text:
\begin{document}

\title[ON BUNGEE SETS OF COMPOSITE TRANSCENDENTAL ENTIRE FUNCTIONS]{ON BUNGEE SETS OF COMPOSITE TRANSCENDENTAL ENTIRE FUNCTIONS }

\author[Anand P. Singh ]{ Anand P. Singh}
\address{Department of Mathematics, Central University of Rajasthan,  Bandarsindri, Kishangarh-305817, Rajasthan, India}
\email{ singhanandp@rediffmail.com }

\date{}

\bigskip
\begin{abstract} Let $f$ be  a transcendental entire function. For $n \in \mathbb{N},$ let $ f^{n}$ denote the $n^{th}$ iterate of $f$. Let
$ I(f) = \{z \in \mathbb{C} : f^n \rightarrow \infty $  as $  n \rightarrow \infty \} $ and
$ K(f) = \{z: \textrm{ there exists } R > 0 \textrm{ such that } | f^n(z) | \leq R  \textrm{ for  }   n \geq 0 \}. $ Then the set $ \mathbb{C}\ \setminus (I(f) \cup K(f)) $  denoted by $ BU(f) $ is called Bungee set of $f$.  In this paper we study some properties of Bungee sets of composite transcendental entire functions and also of Bungee sets of permutable transcendental entire functions.

\end{abstract}

\subjclass[2010]{30D05, 37F10}
\keywords{Transcendental Entire Function, Escaping set, Julia set, Bungee set \\
 Sponsored by SERB grant No. EMR/2016/001645}

{\Large
\bigskip
\maketitle

\section{Introduction}
\setcounter{equation}{0}

Let $f$ be a rational or a transcendental entire function. For $n \in \mathbb{N} $, let $ f^{n}$ denote the $n^{th}$ iterate of $f$. The set \\
$ F(f) = \{ z  : (f^n) $ is normal in some neighbourhood of $ z \}$  is called Fatou set of $f$ and its complement denoted by $J(f)$ is called Julia set of $f$. For properties of these two sets one can refer for instance \cite{Beardon}, \cite{Bergweiler}, \cite{Morosawa}. Here we observe that  $\hat{\mathbb{C} }= \mathbb{C} \cup \{ \infty \}$ has been partitioned in two different classes of sets, viz. the Fatou set and the Julia set with  $ \infty  \in J(f) $ when $f$ is transcendental entire. There are other ways also that $\hat{\mathbb{C} }$ can be partitioned, which is also linked with the Fatou and Julia sets.

For a transcendental entire function, Eremenko \cite{Eremenko} considered  the Escaping set
$ I(f) = \{z \in \mathbb{C} : f^n(z) \rightarrow \infty \textrm{ as } n \rightarrow \infty \}  $
and showed that $ I(f) \cap J(f) \not= \phi $, $ \partial I(f) = J(f) $ and all components of $ \overline{I(f)}$ are unbounded. He further conjectured that all the components of $I(f)$ should be unbounded. This conjecture though not yet solved has given rise to a rich development in the field. The points which tend to infinity with different ``speeds"  such as fast escaping points, slow escaping points, relatively fast escaping points etc. have been studied by various authors,  (see for instance \cite{Rippon}, \cite{Rippon 1}, \cite{ Singh}, \cite{ Wang},  \cite{Zheng}).  In contrast, there are points whose iterates under $f$ remain bounded. The set of such points is denoted by  $ K(f) $ and is defined by

    $ K(f) = \{z: \textrm{ there exists } R > 0 \textrm{ such that } | f^n(z) | \leq R  \textrm{ for  }   n \geq 0 \}. $

 For non linear polynomial $P,$ $ K(P) $ is known as filled Julia set, which has been extensively studied. For transcendental entire function, $K(f) $ has been studied for instance  by Bergweiler \cite{ Bergweiler 1} and Osborne \cite{Osborne}. There are points whose iterate are neither bounded nor tending to $\infty$. Such sets have been considered by Osborne and Sixsmith. They called it Bungee set. More specifically Osborne and Sixsmith \cite{Osborne 1}, defined the Bungee set $ BU(f) $ as $ BU(f) = \mathbb{C} \setminus  ( I(f) \cup K(f) )$. Note that as each of   $I(f)$ and $K(f)$ are  completely invariant, it follows that $ BU(f)$ is also completely invariant, and the sets $I(f)$, $K(f)$, and $ BU(f)$ partition the complex  plane $ \mathbb{C} $  into three classes. Rather than studying the Bungee set as complement of $ I(f) \cup K(f) $  we give here an alternate  definition for Bungee set, which is very easy to use and prove some results using this definition.

\begin{definition} Let $f$ be rational or transcendental entire function. We define the Bungee set of $f$ denoted by $BU(f)$ by:
 $ BU(f) = \{ z :  \textrm{ there exist atleast two subsequences } \{f^{n_k}\},  \{f^{m_k}\}  \\ \textrm{ with } {n_k} \rightarrow \infty \textrm{ and  } {m_k} \rightarrow \infty  \textrm{  as } k   \rightarrow \infty  \textrm{ and a constant  } R > 0 \textrm{    such }  \textrm{that } |f^{n_k}(z) | \leq R  \textrm{ for } k =1, 2, \dots,  \textrm{ and }$  $   f^{m_k}(z)  \rightarrow \infty  \textrm{ as } m_k \rightarrow \infty  \}. $
 \end{definition}

 Note that the  subsequences in the above definition  cannot have a ``pattern" in the sense that for the point $z_o \in BU(f) $ there do not exist sequences  $ {n_k}$ and  $ {m_k} $    such that $ n_k = k n_o, m_k = k m_o, $ where  $ n_o, m_o \geq 1, k = 1, 2, \dots $  and such that  $ |f^{n_k}(z_o) | \leq R  \textrm{ for } k =1, 2, \dots,  \textrm{ and }   f^{m_k}(z_o )  \rightarrow \infty  \textrm{ as } k \rightarrow \infty $, for  if they do exist, then choose $k$ sufficiently large say $l$ such that $  |f^{m_k}(z_o) | > R$  for all $ k \geq l $, and so $  |f^{k m_o}(z_o) | > R$ for all $ k \geq l $, and in particular   $  |f^{k m_o n_o}(z_o ) | > R $ for   $ k \geq l $, contradicting $  |f^{t n_o}(z_o ) | =  |f^{n_t}(z_o ) | \leq R$ for all $t$.  Also, note that  for $z_o \in BU(f)$ if $ ( n_k, m_k, R, l_o )$ denote the sequences $ \{n_k\}$ and  $ \{m_k\} $    such that  $ |f^{n_k}(z_o) | \leq R  \textrm{ for } k =1, 2, \dots,  \textrm{ and }  |  f^{m_k}(z_o ) | > R  $, for all $ k \geq l_o $, then if
 $ n_k = p  n_{k-1} $ and  $ m_k  = q  m_{k-1},  $ then $ p \not= (  \frac{m_o}{n_o} q^l )^{\frac{1}{k}}$ for all $ l \geq l_o. $ For if  for some $ l \geq l_o $ $ p = (  \frac{m_o}{n_o} q^l )^{\frac{1}{k}}$, then $ n_k = p^k  n_o = q^l m_o = m_l, $ and so  $ |f^{n_k}(z_o) | = |f^{m_l}(z_o) | $ which is not possible as left equality is bounded by $R$ and the right equality is greater than $R$.
  \\

    For a non-linear polynomial $P, BU(P) = \phi $, and for a transcendental entire function $f$,  $BU(f) \cap J(f) \not= \phi $ \cite{Baker 1}, and also there are examples where  $BU(f) \cap F(f) \not= \phi $ (see for instance \cite{Bergweiler 1}, \cite{Bishop}). For rational functions, the Bungee set may coincide with Fatou set, for instance if $R(z) = \frac{1}{z^2}$ then $BU(R)
= (|z|<1)\cup (|z|>1) $ and $J(R) = \partial (BU(R))$. It is interesting to note that if $R$ and $S$ are permutable rational functions  ( i.e., $ R \circ S = S \circ R$), then $J(R)= J(S)$, where as this is still an open question for transcendental entire function, though quite some progress has been made (see for instance \cite{Baker}, \cite{Singh}). However the corresponding result on Bungee set   is not true for rational functions. For instance if $R(z) = \frac{1}{z^2} $ and $S(z) = z^2 $ then clearly $ R(S(z)) = S(R(z)) $, and $ BU(R)= (|z|<1) \cup (|z|>1)  \textrm{ whereas } BU(S) = \phi $. It looks, the same type of statement would be true if $f$ is transcendental entire, though we do not have an example for it. Thus the Bungee set may behave quite differently  from Julia sets for rational functions and transcendental entire functions. It is also well known that for any $n$, $J(f^n) = J(f),  F(f^n) = F(f).$ However this is not true for Bungee sets. We only have $ BU(R^n) \subset BU(R)$, since for instance, if $R(z) = \frac{1}{z^2}$, then  $ BU(R^2) = \phi , \textrm{ whereas }   BU(R)= (|z|<1) \cup (|z|>1)$. We are not sure whether  the equality  holds for transcendental entire function. But we can  atleast show $ BU(f^n) \subset BU(f)$. For this observe that if $z_o \in BU(f^n)$, then there exist $\{n_k\}$ and $\{m_k\} $ tending to $ \infty$ such that $|(f^n)^{n_k}(z_o)| \leq R $ for some positive $R$, $ k = 1, 2, \dots$ and $(f^n)^{m_k}(z_o) \rightarrow \infty $
as  $ m_k \rightarrow \infty $. Let $ t_k = n \cdot n_k $ and $p_k = n \cdot m_k$, then  $|(f)^{t_k}(z_o)| \leq R ,  k = 1, 2, \dots$ and $f^{p_k}(z_o) \rightarrow \infty $ as  $ p_k \rightarrow \infty $, and so $z_o \in BU(f)$.\\
\newpage

\section{Main results and their proofs.}
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  In this paper we concentrate on Bungee sets of transcendental entire functions, and more specifically with  composition of transcendental entire functions.\\

If $U$ is a component of  $F(f)$, then by complete invariance, $f(U)$ lies in some component of $F(f)$. If $U_n \cap U_m = \phi $ for $ n \not= m$ where $U_n$ denotes the component of $F(f)$ which contains $f^n(U)$, then $U$ is called a wandering domain, else it is either a periodic or a pre-periodic domain. A complete classification for periodic domain is well known ( see for instance \cite{Morosawa} ). Also it is well known \cite{Sullivan} that rational functions have no wandering domain,  though the same is not true for transcendental entire functions. Infact transcendental entire functions may  have wandering domains. Also there are transcendental entire functions, such as the functions in Speiser class ( i.e. entire functions whose singularities are finite in number)
which do not have wandering domain. There are other classes of transcendental entire functions which do not have wandering domain (see for instance  \cite{Bergweiler}, \cite{Morosawa}).

