \Gamma_0$ satisfying $\diam_0(E)\leq r$, the $r$-components of $F^{-1}(E)$ have $d_\Delta$-diameter bounded by $C'r$, for $C':=\max\{C,8\}$.
\end{lemma}

\begin{proof}
Let $r>0$, and let $E\subset \Gamma_0$ be such that $\diam_0(E)\leq r$. We may assume that $E$ is compact, and that $\diam_0(E)<r$. If $r\geq 1/8$, then we note that $\diam_0(F^{-1}(E))\leq \diam_\Delta(\Gamma)\leq 1\leq 8r$. Thus, we may assume that $r<1/8$.

We claim that $E$ is contained in a subset $E'\subset \Gamma_0$ such that $\diam_0(E')=r$. To see this, we modify the argument employed in \cite[Remark 1.9]{David19}. We first note that $d_0=\lambda$ (the normalized length distance defined in \rf{S:prelims}). Therefore, given $x\in\Gamma_0$, the subset $I(x):=\{y\in \Gamma_0\,|\,d_0(x,y)\leq 1/8\}$ is isometric to some interval in $\mathsf{R}$ of length $1/4$. If $x\in E$, then $E\subset I(x)$. Let $a$ and $b$ denote the first and last points in $E$ along the interval $I(x)$. Thus $d_0(a,b)=\diam_0(E)<r$. Let $c\in I(x)$ be such that $d_0(a,c)=r<1/8$ and $\Gamma_0[a,b]\subset\Gamma_0[a,c]=:E'$. Then $E\subset E'$ and $\diam_0(E')=r$. Clearly, $r$-components of $F^{-1}(E)$ are contained in $r$-components of $F^{-1}(E')$, which (by assumption) have $d_\Delta$-diameter bounded by $Cr$.
\end{proof}

With the above lemma in hand, we begin our proof that $F_0:\Gamma\to\Gamma_0$ is Lipschitz light. Fix $E\subset \Gamma_0$ such that $\diam_0(E)>0$, and let $M^*\in\mathsf{N}$ be maximal such that
\begin{equation}\label{E:delta_size}
2^{-M^*-1}\leq\diam_0(E)<2^{-M^*}.
\end{equation}
We may assume that $M^*\geq 3$, else $\diam_\Delta(F_0^{-1}(E))\leq\diam_\Delta(\Gamma)\leq8\diam_0(E)$. By definition of $M^*$, there exist two adjacent dyadic subintervals $I,J\in\mathcal{I}_{M^*}$ such that $E\subset I\cup J$. In fact, $E$ may be contained in a single element of $\mathcal{I}_{M^*}$, but it will do no harm to assume $E$ is contained in the union of two such intervals.

We claim that it is sufficient to examine pre-images of $H:=I\cup J$. Indeed, given any $\delta>0$, the $\delta$-components of $F^{-1}(E)$ are contained in $\delta$-components of $F^{-1}(H)$.

For the remainder of this section, we set $\delta:=\diam_0(E)$.

\begin{lemma}\label{L:projection}
Given $n\in\mathsf{N}$ and $U\subset \Gamma_0$, we have $F_{n+1}(F_0^{-1}(U))=F_{0,n}^{-1}(U)$.
\end{lemma}

\begin{proof}
Suppose $x\in F_{n+1}(F_0^{-1}(U))$, so $x=F_{n+1}(w)$ for some $w\in F_0^{-1}(U)$. Then
\[F_{0,n}(x)=F_{0,n}(F_{n+1}(w))=\lim_{m\to\infty}F_{0,n}(F_{n+1,m}(w))=\lim_{m\to\infty}F_{0,m}(w)=F_0(w).\]
Since $F_0(w)\in U$, it follows that $F_{n+1}(F_0^{-1}(U))\subset F_{0,n}^{-1}(U)$.

Next, let $x\in F_{0,n}^{-1}(U)$. Write $z_{n+1}:=x$, and choose a point $z_{n+2}\in f_{n+1}^{-1}(z_{n+1})$ so that $f_{n+1}(z_{n+2})=z_{n+1}$. Inductively, for each $k\geq2$, define $z_{n+k}$ such that
\[f_{n+k-1}(z_{n+k})=z_{n+k-1}.\]
We claim that the sequence $\{z_{n+k}\}_{k=1}^\infty$ is Cauchy with respect to  $d_\Delta$, and thus convergent to some point $z\in \Gamma$. Indeed, for any $1\leq i< j$, we note that
\[z_{n+i}=f_{n+i}\circ\dots\circ f_{n+j-1}(z_{n+j}).\]
Let $I\in\mathcal{I}_{n+i}$ denote an interval containing $z_{n+j}$. For all $k\in\mathsf{N}$, we have $f_{n+i+k}(I)=I$. Therefore, $z_{n+i}\in I$, and so
\[d_\Delta(z_{n+i},z_{n+j})\leq\diam_\Delta(I)\leq\max\{\Delta(J)\,|\,J\in\mathcal{I}_{n+i}\}.\]
Since $\max\{\Delta(J)\,|\,J\in\mathcal{I}_{n+i}\}\to0$ as $i\to\infty$, our claim follows.

Next, we claim that $z=\lim_{k\to+\infty}z_{n+k}\in F_0^{-1}(U)$. Via \rf{L:uniform}, we have
\begin{align*}
F_0(z)&=\lim_{m\to+\infty}F_{0,n+m-1}(z_{n+m})\\
&=\lim_{m\to+\infty}F_{0,n}(F_{n+1,n+m-1}(z_{n+m}))=\lim_{m\to+\infty}F_{0,n}(z_{n+1})=F_{0,n}(x)\in U.
\end{align*}
Finally, we claim $F_{n+1}(z)=x$. Again via \rf{L:uniform}, we note that
\[F_{n+1}(z)=\lim_{m\to\infty}F_{n+1,m}(z_{m+1})=\lim_{m\to\infty}f_{n+1}\circ\dots\circ f_{m}(z_{m+1})=z_{n+1}=x\]
Therefore, $x\in F_{n+1}(F_0^{-1}(U))$, and so $F_{0,n}^{-1}(U)\subset F_{n+1}(F_0^{-1}(U))$.
\end{proof}

\begin{lemma}\label{L:component_proj}
Given $m\leq n\in\mathsf{N}$ and $x\in \Gamma$, we have $F_{m,n}(F_{n+1}(x))= F_m(x)$.
\end{lemma}

\begin{proof}
$F_{m,n}(F_{n+1}(w))=\lim_{k\to\infty}F_{m,n}(F_{n+1,k}(w))=\lim_{k\to\infty}F_{m,k}(w)$.
\end{proof}

Let $W$ denote any fixed $\delta$-component of $F_0^{-1}(H)$. By \rf{L:projection}, we have $F_{n+1}(W)\subset F_{0,n}^{-1}(H)$. Given any $n\in\mathsf{N}$, via \rf{L:uniform}, the set $F_{n+1}(W)$ is $\delta$-connected in $\Gamma_{n+1}$. In particular, it is contained in a single $\delta$-component of $F_{0,n}^{-1}(H)$ in $\Gamma_{n+1}$. We denote this $\delta$-component by $V_{n+1}$. Thus, for every $n\geq1$, write $V_n$ to denote the $\delta$-component of $F_{0,n-1}^{-1}(H)$ containing $F_{n}(W)$. We also write $V_0:=H$.

\begin{lemma}\label{L:Vs}
For $n\in\mathsf{N}$, we have $f_n(V_{n+1})\subset V_n$. Furthermore, the set $V_{n+1}$ is a $\delta$-component of $f_n^{-1}(V_n)$.
\end{lemma}

\begin{proof}
If $n=0$, then $V_1\subset F_0^{-1}(H)=f_0^{-1}(H)$, and so $f_0(V_1)\subset H=V_0$. We assume $n\geq1$. Via \rf{L:component_proj}, we have $F_n(W)= f_n(F_{n+1}(W))\subset f_n(V_{n+1})\subset F_{0,n-1}^{-1}(H)$. By definition, $F_n(W)\subset V_n\subset F_{0,n-1}^{-1}(H)$. Therefore, the sets $V_n$ and $f_n(V_{n+1})$ are both subsets of $F_{0,n-1}^{-1}(H)$ and have  non-trivial intersection. Since $f_n:\Gamma_{n+1}\to\Gamma_n$ is $1$-Lipschitz, the set $f_n(V_{n+1})$ is $\delta$-connected. Since $V_n$ is a maximal $\delta$-connected subset of $F_{0,n-1}^{-1}(H)$, we must have $f_n(V_{n+1})\subset V_n$.

Since $V_{n+1}\subset f_n^{-1}(V_n)$ and $V_{n+1}$ is $\delta$-connected,  $V_{n+1}$ is contained in a single $\delta$-component of $f_n^{-1}(V_n)\subset F_{0,n}^{-1}(H)$. Since $V_{n+1}$ is a maximal $\delta$-connected subset of $F_{0,n}^{-1}(H)$, the set $V_{n+1}$ is equal to a single $\delta$-component of $f_n^{-1}(V_n)$.
\end{proof}

\begin{lemma}\label{L:goal}
There exists $N^*\in\mathsf{N}$ such that, if $n\geq N^*$, then
\[\diam_\Delta(W)\leq \diam_n(V_n)+\delta.\]
\end{lemma}

\begin{proof}
Choose $N^*\in\mathsf{N}$ such that, for $n\geq N^*$, we have $\max\{\Delta(I)\,|\,I\in\mathcal{I}_n\}<\delta/4$. Let $x,y\in W$. Since, for $n\in\mathsf{N}$, the map $F_n$ fixes elements of $\mathcal{I}_n$, we note that
\[d_\Delta(x,y)\leq d_\Delta(F_n(x),F_n(y))+2\max\{\Delta(I)\,|\,I\in\mathcal{I}_n\}<d_\Delta(F_n(x),F_n(y))+\delta/2.\]
Furthermore, via \rf{L:distconverge}, we also have (for $n\geq N^*$)
\[d_\Delta(F_n(x),F_n(y))\leq \diam_\Delta(F_n(W))\leq \diam_n(F_n(W))+\delta/2.\]
Since $F_n(W)\subset V_n$, we conclude that, for any $n\geq N^*$ and any $x,y\in W$, we have
\[d_\Delta(x,y)\leq \diam_n(V_n)+\delta.\]
It follows that $\diam_\Delta(W)\leq \diam_n(V_n)+\delta$.
\end{proof}

