text:
\begin{document}

\maketitle

\noindent{\bf Abstract.} \emph{We construct, in an essentially explicit way, various composition
operators on $H^2$ and study their compactness or their membership in the Schatten classes. We construct:
non-compact composition operators on $H^2$ whose symbols have the same modulus on the boundary of
$\D$ as symbols whose composition operators are in various Schatten classes $S_p$ with $p > 2$; compact
composition operators which are in no Schatten class but whose symbols have the same modulus on the
boundary of $\D$ as symbols whose associated composition operators are in $S_p$ for every $p>2$.}
\medskip

\noindent{\bf Mathematics Subject Classification.} Primary: 47B33; 47B10
-- Secondary: 30C80
\medskip

\noindent{\bf Key-words.}  Carleson function -- Carleson measure -- composition operator -- Hardy spaces --
Luecking's criterion -- MacCluer's criterion -- Schatten class

\section{Introduction}

Compactness of composition operators on $H^2$ was first studied in 1968 by H. Schwartz in his
doctoral dissertation \cite{Schw}, and refined in 1973 by J.~Shapiro and P. Taylor \cite{Shap-Tay}, who
discovered the role played by the classical angular derivative, and refined the compactness problem by
asking which composition operators belong to various Schatten classes $S_p$. In particular, they showed
in \cite{Shap-Tay} that $C_\phi \in S_2$, the Hilbert-Schmidt class, if and only if
$\int_{\partial \D} (1 - |\phi^\ast|)^{-1} < +\infty$, where $\phi^\ast$ denotes the radial limit function
of $\phi$. In this paper, we show that for the larger class $S_p$ with $p > 2$, the situation is
completely different: we prove in Theorem~\ref{meme module Schatten}, that for every $p > 2$, there exist
two symbols $\phi_1$ and $\phi_2$ having the same modulus on $\partial \D$ and such that $C_{\phi_1}$
is not compact on $H^2$, but $C_{\phi_2}$ is in the Schatten class $S_p$.\par
\medskip

An amusing feature of the theory of composition operators is that, whereas sophisticated necessary and
sufficient conditions for the composition operators $C_\phi \colon H^2 \to H^2$ to belong to the
Schatten classes $S_p = S_p (H^2)$ have been known for more than twenty years (\cite{Lue}, \cite{Lue-Zhu}),
either in terms of the Nevanlinna counting function, or in terms of the pull-back measure $m_\phi$,  yet
explicit and concrete examples are lacking. For example, D. Sarason posed in 1988 the question
whether there existed a compact composition operator $C_\phi \colon H^2 \to H^2$ which was in \emph{no}
Schatten class, and the (affirmative) answer only came in 1991, by T. Carroll and C. Cowen (\cite{Carroll-Cowen}).
Their example, based on the Riemann mapping Theorem was not completely explicit. Moreover, their
construction used a difficult and delicate argument, with estimates of the hyperbolic metric for certain domains,
due to Hayman (see however \cite{Zhu} and \cite{Jones}). We shall see in this paper that Luecking's criterion
\cite{Lue} for pullback measures leads to very concrete examples of composition operators in various Schatten
classes.\par

In Section~\ref{Preliminaries}, we give a necessary condition,
Proposition~\ref{condition necessaire Schatten}, on the Carleson function $\rho_\phi$ in order for the
composition operator $C_\phi$ to be in $S_p$, as well as a general construction of symbols $ \phi$ with
control on their Carleson function.\par

In Section~\ref{section same modulus}, we construct, for every $p > 2$, symbols $\phi_1$ and $\phi_2$
having the same modulus on $\partial \D$ such that $C_{\phi_1}$ is not compact on $H^2$, but
$C_{\phi_2}$ is in the Schatten class $S_p$ (Theorem~\ref{meme module Schatten}). \par