If $P$ is a bounded periodic component of $F(f)$ then obviously every point $z_0 \in P$ does not belong to $BU(f)$. But what can be said about points on $\partial P$ ? We shall show that even $ \partial P$ does not contain points of $BU(f)$. Thus we shall prove

\begin{Theorem}\label{Theorem 1}
  Let $f$ be transcendental entire function. Let   $P$ be a bounded  periodic component of $ F(f)$. Then $ \partial P \cap BU(f) =  \phi $.
 \end{Theorem}

\noindent \textit{Proof of Theorem} \ref{Theorem 1}. Note that as $P$ is a bounded periodic Fatou component, there exists a constant $M$ such that $ |f^n(P)| \leq M $ for all $ n \in \mathbb{N}$.
Suppose $ \xi_o \in  \partial P \cap BU(f).  $
Then  there exists a sequence $\{ m_k\}$ such that $ f^{m_k}(\xi_o) \to \infty $ as $ m_k \rightarrow \infty. $
Hence for sufficiently large $m_k$ say $ M_k$, $ | f^{M_k}(\xi_o)\mid  >  1+M.$ Also $ f^{M_k}$ is continuous at
$\xi_o$  and hence there exists $ \delta > 0$ such that $ | z - \xi_o |  <  \delta $ implies
$ | f^{M_k}(z) - f^{M_k}(\xi_o) |  < 1$. In particular for $ t  \in  ( | z - \xi_o |  < \delta )\cap P $ we have $ | f^{M_k}(t) - f^{M_k}(\xi_o) |  < 1$, and on the other hand
$ | f^{M_k}(t) - f^{M_k}(\xi_o) | \geq  | f^{M_k}(\xi_o) |- |f^{M_k}(t) | >1. $ This contradiction proves the theorem.\\

{\bf Question}: Is $ \partial P \cap BU(f) = \phi $ if $P$ is unbounded Fatou component ? \\

We next show that  atleast for  a transcendental entire function with no wandering domain, the bungee set cannot be a closed set.

\begin{Theorem}\label{Theorem 2}
  Let $f$ be transcendental entire function without wandering domain. Then  $  BU(f) $ cannot be a closed subset  of $ \mathbb{C} $.
 \end{Theorem}

 For the proof of the theorem we shall need the following theorem of
 Osborne and Sixsmith  \cite{Osborne}

\begin{Theorem}\label{Theorem 3}
Let $f$ be a transcendental entire function.\\
(a) If $U$ is a Fatou component of $f$ and $U \cap BU(f) \not= \phi,$ then $ U \subset BU(f) $ and $U$ is a wandering domain.\\
(b) $J(f) = \partial BU(f).$\\

\end{Theorem}

\noindent \textit{Proof of Theorem} \ref{Theorem 2}. Suppose $  BU(f) $ is a closed subset  of $ \mathbb{C}. $  As $  BU(f) \not= \phi $ it contains infinitely many points. Also clearly $ BU(f)  $ is completely invariant set. Also as  $J(f)$ is the smallest closed completely  invariant set having atleast three points it follows that $ J(f) \subset BU(f)$. If $ J(f) \not=  BU(f)$, then there exists $ z_o \in BU(f) \cap  F(f) $, and consequently there exists $ U \subset F(f) $ which intersects $BU(f)$. By Theorem \ref{Theorem 3}, $U$ must be a wandering domain. This contradiction proves $ J(f) =  BU(f).$ But then this result is also not possible as repelling periodc points lie in $J(f) $ and periodic points obviously do not lie in $BU(f)$. This completes the proof of the  Theorem.\\

Our next two theorems deal with pre-images of bungee sets corresponding to permutable transcendental entire functions.

\begin{Theorem}\label{Theorem 4}
 Let $f$ and $g$ be permutable transcendental entire functions. Let $U \subset BU(f).$ If
$g^{-1} (U) \not= \phi $, then  $g^{-1} (U) \cap ( I(f) \cup BU(f) ) \not= \phi $\\
 \end{Theorem}

\noindent \textit{Proof of Theorem} \ref{Theorem 4}. Suppose $ g^{-1} (U) \cap ( I(f) \cup BU(f) ) = \phi $. Let $z_o \in g^{-1}(U) $. Then there exists a constant $A$ such that $  |f^n(z_o))| \leq A $ for all  $n \in \mathbb{N} $, and consequently $ | g(f^n(z_o))| \leq M(A, g) $ for all $ n \in \mathbb{N}$. Also there exists $\xi_o \in U$ such that $ g(z_o) = \xi_o$, and as
$ U \subset BU(f)$ there exists a sequence $n_k$ such that $f^{n_k}(\xi_o) \rightarrow \infty $ as $ n_k \rightarrow \infty$. So choose $n_k$ sufficiently large so that $ |f^{n_k}(\xi_o) | > M(A,g) +1$. Then for such $n_k$, $ |f^{n_k}(g(z_o) )| > M(A,g) +1$ and where as $ |f^{n_k}(g(z_o) )| = |g(f^{n_k}(z_o) ) | \leq  M(A,g) .$
 This contradiction proves the theorem.\\

 \begin{Theorem}\label{Theorem 5}
 Let $f$ and $g$ be permutable transcendental entire functions. Further let there exist a non linear polynomial $P$ such that $ P\circ f = f \circ g $.  If  $U \subset BU(f), $
 then  $g^{-1} (U)   \subset BU(f).$

  \end{Theorem}

\noindent \textit{Proof of Theorem} \ref{Theorem 5}. Following as in Theorem   \ref{Theorem 4},  if $z_o \in g^{-1}(U) $ and if $  |f^n(z_o))| \leq A $ for all  $n \in \mathbb{N} $, then we have obtained a contradiction. We next show that $ f^n(z_o) \not \rightarrow \infty $ as $ n \rightarrow \infty$, and so $z_o \in BU(f)$. So, let $ f^n(z_o) \rightarrow \infty $ as $ n \rightarrow \infty$. Then there exists $\xi_o \in U $ and a constant $K$  and a sequence $n_k \rightarrow \infty $ such that $  | f^{n_k}(\xi_o)| < K $ for all $ k=1,2,
\dots $. Let $ p_k = n_{k+1}-n_k $ for all $k$. As $P(z)$ is a non linear polynomial, there exists $R$ sufficiently large such that $|P(z)| >  |z| $ for all $z \in (|z| >R) $.As $ f^n(z_o) \rightarrow \infty $ as $ n \rightarrow \infty$, we can choose $n_k $ sufficiently  large so that\\
 $  |f^{n_k}(z_o) | > Max \{R, K+1\} , k=1,2,\dots .$  Let $ f^{n_k}(z_o) = \eta $. Then $ f^{n_k}(g(z_o)) = f^{n_k}(\xi_o)= t $ say where $|t|< K$. And so by permutability, $ t = g(f^{n_k}(z_o))= g(\eta) $, and so
 $ | f^{p_k}(g(\eta ))| = | f^{p_k}(g(f^{n_k}(z_o)))| = | f^{p_k}(f^{n_k}g(z_o))|= | f^{n_{k+1}}(g(z_o))| = | f^{n_{k+1}}(\xi_o))| < K $.\\
 On the other hand, $| f^{p_k}(g(\eta))| =  | f^{n_{k+1}}(g(z_o))|  = | P(f^{n_{k+1}}(z_o))| \geq | (f^{n_{k+1}}(z_o)| > K + 1$, and so  $ z_o \in BU(f)$.  Thus $g^{-1}(U) \subset BU(f)$.\\

  If $f$ is a transcendental entire function and  $z_o \in BU(f)$, then there exists $R > 0 $ and sequences
$\{n_k\}$ and $\{m_k\} $ tending to $ \infty$ such that $ | f^{n_k}(z_o)| \leq R $ for some positive $R$, $ k = 1, 2, \dots$ and $ f^{m_k}(z_o) \rightarrow \infty $
as  $ m_k \rightarrow \infty $. We shall denote such domain $  ( |z| \leq R ) $ by $D_R(z_o)$. (Note that such $D_R(z_o)$ is not unique).\\

\begin{Theorem}\label{Theorem 6}
 Let $f$ and $g$ be permutable transcendental entire functions. Let $P$ be a polynomial of degree $\geq 2$ and let $h$ be  a transcendental entire function such that $ P\circ f = h \circ g$. Let $z_0 \in BU(f)$. Then $ g(D_R(z_o))$ cannot be a periodic domain of $f$.

 \end{Theorem}

\noindent \textit{Proof of Theorem} \ref{Theorem 6}. Let $z_o \in BU(f)$. Then there exists $R > 0 $ and sequences
$\{n_k\}$ and $\{m_k\} $ tending to $ \infty$ such that $| f^{n_k}(z_o)| \leq R $ for positive $R$, $ k = 1, 2, \dots$ and $ f^{m_k}(z_o) \rightarrow \infty $
as  $ m_k \rightarrow \infty.$\\
First suppose $ |f^n(g(D_R(z_o)))| \leq A $ for some constant $A$ and all $ n \in \mathbb{N}$. Then
$  |h(f^n(g(D_R(z_o))))| \leq M(A,h) < B $ say, for all $n \in \mathbb{N}$ where $ M(A,h) = \max _{|z|=A}|h(z)|$.
As $P$ is a polynomial  of degree $ \geq 2 $, we can choose $S$ sufficiently large so that $ |P(z)|>|z|$ for all $|z|>S$.\\
Now choose $M_k$ sufficiently large so that $ | f^{m_k}(z_o)| > \max \{S, B \}  $ for all $ m_k \geq M_k.$\\
Thus  for  $ m_k \geq M_k, $ $ | P( f^{m_k}(z_o))| > | f^{m_k}(z_o)| > B.$\\
On the other hand $ | P( f^{m_k}(z_o))| =  | P( f^{m_k- n_1}(f^{n_1}(z_o)))| = | P( f^{m_k-n_1}(t))| = | h( f^{m_k-n_1-1}g(t))| \leq M(A,h) < B $ where $t= f^{n_1}(z_o)$. This contradiction shows that $ f^n(g(D_R(z_o))) $ cannot be bounded for all $n \in \mathbb{N}$ and hence $g(D_R(z_o))$ cannot be a periodic Fatou component of $f$ unless it is a periodic Baker domain, which also is not possible as  $g(D_R(z_o))$ is bounded.\\

If $f$ and $g$ are permutable transcendental entire functions and if $z_0 \in BU(f)$ as well as in $F(f)$ then we have the following result.

\begin{Theorem}\label{Theorem 7}
 Let $f$ and $g$ be permutable transcendental entire functions. Let $h$ be a trascendental entire function and $P$ be a non-linear polynomial such that $ P \circ f = h \circ g $. If $ z_0 \in F(f) \cap BU(f)$ then $ g^{-1}(z_0) \in F(f) \cap BU(f)$.

 \end{Theorem}

\noindent \textit{Proof of Theorem} \ref{Theorem 7}. Since $ z_0 \in F(f) $ and $F(f)$ is open, there exists a neighbourhood and consequently a component $U$ of Fatou set of $f$ such that $U \cap BU(f) \neq \phi$, and so by Theorem \ref{Theorem 3}, $U \subset BU(f)$ and $U$ is a wandering domain and $\partial BU(f) = J(f)$. If $ g^{-1}(z_0) \in J(f)$, then by complete invariance $ g( g^{-1}(z_0)) \in J(f)$. Thus $z_0 \in J(f)$, contradicting $z_0 \in F(f).$

We now show that $ g^{-1}(z_0) \in BU(f)$. For suppose $ |f^n(g^{-1}(z_0)| \leq A  $ for some constant $A$ and for all $n \in \mathbb{N}$, then $  |g(f^n(g^{-1}(z_0)) | \leq M(A, g) $, and so as $f$ and $g$ are permutable,    $  |(f^n(z_0)) | \leq M(A, g) $ for all $n\in \mathbb{N}$ contradicting $ z_0 \in  BU(f)$. Next suppose  $ | f^n(g^{-1}(z_0) | \rightarrow \infty,$ as $ n \rightarrow \infty. $ Now, as $ z_0 \in  BU(f)$, there exist  a subsequence
 $\{n_k \} \textrm { tending to }\infty $  and a constant  $ c > 0 $,  such that $ | f^{n_k}(z_0)| \leq c, k=1,2,\dots. $
Let $ D = M(c, h)$. Since $ P $ is a nonlinear polynomial, we  can choose $ r > D+1$ sufficiently large so that  for all $ z \in (|z|>r),   |P(z)| > |z|$. Since $ | f^n(g^{-1}(z_0) | \rightarrow \infty $  as $ n \rightarrow \infty $,  we select $n_t$ from the subsequence so that

   $ | f^{n_t}(g^{-1}(z_0) | > r $ and   $ | f^{n_t+1}(g^{-1}(z_0) | > r $.

  Now$ (f^{n_t}\circ g)(g^{-1}(z_0) = f^{n_t}(z_0) = \xi_0 $ say, where $ |\xi_0 | \leq c $.\\
  And so\\
  $ \xi_0 = (f^{n_t}\circ g)(g^{-1}(z_0) = g(f^{n_t}(g^{-1}(z_0)) = g(\eta ) $  say, where $|\eta| = |f^{n_t}g^{-1}(z_0)| > r$.
  Now $ | (h \circ g) (\eta )| = |h(\xi_0)|  \leq M(c, h) = D$, and also  as $ |f^{n_t+1}(g^{-1}(z_0) | > r $, we have\\
  $|(h \circ g) (\eta )| = | (P\circ f)(\eta) | = | P( f^{n_t+1}(g^{-1}(z_0)) | > | f^{n_t+1}(g^{-1}(z_0)) | > r > D+1.
   $  This contradiction proves the theorem.\\

\begin{Theorem}\label{Theorem 8}

 Let $f$ be a transcendental entire function. If $ z_0 \in BU(f \circ g)$ then $ g(z_0) \in BU(g \circ f)$.