\begin{lemma}\label{L:big}
If, for some $n\in\mathsf{N}$, the set $V_n$ is contained in an interval $I_n\in\mathcal{I}_n$ and $\diam_n(V_n)\geq\frac{1}{4}\diam_n(I_n)$, then, for any $k\in\mathsf{N}$, we have $\diam_{n+k}(V_{n+k})\leq4\diam_n(V_n)$.
\end{lemma}

\begin{proof}
We first note that $V_{n+1}\subset I_n$, since, by \rf{L:Vs}, $f_n(V_{n+1})\subset V_n\subset I_n$, and  $f_n(I_n)=I_n$. Via induction, for all $k\in\mathsf{N}$, we have $V_{n+k}\subset I_n$. Therefore, via \rf{E:intn}, we have $\diam_{n+k}(V_{n+k})\leq\diam_{n+k}(I_n)=\diam_n(I_n)\leq 4\diam_n(V_n)$.
\end{proof}

\begin{lemma}\label{L:cases}
Suppose that, for some $n\in\mathsf{N}$, we have
\begin{enumerate}
	\item{$\diam_n(V_n)=\diam_0(V_0)$}
	\item{$V_n$ is the union of two adjacent intervals from $\mathcal{I}_m$, for some $m\geq M^*$,}
	\item{$V_n$ is not symmetric about a point in $\mathcal{D}_n$,}
	\item{$V_n$ is contained in a single interval $I_n\in\mathcal{I}_n$, and}
	\item{$\diam_n(V_n)\leq\frac{1}{4}\diam_n(I_n)$.}
\end{enumerate}
Under these assumptions, $\diam_{n+1}(V_{n+1})\leq 2\diam_n(V_n)$. If $\diam_{n+1}(V_{n+1})>\diam_n(V_n)$, then $\diam_{n+1}(V_{n+1})=2\diam_n(V_n)$ and $V_{n+1}$ is symmetric about a point in $\mathcal{D}_{n+3}\setminus\mathcal{D}_{n+1}$. If, on the other hand, $\diam_{n+1}(V_{n+1})<\diam_n(V_n)$, then $\diam_{n+1}(V_{n+1})=0$ and $V_{n+1}$ is a point in $\mathcal{D}_{n+3}\setminus\mathcal{D}_{n+2}$.
\end{lemma}

\begin{proof}
Note that Assumption (3) follows from Assumption (4) when $n\geq1$; we list Assumption (3) to address the case that $n=0$.

We use binary superscripts to index the four second-generation dyadic sub-intervals $I_n^{00}$, $I_n^{01}$, $I_n^{10}$, and $I_n^{11}$ in $I_n$ such that they proceed consecutively along the positive orientation in $I_n$.

If $f_n$ is the identity on $I_n$, then the lemma is trivial. Therefore, we assume that $f_n$ is a folding map on $I_n$, and we consider the cases below. We preface this case analysis with the reminder that
\[\delta<\frac{1}{2^{M^*}}=\frac{1}{2}\diam_0(V_0)=\frac{1}{2}\diam_n(V_n)\leq\frac{1}{8}\diam_n(I_n).\]

Case 1: $V_n\subset I_n^{00}$. In this case, $f_n^{-1}(V_n)$ consists of either one $\delta$-component or (if $V_n$ contains the right endpoint of $I^{00}_n$) it consists of two. If one, then, via \rf{L:Vs}, we have $V_{n+1}=f_n^{-1}(V_n)\subset I_n^{00}$ and $\diam_{n+1}(V_{n+1})=\diam_n(V_n)$. If two, then one $\delta$-component is contained in $I^{00}_n$ and satisfies $\diam_{n+1}(V_{n+1})=\diam_n(V_n)$ while the other is a single point located at the midpoint of $I^{10}_n$.

Case 2: $V_n\subset I_n^{01}$. In this case, there are at most three $\delta$-components of $f_n^{-1}(V_n)$: one in $I_n^{00}$ and either one or two in $I_n^{10}$. The component in $I_n^{00}$ has $d_{n+1}$-diameter equal to $\diam_n(V_n)$. If there are two components in $I_n^{10}$, then they each have $d_{n+1}$-diameter equal to $\diam_n(V_n)$. If there is one component in $I_n^{10}$, then it has $d_{n+1}$-diameter equal to $2\diam_n(V_n)$, and it is symmetric about the midpoint of $I_n^{10}$. Via \rf{L:Vs}, if $\diam_{n+1}(V_{n+1})>\diam_n(V_n)$, then $V_{n+1}$ is symmetric about a point in $\mathcal{D}_{n+3}\setminus\mathcal{D}_{n+1}$ and $\diam_{n+1}(V_{n+1})=2\diam_n(V_n)$.

Case 3: $V_n\subset I_n^{10}$. By symmetry, we can apply an argument parallel to that used in Case 2 to conclude that $\diam_{n+1}(V_{n+1})\leq 2\diam_n(V_n)$. Furthermore, if $\diam_{n+1}(V_{n+1})>\diam_n(V_n)$, then $V_{n+1}$ is symmetric about a point in $\mathcal{D}_{n+3}\setminus\mathcal{D}_{n+1}$ and $\diam_{n+1}(V_{n+1})=2\diam_n(V_n)$.

Case 4. $V_n\subset I_n^{11}$. By symmetry, we can apply an argument parallel to that used in Case 1 to conclude that, either $V_{n+1}\subset I_n^{11}$ has diameter equal to $\diam_n(V_n)$, or $V_{n+1}$ is a single point at the midpoint of $I_n^{01}$.

Case 5: $V_n$ is symmetric about a point in $\mathcal{D}_{n+2}\setminus\mathcal{D}_{n+1}$. In this case, $f_n^{-1}(V_n)$ consists of two $\delta$-components. One is contained in $I_n^{00}$ (or $I_n^{11}$), and the other is contained in $I_n^{10}$ (or $I_n^{01}$). Each component has $d_{n+1}$-diameter equal to $\diam_n(V_n)$. Via \rf{L:Vs}, $\diam_{n+1}(V_{n+1})=\diam_n(V_n)$.

Case 6: $V_n$ is symmetric about a point in $\mathcal{D}_{n+1}\setminus\mathcal{D}_n$. In this case, there are three $\delta$-components of $f_n^{-1}(V_n)$, and each has $d_{n+1}$-diameter equal to $\diam_n(V_n)$. We note that one of these $\delta$-components is symmetric about a point in $\mathcal{D}_{n+1}\setminus\mathcal{D}_n$. In particular, this is the only case in which $V_{n+1}$ might not be contained in a single interval from $\mathcal{I}_{n+1}$. Via \rf{L:Vs}, $\diam_{n+1}(V_{n+1})=\diam_n(V_n)$.

Having exhausted the possible cases, we conclude the proof of the lemma.
\end{proof}

\begin{lemma}\label{L:none}
If there exists $K\in\mathsf{N}$ such that, for all $k\leq K$, the set $V_k$ is not symmetric about a point in $\mathcal{D}_{k+2}$, then, either there exists $n\geq N^*$ (for $N^*$ as in \rf{L:goal}) such that $\diam_n(V_n)\leq 16\delta$, or, for all $k\leq K$,
\begin{enumerate}
	\item[$(k.1)$]{$\diam_k(V_k)=\diam_0(V_0)$,}
	\item[$(k.2)$]{$V_k$ is the union of two adjacent intervals from $\mathcal{I}_{m}$ for some $m\geq M^*$,}
	\item[$(k.3)$]{$V_k$ is contained in a single interval $I_k\in\mathcal{I}_k$, and}
	\item[$(k.4)$]{$\diam_k(V_k)\leq \frac{1}{4}\diam_k(I_k)$.}
\end{enumerate}
\end{lemma}

\begin{proof}
Suppose $K\in\mathsf{N}$ is such that, for all $k\leq K$, no set $V_k$ is symmetric about a point in $\mathcal{D}_{k+2}$. In preparation for an inductive argument, we affirm the base case  $k=0\leq K$. Indeed, for $V_0=H$, we have
\begin{itemize}
	\item[(0.1)]{$\diam_0(V_0)=\diam_0(V_0)$,}
	\item[(0.2)]{$V_0$ is the union of two adjacent intervals from $\mathcal{I}_{M^*}$, and}
	\item[(0.3)]{$V_0$ is contained in a single interval $I_0\in\mathcal{I}_0$.}
	\item[(0.4)]{$\diam_0(V_0)\leq \frac{1}{4}\diam_0(I_0)$.}
\end{itemize}
Here we recall that $M^*\geq 3$. To proceed, we assume that, either there exists $n\geq N^*$ such that $\diam_n(V_n)\leq 16\delta$, or, for all $n\leq k-1\leq K-1$, we have
\begin{itemize}
	\item[$(n.1)$]{$\diam_n(V_n)=\diam_0(V_0)$,}
	\item[$(n.2)$]{$V_n$ is the union of two adjacent intervals from $\mathcal{I}_{m}$, for some $m\geq M^*$,}
	\item[$(n.3)$]{$V_n$ is contained in a single interval $I_n\in\mathcal{I}_n$, and}
	\item[$(n.4)$]{$\diam_n(V_n)\leq \frac{1}{4}\diam_n(I_n)$.}
\end{itemize}
Therefore, either there exists $n\geq N^*$ such that $\diam_n(V_n)\leq 16\delta$, or we satisfy the assumptions of \rf{L:cases} for $V_{k-1}$. Since $V_k$ is not symmetric about $\mathcal{D}_{k+2}$, \rf{L:cases} tells us that
\begin{itemize}
	\item[$(k.1)$]{$\diam_k(V_k)=\diam_{k-1}(V_{k-1})$.}
\end{itemize}
We note that, if $f_{k-1}$ is the identity on $I_{k-1}$, then $V_k=V_{k-1}$. If $f_{k-1}$ is a folding map on $I_{k-1}$, then $V_k$ is the union of two adjacent intervals in $\mathcal{I}_{m+1}$ (here we are using (k.1)). In either case,
\begin{itemize}
	\item[$(k.2)$]{$V_k$ is the union of two adjacent intervals in $\mathcal{I}_{m}$, for some $m\geq M^*$.}
\end{itemize}
Furthermore, since $V_{k-1}$ is not symmetric about a point in $\mathcal{D}_{k}$, Case 6 (in the proof of \rf{L:cases}) cannot occur. It follows that
\begin{itemize}
	\item[$(k.3)$]{$V_k$ is contained in a single interval $I_k\in\mathcal{I}_k$.}
\end{itemize}
Furthermore, if $\diam_k(V_k)>\frac{1}{4}\diam_k(I_k)$, then, (since $V_k$ is the union of two adjacent dyadic intervals) we must have either $\diam_k(V_k)=\frac{1}{2}\diam_k(I_k)$ or $\diam_k(V_k)=\diam_k(I_k)$. Since, by assumption, $V_k$ is not symmetric about a point in $\mathcal{D}_{k+2}$, neither case can occur. Therefore,
\begin{itemize}
	\item[$(k.4)$]{$\diam_k(V_k)\leq\frac{1}{4}\diam_k(I_k)$.}
\end{itemize}