In Section~\ref{log puissance theta}, we revisit an example of J. Shapiro and P. Taylor (\cite{Shap-Tay}, \S~4)
to show that for every $p_0 > 0$, there exists a symbol  $\phi $ such that the composition
operator $C_\phi \colon H^2 \to H^2$ is in the Schatten class $S_p$ for every $p > p_0$, but not in
$S_{p_0}$ (Theorem~\ref{theo Schatten}), and also that for every $p_0 >0$, there exists a symbol $\phi$
such that $C_\phi \colon H^2 \to H^2$ is in the Schatten class $S_{p_0}$, but not in $S_p$, for $p < p_0$
(Theorem~\ref{theo Schatten bis}). Moreover, there exists a symbol $\phi$ such that
$C_\phi \colon H^2 \to H^2$ is compact but in no Schatten class $S_p$ for $p < \infty$
(Theorem~\ref{no Schatten}) and there exists a symbol $\psi$, whose boundary values $\psi^\ast$ have the
same modulus as those $\phi^\ast$ of $\phi$ on $\partial \D$, but for which $C_\psi \colon H^2 \to H^2$ is
in $S_p$ for every $p>2$ (Theorem~\ref{no Schatten meme module}).\par
After our work was completed, we became aware of the papers \cite{Zhu} and \cite{Jones}; in \cite{Jones},
M. Jones gives another proof of the theorem of Carroll and Cowen (our Theorem~\ref{no Schatten}), and Y. Zhu
gives also another proof of this theorem, as well as a proof of our Theorem~\ref{theo Schatten}. However, our
proofs are different and lead to further results: see Theorem~\ref{meme module Schatten}.\par
\bigskip

\noindent{\bf Acknowledgements.} We thanks the referee for a very careful reading of this paper and
many suggestions to improve its writing.

\section{Notation}

Throughout this paper, the notation $f\approx g$ will mean that there are two constants $0 < c < C < +\infty$
such that $ c f(t) \leq g(t) \leq C f(t)$ (for $t$ sufficiently near of a specified value), and the notation
$ f (t) \lesssim g (t)$, when $t$ is in the neighbourhood of some value $t_0$, will have the same meaning as
$g = O (f)$.\par
\medskip

We shall denote by $\D$ the open unit disc of the complex plane: $\D = \{ z\in \C\,; \ |z| < 1\}$, and by
$\T = \partial \D$ its boundary: $\T =\{ z\in \C\,;\ |z | = 1\}$. We shall denote by $m$ the normalized
Lebesgue measure on $\T$.\par
For every analytic self-map $\phi \colon \D \to \D$, the composition operator $C_\phi$ is the map
$f \mapsto f\circ \phi$. By Littlewood's subordination principle (see \cite{Duren}, Theorem~1.7), every
composition operator maps every Hardy space $H^p$ ($p >0$) into itself, and is continuous on $H^p$.\par
\smallskip

For every $\xi \in \T$ and $0 < h < 1$, the Carleson window $W (\xi, h)$ is the set
\begin{displaymath}
W (\xi, h) = \{ z \in \D\,;\ |z| \geq 1 - h\quad \text{and} \quad |\arg ( z \bar{\xi})| \leq h\}.
\end{displaymath}
For every finite positive measure $\mu$ on $\D$, one sets:
\begin{displaymath}
\rho_\mu (h) = \sup_{\xi \in \T} \mu[W (\xi, h)].
\end{displaymath}
We shall call this function $\rho_\mu$ the \emph{Carleson function} of $\mu$.\par
When $\phi \colon \D \to \D$ is an analytic self-map of $\D$, and $\mu =m_\phi$ is the
measure defined on $\D$, for every Borel set $B \subseteq \D$, by:
\begin{displaymath}
m_\phi (B) = m (\{\xi\in \T\,;\ \phi^\ast (\xi) \in B \}),
\end{displaymath}
where $\phi^\ast$ is the boundary values function of $\phi$, we shall denote $\rho_{m_\phi}$ by
$\rho_\phi$. In this case, we shall say that $\rho_\phi$ is the Carleson function of $\phi$.\par
\smallskip

For $\alpha \geq 1$, we shall say that $\mu$ is an \emph{$\alpha$-Carleson measure} if
$\rho_\mu (h) \lesssim h^{\alpha}$. For $\alpha =1$, $\mu$ is merely said to be a \emph{Carleson measure}.
\par\smallskip

The Carleson Theorem (see \cite{Duren}, Theorem~9.3) asserts that, for $1 \leq p < \infty$ (actually, for
$ 0 < p < \infty$), the canonical inclusion $j_\mu \colon H^p \to L^p (\mu)$ is bounded if and only if
$\mu$ is a Carleson measure. Since every composition operator $C_\phi$ is continuous on $H^p$, it defines a
continuous map $j_\phi \colon H^p \to L^p (\mu_\phi)$; hence every pull-back measure $\mu_\phi$ is a
Carleson measure.\par
When $C_\phi \colon H^2 \to H^2$ is compact, one has, as it is easy to see:
\begin{equation}\label{condition phi plus petit que un au bord}
|\phi^\ast| < 1 \qquad \text{\emph{a.e.} on } \partial \D.
\end{equation}
Hence, we shall only consider in this paper symbols $\phi$ for which
\eqref{condition phi plus petit que un au bord} is satisfied (which is the case, as we said, when $C_\phi$ is
compact on $H^2$).\par
\smallskip