 \end{Theorem}
 \noindent \textit{Proof of Theorem} \ref{Theorem 8}. Suppose $ g(z_0) \notin BU(g \circ f)$. Then  either (i) $ | (g \circ f )^n (g(z_0)) | \leq A $ for all
$n \in \mathbb{N} $ and some $ A > 0$ or (ii) $ \lim_{n \rightarrow \infty }(g\circ f)^n (g(z_0)) = \infty $. \\
Now if (i) holds, then $ | g(f\circ g)^n(z_0) | \leq A $ for all  $n \in \mathbb{N} $, and so $ | f(g(f\circ g)^n(z_0)|\leq M(A, f) $ for all $ n \in \mathbb{N},$ where $ M(A, f) = \max_{|z|=A}|f(z)| $. Thus $ | (f\circ g)^{n+1}(z_0) | \leq M(A, f) $ for all $n\in \mathbb{N}$, contradicting $ z_0 \in  BU(f \circ g).$\\
Next, if (ii) holds, then since $z_0 \in BU(f \circ g)$, there exists a sequence ${n_k}$ such that $ |(f \circ g)^{n_k}(z_0) | \leq \beta,$ for some  $\beta > 0 $. And so $ | g(f\circ g)^{n_k}(z_0)| \leq M( \beta, g), k = 1, 2, \dots $. Thus $ | (g\circ f)^{n_k}(g(z_0))| \leq M( \beta , g)$, contradicting (ii). This proves the theorem.\\

config: arxiv


TITLE: Decomposition of 4x4 or larger affine transformation matrix to individual variables per degree of freedom
QUESTION [3 upvotes]: There are a couple of problems and solutions where affine matrices are decomposed into their separate transformations. However, they are all for the 2D case and I`m finding it difficult to generalise it to higher dimensions. Specifically, I have to decompose them for the 3D case, to variables that represent the individual degrees of freedom such as rotations, translations, scalings, and shears. I don't understand how to do that. Should there be a general approach for the nD case? Should I write out the formulas?
So I think I could write my problem for 3D like this, where I know all entries of A. With unknowns $v$ the scaling vector, $s$ the information needed for shearing which I don't know how to represent, $p$ the translation vector and $\theta, \psi, \phi$ the rotations: $$ A(\theta, \psi, \phi, v, s, p) = R_z(\phi) R_y(\psi) R_x(\theta) Sc(v) Sh(s) + T(p) $$
Maybe I should formulate this differently?
User7530 shows a nice decomposition here but does not explain (enough for my understanding) why and how it works. Especially why the QR decomposition does the trick here and why $\theta = \arctan\left(\frac{d}{a}\right).$
John does domething similar but with less linear algebra and more algebra.
Michael Albanese shows a nice group theory approach to explain what an affine matrix is made up of, but does not say anything about decomposition. Also I don't really understand this.

REPLY [1 votes]: Factoring out the translation, assume that that the remaining $N$ by $N$ matrix is $A=QKS$ where $Q$ is a rotation matrix, $K$ is a scaling matrix and $S$ is a shear matrix. Then $A$ can be decomposed using QR decomposition (say by using Gram–Schmidt) into a rotation matrix $Q$ and an upper triangular matrix $R$ such that $A=QR$.
Next, decompose $R$ by setting the scaling matrix $K$ to the diagonal elements of $R$ and then computing $S=K^{-1}R$. Note that the inverse of $K$ can be computed by just taking the reciprocal of its diagonal elements.
Now the determinant of $Q$ is $1$, thus it is a rotation matrix. $K$ is a scaling matrix since all elements off the main diagonal are zero. $S$ is a shear matrix since all main diagonal elements are $1$ and its rank and trace are equal to $N$, its determinant is $1$ and all its eigenvalues are equal to $1$.
I believe this factorization works for any square matrix $A$ that has a non-zero determinant. For small $N$, it may be practical to find algebraic solutions for $Q$, $K$ and $S$. See this for computing Euler angles from $Q$.


TITLE: What is the probability of a pen touching a bar given that the length of the pen is $10$ cm and the bars are regularly spaced at $15$ cm?
QUESTION [6 upvotes]: Problem:
If a pen of length $10$ cm is thrown out of infinitely large window having vertical bars regularly spaced at $15$ cm, then find the probability that it will touch any of the bars. (Assume that the vertical bars are infinitely long)

I don't know how to even begin with the question, so any help will be appreciated. All I know is that integration is involved in the calculation of this probability, but I don't know what to integrate, since the pen can be thrown anywhere, and at any angle.

REPLY [0 votes]: I will interpret the problem in the simplest terms first, the classic problem assuming the pen is always crossing with a length of 10cm and not rotating.
A simple and elegant solution:
Consider a circle of diameter 15 cm. This circle will ALWAYS touch the bars at 2 distinct points.
Since the circle touches the bars at 2 points, and has a circumference of 15pi, we can say that a part of the circle with length x has a 2x/(15pi) chance of touching the line.
As the circle gets smaller, it is noticably less circle- like, indeed, approaching that of a line. This gets us thinking, and soon we realize that, independent of shape, any given point should have the same probability of touching the bars. Furthermore, we know that probability!
We can therefore say that, for any segment, of any shape, of length X, the expected number of times that the segment touches the bars is 2x/(15pi).
Since the pen can't touch the bars more than once, the expected number of times that the segment touches the bars is equal to the chance that the segment touches the bars!
Therefore, the solution is simply 2*10/(15*pi)= 20/(15*pi)
From here, it isn't so hard to consider the pen falling at any angle (Assuming no spin).
Since we've just proved this relationship to be linear, we realize that we can just take the average length of the pen as it spins. As it spins, the length of the pen is just sin(theta)*10. The sin function can be found to have an average of 1/sqrt(2) through calculus, so the answer to that question must be 20/(15*pi*sqrt(2))


TITLE: Does universal quantification order matter?
QUESTION [3 upvotes]: For example if I have $$\forall a \in A \forall b_2 \in B \forall b_1 \in B((a,b_1)\notin R \lor (b_1, b_2)\notin R)$$ is is the same as
$$\forall b_2 \in B   \forall b_1 \in B \forall a \in A ((a,b_1)\notin R \lor (b_1, b_2)\notin R)$$
and other variations thereof? Is this true in general? I ask because I think it is the same thing since there is not a "dependency" between these 3 variables but I might be wrong.

REPLY [5 votes]: Universal quantifiers if not separated by other types of quantifiers (existential for example) are interchangeble IF they do not depend on each other.
Example where the exchange is possible:
$$\forall a \in A,\, \forall b \in B : p(a,b) = T$$
$$\forall b \in B,\, \forall a \in A : p(a,b) = T$$
Example where the exchange cannot be performed:
$$\forall a \in A,\, \forall b \in (B \setminus \{a\}) : p(a,b) = T$$
in the second example you can observe that the second universal quantifier depends on the first since the set $(B \setminus \{a\})$ does not contain the element chosen by the first universal quantifier.
indeed like you said in your proposition the universal quantifiers are all togheter and they are independent so you can swap them the way you like.


TITLE: Isn't my book doing this math about differentiation wrongly?
QUESTION [13 upvotes]: Problem:
Differentiate with respect to $x$: $\ln(e^x(\frac{x-1}{x+1})^{\frac{3}{2}})$
My attempt:
Let,
$$y=\ln(e^x(\frac{x-1}{x+1})^{\frac{3}{2}})$$
Both $e^x$ and $\frac{x-1}{x+1}$ are positive: $e^x$ can never be negative, and you can take the square root of only positive numbers, so $\frac{x-1}{x+1}$ is positive as well. So, we can apply logarithm properties:
$$=\ln e^x+\ln(\frac{x-1}{x+1})^{\frac{3}{2}}$$
$$=x+\frac{3}{2}\ln(\frac{x-1}{x+1})$$
Now,
$$\frac{dy}{dx}=1+\frac{3}{2}.\frac{x+1}{x-1}.\frac{d}{dx}(\frac{x-1}{x+1})$$
$$=1+\frac{3}{2}.\frac{x+1}{x-1}.\frac{x+1-x+1}{(x+1)^2}$$
$$=1+\frac{3}{2}.\frac{x+1}{x-1}.\frac{2}{(x+1)^2}$$
We can cancel $(x+1)$ and $(x+1)$ in the numerator and denominator because the graph doesn't change.
$$=1+\frac{3}{x^2-1}$$
$$=\frac{x^2+2}{x^2-1}$$
My book's attempt:
Let,
$$y=\ln(e^x(\frac{x-1}{x+1})^{\frac{3}{2}})$$
$$=\ln e^x+\ln(\frac{x-1}{x+1})^{\frac{3}{2}}$$
$$=x+\frac{3}{2}\ln(\frac{x-1}{x+1})$$
$$=x+\frac{3}{2}(\ln(x-1)-\ln(x+1))\tag{1}$$
$$...$$
$$\frac{dy}{dx}=\frac{x^2+2}{x^2-1}$$
Question:

In my book's attempt, is line $(1)$ valid? I think my book's assumption in $(1)$ that $(x-1)$ and  $(x+1)$ must be positive is unfounded. I deduced that $\frac{x-1}{x+1}$ is positive; $\frac{x-1}{x+1}$ can be positive even when both $(x-1)$ and $(x+1)$ is negative. So, is line $(1)$ of my book valid?

REPLY [5 votes]: My book's attempt:
$$y=x+\frac{3}{2}\ln(\frac{x-1}{x+1})$$
$$=x+\frac{3}{2}(\ln(x-1)-\ln(x+1))\tag{1}\\\ldots$$
$$\frac{dy}{dx}=\frac{x^2+2}{x^2-1}$$
is line $(1)$ valid? I think my book's assumption that $(x-1)$ & $(x+1)$ must be positive is unfounded. $\frac{x-1}{x+1}$ can be positive even when both $(x-1)$ & $(x+1)$ are negative.

The author is hand-waving; pretending that $x-1$ and $x+1$ are nonnegative doesn't affect the result.
To be rigorous: on our domain of interest, $$\frac{\mathrm d}{\mathrm dx}\left(x+\frac32\ln \left(\frac{x-1}{x+1}\right)\right)\\ =\frac{\mathrm d}{\mathrm dx}\left(x+\frac32\ln \left|\frac{x-1}{x+1}\right|\right)\\ =\frac{\mathrm d}{\mathrm dx}\left(x+\frac32\Big(\ln \left|{x-1}\right|-\ln\left|{x+1}\right| \Big)\right),$$ noting that $$\frac{\mathrm d}{\mathrm dx}\ln|f(x)|=\frac{f’(x)}{f(x)}.$$
This hand-wavy issue of implicit modulus signs when there are logarithmic functions, is also common when solving differential equations: see here and here.
And here is a related answer: How is taking $\ln$ on both sides justified?

Addendum
For negative $f(x),$ $$\frac{\mathrm d}{\mathrm dx}\ln\left|f(x)\right|\\=\frac{\mathrm d}{\mathrm dx}\ln\left(-f(x)\right)\\=\frac{-f'(x)}{-f(x)}\\=\frac{f'(x)}{f(x)};$$ for positive $f(x),$ again $$\frac{\mathrm d}{\mathrm dx}\ln\left|f(x)\right|=\frac{f'(x)}{f(x)}.$$


text:
\begin{document}

\title[Tessellating Quantum Graphs]{Strong Isoperimetric Inequality \\
For Tessellating Quantum Graphs}

\author[N. Nicolussi]{Noema Nicolussi}
\address{Faculty of Mathematics\\ University of Vienna\\
Oskar-Morgenstern-Platz 1\\ 1090 Vienna\\ Austria}
\email{\mailto{noema.nicolussi@univie.ac.at}}

\thanks{{\it Research supported by the Austrian Science Fund (FWF)
under Grants P28807 and W1245.}}

\keywords{Quantum graph, tilling, isoperimetric inequality}
\subjclass[2010]{Primary \msc{34B45}; Secondary \msc{35P15}; \msc{81Q35}}

\begin{abstract}
We investigate isoperimetric constants of infinite tessellating metric graphs. We introduce a curvature-like quantity, which plays the role of a metric graph analogue of discrete curvature notions for combinatorial tessellating graphs. Based on the definition in \cite{kn17}, we then prove a lower estimate and a criterium for positivity of the isoperimetric constant.
\end{abstract}

\maketitle

\section{Introduction}
	{\em Isoperimetric constants}, which relate surface area and volume of sets, are among the most fundamental tools in spectral geometry of manifolds and graphs. They first appeared in this context in \cite{che}, where Cheeger obtained a lower bound on the spectral gap of Laplace--Beltrami operators. For discrete Laplacians on graphs, versions of Cheeger's inequality are known in various settings, e.g. \cite{al, am, bkw15, dod, dk86, fuj, km, moh88, moh92}. They find application in many fields (such as the study of expander graphs and random walks on graphs, we refer to \cite{lub} and \cite{woe} for more information) and consequently, there is a very large interest in graph isoperimetric constants.