Thus we conclude our inductive argument, and the proof of the lemma.
\end{proof}

\begin{lemma}\label{L:symm_next}
Suppose there exist $n\leq K\in\mathsf{N}$ such that, for all $n\leq k\leq K$, the set $V_{k}$ is not symmetric about a point in $\mathcal{D}_{k+1}\setminus\mathcal{D}_{k}$, and $n\in\mathsf{N}$ is such that
\begin{enumerate}
	\item[$(n.1)$]{Either $\diam_n(V_n)=0$ or $\diam_0(V_0)\leq\diam_n(V_n)\leq2\diam_0(V_0)$}
	\item[$(n.2)$]{$V_n$ is symmetric about a point in $\mathcal{D}_{n+2}\setminus\mathcal{D}_{n+1}$,}
	\item[$(n.3)$]{$V_n$ is contained in a single interval $I_n\in\mathcal{I}_n$, and}
	\item[$(n.4)$]{$\diam_n(V_n)\leq \frac{1}{4}\diam_n(I_n)$.}
\end{enumerate}
Under these assumptions on $n$ and $K$, for all $n\leq k\leq K$, it is true that
\begin{enumerate}
	\item[$(k.1)$]{$\diam_k(V_k)=\diam_n(V_n)$.}
	\item[$(k.2)$]{$V_{k}$ is symmetric about a point in $\mathcal{D}_{k+2}\setminus\mathcal{D}_{k+1}$,}
	\item[$(k.3)$]{$V_{k}$ is contained in a single interval $I_{k}\in\mathcal{I}_{k}$, and}
	\item[$(k.4)$]{$\diam_{k}(V_{k})\leq \frac{1}{4}\diam_n(I_{k})$.}
\end{enumerate}
\end{lemma}

\begin{proof} By way of induction, we first note that the base case $k=n$ is included in our assumptions. Thus, we assume that $K>n$ and, for all $n\leq j\leq k-1\leq K-1$,
\begin{itemize}
	\item[$(j.1)$]{$\diam_j(V_j)=\diam_n(V_n)$,}
	\item[$(j.2)$]{$V_{j}$ is symmetric about a point in $\mathcal{D}_{j+2}\setminus\mathcal{D}_{j+1}$,}
	\item[$(j.3)$]{$V_{j}$ is contained in a single interval $I_{j}\in\mathcal{I}_{j}$, and}
	\item[$(j.4)$]{$\diam_{j}(V_{j})\leq \frac{1}{4}\diam_{j}(I_{j})$.}
\end{itemize}
We prove that the analogous conclusions hold for $V_{k}$. Indeed, if $f_{k-1}$ is the identity on $I_{k-1}$, then $V_{k}=V_{k-1}$ is symmetric about a point in $\mathcal{D}_{k+1}\setminus\mathcal{D}_k$. Since, by assumption, this cannot occur, we only need to consider the case that $f_{k-1}$ is a folding map on $I_{k-1}$. Since $\delta<\frac{1}{2}\diam_0(V_0)\leq\frac{1}{8}\diam_{k-1}(I_{k-1})$, it is clear that
\begin{enumerate}
	\item[$(k.1)$]{$\diam_k(V_k)=\diam_{k-1}(V_{k-1})$,}
	\item[$(k.2)$]{$V_{k}$ is symmetric about a point in $\mathcal{D}_{k+2}\setminus\mathcal{D}_{k+1}$,}
	\item[$(k.3)$]{$V_{k}$ is contained in a single interval $I_{k}\in\mathcal{I}_{k}$, and}
	\item[$(k.4)$]{$\diam_{k}(V_{k})\leq \frac{1}{4}\diam_{k}(I_{k})$.}
\end{enumerate}
This completes the inductive argument, and the proof of the lemma.
\end{proof}

\begin{lemma}\label{L:not_middle}
Suppose there exist $n\leq K\in\mathsf{N}$ such that, for all $n\leq k\leq K$, the set $V_{k}$ is not symmetric about a point in $\mathcal{D}_{k}$, and $n$ is such that
\begin{enumerate}
	\item[$(n.1)$]{Either $\diam_n(V_n)=0$ or $\diam_0(V_0)\leq\diam_n(V_n)\leq2\diam_0(V_0)$,}
	\item[$(n.2)$]{$V_n$ is symmetric about a point in $\mathcal{D}_{n+1}\setminus\mathcal{D}_n$,}
	\item[$(n.3)$]{$V_n$ is contained in a single interval $I_n\in\mathcal{I}_n$, and}
	\item[$(n.4)$]{$\diam_n(V_n)\leq\frac{1}{4}\diam_n(I_n)$.}
\end{enumerate}
Under these assumptions, for all $n\leq k\leq K$,
\begin{enumerate}
	\item[$(k.1)$]{$\diam_k(V_k)=\diam_n(V_n)$.}
	\item[$(k.2)$]{$V_k$ is symmetric about a point in $\mathcal{D}_{k+1}$,}
	\item[$(k.3)$]{$V_k$ is contained in a single interval $I_k\in\mathcal{I}_k$, and}
	\item[$(k.4)$]{$\diam_k(V_k)\leq\frac{1}{4}\diam_n(I_k)$.}
\end{enumerate}
\end{lemma}

\begin{proof}
The proof consists of a straightforward inductive argument similar to that used to prove \rf{L:symm_next}. For the sake of brevity, we omit the details.
\end{proof}

We are now ready to prove the following, which, via \rfs{L:goal} and \ref{L:def}, will be sufficient to prove that $F_0:\Gamma\to\Gamma_0$ is Lipschitz light.

\begin{lemma}\label{L:final}
There exists $n\in\mathsf{N}$ such that $n\geq N^*$ and $\diam_n(V_n)\leq 128\delta$.
\end{lemma}

\begin{proof}
If there is no index $n_1$ for which $V_{n_1}$ is symmetric about a point in $\mathcal{D}_{n_1+2}$, then, by \rf{L:none}, we conclude that $\diam_N(V_N)=\diam_0(V_0)\leq 4\delta$. Therefore, we may assume $n_1$ is the minimal such index. We first consider the case that $n_1\geq1$. By the definition of $n_1$ and \rf{L:none}, we may assume that
\begin{enumerate}
	\item{$\diam_{n_1-1}(V_{n_1-1})=\diam_0(V_0)$,}
	\item{$V_{n_1-1}$ is the union of two adjacent intervals from $\mathcal{I}_{m}$, for some $m\geq M^*$,}
	\item{$V_{n_1-1}$ is contained in a single interval $I_{n_1-1}\in\mathcal{I}_{n_1-1}$, and}
	\item{$\diam_{n_1-1}(V_{n_1-1})\leq\frac{1}{4}\diam_{n_1-1}(I_{n_1-1})$.}
\end{enumerate}
Via \rf{L:Vs}, it follows from the definition of $f_{n_1-1}$ and the minimality of $n_1$ that $V_{n_1}$ is symmetric about a point in $\mathcal{D}_{n_1+2}\setminus\mathcal{D}_{n_1+1}$, and, either $\diam_{n_1}(V_{n_1})=0$, or
\[\diam_0(V_0)\leq\diam_{n_1}(V_{n_1})\leq2\diam_{n_1-1}(V_{n_1-1})=2\diam_0(V_0).\]
Furthermore, $V_{n_1}$ is contained in a single interval $I_{n_1}\in\mathcal{I}_{n_1}$. If $\diam_{n_1}(V_{n_1})>\frac{1}{4}\diam_{n_1}(I_{n_1})$, then, via \rf{L:big} and \rf{E:delta_size}, there exists $n\geq N^*$ such that
\[\diam_n(V_n)\leq4\diam_{n_1}(V_{n_1})\leq 8\diam_0(V_0)\leq 32\delta.\] Therefore, we assume that $\diam_{n_1}(V_{n_1})\leq \frac{1}{4}\diam_{n_1}(I_{n_1})$.

If there is no index $n>n_1$ such that $V_n$ is symmetric about a point in $\mathcal{D}_{n+1}\setminus\mathcal{D}_{n}$, then, by \rf{L:symm_next}, we  conclude that there exists $n\geq N^*$ such that $\diam_n(V_n)\leq 2\diam_0(V_0)\leq8\delta$. Therefore, we may assume that there exists $n_2>n_1$ minimal such that $V_{n_2}$ is symmetric about a point in $\mathcal{D}_{n_2+1}\setminus\mathcal{D}_{n_2}$. Furthermore, the proof of \rf{L:symm_next} makes it clear that $f_{n_2-1}$ is the identity on $I_{n_2-1}$ and so $V_{n_2}=V_{n_2-1}$. Therefore, $V_{n_2}$ is contained in an interval $I_{n_2}\in\mathcal{I}_{n_2}$ and, either $\diam_{n_2}(V_{n_2})=0$, or $\diam_0(V_0)\leq \diam_{n_2}(V_{n_2})\leq2\diam_0(V_0)$. If $\diam_{n_2}(V_{n_2})>\frac{1}{4}\diam_{n_2}(I_{n_2})$, then, via \rf{L:big} and \rf{E:delta_size}, there exists $n\geq N^*$ such that $\diam_n(V_n)\leq 32\delta$. Therefore, we may assume that $\diam_{n_2}(V_{n_2})\leq \frac{1}{4}\diam_{n_2}(I_{n_2})$,

If there is no index $n>n_2$ such that $V_n$ is symmetric about a point in $\mathcal{D}_n$, then, via \rf{L:not_middle} and \rf{E:delta_size}, we conclude that there exists $n\geq N^*$ such that $\diam_n(V_n)\leq 2\diam_0(V_0)\leq 8\delta$. Thus, we assume that there exists $n_3>n_2$ minimal such that $V_{n_3}$ is symmetric about a point in $\mathcal{D}_{n_3}$.