B. MacCluer (\cite{McCluer}, Theorem~1.1) has shown (assuming
condition~\eqref{condition phi plus petit que un au bord}) that $C_\phi$ is compact on $H^p$ if and
only if $\rho_\phi (h) = o\, (h)$, as $h$ goes to $0$.
\medskip

Note that, in this paper, we shall not work, most often, with exact inequalities, but with
inequalities up to constants. It follows that we shall not actually work with true Carleson windows
$W (\xi, h)$ (or Luecking sets, defined below), but with distorted Carleson windows:
\begin{displaymath}
\tilde W (\xi, h) =\{ z\in \D\,;\ |z| \geq 1 - ah \quad \text{and} \quad | \arg (z \bar{\xi} )|\leq bh\},
\end{displaymath}
where $a,b > 0$ are given constants. Since, for a given symbol $\phi$, one has:
\begin{displaymath}
m_\phi \big(W (\xi, c\,h) \big) \leq m_\phi \big(\tilde W (\xi, h) \big) \leq m_\phi \big(W (\xi, C\,h) \big)
\end{displaymath}
for some constants $c= c(a,b)$ and $C = C (a,b)$ which only depend on $a$ and $b$, that will not matter
for our purpose.

\section{Preliminaries}\label{Preliminaries}

\subsection{Luecking sets and Carleson windows}

We shall begin by recalling the characterization, due to D. Luecking (\cite{Lue}), of the composition operators
on $H^2$ which belong to the Schatten classes. Let, for every integer $n \geq 1$ and $0 \leq j \leq 2^n -1$:
\begin{displaymath}
R_{n,j} = \Big\{z\in \D\,;\ 1 - 2^{-n} \leq |z| < 1 - 2^{-n - 1}\quad \text{and}\quad
\frac{2j\pi}{2^n} \leq \arg z < \frac{2(j+1)\pi }{2^n}\,\Big\}
\end{displaymath}
be the \emph{Luecking sets}.\par
\smallskip

The result is:

\begin{theoreme}[Luecking \cite{Lue}]\label{theo Luecking}
For every $p > 0$,  the composition operator $C_\phi$, assuming
condition~\eqref{condition phi plus petit que un au bord}, is in the Schatten class $S_p$ if and only if:
\begin{equation}\label{critere Luecking}
\sum_{n\geq 0} 2^{np/2} \bigg( \sum_{j=0}^{2^n-1} \big[m_\phi (R_{n,j})\big]^{p/2} \bigg) <+\infty.
\end{equation}
\end{theoreme}

Majorizing $m_\phi (R_{n,j})$ by
$m_\phi \big( W (\e^{2^{-n}(2j+1)i \pi}, 2^{-n})\big) \leq \rho_\phi (2^{-n})$, one gets:

\begin{corollaire}\label{coro Luecking}
Let $\phi \colon \D \to \D$ be an analytic self-map, with
condition~\eqref{condition phi plus petit que un au bord}, and assume that $m_\phi$ is an $\alpha$-Carleson
measure, with $\alpha > 1$. Then $C_\phi \in S_p$ for every $p > \frac{2}{\alpha - 1}\,\cdot$
\end{corollaire}

\noindent{\bf Proof.} Since $\rho_\phi (h) \lesssim h^\alpha$, one gets:
\begin{displaymath}
\sum_{n\geq 0} 2^{np/2} \bigg( \sum_{j=0}^{2^n-1} \big[m_\phi (R_{n,j})\big]^{p/2} \bigg)
\lesssim \sum_{n \geq 0} 2^{np/2}.2^n.(2^{-n\alpha})^{p/2}
= \sum_{n \geq 0} 2^{n [ 1 - (\alpha - 1)p/2]},
\end{displaymath}
which is $< +\infty$ since $1 - (\alpha - 1) \frac{p}{2} <0$. \qed
\bigskip