	In the case of {\em tessellating graphs} (i.e.\ edge graphs of tessellations of $\R^2$), they have been investigated using certain notions of {\em discrete curvature} (see for example \cite{h01, kp11, oh, woe, zuk}). On the other hand, the idea of plane graph curvature already appears earlier in several unrelated works \cite{g87, ish, s76} and was also employed to describe other geometric properties, for instance discrete analogues of the Gauss--Bonnet formula and the Bonnet--Myers theorem, e.g. \cite{bp01, dVm, HJL, k11, kp11, s76}.

	Another framework for isoperimetric constants are {\em metric graphs} $\cG$, i.e.\ combinatorial graphs $\cG_d = (\cV, \cE)$ with vertex set $\cV$ and edge set $\cE$, where each edge $e \in \cE$ is identified with an interval $I_e = (0, |e|)$ of length $0 < |e|< \infty$ . Topologically, $\cG$ may be considered as a ``network" of intervals glued together at the vertices.  The analogue of the Laplace--Beltrami operator for metric graphs is the {\em Kirchhoff--Neumann Laplacian} $\bH$ (also known as a {\em quantum graph}). It acts as an edgewise (negative) second derivative $f_e \mapsto -\frac{d^2}{dx_e^2} f_e$, $e \in \cE$, and is defined on edgewise $H^2$-functions satisfying continuity and Kirchhoff conditions at the vertices  (we refer to \cite{bcfk06, bk13, ekkst08, post} for more information and references; see also \cite{ekmn, kn17} for the case that $\cG$ is infinite). A well-known result for {\em finite metric graphs} (i.e.\ $\cV$ and $\cE$ are finite sets) is a spectral gap estimate for $\bH$ in terms of an isoperimetric constant due to Nicaise \cite{nic} (see also \cite{kkmm16, kur}).

	In this work, we are interested in {\em infinite metric graphs} (infinitely many vertices and edges). A notion of an isoperimetric constant $\alpha(\cG)$  in this context  was introduced recently in \cite{kn17} (see \eqref{eq:a_G} below for a precise definition) together with the following estimate
		\begin{align}\label{eq:cheeger}
		\frac{1}{4}\alpha(\cG)^2 \leq \lambda_0(\bH) \leq  \frac{\pi^2}{2\, \inf_{e\in \cE} |e| }\alpha(\cG),
		\end{align}
	which holds for every connected, simple, locally finite, infinite metric graph.
	Here $\lambda_0(\bH) := \inf\, \sigma(\bH)$ is the bottom of the spectrum of $\bH$.

	However, let us stress that explicit computation of isoperimetric constants in general is a difficult problem (known to be NP-hard for combinatorial graphs \cite{moh89}). Hence the question arises, whether one can find bounds on $\alpha(\cG)$ in terms of less complicated quantities. On the other hand, the definition of $\alpha(\cG)$ is purely combinatorial and moreover $\alpha(\cG)$ is related to the isoperimetric constant $\alpha_{\comb} (\cG_d)$ of the combinatorial graph $\cG_d$ (see \cite{kn17} for further details). This strongly suggests to use discrete methods for further study. For combinatorial tessellating graphs, such tools are available in the form of discrete curvature and it is natural to ask whether similar techniques also apply to metric graphs.  Moreover, the class of plane graphs contains important examples such as trees and edge graphs of regular tessellations.

	Motivated by this, the subject of our paper are isoperimetric constants of {\em infinite tessellating metric graphs} (see Definition \ref{def:tess}). Our main contribution is the definition of a {\em characteristic value} $c(\cdot)$ of the edges of a given metric graph (see \eqref{eq:charval}), which takes over the role of the classical discrete curvature  (up to sign convention; as opposed to e.g. \cite{ h01, k11,  kp11}, our results on $\alpha(\cG)$ are formulated in terms of positive curvature). In the simple case of equilateral metric graphs (i.e.\ $|e| = 1$ for all $e \in \cE$), $c$ coincides with the characteristic edge value introduced by Woess in \cite{woe98}. Moreover, for a finite tessellating metric graph (Corollary \ref{cor:GB}),
	\begin{equation}
	\sum_{e \in \cE} - c(e)|e| = 1,
\end{equation}
which can be interpreted as a metric graph analogue of the combinatorial Gauss--Bonnet formula known in the discrete case (see e.g. \cite{k11}).

	In terms of these characteristic values, we then formulate our two main results: Theorem \ref{prop:average} contains a criterium for positivity of $\alpha(\cG)$ based on the averaged value of $c(\cdot)$ on large subgraphs $\wt \cG \subset \cG$. In Theorem \ref{prop:est}, we obtain explicit lower bounds on $\alpha(\cG)$. A simplified version of this estimate is the following inequality:
	\begin{equation}
	\alpha(\cG) \geq \inf_{e \in \cE} c(e).
	\end{equation}
	Theorem \ref{prop:est} can be interpreted as a metric graph analogue of the estimate in \cite[Theorem 1]{kp11} and a result by McKean in the manifold case \cite{mckean}.

	Finally, we demonstrate the use of our theory by examples. First, we consider the case of equilateral  $(p,q)$-regular graphs. Here, $\alpha(\cG)$ is closely related to $\alpha_{\comb} (\cG_d)$  and hence can be computed explicitly. It turns out for large $p$ and $q$,  the estimate in Theorem \ref{prop:est} is quite close to the actual value. Second, we show how to construct an example where $\alpha(\cG)$ and $\alpha_{\comb} (\cG_d)$ behave differently.

	Let us finish the introduction by describing the structure of the paper.
	In Section 2, we recall a few basic notions and give a precise definition of infinite tessellating metric graphs. Moreover, we review the definition of $\alpha(\cG)$ and define the characteristic values. Section 3 contains our main results and proofs. In the final section, we consider examples.

	\section{Preliminaries}\label{sec:Prelim}

	\subsection{(Combinatorial) planar graphs}\label{ss:II.01}
	Let $\cG_d = (\cV, \cE)$ be an infinite, unoriented graph with countably infinite sets of vertices $\cV$ and edges $\cE$.  For a vertex $v \in \cV$, we set
	\be\label{eq:E_v}
	\cE_v := \{e \in \cE | \; e \text{  is incident to } v\}
	\ee
and denote by $\deg(v) := \# \cE_v$ the {\em combinatorial degree} of $v \in \cV$. Throughout the paper, $\# A$ denotes the number of elements of a given  set $A$. We will always assume that $\cG_d$ is connected, simple (no loops or multiple edges) and  locally finite ($\deg(v) < \infty$ for all $v \in \cV$).

Moreover, we suppose that $\cG_d$ is {\em planar} and consider a fixed planar embedding. Hence $\cG_d$ can be regarded as a subset of the plane $\R^2$, which we assume to be closed. Denote by $\cT$ the set of {\em faces} of $\cG_d$, i.e.\ the closures of the connected components of $\R^2 \setminus \cG_d$. Note that unbounded faces of $\cG_d$ are included in $\cT$ as well. Motivated by the next definition, we will refer to the elements $T \in \cT$ as the {\em tiles} of $\cG_d$.

\begin{definition}  \label{def:tess}
A planar graph $ \cG_d$ is {\em tessellating} if the following additional conditions hold:
\begin{enumerate}[label=(\roman*), ref=(\roman*), leftmargin=*, widest=iiii]
		\item $\cT$ is locally finite, i.e.\ each compact subset $K$ in $\R^2$ intersects only finitely many tiles.
		\item Each bounded tile $T \in \cT$ is a closed topological disc and its boundary $\partial T$ consists of a finite cycle of at least three edges.
		\item Each unbounded tile $T \in \cT$ is a closed topological half-plane and its boundary $\partial T$ consists of a (countably) infinite chain of edges.
		\item Each edge $e \in \cE$ is contained in the boundary of precisely two different tiles.
		\item Each vertex $v \in \cV$ has degree $\geq 3$.
	\end{enumerate}
\end{definition}
Here, a subset $A \subseteq \R^2$ is called a closed topological disc (half-plane) if it is the image of the closed unit ball in $\R^2$ (the closed upper half-plane) under a homeomorphism $\phi \colon \R^2 \to \R^2$. For a tile $T \in \cT$, we define
\begin{align}\label{eq:E_T}
\cE_T & := \{e \in \cE | \;  e \subseteq \partial T\}, & d_{\cT}(T) & := \# \cE_T,
\end{align}
where the latter is called the {\em degree of a tile} $T\in\cT$. Notice that according to Definition \ref{def:tess}(ii), $d_{\cT}(T) \geq 3$ for all $T \in \cT$.

The above assumptions (i)--(v) imply that $\cT$ is a {\em locally finite tessellation} of $\R^2$, i.e.\ a locally finite, countable family of closed subsets $T \subset \R^2$ such that the interiors are pairwise disjoint and $\bigcup_{T \in \cT} T = \R^2$. In addition, $\cG_d = (\cV, \cE)$ coincides with the edge graph of the tessellation in the following sense: by calling a connected component of the intersection of at least two tiles $T \in \cT$ a $\cT$-vertex, if it has only one point and a $\cT$-edge otherwise, we recover the vertex and edge sets $\cV$ and $\cE$.

For a finite subgraph $\wt \cG \subset \cG_d$, let $ \cF(\wt \cG)$ be the set of {\em bounded faces} of $\wt \cG$, i.e.\ the closures of all bounded, connected components of $\R^2 \backslash \wt \cG$. By local finiteness, each bounded face of $\wt \cG$ is a finite union of bounded tiles $T \in \cT$. Moreover, define $ \cP(\wt \cG)$ as the set of tiles $T \in \cT$ with $\partial T \subseteq  \wt   \cG$. Note that always
	\[
	\cP(\wt \cG) \subseteq \cF(\wt \cG).
	\]

\subsection{Metric graphs} \label{ss:II.02}
After assigning each edge $e \in \cE$ a finite length $|e|\in (0,\infty)$, we obtain a {\em metric graph} $\cG:=(\cV,\cE,|\cdot| )= (\cG_d,|\cdot|)$. For a subgraph $\wt \cG = (\wt \cV, \wt \cE) \subset \cG$, define its {\em boundary vertices} by
	\begin{equation}\label{eq:bdG}
		\partial_\cG \wt{\cG} 	:= \big\{v\in \widetilde{\cV} |\  \deg_{\wt\cG}(v)<\deg(v)\big\},
	\end{equation}
	where $\deg_{\wt \cG}(v)$ denotes the degree of a vertex $v \in \wt \cV$ with respect to $\wt \cG$. For a given finite subgraph $\wt\cG\subset \cG$ we then set
\be\label{eq:deg_wtG}
\deg(\partial_\cG \wt{\cG}) := \sum_{v\in\partial_\cG \wt{\cG}} \deg_{\wt\cG}(v).
\ee
Following \cite{kn17}, the {\em isoperimetric constant}  of a metric graph $\cG$ is then defined by
	\be\label{eq:a_G}
	\alpha(\cG) := \inf_{\wt{\cG}} \; \frac{ \deg (\partial_\cG \wt{\cG}) }{\mes ( \wt{\cG} ) } \in [0, \infty),
	\ee
	where the infimum is taken over all finite, connected subgraphs $\wt \cG \subset \cG$ and $\mes(\wt\cG)$ denotes the Lebesgue measure of $\wt\cG$, $ \mes(\wt\cG) := \sum_{e\in\wt\cE} |e|$. We will say that the metric graph $\cG$ satisfies the {\em strong isoperimetric inequality} if $\alpha(\cG)>0$.

	Recall that for a combinatorial graph $\cG_d= (\cV, \cE)$ the {\em (combinatorial) isoperimetric constant} $\alpha_{\comb}(\cG_d)$ is defined by (see, e.g., \cite{dk86})
\be\label{eq:a_comb}
	\alpha_{\comb}(\cG_d) = \inf_{U \subset \cV} \frac{\# \{e \in \cE| \; e \text{ connects } U \text{ and } \cV \setminus U \}}{ \sum_{v \in U}\deg(v)},
\ee
where the infimum is taken over all finite subsets $U \subset \cV$. There is a close connection between $\alpha_{\comb}(\cG_d)$ and $\alpha(\cG)$ and we refer for further details to \cite{kn17}.