If $V_{n_3}$ is a single point, then we note that, for all $n\geq n_3$, we have $f_n^{-1}(V_{n_3})=V_{n_3}$ (since $f_n$ fixes points in $\mathcal{D}_{n_3}$). Therefore, there exists $n\geq N^*$ such that $\diam_n(V_n)=0<\delta$. Thus we may assume that $\diam_{n_3}(V_{n_3})>0$. In this case, we note that $n_3$ is minimal such that $V_{n_3}$ is not contained in a single interval from $\mathcal{I}_{n_3}$. It is also easy to verify (via \rf{L:not_middle}) that $\diam_{n_3}(V_{n_3})=\diam_{n_2}(V_{n_2})$. Indeed, $V_{n_3}$ is contained in the interior of the union of two adjacent intervals from $\mathcal{I}_{n_3}$ whose union forms $I_{n_3-1}\in\mathcal{I}_{n_3-1}$.

Write $I_{n_3}'$ to denote the left dyadic child of $I_{n_3-1}$, and write $V_{n_3}':=V_{n_3}\cap I_{n_3}'$. Inductively, for each $k\geq1$, write $I_{n_3+k}'$ to denote the right dyadic child of $I_{n_3+k-1}'$. Since $\diam_{n_3}(V_{n_3}')\leq\frac{1}{4}\diam_{n_3}(I_{n_3}')$, we have $V_{n_3}'\subset I'_{n_3+2}$. If $\diam_{n_3}(V_{n_3}')=\frac{1}{4}\diam_{n_3}(I_{n_3}')$, then
\[\diam_{n_3-1}(V_{n_3-1})=2\diam_{n_3}(V_{n_3}')=\frac{1}{2}\diam_{n_3}(I_{n_3}')\geq\frac{1}{4}\diam_{n_3-1}(I_{n_3-1}).\]
Therefore, by \rf{L:big} and \rf{E:delta_size}, there exists $n\geq N^*$ such that
\[\diam_n(V_n)\leq 4\diam_{n_3-1}(V_{n_3-1})\leq8\diam_0(V_0)\leq32\delta.\]
Therefore, we may assume that $\diam_{n_3}(V_{n_3}')<\frac{1}{4}\diam_{n_3}(I_{n_3}')$.

For each $n\geq n_3$, we define $V_n':=V_n\cap I_{n_3}'$, and we define the ratio
\[R(n):=\frac{\diam_n(V'_n)}{\diam_n(I_n')}.\]
Since $R(n_3)<\frac{1}{4}$, we have $\diam_{n_3+1}(V_{n_3+1}')=\diam_{n_3}(V_{n_3}')$. If $f_{n_3}$ is a folding map on $I_{n_3}'$, then $V_{n_3+1}'\subset I_{n_3+3}'$ and $R(n_3+1)=R(n_3)<\frac{1}{4}$. If $f_{n_3}$ is the identity on $I_{n_3}$, then $R(n_3+1)=2R(n_3)$. If $R(n_3+1)\geq\frac{1}{4}$, then define $n_4:=n_3+1$. If not, then we proceed inductively, and assume that, for all $n_3+1\leq j\leq k-1$, we have $V_{n_3+j}'\subset I_{n_3+j+2}'$, $\diam_{n_3+j}(V_{n_3+j}')=\diam_{n_3}(V_{n_3}')$, and $R(n_3+j)<\frac{1}{4}$.

Under this inductive hypothesis, we examine $V_{n_3+k}'$. Either $f_{n_3+k-1}$ is a folding map on $I_{n_3+k-1}'$, and $R(n_3+k)=R(n_3+k-1)<\frac{1}{4}$, or $f_{n_3+k-1}$ is the identity on $I_{n_3+k-1}'$, and $R(n_3+k)=2R(n_3+k-1)$. If  $R(n_3+k)\geq\frac{1}{4}$, then write $n_4:=n_3+k$.

Via induction, we are faced with two possibilities: either there exists $n_4>n_3$ minimal such that $V_{n_4}'\subset I_{n_4}'$ and $R(n_4)\geq\frac{1}{4}$, or, for all $n> n_3$, we have $V_n'\subset I_{n+2}'$ and $R(n)< \frac{1}{4}$. We claim this latter case cannot occur. Indeed, we note that, for all $n>n_3$, we have $R(n+1)\geq R(n)$. Moreover, $R(n+1)>R(n)$ if and only if $f_n$ is the identity on $I_{n}'$ and $R(n+1)=2R(n)$. Since $\diam_n(I_n')\to 0$, the map $f_n$ must be the identity on $I_n'$ infinitely often, and thus $R(n+1)=2R(n)$ infinitely often. This would imply that $R(n)\to+\infty$, and this contradiction proves our claim.

Thus we have $V_{n_4}'=V_{n_4}\cap I_{n_3}'\subset I_{n_4}'$ such that $\diam_{n_4}(V_{n_4}')\geq\frac{1}{4}\diam_{n_4}(I_{n_4}')$. Recall that, for any $n\geq n_4$, the map $f_n$ fixes elements of $\mathcal{I}_{n_4}$ and $\mathcal{I}_{n_3}$. Therefore, for any $n\geq n_4$, we have $V_n'\subset I_{n_4}'$, and so
\begin{align*}
\diam_n(V_n')&\leq \diam_n(I_{n_4}')=\diam_{n_4}(I_{n_4}')\\
&\leq4\diam_{n_4}(V_{n_4}')=8\diam_{n_3}(V_{n_3})\leq 16\diam_0(V_0)\leq64\delta.
\end{align*}

An analogous argument applies to the set $V_{n_3}'':=V_{n_3}\cap I_{n_3}''$, where $I_{n_3}''$ denotes the right dyadic child of $I_{n_3-1}$. In particular, there exists $n_5>n_3$ such that, if $n\geq n_5$, then $\diam_n(V_n'')\leq 64\delta$. Therefore, there exists $n\geq N^*$ such that
\[\diam_n(V_n)\leq \diam_n(V_n\cap I_{n_3}')+\diam_n(V_n\cap I_{n_3}'')\leq 128\delta.\]

We finish by briefly considering the case that $n_1=0$. If $V_0$ is symmetric about a point in $\mathcal{D}_2\setminus \mathcal{D}_1$, then we argue as in the case that $n_1\geq1$. If $V_0$ is symmetric about a point in $\mathcal{D}_1\setminus\mathcal{D}_0$, then we apply the argument utilized in our above analysis of $V_{n_2}$. If $V_0$ is symmetric about the point in $\mathcal{D}_0$, then we apply (a simple modification of) the argument used in our above analysis of $V_{n_3}$.
\end{proof}

\bibliographystyle{amsalpha}
\bibliography{bib}

\end{document}

config: arxiv


TITLE: Show that the polynomial $x^2+y^2-1$ is irreducible over $\mathbb{Q}[x,y]$
QUESTION [1 upvotes]: Show that the polynomial $x^2+y^2-1$ is irreducible over $\mathbb{Q}[x,y]$.

Theorem : Let $A$ an integral domain and $I$ a proper ideal of $A$. If $f(x) \not \equiv a(x)b(x) \pmod I$ for any polynomials $a(x)$, $b(x)$ $\in A[x]$ of degree $\in [1, \deg(f))$, then $f(x)$ is irreducible in $A[x]$
I think I have to use this theorem, maybe in using the ideal $(y^2)$
The fact that $\mathbb{Q}[x,y]=\mathbb{Q}[x][y]$ and $\mathbb{\mathbb{Q}[y]}$ is a unique factorization domain could it help me here? Is anyone could help me at this point?

REPLY [1 votes]: Hint: you're on the right track in considering the isomorphism $\mathbb{Q}[x, y] \cong (\mathbb{Q}[y])[x]$. Viewing $x^{2}+y^{2}-1$ as a polynomial in $x$ with coefficients in $\mathbb{Q}[y]$, observe that $y^{2}-1$ belongs to the ideal $\langle y-1\rangle \in \mathbb{Q}[y]$. Is this ideal prime? Does $y^{2}-1$ belong to $\langle (y-1)^{2} \rangle$?

REPLY [1 votes]: As you observed, $\mathbf{Q}[x,y] = \mathbf{Q}[x][y]$ so viewing $x^2 + y^2 - 1$ as a polynomial with coefficients in $\mathbf{Q}[x]$, we can apply Eisenstein with, for example, $(x+1)$, which is a prime ideal in $\mathbf{Q}[x]$.


TITLE: Choosing path to show limit does not exist
QUESTION [1 upvotes]: I'm trying to show that the limit as $(x,y)$ go to $(0,0)$ for the function $f(x,y) = sin( x + y )/( |x| + |y|)$ does not exist. I initially tried the path $y=2$ and $y=1$, but I don't think I can use these paths because they don't go through the origin. Any thoughts on which path to choose to show that this limit doesn't exist?

REPLY [2 votes]: Try the paths

$y=x$ with $x>0$; and
$y=-x$ with $x>0$.