	We also need the following quantities. The {\em weight} $m(v)$ of a vertex $v \in \cV$ is given by
	\be\label{eq:def_m}
		m(v) = \sum_{e \in \cE_v} |e|.
	\ee
 Clearly, $m(v)$ equals the Lebesgue measure (the ``length") of the star $\cE_v$.
 	The {\em perimeter} $p(T)$ of a tile $T \in \cT$ is defined as
	\begin{align}\label{eq:def_p}
		p(T) := \begin{cases} \sum_{e \in \cE_T} |e|, &   T \in \cT\ \text{is  bounded} \\  \infty, & T \in \cT \ \text{is unbounded} \end{cases}.
	\end{align}

For every $e\in\cE$, we define its {\em characteristic value} $c(e)$  by
	\begin{align}\label{eq:charval}
		c(e) := \frac{1} { |e| }- \sum_{v: v \in e} \frac{1} {m(v)} - \sum_{T: e \subseteq \partial T} \frac{1}{p(T)}.
	\end{align}
Here we employ the convention that whenever $\infty$ appears in a denominator, the corresponding fraction $1/p$ has to be interpreted as zero if $p$ is infinite. Let us mention that for equilateral metric graphs $\cG$ (i.e.\ $|e| \equiv 1$ for all $e \in \cE$), the characteristic value $c(e)$ coincides with the characteristic edge value introduced in \cite{woe98} in the context of combinatorial graphs.

Finally, we need the following class of subgraphs of $\cG$.
A subgraph $\wt \cG =(\wt \cV, \wt \cE) \subset \cG$ is called {\em star-like}, if it can be written as a finite, connected union of stars. More precisely,
	\[ \wt \cE = \bigcup_{v \in \wt U}\cE_v  \]
	for some finite, connected vertex set $\wt U \subseteq \wt{\cV}$.

	Also, for a finite subgraph $\wt \cG \subset \cG$, we define its {\em interior graph} $\wt \cG_{\Int} = (\wt \cV_{\Int},\wt  \cE_{\Int} ) $ as the set of interior vertices $v \in \wt \cV_{\Int} := \wt \cV \setminus \partial \wt \cG$ together with all edges between such vertices.  We say that $\wt \cG$ is {\em complete}, if $ \cF(\wt \cG_{\Int}) = \cP(\wt \cG_{\Int})$, or equivalently if every bounded face of $\wt \cG_{\Int}$ consists of exactly one tile $T \in \cT$. Let us denote the class of star-like complete subgraphs by $\cS(\cG)$.

\section{Strong isoperimetric inequality for tessellating quantum graphs}\label{sec:main}

Now we are in position to formulate our main results. Our first theorem relates the positivity of the isoperimetric constant with the positivity of the characteristic values of a metric graph.

\begin{theorem} \label{prop:average}
	Let $\cG = (\cV, \cE, |\cdot|)$ be a tessellating metric graph having infinite volume, $\mes(\cG) = \sum_{e\in\cE}|e|=\infty$. If
	\be\label{eq:ellast}
	\ell^\ast(\cG):= \sup_{e \in \cE} |e| < \infty
	\ee
	 and
		\be\label{eq:aver_c}
		\liminf_{\substack{\mes(\wt \cG) \to \infty }} \frac{1}{ \mes(\wt \cG)}\sum_{e \in \wt \cE} c(e) |e|
		= \liminf_{\substack{\mes(\wt \cG) \to \infty }} \frac{\sum_{e \in \wt \cE} c(e) |e|}{\sum_{e \in \wt \cE} |e|}
		    > 0,
		\ee
	then $\alpha(\cG) >0$. Here $\liminf$ is taken over all star-like complete subgraphs $\wt \cG\in\cS(\cG)$.
\end{theorem}

\begin{remark}
A few remarks are in order.
\begin{enumerate}[label=(\roman*), ref=(\roman*), leftmargin=*, widest=iiii]
\item[(i)] Let us mention that \eqref{eq:ellast} is necessary for the strong isoperimetric inequality to hold for an arbitrary metric graph since (see, e.g., \cite[Remark 3.3]{kn17})
\be\label{eq:estSol}
\alpha(\cG)  \le \frac{2}{\ell^\ast(\cG)}.
\ee
\item[(ii)] If $\mes(\cG) = \sum_{e\in\cE}|e| < \infty$, then the lower bound
\begin{equation} \label{eq:estvol}
	\alpha(\cG) \geq \frac{2}{\mes(\cG)}>0
\end{equation}
holds. In fact, if $\cG$ is tessellating, then $\deg (\partial \wt \cG) \ge 2$ for every finite subgraph $\wt \cG \subset \cG$ and \eqref{eq:estvol} follows immediately from \eqref{eq:a_G}.
\item[(iii)] Theorem \ref{prop:average} can be seen as the analogue of \cite[Theorem 1]{woe98}.
\item[(iv)] As we will see below, the proof of Theorem \ref{prop:average} implies the explicit estimate
\begin{equation} \label{eq:est01}
	\alpha (\cG) \geq \min \Big\{\frac{2}{\ell^\ast(\cG)}, \; \ \inf_{\wt \cG \in \cS}  \frac{1}{ \mes(\wt \cG)}\sum_{e \in \wt \cE} c(e) |e|  \Big\},
\end{equation}
however, the conditions in Theorem \ref{prop:average} are weaker than positivity of the right-hand side in \eqref{eq:est01}.
\end{enumerate}
\end{remark}

The next result shows that pointwise estimates for the characteristic values also yield lower estimates for the isoperimetric constant. To this end, introduce
the following quantities
\begin{align}
M(\cG) & := \sup_{v\in\cV} \frac{m(v)}{\inf_{e\in\cE_v}|e|}, & P(\cG) & := \sup_{T \in \cT} \frac{p(T)}{\inf_{e\in \cE_T}|e|},
\end{align}
and set
\be\label{eq:K_G}
	 K  (\cG) := 1-  \frac{1}{M(\cG)}  -  \frac{2}{P(\cG)} - \frac{1}{(M(\cG) -2) P(\cG)}.
	\ee

	\begin{theorem} \label{prop:est}
	Let $\cG = (\cV, \cE, |\cdot|)$ be a tessellating metric graph. Suppose
	\be\label{eq:ce_ast}
	c_\ast(\cG) := \inf_{e \in \cE} c(e) > 0.
        \ee
        Then
	\be\label{eq:iso_02}
		\alpha(\cG) \geq    \frac{c_\ast (\cG)}{K (\cG)}  > 0.
	\ee
\end{theorem}
Theorem \ref{prop:est}  can be considered as the metric graph analogue of the corresponding estimate for combinatorial graphs in \cite[Theorem 1]{kp11}.
\begin{remark} \label{rem:K}
The following obvious estimates
\begin{align} \label{est:Mdeg}
M(\cG) & \ge \sup_{v\in\cV} \deg(v) \ge 3, & P(\cG) & \ge \sup_{T\in\cT} d_\cT(T) \ge 3,
\end{align}
imply that $K(\cG) \le 1$.
Moreover, one can show that $K(\cG) > 0$ if $c_\ast(\cG) > 0$. Indeed, noting that
\begin{align*}
m(v) & \le  \deg(v)\ell^\ast(\cG), & p(T) & \le  d_\cT(T)\ell^\ast(\cG),
\end{align*}
we easily get the following rough estimate
\begin{align}\label{eq:c_ast_est1}
	c_\ast (\cG) &\leq
		 \frac{1}{\ell^\ast (\cG)}   \Big( 1 - \frac{2}{ \deg^\ast (\cG) } -  \frac{2}{   d_\cT^\ast (\cG) }    \Big),
\end{align}
where $\deg^\ast (\cG)  :=  \sup_{v\in\cV} \deg(v)$ and $  d_\cT^\ast (\cG)   :=  \sup_{T\in\cT} d_\cT(T)$.
On the other hand,
\begin{align*}
	K(\cG) & =  \frac{1}{2}\Big(1 -  \frac{2}{M(\cG)}  -  \frac{2}{P(\cG)}\Big)+ \frac{1}{2} -  \frac{1}{P(\cG)} - \frac{1}{(M(\cG) -2) P(\cG)} \\
	&\ge \frac{1}{2}\Big( 1 - \frac{2}{ \deg^\ast (\cG) } -  \frac{2}{   d_\cT^\ast (\cG)  }    \Big)  + \frac{1}{2} - \frac{1}{   d_\cT^\ast (\cG) } - \frac{1}{( \deg^\ast (\cG)- 2)  d_\cT^\ast (\cG) }.
\end{align*}
If $c_\ast(\cG) > 0$, then so is the right-hand side in \eqref{eq:c_ast_est1} which implies
\begin{align*}
	K(\cG)  > & \frac{1}{2} - \frac{1}{  d_\cT^\ast (\cG) } - \frac{1}{( \deg^\ast (\cG) - 2)   d_\cT^\ast (\cG) }\\
	 > & \frac{1}{ \deg^\ast (\cG) } - \frac{1}{( \deg^\ast (\cG) - 2)  d_\cT^\ast (\cG) } \ge 0.
\end{align*}
\end{remark}

To prove Theorems \ref{prop:average} and \ref{prop:est}, we first show that we can restrict in \eqref{eq:a_G} to star-like complete subgraphs.
\begin{lemma} \label{lem:reduction}
	Let $\cG = (\cV, \cE, |\cdot|)$ be a tessellating metric graph. Then
	\begin{align} \label{eq:minalpha}
		\alpha(\cG) = \min \Big\{\frac{2}{\ell^\ast(\cG)}, \alpha_\cS(\cG)  \Big\},
	\end{align}
	where
	\be\label{eq:aS_G}
		\alpha_\cS(\cG) := \inf_{\wt \cG \in \cS }  \frac{\deg ( \partial \wt \cG)}{ \mes(\wt \cG)}.
	\ee
	\end{lemma}
	\begin{proof}
	(i) First, we show that it suffices to consider subgraphs that are either star-like or consist of a single edge. Let $\wt \cG = (\wt \cV, \wt \cE)$ be a finite, connected subgraph of $\cG$ and $\wt \cV_{\Int} = \wt \cV \setminus \partial \wt \cG$. We split $ \wt \cV_{\Int} = \bigcup_{i=1}^n \cV_i$ into a finite, disjoint union of connected vertex sets $\cV_i$. Let $ \cG_i =(\cV_i, \cE_i) \subset \cG$ be the subgraph with edge set
	\[
		 \cE_i = \bigcup_{v \in \cV_i} \cE_v.
	\]
	 By construction, each $ \cG_i$ is star-like and each edge $e \in  \wt \cE$ belongs to at most one $ \cG_i$. Let
	 $\wt\cE_r = \wt \cE \setminus \bigcup_{i=1}^n \cE_i$ be the remaining edges. Then
	 \[
	 \mes(\wt \cG) = \sum_{i=1}^n \mes(\cG_i)  + \sum_{e \in  \wt\cE_r} |e|.
	 \]
	Moreover, both vertices of an edge $e \in \wt \cE_r$ are in $\partial \wt \cG$ and $\partial \cG_i = \partial \wt \cG \cap \cV_i$. Hence
	\begin{align*}
	\deg(\partial \wt{\cG}) = \sum_{v \in \partial \wt \cG}  \sum_{i=1}^n \deg_{\cG_i} (v) +  2 \# \cE_r = \sum_{i=1}^n \deg(\partial \cG_i) + 2 \# \cE_r.
	\end{align*}
	This finally implies
	\begin{align*}
		\frac{ \deg(\partial \wt{\cG} )}{\mes(\wt \cG)} = \frac { \sum_{i=1}^n \deg(\partial \cG_i) + 2 \# \cE_r } { \sum_{i=1}^n \mes(\cG_i)  + \sum_{e \in  \cE_r} |e|} \geq \min_{\substack{i=1,...,n, \\ e \in \cE_r}} \left \{  \frac{ \deg(\partial \cG_i  )}{ \mes(\cG_i)}, \frac{2}{ | e |} \right \}.
	\end{align*}