Both paths get arbitrarily close to the origin. Compute the limit restricted to each of those paths (you may need to apply L'Hôpital's rule). If you get two different results, the limit does not exist (because the limit can't depend on the path).


text:
\begin{document}
\maketitle
{\bf Abstract:}
In this article, we study regularity criteria for the 3D  micropolar fluid equations in terms of one partial derivative of the velocity. It is proved that if
\begin{equation*}
\int^{T}_{0}\|\partial_{3}u\|^{\frac{2}{1-r}}_{\dot{B}^{-r}_{\infty,\infty}} dt<\infty \quad \text{with} \quad  0< r<1,
\end{equation*}
then, the solutions of the micropolar fluid equations actually are smooth on $(0, T)$. This improves and extends many previous results.
\medskip

{\bf Mathematics Subject Classification (2010):} \  35Q35, 35B65, 76N10
\medskip

{\bf Keywords:} \ Micropolar fluid equations; Regularity criteria; Besov spaces
\section{Introduction}
We are interested in the regularity of weak solutions to the incompressible micropolar fluid equations in $\mathbb{R}^3$:
\begin{spacing}{1.5}
\begin{equation}
\left\{
             \begin{array}{lr}
             \partial_t u + (u\cdot\nabla) u +\nabla \pi
 =  \Delta u +  \nabla \times \omega,&\\
\partial_t \omega + (u\cdot \nabla) \omega + 2 \omega
 =  \Delta \omega+  \nabla \nabla\cdot \omega +  \nabla \times u,&\\
             \nabla\cdot u=0,&\\
             u(x,0)=u_{0}(x),\omega(x,0)=\omega_{0}(x),&
\end{array}
\right.\tag{1.1}
\end{equation}
\end{spacing}
where, for $\mathbf{x}\in \mathbb{R}^3$ and $t\ge 0$,
$u=u(\mathbf{x},t), \omega=\omega(\mathbf{x},t)$  and $\pi=\pi(\mathbf{x},t)$
denote the velocity field, the micro-rotation field and the pressure,
respectively.
\par Micropolar fluids represent a class of fluids with nonsymmetric stress tensor (called polar fluids) such as fluids consisting of suspending particles, dumbbell molecules, (see, e.g.,
\cite{ Erdo, Er, Er2}). Mathematically, many authors \cite{RM,YK,DZ} treated the well-posedness
and large-time behaviour of solutions to system (1.1). However, the issue of global regularity
of weak solutions to (1.1) remains an open problem. In the absence of micro-rotational
effects ($\omega=0$), this system reduces to well-known incompressible Navier Stokes equations. Therefore, it is an interesting thing that regularity of a given weak solution of the 3D micropolar fluids or the 3D Navier-Stokes equations can be shown under some additional conditions, and over the years different criteria for regularity of the weak solutions have been proposed. Some fundamental Serrin-type regularity criteria for system $(1.1)$ in terms of the velocity only were carried out in \cite{YB,YB1,DZ1} independently. Recently, some
improvement and extension were made on the basis of the present paper (see e.g. \cite{DZ2,GS,ZZ,WY} and the references therein) were derived to guarantee the regularity of the weak solution.
\par In this paper, we are concerned with the regularity conditions of partial components of three-dimensional micropolar fluid equations. In this respect, Jia et al. \cite{JZZ}
showed that if
\begin{equation}
\partial_3u\in L^{\frac{2p}{p-3}}(0,T;L^{p,\infty}) \quad  \text{with} \quad 3<p\leq\infty ,\tag{1.2}
\end{equation}
or
\begin{equation}
\partial_1u_1, \partial_2u_2\in L^{\frac{2p}{2p-3}}(0,T;L^{p,\infty}) \quad \text{with}\quad \frac{3}{2}<p\leq\infty,\tag{1.3}
\end{equation}
then the solution is smooth on (0, T). Later, in \cite{GS1}, the authors extended the regularity criterion $(1.2)$ to the Morrey-Campanato spaces. More precisely, they obtained following interesting regularity criterion
 \begin{equation}
\partial_3u\in L^{\frac{2}{1-r}}(0,T;\dot{M}_{2,\frac{3}{r}}) \quad \text{with} \quad  0< r<1.\tag{1.4}
\end{equation}
Recently, many authors refine earlier results from the Lebesgue
spaces to Besov spaces. Zhang \cite{ZZ} and Yuan \cite{YB2} refined the BKM criterion
\begin{equation}
u\in L^{\frac{2}{1+r}}(0,T;\dot{B}^{r}_{\infty,\infty}) \quad \text{with} \quad 0<r<1,\tag{1.5}
\end{equation}
and
 \begin{equation}
\nabla_{h}u,\nabla_{h}\omega\in L^{1}(0,T;\dot{B}^{0}_{\infty,\infty}), \tag{1.6}
\end{equation}
or
\begin{equation}
\nabla_{h}u,\nabla_{h}\omega\in L^{\frac{8}{3}}(0,T;\dot{B}^{-1}_{\infty,\infty}), \tag{1.7}
\end{equation}
respectively. Readers interested in regularity criteria for other incompressible models may refer to \cite{GLR,GLR1,GLR2,LQ,GS,CW,GS2,JL,JX,ZY}.
\par Motivated by the references mentioned above, the purpose of the present paper
is to extend the regularity criterion of weak solutions to the micropolar fluid equations $(1.1)$ in terms of one partial derivative of the velocity in the framework of the homogeneous Besov space, which improves the recently results of \cite{JZZ,GS1}.
\medskip
\par Now, we state our results as follows.
\medskip
\\\textbf{Theorem 1.1}\quad Assume that the initial velocity and micro-rotation field $(u_{0},\omega_0)\in H^{1}(R^{3})$ with $\nabla\cdot u_{0}=0$. Let $(u,\omega)$ be a weak solution to the system $(1.1)$ on some interval $[0,T]$ with $0<T< \infty$. If additionally
\begin{equation}
\int^{T}_{0}\|\partial_{3}u\|^{\frac{2}{1-r}}_{\dot{B}^{-r}_{\infty,\infty}} dt<\infty \quad \text{with} \quad  0< r<1\tag{1.8}
\end{equation}
then, the solution $(u,\omega)$ smooth on $[0,T]$.
\medskip
\\\textbf{Remark 1.1}\quad For $2\leq p\leq\frac{3}{r}$ and $0\leq r<\frac{3}{2}$, we have following inclusion relations:
\[L^{\frac{3}{r}}(\mathbb{R}^3)\hookrightarrow L^{\frac{3}{r},\infty}(\mathbb{R}^3)\hookrightarrow\dot{M}_{p,\frac{3}{r}}(\mathbb{R}^3)\hookrightarrow\dot{X}_r(\mathbb{R}^3)\hookrightarrow\dot{M}_{2,\frac{3}{r}}(\mathbb{R}^3)\hookrightarrow\dot{B}^{-r}_{\infty,\infty}(\mathbb{R}^3).\]
Therefore, Theorem 1.1 can be regarded as an extension of condition $(1.2)$ and $(1.4)$.
 \medskip
\\\textbf{Remark 1.2}\quad  If let $\omega= 0$ in system $(1.1)$, it is clear that Theorem 1.1 improve and extend the results in \cite{GS,LQ} for the 3D incompressible Navier-Stokes equations.
 \medskip
\\\textbf{Remark 1.3}\quad  Different from the literatures \cite{JZZ,GS1}, we estimate the first equation of $(1.1)$ to obtain the priori estimate of the velocity field $u$, and then use the iterative method to estimate the micro-rotation field $\omega$ to obtain the regularity condition of the solution for the three-dimensional micropolar fluid equation.
\medskip
\par Next, in order to derive the regularity criterion of weak solutions to the micropolar
fluid equations $(1.1)$, we introduce the definition of a weak solution.
\medskip
\\\textbf{Definition 1.1}\quad Let $u_0\in L^2(\mathbb{R}^3)$ with $\nabla\cdot u_0=0$, $\omega_0=L^2(\mathbb{R}^3)$. A measurable $\mathbb{R}^3$-valued pair $(u,\omega)$ is called a weak solution to system $(1.1)$ on $(0,T)$, provided that following two conditions hold,
\medskip
\par $(1)$\quad $(u,\omega)\in L^{\infty}(0,T;L^2(\mathbb{R}^3))\cap L^{2}(0,T;H^1(\mathbb{R}^3))$;
\medskip
\par $(2)$\quad $(u,\omega)$ verifies system $(1.1)$ in the distributions sense.
\medskip
\par The rest of this paper is organized as follows. We shall present some preliminary
on functional settings and some useful lemmas in Section 2, while the proof of Theorem 1.1 will be presented in Section 3.
\section{Preliminaries}
In this section, we give some lemmas and introduce some basic facts on Littlewood-Paley theory, which will be used in the proof of our main results.
\par \par Let $\mathcal{S}(\mathbb{R}^3)$ be the Schwartz class of rapidly decreasing functions. Given $f\in \mathcal{S}(\mathbb{R}^3)$, its Fourier transform $\mathcal{F}f=\hat{f}$ is defined as
\[\hat{f}(\xi)=\int_{\mathbb{R}^3}f(x)e^{-ix\cdot\xi}dx.\]
Let $(\chi,\varphi)$ be a couple of smooth functions valued in $[0,1]$ such that $\chi$ is supported in $B=\{\xi \in \mathbb{R}^3:|\xi|\leq\frac{4}{3}\}$,
$\varphi$ is supported in $\mathcal{C}=\{\xi \in \mathbb{R}^3:\frac{3}{4}\leq|\xi|\leq\frac{8}{3}\}$ such that
\[\chi(\xi)+\sum\limits_{j\geq 0}\varphi(2^{-j}\xi)=1,\quad \forall\xi\in \mathbb{R}^3,\]
\[\sum\limits_{j\in Z}\varphi(2^{-j}\xi)=1, \quad \forall\xi\in \mathbb{R}^3\setminus \{0\}.\]
Denoting $\varphi_{j}=\varphi(2^{-j}\xi)$, $h=\mathcal{F}^{-1}\varphi$ and $\widetilde{h}=\mathcal{F}^{-1}\chi$, the
dyadic blocks are defined as follows, respectively.
\[\dot{\Delta}_{j}f=\varphi(2^{-j}D)f=2^{3j}\int_{\mathbb{R}^3}h(2^{j}y)f(x-y)dy,\]
\[\dot{S}_{j}f=\sum \limits_{k\leq j-1}\Delta_{j}f=\chi(2^{-j}D)f=2^{3j}\int_{\mathbb{R}^3}\widetilde{h}(2^{j}y)f(x-y)dy.\]
Furthermore, the above dyadic decomposition has nice properties of quasi-orthogonality,
namely,
\begin{center}
$\dot{\Delta}_{j}\dot{\Delta}_{q}f\equiv0$ \quad with $|j-q|\geq2.$
\end{center}
We have the following formal decomposition:
\[f=\sum \limits^{\infty}_{-\infty}\dot{\Delta}_{j}f\quad for \quad f\in\mathcal{S'}(\mathbb{R}^3)\setminus \mathcal{P}(\mathbb{R}^3),\]
where $\mathcal{P}(\mathbb{R}^3)$ is the set of polynomials, which can be found in \cite{BC} about the details of the Littlewood-Paley decomposition theory.
\par Lets now recall the homogeneous Besov spaces.
\\\\\textbf{Definition 2.1}\quad Let $s\in \mathbb{R}, (p,q)\in [1,\infty]^{2}$, the homogeneous Besov space $\dot{B}_{p,q}^{s}$ is defined by
\[\dot{B}_{p,q}^{s}=\{f\in \mathcal{Z^{'}}(\mathbb{R}^3);\|f\|_{\dot{B}_{p,q}^{s}}<\infty\},\]
where
$$\|f\|_{\dot{B}_{p,q}^{s}}=
\begin{cases}
(\sum\limits_{j\in Z}2^{jsq}\|\Delta_{j}f\|_{p}^{q})^{\frac{1}{q}}& \text{$q<\infty$}, \\
\sup\limits_{j\in Z}2^{js}\|\Delta_{j}f\|_{p}& \text{$q=\infty$.}
\end{cases}$$
Here $ \mathcal{Z^{'}}(\mathbb{R}^3)$ is the dual space of
\[\mathcal{Z}(\mathbb{R}^3)=\{f\in\mathcal{S}(\mathbb{R}^3);D^{\alpha}\widehat{f}(0)=0, \forall\alpha\in \mathbb{N}^3\} .\]
Finally, for completeness, we give following Bernstein's inequalities and interpolation inequality (see \cite{BC}), which plays an important role in the proof of main results.
\medskip
\\\textbf{Lemma 2.1}\quad Let $1\leq q<p<\infty$ and $\alpha$ be a positive real number. A constant $C$ exists such that
\begin{equation}
\|f\|_{L^{p}}\leq C\|f\|^{1-\theta}_{\dot{B}^{-\alpha}_{\infty,\infty}}\|f\|^{\theta}_{\dot{B}^{\beta}_{q,q}},\tag{2.1}
\end{equation}
where $\beta=\alpha(\frac{p}{q}-1)$ and $\theta=\frac{q}{p}$.
\medskip
\\\textbf{Lemma 2.2} \cite{CW}\quad Let $\mu,\theta,\lambda$ and $\kappa$ be four numbers satisfying the following relations
\begin{displaymath}
1\leq\mu,\theta,\lambda,\kappa<\infty,\quad\frac{1}{\theta}+\frac{1}{\lambda}+\frac{1}{\kappa}>1
\quad \text{and} \quad 1+\frac{3}{\mu}=\frac{1}{\theta}+\frac{1}{\lambda}+\frac{1}{\kappa}.
\end{displaymath}
Assume that $\varphi(x)=\varphi(x_{1},x_{2},x_{3})$ with $\partial_{1}\varphi\in L^{\theta}(\mathbb{R}^3),\partial_{2}\varphi\in L^{\lambda}(\mathbb{R}^3)$ and $\partial_{3}\varphi\in L^{\kappa}(\mathbb{R}^3)$.
Then, there exists a constant $C=C(\theta,\lambda,\kappa)$ such that
\begin{equation}
\|\varphi\|_{L^{\mu}}\leq C\|\partial_{1}\varphi\|^{\frac{1}{3}}_{L^{\theta}}
\|\partial_{2}\varphi\|^{\frac{1}{3}}_{L^{\lambda}}
\|\partial_{3}\varphi\|^{\frac{1}{3}}_{L^{\kappa}}.\tag{2.2}
\end{equation}
Especially, when $\theta=\lambda=2$ and $1\leq\kappa<\infty$, there exists a constant $C=C(\kappa)$ such that
\begin{equation}
\|\varphi\|_{L^{3\kappa}}\leq C\|\partial_{1}\varphi\|^{\frac{1}{3}}_{L^{2}}
\|\partial_{2}\varphi\|^{\frac{1}{3}}_{L^{2}}
\|\partial_{3}\varphi\|^{\frac{1}{3}}_{L^{\kappa}},\tag{2.3}
\end{equation}
which holds for any $\varphi$ with $\partial_{1}\varphi\in L^{2}(\mathbb{R}^3), \partial_{2}\varphi\in L^{2}(\mathbb{R}^3)$ and $\partial_{3}\varphi\in L^{\kappa}(\mathbb{R}^3)$.
\section{Proof of Theorem 1.1}
This section is devoted to the proof of Theorem 1.1. It is based on the establishment of a priori estimates under condition $(1.8)$.
\par Multiplying the first equation of $(1.1)$ by $u$, and the second equation of $(1.1)$ by $\omega$, integrating over $\mathbb{R}^3$, then we add the resulting equations, it yields
\begin{equation}
\frac{1}{2}\frac{d}{dt}(\|u\|^{2}_{L^{2}}+\|\omega\|^{2}_{L^{2}})+\|\nabla u\|^{2}_{L^{2}}+\|\nabla \omega\|^{2}_{L^{2}}+2\|\omega\|^{2}_{L^{2}}+\|\nabla\cdot\omega\|^{2}_{L^{2}}=0.\tag{3.1}
\end{equation}
Integrating in time, we get the fundamental energy estimate
 \begin{equation}
\frac{1}{2}\|(u,\omega)(t)\|^{2}_{L^{2}}+\int^{t}_{0}\|(\nabla u,\nabla \omega)(s)\|^{2}_{L^{2}}ds+\int^{t}_{0}2\|\omega(s)\|^{2}_{L^{2}}ds
+\int^{t}_{0}\|\nabla\cdot\omega(s)\|^{2}_{L^{2}}ds\leq\frac{1}{2}\|(u_{0},\omega_{0})\|^{2}_{L^{2}}.\tag{3.2}
\end{equation}
\medskip
\\\textbf{Lemma 3.1}\quad Assume that $u_0,\omega_0\in H^1(\mathbb{R}^3)$ with $\nabla\cdot u_0$, let $T>0$, $(u,\omega)$ is the weak solution to the system $(1.1)$. If there exists a constant $K>0$ such that the following conditions holds,
\begin{equation}
\int^{T}_{0}\|\partial_{3}u\|^{\frac{2}{1-r}}_{\dot{B}^{-r}_{\infty,\infty}} dt<K \quad \text{with} \quad  0< r<1,\tag{3.3}
\end{equation}
then, we have
 \begin{equation}
\begin{split}
\sup\limits_{0\leq t\leq T}\|\nabla u\|^{2}_{L^{2}}+\int^T_0\|\Delta u\|^2_{L^2}dt\leq C,
\end{split}\tag{3.4}
\end{equation}
where constant $C$ depending only on $u_0$, $\omega_0$, $T$ and $K$.\\
\\\textbf{Proof:}
Differentiating the first equation of system $(1.1)$ about space variable $x_{3}$, then multiplying the resulting equation by $\partial_{3}u$, and integrating it we get
\begin{equation}
\begin{split}
\frac{1}{2}\frac{d}{dt}\|\partial_{3} u\|^{2}_{L^{2}}+\|\nabla \partial_{3}u\|^{2}_{L^{2}}&=-\int_{\mathbb{R}^3}\partial_{3}(u\cdot\nabla u)\cdot\partial_{3} u dx+\int_{\mathbb{R}^3}\partial_{3}(\nabla\times\omega)\cdot\partial_{3} u dx\\
&=I_{1}+I_{2}.
\end{split}\tag{3.5}
\end{equation}
By Young inequality, standard multiplicative inequality, Lemma 2.1 and Lemma 2.2, we can estimate $I_{1}$ and $I_{2}$ as follows:
\begin{equation}
\begin{split}
|I_{1}|&=\int_{\mathbb{R}^3}\partial_{3}u\cdot\nabla u\cdot\partial_{3} u dx\\
&\leq\|\partial_{3}u\|^{2}_{L^{4}}\|\nabla u\|_{L^{2}}\\
&\leq\|\partial_{3}u\|_{\dot{B}^{-r}_{\infty,\infty}}\|\partial_{3}u\|_{\dot{H}^{r}}\|\nabla u\|_{L^{2}}\\
&\leq\|\partial_{3}u\|_{\dot{B}^{-r}_{\infty,\infty}}\|\partial_{3}u\|^{1-r}_{L^{2}}
\|\nabla\partial_{3}u\|^{r}_{L^{2}}\|\nabla u\|_{L^{2}}\\
&\leq C(\|\nabla\partial_{3}u\|^{2}_{L^{2}})^{\frac{r}{2}}(\|\partial_{3}u\|^{\frac{2}{2-r}}_{\dot{B}^{-r}_{\infty,\infty}}
\|\partial_{3}u\|^{2\cdot\frac{1-r}{2-r}}_{L^{2}}\|\nabla u\|^{\frac{2}{2-r}}_{L^{2}})^{\frac{2-r}{2}}\\
&\leq \frac{1}{4}\|\nabla\partial_{3}u\|^{2}_{L^{2}}+C\|\partial_{3}u\|^{\frac{2}{2-r}}_{\dot{B}^{-r}_{\infty,\infty}}
\|\partial_{3}u\|^{2\cdot\frac{1-r}{2-r}}_{L^{2}}\|\nabla u\|^{\frac{2}{2-r}}_{L^{2}}\\
&\leq \frac{1}{4}\|\nabla\partial_{3}u\|^{2}_{L^{2}}+C[(\|\partial_{3}u\|^{\frac{2}{1-r}}_{\dot{B}^{-r}_{\infty,\infty}})^{\frac{1-r}{2-r}}
(\|\nabla u\|^{2}_{L^{2}})^{\frac{1}{2-r}}]
\|\partial_{3}u\|^{2\cdot\frac{1-r}{2-r}}_{L^{2}}\\
&\leq \frac{1}{4}\|\nabla\partial_{3}u\|^{2}_{L^{2}}+C(1+\|\partial_{3}u\|^{2}_{L^{2}})(\|\partial_{3}u\|^{\frac{2}{1-r}}_{\dot{B}^{-r}_{\infty,\infty}}
+\|\nabla u\|^{2}_{L^{2}}),
\end{split}\tag{3.6}
\end{equation}
and
\begin{equation}
\begin{split}
|I_{2}|=|\int_{\mathbb{R}^3}\partial_{3}(\nabla\times\omega)\cdot\partial_{3} u dx|\leq \|\partial_{33}u\|_{L^{2}}\|\nabla\omega\|_{L^{2}}\leq \frac{1}{4}\|\nabla\partial_{3} u\|^{2}_{L^{2}}+C\|\nabla\omega\|^{2}_{L^{2}}.
\end{split}\tag{3.7}
\end{equation}
Combining $(3.5)$, $(3.6)$ and $(3.7)$ together, then we deduce that
\begin{equation}
\begin{split}
\frac{d}{dt}(1+\|\partial_{3} u\|^{2}_{L^{2}})+\|\nabla \partial_{3}u\|^{2}_{L^{2}}&\leq C(1+\|\partial_{3} u\|^{2}_{L^{2}})(1+\|\nabla u\|^{2}_{L^{2}}+\|\nabla\omega\|^{2}_{L^{2}}+\|\partial_{3}u\|^{\frac{2}{1-r}}_{\dot{B}^{-r}_{\infty,\infty}}).
\end{split}\tag{3.8}
\end{equation}
Applying the Gronwall inequality to $(3.8)$ yields that
\begin{equation}
\begin{split}
(1+\|\partial_{3} u\|^{2}_{L^{2}})&\leq (1+\|\partial_{3} u_{0}\|^{2}_{L^{2}})\exp\{C\int^{T}_{0}(1+\|\nabla u\|^{2}_{L^{2}}+\|\nabla\omega\|^{2}_{L^{2}}+\|\partial_{3}u\|^{\frac{2}{1-r}}_{\dot{B}^{-r}_{\infty,\infty}})d\tau\},
\end{split}\tag{3.9}
\end{equation}
which implies that
\begin{equation}
\begin{split}
\sup\limits_{0\leq t\leq T}\|\partial_{3} u(t)\|^{2}_{L^{2}}+\int^{T}_{0}\|\nabla\partial_{3} u(t)\|^{2}_{L^{2}}dt&\leq C.
\end{split}\tag{3.10}
\end{equation}
Next, multiplying the first equation of system $(1.1)$ by $\Delta u$, integrating over $\mathbb{R}^3$, we get
 \begin{equation}
 \begin{split}
\frac{1}{2}\frac{d}{dt}\|\nabla u\|^{2}_{L^{2}}+\|\Delta u\|^{2}_{L^{2}}&=\int_{\mathbb{R}^3}u\cdot\nabla u\cdot\Delta u dx-\int_{\mathbb{R}^3}(\nabla\times\omega)\cdot\Delta u dx\\
&=J_{1}+J_{2}.
\end{split}\tag{3.11}
\end{equation}
For $J_{1}$, by using the interpolation inequality and Lemma 2.2 (with $\kappa=2$ in (2.3)), it is not difficult to see that
\begin{equation}
\begin{split}
J_{1}&\leq C\|\nabla u\|^{3}_{L^{3}}\leq C\|\nabla u\|^{\frac{3}{2}}_{L^{2}}\|\nabla u\|^{\frac{3}{2}}_{L^{6}}\\
&\leq C\|\nabla u\|^{\frac{3}{2}}_{L^{2}}\|\nabla\partial_{1}u\|^{\frac{1}{2}}_{L^{2}}
\|\nabla\partial_{2}u\|^{\frac{1}{2}}_{L^{2}}\|\nabla\partial_{3}u\|^{\frac{1}{2}}_{L^{2}}\\
&\leq C\|\nabla u\|^{\frac{3}{2}}_{L^{2}}\|\nabla^{2}u\|_{L^{2}}
\|\nabla\partial_{3}u\|^{\frac{1}{2}}_{L^{2}}\\
&=C(\|\nabla^{2}u\|^{2}_{L^{2}})^{\frac{1}{2}}(\|\nabla u\|^{3}_{L^{2}}\|\nabla\partial_{3}u\|_{L^{2}})^{\frac{1}{2}}\\
&\leq\frac{1}{4}\|\Delta u\|^{2}_{L^{2}}+C\|\nabla u\|^{3}_{L^{2}}\|\nabla\partial_{3}u\|_{L^{2}}\\
&\leq\frac{1}{4}\|\Delta u\|^{2}_{L^{2}}+C\|\nabla u\|^{2}_{L^{2}}(\|\nabla u\|^2_{L^{2}}+\|\nabla\partial_{3}u\|^2_{L^{2}}).
\end{split}\tag{3.12}
\end{equation}
For $J_{2}$, we have
\begin{equation}
J_{2}\leq\frac{1}{4}\|\Delta u\|^{2}_{L^{2}}+C \|\nabla\omega\|^{2}_{L^{2}}.\tag{3.13}
\end{equation}
Substituting $(3.12),(3.13)$ in $(3.11)$, we obtain
 \begin{equation}
 \begin{split}
\frac{d}{dt}\|\nabla u\|^{2}_{L^{2}}+\|\Delta u\|^{2}_{L^{2}}
&\leq C(\|\nabla u\|^{2}_{L^{2}}+e)(\|\nabla\partial_{3}u\|^{2}_{L^{2}}+\|\nabla u\|^{2}_{L^{2}}+\|\nabla \omega\|^{2}_{L^{2}}+1).\\
\end{split}\tag{3.14}
\end{equation}
Applying the Gronwall inequality yields that
\begin{equation}
\begin{split}
\sup\limits_{0\leq t\leq T}&\|\nabla u\|^{2}_{L^{2}}\leq (e+\|\nabla u_{0}\|^{2}_{L^{2}})\exp\{\int^{T}_{0}(\|\nabla\partial_{3}u\|^{2}_{L^{2}}+\|\nabla u\|^{2}_{L^{2}}+\|\nabla \omega\|^{2}_{L^{2}}+1)d\tau\},
\end{split}\tag{3.15}
\end{equation}
which gives that
\begin{equation}
u\in L^{\infty}\big(0,T;H^{1}(\mathbb{R}^3)\big)\cap L^{2}\big(0,T;H^{2}(\mathbb{R}^3)\big).\tag{3.16}
\end{equation}
The proof of Lemma 3.1 is completed.
\medskip
\par \textbf {Proof of Theorem 1.1}\quad Taking the inner product of system $(1.1)_1$ with $-\Delta u$ in $L^2(\mathbb{R}^3)$, we obtain
\begin{equation}
\begin{split}
\frac{1}{2}\frac{d}{dt}\|\nabla u\|^{2}_{L^2}+\|\Delta u\|^{2}_{L^2}=\int_{\mathbb{R}^3}u\cdot\nabla u\cdot\Delta u dx-\int_{\mathbb{R}^3}\nabla \times\omega\cdot\Delta u dx.
\end{split}\tag{3.17}
\end{equation}
On the other hand, testing $(1.1)_2$ by $-\Delta\omega$, we get
\begin{equation}
\begin{split}
\frac{1}{2}\frac{d}{dt}\|\nabla \omega\|^{2}_{L^2}+\|\Delta \omega\|^{2}_{L^2}+2\|\nabla \omega\|^{2}_{L^2}+\|\nabla\nabla\cdot\omega\|^{2}_{L^2}
=\int_{\mathbb{R}^3}u\cdot\nabla \omega\cdot\Delta \omega dx-\int_{\mathbb{R}^3}\nabla\times u\cdot\Delta \omega dx.
\end{split}\tag{3.18}
\end{equation}
Plugging (3.17) and (3.18) together, it yields
\begin{equation}
\begin{split}
&\frac{1}{2}\frac{d}{dt}(\|\nabla u\|^{2}_{L^2}+\|\nabla \omega\|^{2}_{L^2})+\|\Delta u\|^{2}_{L^2}+\|\Delta \omega\|^{2}_{L^2}+2\|\nabla \omega\|^{2}_{L^2}+\|\nabla\nabla\cdot\omega\|^{2}_{L^2}\\
=& \int_{\mathbb{R}^3}u\cdot\nabla u\cdot\Delta u dx-\int_{\mathbb{R}^3}\nabla \times\omega\cdot\Delta u dx+\int_{\mathbb{R}^3}u\cdot\nabla \omega\cdot\Delta \omega dx-\int_{\mathbb{R}^3}\nabla\times u\cdot\Delta \omega dx\\
=&K_1+K_2+K_3+K_4.
\end{split}\tag{3.19}
\end{equation}
By using H\"{o}lder and Young inequalities, it is clear that
\begin{equation}
\begin{split}
K_{2}+K_{4}&=-\int_{\mathbb{R}^3}\nabla \times\omega\cdot\Delta u dx-\int_{\mathbb{R}^3}\nabla\times u\cdot\Delta u dx\\
&\leq C\|\nabla \times \omega\|_{L^{2}}\|\Delta u\|_{L^{2}}+C\|\nabla \times u\|_{L^{2}}\|\Delta \omega\|_{L^{2}}\\
&\leq 2C\|\nabla u\|^{2}_{L^{2}}+2C\|\nabla \omega\|^{2}_{L^{2}}+\frac{1}{2}\|\Delta u\|^{2}_{L^{2}}+\frac{1}{2}\|\Delta \omega\|^{2}_{L^{2}}.
\end{split}\tag{3.20}
\end{equation}
Combining $(3.19)$ and $(3.20)$, one obtains
\begin{equation}
\begin{split}
&\frac{d}{dt}(\|\nabla u\|^{2}_{L^2}+\|\nabla \omega\|^{2}_{L^2})+\|\Delta u\|^{2}_{L^2}+\|\Delta \omega\|^{2}_{L^2}+\|\nabla\nabla\cdot\omega\|^{2}_{L^2}\\
\leq& 4C\|\nabla u\|^{2}_{L^2}+4C\|\nabla \omega\|^{2}_{L^{2}}+2\int_{\mathbb{R}^3}u\cdot\nabla u\cdot\Delta u dx+2\int_{\mathbb{R}^3}u\cdot\nabla \omega\cdot\Delta \omega dx
\end{split}\tag{3.21}
\end{equation}
By the H\"{o}lder inequality, Gagliardo-Nirenberg inequality and Sobolev inequality,
the last two terms of $(3.21)$ can be estimated as
\begin{equation}
\begin{split}
&2\int_{\mathbb{R}^3}u\cdot\nabla u\cdot\Delta u dx+2\int_{\mathbb{R}^3}u\cdot\nabla \omega\cdot\Delta \omega dx\\
\leq& 2\|u\cdot\nabla u\|_{L^2}\|\Delta u\|_{L^2}+2\|u\cdot\nabla \omega\|_{L^2}\|\Delta \omega\|_{L^2}\\
\leq& C\|u\cdot\nabla u\|^{2}_{L^2}+C\|u\cdot\nabla \omega\|^{2}_{L^2}+\epsilon\|\Delta u\|^{2}_{L^2}+\epsilon\|\Delta \omega\|^{2}_{L^2}\\
\leq& C\| u\|^{2}_{L^\infty}(\|\nabla u\|^{2}_{L^2}+\|\nabla \omega\|^{2}_{L^2})+\epsilon\|\Delta u\|^{2}_{L^2}+\epsilon\|\Delta \omega\|^{2}_{L^2}\\
\leq& C\| \nabla u\|_{L^2}\| \Delta u\|_{L^2}(\|\nabla u\|^{2}_{L^2}+\|\nabla \omega\|^{2}_{L^2})+\epsilon\|\Delta u\|^{2}_{L^2}+\epsilon\|\Delta \omega\|^{2}_{L^2}\\
\leq& C\| \nabla u\|^{2}_{L^2}(\|\nabla u\|^{2}_{L^2}+\|\nabla \omega\|^{2}_{L^2})^2+2\epsilon\|\Delta u\|^{2}_{L^2}+\epsilon\|\Delta \omega\|^{2}_{L^2}.\\
\end{split}\tag{3.22}
\end{equation}
Plugging estimates $(3.22)$ into $(3.21)$, give us to
\begin{equation}
\begin{split}
&\frac{d}{dt}(\|\nabla u\|^{2}_{L^2}+\|\nabla \omega\|^{2}_{L^2})+\frac{1}{2}(\|\Delta u\|^{2}_{L^2}+\|\Delta \omega\|^{2}_{L^2})+\|\nabla\nabla\cdot\omega\|^{2}_{L^2}\\
\leq& 4C\|\nabla u\|^{2}_{L^2}+4C\|\nabla \omega\|^{2}_{L^2}+ C\| \nabla u\|^{2}_{L^2}(\|\nabla u\|^{2}_{L^2}+\|\nabla \omega\|^{2}_{L^2})^2.
\end{split}\tag{3.23}
\end{equation}
Setting $Y(t)=\|\nabla u\|^{2}_{L^{2}}+\|\nabla \omega\|^{2}_{L^{2}}+e$, from $(3.23)$ we see that
\begin{equation}
\frac{d}{dt}\ln Y(t)\leq 4C+ C\| \nabla u\|^{2}_{L^2}(\|\nabla u\|^{2}_{L^2}+\|\nabla \omega\|^{2}_{L^2})\ln Y(t).\tag{3.24}
\end{equation}
Due to the Gronwall inequality and Lemma 3.1, we conclude that
 \begin{equation}
\begin{split}
\ln Y(t)&\leq \bigg(\ln(e+\|\nabla u_{0}\|^{2}_{L^{2}}+\|\nabla \omega_{0}\|^{2}_{L^{2}})+4CT\bigg)\exp \int^{T}_{0}C\| \nabla u\|^{2}_{L^2}(\|\nabla u\|^{2}_{L^2}+\|\nabla \omega\|^{2}_{L^2})dt\\
&\leq \bigg(C+4CT\bigg)\exp C\int^{T}_{0}\|\nabla u\|^{2}_{L^2}+\|\nabla \omega\|^{2}_{L^2}dt.
\end{split}\tag{3.25}
\end{equation}
Combining the basic energy estimate $(3.2)$ and taking the exponential function on both side of $(3.25)$, we have
 \begin{equation}
\begin{split}
\sup\limits_{0\leq t\leq T}&(\|\nabla u\|^{2}_{L^{2}}+\|\nabla \omega\|^{2}_{L^{2}})\leq
\exp\{(C+4CT)\exp\{CT(\| u_0\|^{2}_{L^{2}}+\|\omega_0\|^{2}_{L^{2}})\}\}<\infty,
\end{split}\tag{3.26}
\end{equation}
thus
\[u,\omega\in L^{\infty}\big(0,T;H^{1}(\mathbb{R}^3)\big)\cap L^{2}\big(0,T;H^{2}(\mathbb{R}^3)\big).\]
This completes the proof of Theorem 1.1.
\section*{Acknowledgments}
\par The first author is partially supported by I.N.D.A.M-G.N.A.M.P.A. 2019 and the ``RUDN University Program 5-100''.