	(ii) To complete the proof, it suffices to construct for every star-like subgraph $\wt \cG $ a star-like, complete subgraph $\widehat \cG \in\cS(\cG)$ with $\widehat \cG  \supseteq   \wt \cG  $ and $ \deg ( \partial  \wt \cG)  \geq  \deg ( \partial  \widehat \cG) $.
	Let $\wt\cG_{\Int}=(\wt \cV_{\Int}, \wt \cE_{\Int})$ be the interior graph of $\wt \cG$. Denote by $  \cF_0  $ the set of bounded, open components of  $\R^2 \setminus \wt \cG_{\Int}$ and by $\cF = \{  F = \ol{f} | \; f \in \cF_0\}$ the bounded faces of $\wt \cG_{\Int}$. The idea is to add ``edges contained in bounded faces". Define the  subgraph $\widehat \cG = (\widehat \cV, \widehat \cE)$ by its edge set
	\[
		\widehat \cE =  \wt \cE \cup \bigcup_{v \in f\colon f \in \cF_0} \cE_v   .
	\]
	If an edge $e \in \cE$ is incident to a vertex $v \in f$ with $ f \in \cF_0$, then its other vertex lies in $F = \ol{f}$. Hence $\deg_{\widehat \cG} (v) = \deg_{\wt \cG} (v) $ for every vertex $v$ with $v \notin K := \bigcup_{F \in \cF }F$. On the other hand, every vertex $v \in K$ belongs to $\widehat \cG$ and satisfies $\deg_{\widehat \cG} (v) = \deg_{\cG} (v)$. Indeed, if $v \in  F = \ol{f}$, then either $v \in f$ or $v \in \partial f \subseteq \wt \cG_{\Int} $.
	This implies $\partial \widehat \cG \subseteq \partial \wt \cG$ and $ \deg ( \partial  \wt \cG)  \geq  \deg ( \partial  \widehat \cG) $.
	Moreover, $\widehat \cE = \cup_{v \in \widehat U} \cE_v$, where
	\[
		\widehat U = \wt \cV_{\Int} \cup \bigcup_{f \in \cF_0} \{ v \in \cV | \; v \in f \}.
	\]
	 Also, $\widehat U$ is finite by local finiteness and connected since $\wt \cG$ is star-like and $\partial f \subseteq \wt \cG_{\Int}$ for $f \in \cF_0$. It remains to show that $\widehat \cG$ is complete. Let $\widehat F$ be a bounded face of the interior graph $\widehat \cG_{\Int}$. Suppose $T \in \cT$ with $T \subseteq \widehat F$. Then $T \subseteq \widehat F \subseteq F$ for some bounded face $F$ of $\wt \cG_{\Int}$. In particular, $e \subseteq K$ for every edge $e \subseteq \partial T$. But every vertex $ v \in K$ belongs to $\widehat \cG_{\Int}$, and hence $\partial T \subseteq \widehat \cG_{\Int}$ and $\widehat F = T$
\end{proof}

\begin{remark} \label{rem:min}
Combining \eqref{eq:minalpha} with \eqref{eq:estSol}, one concludes that the inequality
\begin{equation} \label{eq:estmin}
\frac{2}{\ell^\ast(\cG)} \le  \alpha_\cS(\cG) = \inf_{\wt \cG \in \cS }  \frac{\deg ( \partial \wt \cG)}{ \mes(\wt \cG)}
\end{equation}
implies that
\be
\alpha(\cG) = \frac{2}{\ell^\ast(\cG)}.
\ee
In Example \ref{ex:min}, we provide an explicit construction of a graph satisfying \eqref{eq:estmin}.
\end{remark}

The next lemma contains the connection between $c(e)$ and $\alpha(\cG)$.
\begin{lemma} \label{lem:degsum}
		The following inequality
		\be\label{eq:isoStar}
			\sum_{e \in \wt \cE} c(e) |e| \leq \deg (\partial_\cG \wt \cG)
		\ee
		holds for any star-like, complete subgraph $\wt \cG\in \cS(\cG)$.
	\end{lemma}

\begin{proof}
	Let $\wt \cG_{\Int} = (\wt \cV_{\Int}, \wt \cE_{\Int})$ be the interior graph and  $\cE^b := \wt \cE \setminus \wt \cE_{\Int}$ the remaining edges. Then
 		\begin{align*}
		\sum_{e \in \wt \cE} c(e) |e| &= \sum_{ e \in \wt \cE} 1 - \sum_{v \in \wt \cV}   \frac{\mes(\cE_v \cap \wt \cE)}{ m(v)} - \sum_{T \in \cT} \frac{\mes( \cE_T \cap \wt \cE_{\Int})}  {p(T)} - \sum_{T \in \cT} \frac{\mes( \cE_T \cap \cE^b)}{p(T)}    \\
		&= \# \cE^b  +  \# \wt \cE_{\Int}   - \# \wt \cV_{\Int}  -   \# \cP(\wt \cG_{\Int})	\\
		& \; \; -   \sum_{v \in \partial \wt \cG} \frac{\mes(\cE_v \cap \wt \cE )}{ m(v)}- \sum_{T \in \cT, \cE_T \not \subseteq \wt \cE_{\Int}}  \frac{\mes( \cE_T \cap \wt \cE_{\Int})} {p(T)}  - \sum_{T \in \cT} \frac{\mes( \cE_T \cap \cE^b)}{p(T)},
	\end{align*}
	where $\cP(\wt \cG_{\Int})$ is the set of tiles $T \in \cT$ with $\cE_T \subseteq \wt \cE_{\Int}$. By Euler's formula (see, e.g., \cite{har})
	\[
		  \# \wt \cE_{\Int} - \# \wt \cV_{\Int}  - \#  \cF(\wt \cG_{\Int})  = - 1,
	\]
Because $\wt \cG$ is complete,  $\cF(\wt \cG_{\Int}) = \cP(\wt \cG_{\Int})$ and $\sum |e| c(e)$ is equal to
	\begin{align*}
		 & \# \cE^b -1 -  \sum_{v \in \partial \wt \cG} \frac{\mes(\cE_v \cap \wt \cE )}{ m(v)} - \sum_{T \in \cT, \cE_T \not \subseteq \wt \cE_{\Int}}  \frac{\mes( \cE_T \cap \wt \cE_{\Int})} {p(T)}  - \sum_{T \in \cT} \frac{\mes( \cE_T \cap \cE^b)}{p(T)}.
	\end{align*}
 Since $\wt \cG$ is star-like, there are no edges $e \in \wt \cE$ with both vertices in $\partial \wt \cG$. Therefore, $
  \# \cE^b = \deg( \partial \wt \cG)$ and the proof is complete.
\end{proof}

\begin{remark}

For future reference, observe that
	\begin{align*}
  	\sum_{v \in \partial \wt \cG} & \frac{\mes(\cE_v \cap \wt \cE )}{ m(v)} + \sum_{T \in \cT, \cE_T \not \subseteq \wt \cE_{\Int}}  \frac{\mes( \cE_T \cap \wt \cE_{\Int})} {p(T)}  + \sum_{T \in \cT} \frac{\mes( \cE_T \cap \cE^b)}{p(T)} \\
	& \geq \sum_{v \in \partial \wt \cG} \frac{\deg_{\wt \cG} (v) }{ M(\cG)  } +   \sum_{T \in \cT, \cE_T \not \subseteq \wt \cE_{\Int}} \frac{ \# (\cE_T \cap \wt \cE_{\Int}) }{ P(\cG) }  +   \sum_{T \in \cT} \frac{ \# ( \cE_T \cap \cE^b )}{ P(\cG) }.
  \end{align*}
   This implies the following estimate
  \begin{equation} \label{eq:tech}
  \begin{split}
  	\sum_{e \in \wt \cE} c(e) |e| \leq&  \deg (\partial \wt \cG)\Big( 1 - \frac{1}{M(\cG)} - \frac{2}{P(\cG) }  \Big)  \\
	&- \frac{1}{P(\cG) }  \sum_{ e \in \wt \cE_{\Int}} \#  \big\{ T | \; e \in \cE_T \text{ and } \cE_T \not \subseteq \wt \cE_{\Int} \big\}
	\end{split}
  \end{equation}
  for every  star-like, complete subgraph $\wt \cG \in \cS(\cG)$.
\end{remark}

\begin{corollary}\label{cor:GB}
Let  $\cG = (\cV, \cE, |\cdot|)$ be  a finite tessellating metric graph, that is a finite graph satisfying all the assumptions of Section \ref{ss:II.01} except (iii) of Definition \ref{def:tess}.
Then
	\begin{equation} \label{eq:GB1}
		\sum_{e \in \cE} - c(e)|e| = 1.
	\end{equation}
\end{corollary}

\begin{proof}
	By Euler's formula
	\begin{align*}
		\sum_{e \in \cE} c(e)|e|  &= \# \cE - \sum_{v \in \cV} \frac{\mes(\cE \cap \cE_v) }{m(v)} - \sum_{T \in \cT}  \frac{\mes(\cE \cap \cE_T) }{ p(T) } \\
						&= \# \cE - \# \cV - \# \cF(\cG) = -1. \qedhere
	\end{align*}
\end{proof}
\begin{remark} \label{rem:GB}
Formula \eqref{eq:GB1}  can be seen as the analogue of the combinatorial Gauss--Bonnet formula known for combinatorial graphs (see \cite[Proposition 1]{k11}). Let us also mention that the difference in the right-hand side arises from our convention  $p(T) = \infty$ for unbounded $T \in \cT$.
\end{remark}

Theorem \ref{prop:average} now follows from Lemma \ref{lem:reduction} and \ref{lem:degsum} together with the inequality $\deg( \partial \wt \cG) \geq 1$ for $\wt \cG \in \cS(\cG)$. Moreover, we can already deduce (see \eqref{eq:est01} and \eqref{eq:c_ast_est1}) the basic estimate
\begin{equation} \label{eq:estbasic}
	\alpha(\cG) \geq c_\ast(\cG).
\end{equation}
By improving this bound further we finally obtain Theorem \ref{prop:est}.

\begin{proof}[Proof of Theorem \ref{prop:est}]
We start by providing a basic inequality. By \ref{rem:K}, we have $K(\cG) >0$. Let $\deg^\ast(\cG) := \sup_{v \in \cV} \deg(v)$. Then using \eqref{est:Mdeg} and \eqref{eq:c_ast_est1}, a lengthy but straightforward calculation implies
\begin{equation} \label{eq:fundamentalest}
	\frac{c_\ast(\cG)}{K(\cG)}  \leq \frac{\deg^\ast (\cG)-2}{\deg^\ast (\cG)-1} \frac{1}{\ell^\ast(\cG)}.
\end{equation}
Hence by Lemma \ref{lem:reduction}, it suffices to show that
\begin{equation} \label{eq:toprove}
	\frac{\deg(\partial \wt \cG)}{\mes(\wt \cG)} \geq \frac{c_\ast(\cG)}{K(\cG)}
\end{equation}
for every $\wt \cG= (\wt \cV, \wt \cE) \in \cS(\cG)$.

 We will obtain \eqref{eq:toprove} by induction over $\#\wt \cV_{\Int} $. If $\# \wt \cV_{\Int}  = 1$, that is, $\wt \cV_{\Int}  = \{v\}$ for some $v\in\cV$, then $\wt \cG$ ``consists of a single star". More precisely, $\wt \cE = \cE_v$ and \eqref{eq:fundamentalest} implies
\[
	\frac{\deg(\partial \wt \cG)}{\mes(\wt \cG)} \geq \frac{\deg(v)}{\deg(v) \ell^\ast(\cG)} \geq  \frac{c_\ast(\cG)}{K(\cG)} .
\]
Now suppose $\# \wt \cV_{\Int}  = n \geq 2$ and \eqref{eq:toprove} holds for all $\widehat \cG \in \cS(\cG)$ with $ \#\widehat \cV_{\Int}  < n$. We distinguish two cases:

(i) First, assume
\[
\#\{u \in \partial \wt \cG| \; u \text{ is connected to } v \} \leq \deg^\ast (\cG)-2
\]
 for all $v \in \wt \cV_{\Int} $. In view of \eqref{eq:tech}, define
 \[
 \cE_i := \{e \in \wt \cE_{\Int}  |\; \# \{ T | \; e \in \cE_T \text{ and } \cE_T \not \subseteq  \wt \cE_{\Int}\} = i \}
 \]
for $i\in\{1,2\}$. Then
\begin{align*}
\sum_{ e \in \wt \cE_{\Int}} {\#\{ T | \; e \in \cE_T \text{ and } \cE_T \not \subseteq \wt \cE_{\Int} \} } = \# \cE_1 + 2 \# \cE_2 = \sum_{v \in \wt \cV_{\Int} } \delta(v) ,
\end{align*}
where $\delta(v):= \#( \cE_v \cap \cE_1) /2 +  \#( \cE_v \cap \cE_2)$ for all $v \in \cV$.

Now assume that $v \in \wt \cV_{\Int} $ and that $v$ is connected to at least one vertex in $ \partial \wt \cG$. Since $\wt \cG$ is star-like and $\# \wt \cV_{\Int}  \geq 2 $, $v$ is connected to another vertex in $\wt \cV_{\Int} $ and hence there exists an interior edge $e \in \wt \cE_{\Int} $ incident to $v$. Going through the edges incident to $v$ in counter-clockwise direction starting from $e$, denote by $e_+$ the ``last" edge incident to $v$ with $e_+ \in \wt \cE_{\Int} $. Define $e_-$ analogously by going clockwise. Then $e_\pm \in \cE_1 \cup \cE_2$. Moreover, if $e_+ = e_-$, then $e = e_+ = e_- \in \cE_2$. Thus $\delta(v) \geq 1$ for every such $v \in \wt \cV_{\Int} $. Since $\wt \cG$ is star-like,
\begin{align*}
	\sum_{v \in \wt \cV_{\Int} } \delta(v) &\geq \frac{1}{\deg^\ast (\cG)-2} \sum_{v \in \wt \cV_{\Int} }  \# \{u \in \partial \wt \cG|\; u \text{ is connected to } v\} \\
	&\geq \frac{1}{M (\cG)-2} \deg(\partial \wt \cG),
\end{align*}
	and \eqref{eq:toprove} follows from \eqref{eq:tech}.

	(ii) Assume that $\#\{u \in \partial \wt \cG| \; u \text{ is connected to } v \} \geq \deg^\ast (\cG) -1 $ for some vertex $v \in \wt \cV_{\Int} $. Since $\# \wt \cV_{\Int} \geq 2$, this implies $\deg(v) = \deg^\ast (\cG)$ and that $v$ is connected to exactly one $w \in \wt \cV_{\Int} $. We ``cut out" the $\deg^\ast (\cG)-1$ edges between $v$ and $\partial \wt \cG$ and define $\widehat \cG = (\widehat \cV, \widehat \cE)$ by its edge set
	\[
		\widehat \cE = \wt \cE \setminus \{e \in \cE| \; e \text{ connects } v \text{ and } \partial \wt \cG \}.
	\]
	Then $\widehat \cG$ is again star-like and complete. Its interior graph $\widehat \cG_{\Int}  = (\widehat \cV_{\Int} , \widehat \cE_{\Int} )$ is given by $\widehat \cV_{\Int}  = \wt \cV_{\Int}  \setminus \{v\}$ and $\widehat \cE_{\Int}  = \wt \cE_{\Int}  \setminus \{e_{v,w}\}$,
	where $e_{v,w}$ is the edge between $v$ and $w$. In particular, $\widehat \cG$ satisfies \eqref{eq:toprove}.

	Now assume $\eqref{eq:toprove}$ fails for $\wt \cG$. Then
	\[
	\deg(\partial \wt \cG) (\mes(\wt \cG) - \mes(\widehat \cG)) \leq (\deg^\ast (\cG)-2)  \mes(\wt \cG)
	\]
	by \eqref{eq:fundamentalest}. Consequently,
	\begin{align*}
		\frac{\deg(\partial \widehat \cG)}{\mes(\widehat \cG)} = \frac{ \# \widehat \cE - \# \widehat \cE_{\Int}  }{\mes(\widehat \cG)} &= \frac{ \# \wt \cE - \# \wt \cE_{\Int}  - \deg^\ast (\cG) + 2 }{\mes(\widehat \cG)}  \\
		&= \frac{\deg(\partial \wt \cG) -\deg^\ast (\cG) +2}{\mes(\widehat \cG)} \leq \frac{\deg(\partial \wt \cG)}{\mes(\wt \cG)} < \frac{c_\ast(\cG)}{K(\cG) },
	\end{align*}
	which is a contradiction.
\end{proof}

\section{Examples}\label{sec:examples}

In this section, we illustrate the use of our results in three examples.

\subsection{(p,q)-regular graphs} \label{ex:pq}
Let $p\in \Z_{\ge 3}$ and  $q\in \Z_{\ge 3}\cup \{\infty\}$. A tessellating (combinatorial) graph $\cG_d=(\cV, \cE)$ is called {\em $(p,q)$-regular}, if $\deg(v) =p$ for all $v \in \cV$ and $d_{\cT}(T)  = q$ for all $T \in \cT$. Let $\cG_{p,q}$ denote both the corresponding combinatorial graph and the associated equilateral  metric graph, that is,  we put $|e| \equiv 1$ for all $e\in\cE_{p,q} = \cE(\cG_{p,q})$. Notice that $\cG_{p,\infty}$ is an infinite $p$-regular tree $\T_p$ (also known as a {\em Cayley tree} or a {\em Bethe lattice}).

Next, by \eqref{eq:charval},  we get
\begin{equation}\label{eq:c_pq}
	c(e) = 1- \frac{2}{p} - \frac{2}{q} =: c_{p,q},
\end{equation}
for all $e \in \cE_{p,q}$, and the {\em vertex curvature} of the combinatorial graph $\cG_{p,q}$ (see for example \cite{h01, dVm, woe98}) is given by
\begin{align}\label{eq:kappa_pq}
	 \kappa(v) = 1 - \frac{\deg(v)}{2} + \sum_{T: v \in T} \frac{1}{  d_{\cT}(T) } = 1 - \frac{p}{2} + \frac{p}{q} = -\frac{p}{2}\,c_{p,q},
\end{align}
for all $ v \in \cV$.

Since strictly positive vertex curvature implies that $\cG_d$ has only finitely many vertices (see \cite[Theorem 1.7]{dVm}), the characteristic value should satisfy $c_{p,q} \geq 0$.
 Clearly, $c_{p,q}=0$ exactly when $(p,q) \in \{ (4,4), (3,6), (6,3)\}$ and in these cases $\cG_{p,q}$ is isomorphic to the square, hexagonal or triangle lattice in $\R^2$. If $c_{p,q} >0$, then $\cG_{p,q}$ is isomorphic to the edge graph of a tessellation of the Poincar\'e disc ${\rm H}^2$ with regular $q$-gons of interior angle $2 \pi/p$ (see \cite[Remark 4.2.]{hjl} and \cite{iv}). In the latter case, Theorem \ref{prop:est} implies $\alpha(\cG_{p,q})> 0$ and the estimate
\begin{align} \label{eq:estpq}
	\alpha(\cG_{p,q}) \geq  \frac{  q(p-2)c_{p,q} }{q(p-1)c_{p,q} + 1 }
	=  \frac{p-2}{p-1}\times \begin{cases} \frac{1}{1+ (q(p-1)\,c_{p,q})^{-1}}, &q < \infty \\[2mm] 1, &q =  \infty\end{cases}.
\end{align}
 Notice that in the case $q = \infty$, equality holds true in \eqref{eq:estpq} (see, e.g., \cite[Example 8.3]{kn17}).

 It is well-known that (see \cite{hs03}, \cite{hjl}),
\begin{equation} \label{eq:alphacomb}
	\alpha_{\comb} (\cG_{p,q}) = \frac{p-2}{p} \sqrt{1 - \frac{4}{(p-2)(q-2)}}.
\end{equation}
By (a slight modification of)  \cite[Lemma 4.1]{kn17},
\begin{align} \label{eq:combmetric}
	\alpha(\cG) = \frac{2 \alpha_{\comb} (\cG_d) }{\alpha_{\comb}( \cG_d )+1}
\end{align}
for every equilateral metric graph $\cG=(\cV, \cE, | \cdot |)$ with underlying combinatorial graph $\cG_d = (\cV, \cE)$. Hence
\be\label{eq:a_Gpg}
	\alpha(\cG_{p,q}) = \frac{p-2}{p-1 + \frac{p}{2} \left( \sqrt{ \frac{(p-2)(q-2)}{pq-2(p+q) }} - 1 \right)}
	=  \frac{p-2}{p-1} \times
	\begin{cases} \; \frac{  1 }{1 +  \delta (q(p-1)\, c_{p,q}) ^{-1}   },  &q < \infty \\[2 mm]1, &q =  \infty \end{cases},
	\ee
where
\[
\delta := \frac{pq-2(p+q)}{2} \left( \sqrt{ 1 + \frac{4}{pq-2(p+q)}} - 1 \right) \leq 1.
\]
Comparing \eqref{eq:a_Gpg} with \eqref{eq:estpq}, we conclude that the error in the estimate \eqref{eq:estpq} is uniformly of order $\frac{1}{(pq)^2}$.

Finally, let us mention that using \eqref{eq:combmetric}, we can turn \eqref{eq:estpq} into a lower estimate for $\alpha_{\comb}(\cG_{p,q})$ as well. After a short calculation, we recover Theorem 1 from \cite{kp11},
\begin{equation}
	\alpha_{\comb}(\cG_{p,q}) \geq \frac{p-2}{p} \left( 1 - \frac{2}{(p-2)(q-2)-2} \right).
\end{equation}

\subsection{Another example} \label{ex:c}
Denote by $\Z^2_+ $ the square lattice of the upper half-plane, i.e.\ the combinatorial graph with vertex set $\Z \times  \Z_{\geq 0}$ and two vertices connected if and only if they are connected in the square lattice $\Z^2= \Z \times \Z$. Fix $k \in \Z_{\ge 3}$ and let $\cG_k$ be the graph obtained from $\Z_+^2$ by attaching to each vertex $v \in \Z \times \{ 0 \}$ an infinite $k$-regular tree (see Figure~\ref{fig:G1}).

To assign edge lengths, we first define a partition of the edge set $\cE_{k}$.  We denote by $\cE_{k,{\rm tree}}$ the set of edges $e \in \cE_k$ belonging to one of the attached trees. Also,  let
\[
 \cV_n = \{(z, n)| \; z \in \Z \} = \Z \times \{n\},\qquad n\in\Z,
 \]
be the vertices on the ``$n$-th horizontal line".
For $n \in \Z_{\ge 0}$, we define $ \cE_{k,n}^+$ as the set of ``vertical" edges between the $n$-th horizontal line $\cV_n$ and the $(n+1)$-th horizontal line $\cV_{n+1}$, and $ \cE_{k,n}^-$ as the set of ``horizontal" edges connecting vertices in the $n$-th horizontal line $\cV_n$ (see Figure \ref{fig:G1}).
Finally, we equip $ \cG_k$ with edge lengths in the following way:
\be\label{eq:length2}
|e| = \begin{cases} 1, & e\in \cE_{k,{\rm tree}} \\[1mm] \frac{1}{(2n+2)^2}, & e\in \cE^-_{k,n} \\[1mm] \frac{1}{(2n+3)^2}, & e\in \cE^+_{k,n} \end{cases}.
\ee

\begin{figure} [ht]
	\begin{center}
		\begin{tikzpicture}    [scale=1.3]

		\foreach \x in {-3,...,3}
		\foreach \y in {0,...,2}
		{ \filldraw  ( \x ,\y) circle (1 pt);  }

	\foreach \y in {0,...,2}{
		\draw (-3.5,  \y) -- (3.5 ,  \y) ;
		\draw [ dotted] (-4,  \y) -- (-3.5 ,  \y) ;
		\draw [ dotted] (3.5,  \y) -- (4 ,  \y) ;
	}

	\draw (-4,  0) -- (4 ,  0) ;

	\foreach \x in {-3,...,3}{
		\draw ( \x, 0) -- (\x, 2.5) ;
		\draw [ dotted] ( \x, 2) -- (\x, 3) ;
	}

	\foreach \y in {0,1, 2}{
		\node [right] at (-4.6, \y ) { $  \cE_{k, \y}^-$};
		\node [right] at (-3.8, \y + 0.5 ) { $  \cE_{k, \y}^+$};
	}

	\node [right] at (-4.6, - 1 ) {$  \cE_{k,{\rm tree}}$};

	\foreach \x in {-3,...,3}{
	\foreach \y in {-1,...,1}{
	\foreach \z in {-1,1}{
	\foreach \w in {-1,1}{
		\draw ( \x, 0) -- (\x -1/3*\y, -0.5) ;
		{ \filldraw (\x -1/3*\y, -0.5) circle (1 pt);  }
		\draw (\x -1/3*\y, -0.5) -- (\x -1/3*\y + 1/12*\z, -1) ;
		{ \filldraw (\x -1/3*\y + 1/12*\z, -1) circle (1 pt);  }
		\draw [densely dotted]  (\x -1/3*\y + 1/12*\z, -1) -- (\x -1/3*\y + 1/12*\z + 1/20*\w, -1.5) ;
	}}}}

		\end{tikzpicture}
	\end{center}
	\caption{$\cG_k$ for $k = 3$.} \label{fig:G1}
\end{figure}

Now let us compute the characteristic values. First of all, for all $e\in \cE_{k,{\rm tree}}$ we have the estimate
\[
 \inf_{e\in\cE_{k,{\rm tree}}} c(e) = 1 - \frac{2}{k} = \frac{k-2}{k}.
\]
Next, taking into account that $k\ge 3$, we get
\[
	c(e) =  4 - \frac{2}{ k + 2\frac{1}{4} + \frac{1}{9} } - \frac{1}{\frac{1}{4} + 2\frac{1}{9}+\frac{1}{16} } = \frac{164}{77} - \frac{2}{ k + \frac{11}{18}} > 1
\]
for all $e \in \cE_{k,0}^-$, and
\[
	c(e) = 9 - \frac{1}{ k + 2\frac{1}{4} + \frac{1}{9} } - \frac{1}{\frac{1}{9}+\frac{1}{25} + 2\frac{1}{16}} - \frac{2}{\frac{1}{4} + 2\frac{1}{9}+\frac{1}{16} }= \frac{8955}{ 5467}  -    \frac{1}{ k + \frac{11}{18}} > 1
\]
for $ e \in \cE_{k,0}^+$.