TITLE: Lack of invariance of relativistic doppler effect for inertial frames?
QUESTION [0 upvotes]: The equation for the relativistic doppler effect in a medium where waves travel at speed $c_m$ is
$$\frac{f_r}{f_s} = \frac{c_m - v_r}{c_m-v_s}\frac{\gamma_r}{\gamma_s}$$
My question is whether this value stays invariant under velocity shifts as it does in the galilean case. That is we apply transformation, $v' = \frac{v-u}{1-\frac{uv}{c^2}}$ to the speed of waves, velocity of source and receiver and see if the equation stays the same. This doesn't seem at first glance to be true. If not, why?
Moreover let us consider the case of the doppler effect for light. The equation simplifies to $$\frac{f_r}{f_s} = \sqrt{\frac{(c-v_r)(c+v_s)}{(c+v_r)(c-v_s)}}$$
If we look carefully, we can see that this is equivalent to the standard 'correct' relativistic doppler equation 1 cited on this wikipedia page, compounded twice. To clarify, this is equivalent to a receiver in an intermediate frame transmitting another light beam of the frequency observed by them towards the receiver. Because of the constancy of $c$ this should give us the same value (if the equation were correct), right? But still it doesn't seem like the equation is actually invariant with velocity shifts. Which then implies a contradiction to the postulates of relativity.
Just to make sure I have everything in check, I will post my derivation of the equation below.