Moreover, after lengthy but straightforward calculations one can see that
\[
c(e) > 1
\]
for all $e \in \cE_{k,n}^\pm$ with $n \geq 1$.
Thus we obtain
\be
	c_\ast(\cG_k) = \inf_{e\in \cE_{k,{\rm tree}}} c(e) = \frac{k-2}{k} >0,
\ee
and hence, by Theorem \ref{prop:est}, $\alpha(\cG_k)  >0$.

Now let us compute $K(\cG_k)$. If $v \in \cV_n$ with $n \geq 1$, then
	\begin{align*}
		\sup_{v\in \cup_{n\ge 1}\cV_n} \frac{m(v)}{\inf_{e \in \cE_v}|e|} &= \sup_{n\ge 1}\,(2n+3)^2 \left( \frac{1}{(2n+1)^2} +  \frac{2}{(2n+2)^2} + \frac{1}{(2n +3)^2} \right) \\
								&=\left (1 + \frac{2}{3} \right)^2     + 2     \left (1 + \frac{1}{4} \right)^2   + 1  = \frac{497}{72} = 6.902\dot 7
	\end{align*}
	For $v \in \cV_0$, we obtain	\[
		 \frac{m(v)}{\inf_{e \in \cE_v}|e|}  = 9 \left( k + 2 \frac{1}{4} + \frac{1}{9} \right) = 9k + \frac{11}{2}
	\]
	Moreover, for the remaining vertices $v \in \cV$ belonging to one of the attached trees,
	\[
		\frac{m(v)}{\inf_{e \in \cE_v} |e|} = k.
	\]
	By assumption, $k \geq 3$ and hence $M(\cG_k) = 9k + \frac{11}{2}$. In addition, $P(\cG_k) = \infty$ since $\cT $ contains unbounded tiles. Thus we obtain
	\be\label{eq:K_Gk}
		K(\cG_k) = 1 - \frac{1}{M(\cG_k)} = \frac{18 k + 9}{18k + 11},
	\ee
	and Theorem \ref{prop:est} implies the lower estimate
	\[
		\alpha(\cG_k) \geq \frac{18 k +11}{18 k + 9} \frac{k-2}{k}.
	\]
	Our next goal is to derive an upper estimate. For $l \in \Z_{\ge 2}$, let $\wt \cG_l$ be the subgraph consisting of all edges in one of the attached trees between $\cV_0$ and $\cV_{-l}$.  Then it is straightforward to verify
	\begin{align*}
		& \mes(\wt \cG_l) = \sum_{j=0}^{l-1} k(k-1)^j = \frac{ k ( (k-1)^{l} - 1) }{k-2}  , &&\deg (\partial \wt \cG_l) = k + 3 + k(k-1)^{l-1},
	\end{align*}
	and as a consequence,
	\[
		\lim_{l \to \infty} \frac{ \deg (\partial \wt \cG_l)}{ \mes(\wt \cG_l) }= \frac{k-2}{k-1} = \alpha(\T_k),
	\]
	where $\T_k$ is the equilateral, $k$-regular tree (see Example \ref{ex:pq} or \cite[Example 8.3]{kn17}). This implies the two-sided estimate
	\[
	 	\frac{18 k +11}{18 k + 9} \frac{k-2}{k} \leq \alpha(\cG_k) \leq  \frac{k-2}{k-1}.
	\]
	In particular, $\alpha(\cG_k) \to 1$ for $k \to \infty$.

\begin{remark}
The two above examples demonstrate the use of Theorem \ref{prop:est} in two different situations. First of all, let us mention that by \cite[Corollary 4.4.]{kn17} the metric graph $\cG$ satisfies the strong isoperimetric inequality if $\ell^\ast(\cG)<\infty$ and the combinatorial isoperimetric constant $\alpha_{\comb}(\cG_d)$ is positive,
\[
 \alpha_{\comb}(\cG_d) >0.
\]
In Example \ref{ex:pq}, the positivity of $\alpha_{\comb}(\cG_{p,q})$ is known (see \eqref{eq:alphacomb}) and hence it is a priori clear that $\alpha(\cG) > 0$. However,  Example \ref{ex:pq} shows that in certain situations Theorem \ref{prop:est} gives a good quantitative estimate.

On the other hand, in Example \eqref{eq:alphacomb} we have $\alpha_{\comb}(\cG_k) = 0$ (since clearly $\alpha_{\comb}(\Z^2_+) = 0$), however, $\alpha(\cG) > 0$. In particular, Theorem \ref{prop:est} shows that the isoperimetric constants of the combinatorial and metric graph behave differently.
\end{remark}

\subsection{Non-equilateral $p$-regular trees} \label{ex:min}
We conclude with an example showing the use of Remark \ref{rem:min}.
For $p \in\Z_{\ge 5}$, let $\T_p$ be the equilateral, $p$-regular tree from Example \ref{ex:pq}.  Fix an edge $\hat e \in \cE(\T_{p}) $. In the following, we will consider $\T_p$ equipped with another choice of edge lengths.  Define the metric graph $\cT_p:=(\T_p,|\cdot|)$ by assigning
\[
 |e| := \begin{cases} p, &e = \hat e \\[1mm] 1, &e \in \cE(\T_{p}) \setminus \{ \hat e \} \end{cases}.
\]

Let $\wt \cG \in \cS(\cT_p)$ be a star-like complete subgraph. If $\hat e \notin \wt \cE$, then $\mes (\wt \cG) = \# \wt\cE $. If $\hat e \in \wt \cE$, then $\wt \cV_{\Int} \neq \varnothing$ since $\wt \cG$ is star-like. Hence
\[
	\mes(\wt \cG) =\# \wt\cE+ p-1 \le 2 \# \wt\cE.
\]
 Thus we conclude from \eqref{eq:a_Gpg} and Lemma \ref{lem:reduction} that
\[
	\alpha_\cS( \cT_p)  = \inf_{\wt \cG \in \cS }  \frac{\deg ( \partial \wt \cG)}{ \mes(\wt \cG)} \geq \frac{1}{2}  \inf_{\wt \cG \in \cS }  \frac{\deg ( \partial \wt \cG)}{\#\wt\cE} = \frac{1}{2} \alpha( \T_p) = \frac{1}{2} \frac{p-2}{p-1} \ge \frac{2}{5}
\]
for all $p\ge 6$.
On the other hand, $\ell^\ast(\cT_p) = p \geq 6$ by assumption. Hence Remark \ref{rem:min} implies
\[
	\alpha(  \cT_p) = \frac{2}{\ell^\ast (  \cT_p)} = \frac{2}{p}.
\]

\noindent
\ack I thank Aleksey Kostenko for helpful discussions throughout the preparation of this article and Delio Mugnolo for useful comments and hints with respect to the literature.


TITLE: How to prove this formula involving the push-forward funtion
QUESTION [1 upvotes]: Let $f:M^{n} \to N^{m}$, and suppose that $(x,U)$ and $(y,V)$ are coordinate systems around $p$ and $f(p)$, respectively, then I want to prove that
$$f_{*} (\frac{\partial}{\partial x_i} \big{|}_p )=\sum_{j=1}^{m} \frac{\partial(y^{j} \circ f)}{\partial x^{i}}(p) · \frac{\partial}{\partial y^{j}} \big{|}_{f(p)} $$
So then I think I can get a formula just extending by linearity for
$$f_{*} (\sum_{i=1}^{n} a^{i} \frac{\partial}{\partial x_i} \big{|}_p )$$
in terms of $\frac{\partial}{\partial y^{j}} \big{|}_{p}$ (I don't know if what I am thinking of this is right or this fact involves another proof). So then I am trying to use that
$$\frac{\partial(g \circ f)}{\partial x^{i}}(p)=\sum_{j=1}^{m}\frac{\partial(g )}{\partial y^{j}}(f(p))·\frac{\partial(y^{j} \circ f)}{\partial x^{i}}(p)$$
because I have a similar expression but I can't figure how can I compute $f_{*} (\frac{\partial}{\partial x_i} \big{|}_p )$.
Can someone provide a detailed proof of this result please?

Attempt

We know that $f_{*}v(·)=v(· \circ f)$ then in this particular case we have $v=\frac{\partial}{\partial x_i} \big{|}_p $ so then using that
$$\frac{\partial(g \circ f)}{\partial x^{i}}(p)=\sum_{j=1}^{m}\frac{\partial(g )}{\partial y^{j}}(f(p))·\frac{\partial(y^{j} \circ f)}{\partial x^{i}}(p)$$
we get that
$$\frac{\partial}{\partial x_i} \big{|}_p (· \circ f)=\sum_{j=1}^{m} \frac{\partial(y^{j} \circ f)}{\partial x^{i}}(p) · \frac{\partial}{\partial y^{j}} \big{|}_{f(p)}$$
Am I right? and for the generalization what can be done?
Thanks in advance.

REPLY [1 votes]: If $\phi_M,\ \phi_N$ are charts then we have $G :=\phi_N\circ f \circ \phi_M^{-1} : {\bf R}^n\rightarrow {\bf R}^m$ so that $G_\ast \partial_{x_i}= \frac{\partial G^j}{\partial x_i} \partial_{y_j} $ If $E_i:=d\phi_M^{-1} \partial_{x_i}$ and $F_j:=d\phi_N^{-1}\partial_{y_j}$, then $$ f_\ast E_i= \frac{\partial G^j}{\partial x_i}  F_j $$
For $f_\ast \sum_i a_i\partial_{x_i}$, we need the argument about tangent space of manifold : If $d\phi_M^{-1},\ d\phi_N$ is linear, then since $dG$ is linear, we are done
If $c$ is curve in ${\bf R}^n$, then $v:=d\phi_M^{-1}\ c'(0)\in
T_aM$ where $a=\phi_M^{-1}\circ c(0)$ If $h$ is a function on $M$,
then $$vh := \frac{d}{dt} h\circ \phi_M^{-1}\circ c $$
That is tangent vector $v$ is viewed as a differential operator. And
 $$ w:= \frac{d}{dt} \phi_M^{-1}\circ c_2 \Rightarrow
 Cv+w = \frac{d}{dt} \phi_M^{-1} ( c(Ct) + c_2(t)-c_2(0)  ) $$
$$ Cv+w = d\phi_M^{-1} \ (Cc' + c_2') $$
And $$ (Cv+w)h= d ( h\circ \phi_M^{-1} )\ ( Cc' + c_2')
=  Cd ( h\circ \phi_M^{-1} )\ c'+ d ( h\circ \phi_M^{-1} )\ c_2'
$$
$$ = C vh + w h $$
That is $d\phi_M^{-1} \ (Cc' + c_2') =Cd\phi_M^{-1} \ c'+
d\phi_M^{-1} \ c_2' $