I have tried to derive the doppler effect equation the following way.
  Let there be 3 frames. The source frame $S$, the medium frame $S'$ and
  the receiver frame $S''$. The time period in $S$ is $\Delta t$ and the
  velocity of the source and receiver w.r.t $S'$ is $v_s,v_r$
  respectively. Applying the inverse Lorentz transforms we go from $S$
  to $S'$ to find
$\Delta t'=\gamma_s\Delta t$
$\Delta x'=\gamma_sv_s\Delta t$
The path of the two periodic disturbances in $S'$ is
$x=c_mt$ and
$x-\gamma_sv_s\Delta t = c_m(t-\gamma_s\Delta t)$
and the path of the receiver is $x = v_rt+\alpha$ where $\alpha$ is
  the displacement of the reciever from the source. Solving for the
  difference between the two solutions for the two times as the wave
  intersects the receiver in $S'$ gives us $\tilde{\Delta t'}= \frac{c_m-v_s}{c_m-v_r}\gamma_s\Delta t$, $\tilde{\Delta x'}= v_r\tilde{\Delta t'}$. Applying Lorentz transforms gives us $\Delta t'' = \tilde{\Delta t'}(1-\frac{v_r^2}{c^2})\gamma_r$ so $\Delta t'' = \frac{c_m-v_s}{c_m-v_r}\frac{\gamma_s}{\gamma_r}\Delta t$
So
$$\frac{f_r}{f_s} = \frac{c_m - v_r}{c_m-v_s}\frac{\gamma_r}{\gamma_s}$$

I believe this derivation should apply for light waves in any frame where we have the special case $c_m = c$. If this is the case, then it is necessary that the equation should be invariant under velocity shifts. That is, $v' = \frac{v-u}{1-\frac{uv}{c^2}}$

REPLY [0 votes]: It's also invariant for waves in a medium, just the algebra is extremely tedious.

I think the equation for waves in a medium can be proved invariant in the following fairly simple way. You can rearrange it to give
$$\frac{f_r/\gamma_r}{f_s/\gamma_s}=\frac{c_m-v_r}{c_m-v_s}.$$
If $f_r$ refers to the proper frequency, then $\gamma_r/f_r$ is the period referred to the frame of the medium, and similarly for $\gamma_s/f_s$. Therefore the entire equation is invariant because every quantity is explicitly referred to a particular frame, so they don't transform at all.


TITLE: Effect of the measure on an exponential family
QUESTION [5 upvotes]: This might be a very naive question.
Wikipedia describes an exponential family as a distribution
$f(x \mid \theta) =  h(x)\exp( - \theta x - A(\theta))$
where $A(\theta) = \int h(x)\exp( - \theta x) \, dx$
I want to understand the role of the underlying measure or basically the $h(x)$ in the above equation.
For instance can any function of $\theta$ be written as $A(\theta)$ for some choice of $h(x)$.

REPLY [4 votes]: Every probability density is with respect to a measure.  The Lebesgue measure is the measure that assigns to every interval its length, so that the measure of the interval $(2,6) = \{ x : 2<x<6\}$ is $6-2=4$.  The standard normal density function is $\displaystyle x\mapsto\frac 1 {\sqrt{2\pi}} e^{-x^2/2}$ and that is a density with respect to Lebesgue measure.  In integrals $\text{“}dx\text{''}$ represents the Lebesgue measure, and that means
$$
\int_{(2,6)} 13\,dx = \Big( 13\times\text{the measure of } (2,6) \Big) = 13\times 4 = 52.
$$
If one multiplies the density $\displaystyle \frac 1 {\sqrt{2\pi}} e^{-x^2/2}$ with respect to Lebesgue measure by the Lebesgue measure $dx$ then one gets a measure called $\displaystyle \overbrace{\frac 1 {\sqrt{2\pi}} e^{-x^2/2} \, dx}^\text{This includes “$dx$''.}$, which is the standard normal distribution.  The measure that that assigns to the interval $(2,6)$ is
$$
\int_{(2,6)} \frac 1 {\sqrt{2\pi}} e^{-x^2/2} \, dx = \int_2^6 \frac 1 {\sqrt{2\pi}} e^{-x^2/2} \, dx,
$$
the probability that a random variable with a standard normal distribution is in the interval $(2,6)$.
The measure considered in the definition of "exponential family" is not $h(x)$, but $h(x)\,dx$.  And this is somewhat misleading, in that it suggests that there is always some density function $x\mapsto h(x)$ that gets integrated with respect to Lebesgue measure to get the measure $h(x)\,dx$.  But in fact one often considers discrete measures, made up entirely of point masses, that assign positive measure only to sets of Lebesgue measure zero, so there is no density that can be integrated with respect to Lebesgue measure to obtain that discrete measure.  For example
\begin{align}
x\mapsto {} & \binom n x p^x (1-p)^{n-x} = \binom n x \left( \frac p {1-p} \right)^x (1-p)^n \\[12pt]
= {} & h(x) \exp\left( -x\theta - A(\theta) \right) \\
& \text{where }\theta = -\log \frac p {1-p} \text{ and }A(\theta) = -n\log(1-p) \text{ and }h(x) = \binom n x.
\end{align}
Here one integrates not with respect to Lebesgue measure but with respect to counting measure on the set $\{0,1,2,\ldots,n\}$, so that the measure of any set $A$ is the number of points in $A\cap\{0,\ldots,n\}$, and so the integral is a sum:
$$
\int g(x) \, dc(x) = \sum_{x=0}^n g(x) \text{ where $c$ is the counting measure on } \{0,\ldots,n\}.
$$
As to which functions can be realized as $\theta\mapsto A(\theta)$, that is a matter of inverting a Laplace transform.  There are some inversion theorems, but I'm not sure if we need separate inversion theorems for different measures.