In order to get this corollary, we majorized crudely $m_\phi (R_{n,j})$ by $\rho_{\phi} (2^{-n})$. Actually,
we shall see, through the results of this paper, that we lose too much with this majorization. Nevertheless, if we
replace the Luecking sets by the \emph{dyadic Carleson windows}:
\begin{displaymath}
W_{n,j}=\Big\{ z\in\D \,;\  1 - 2^{-n} \leq |z| < 1\, ,\quad
\frac{2j\pi}{2^n} \leq \arg(z) < \frac{2 (j + 1) \pi}{2^n} \Big\},
\end{displaymath}
($j=0, 1, \ldots, 2^n - 1$, $n=1,2,\ldots$), we have the same behaviour:

\begin{proposition}\label{Luecking vs Carleson}
Let $\mu$ be a finite positive measure on the open unit disk $\D$ and let $\alpha >0$. Then the following
assertions are equivalent:\par
\smallskip
\qquad $(a)$ \quad $\displaystyle{
\sum_{n=1}^\infty \sum_{j=0}^{2^n - 1} 2^{n\alpha} \bigl(\mu(R_{n,j})\bigr)^\alpha < +\infty} $;\par
\smallskip
\qquad $(b)$ \quad $\displaystyle{
\sum_{n=1}^\infty \sum_{j=0}^{2^n - 1} 2^{n\alpha} \bigl(\mu(W_{n,j})\bigr)^\alpha < +\infty} $.
\end{proposition}

\noindent{\bf Proof.} It is clear that $(b)$ implies $(a)$ since $R_{n,j}\subset W_{n,j}$ for all $n$ and $j$
(we already used this in the proof of Corollary~\ref{coro Luecking}). \par
For the proof of the converse implication, we shall need the following sets, for positive integers $l, n$
with $l \geq n$, and $j\in \{0 ,1,\ldots, 2^n - 1\}$:
\begin{displaymath}
H_{l, n, j} = \Bigl\{ k\in \{0, 1, \ldots, 2^l -1\}\,;\
\frac{j}{2^n} \leq \frac{k}{2^l} < \frac{j + 1}{2^n} \Bigr\}\,.
\end{displaymath}

It is clear that we have, for every $n$ and $j$:
\begin{displaymath}
W_{n,j}  = \bigcup_{l \geq n}\; \bigcup_{k\in H_{l, n, j}} R_{l, k},
\end{displaymath}
and
\begin{displaymath}
\mu (W_{n, j}) = \sum_{l \geq n}\; \sum_{k \in H_{l, n, j}} \mu (R_{l, k})\,.
\end{displaymath}
\medskip

We shall first treat the case $\alpha\leq 1$, where we can use, for $x_1, x_2, \ldots, x_N \geq 0$:
\begin{displaymath}
(x_1 + x_2 + \cdots + x_N)^\alpha \leq x_1^\alpha + x_2^\alpha + \cdots + x_N^\alpha\,.
\end{displaymath}

We have:
\begin{align*}
\sum_{n=1}^\infty \sum_{j = 0}^{2^n - 1} 2^{n\alpha} \bigl(\mu(W_{n,j})\bigr)^\alpha
& = \sum_{n=1}^\infty \sum_{j=0}^{2^n - 1} 2^{n\alpha} \bigg( \sum_{l \ge n}\;
\sum_{k \in H_{l,n,j}} \mu (R_{l,k}) \bigg)^\alpha \\
& \leq  \sum_{n=1}^\infty \sum_{j=0}^{2^n - 1} 2^{n\alpha} \sum_{l \geq n}\;
\sum_{k\in H_{l,n,j}} \bigl(\mu(R_{l,k})\bigr)^\alpha \\
& =   \sum_{l =1}^\infty \sum_{k=0}^{2^l - 1} \;
\bigl(\mu(R_{l,k})\bigr)^\alpha
\sum_{{(n,j) \,:\, n\le l \,;\, k\in H_{l,n,j}}} 2^{n\alpha}
\end{align*}

Observe that, for every $n\le l$, there is only one $j$ such that $k\in  H_{l,n,j}$. Since we have
\begin{displaymath}
\sum_{n = 1}^{l} 2^{n\alpha} \le C_\alpha 2^{l\alpha},
\end{displaymath}
we get:
\begin{displaymath}
\sum_{n=1}^\infty \sum_{j=0}^{2^n - 1} 2^{n\alpha} \bigl(\mu(W_{n,j})\bigr)^\alpha
\le C_\alpha
\sum_{l=1}^\infty \sum_{k=0}^{2^l - 1} 2^{l\alpha} \bigl(\mu(R_{l,k})\bigr)^\alpha\,,
\end{displaymath}
and $(a)$ implies $(b)$ in the case $\alpha\le 1$.
\medskip

If $\alpha>1$, we can use H\"older's inequality. Let $\beta$ be the conjugate exponent of $\alpha$. Choose
$a$ such that $1 < a < 2 < a^\beta$. Then:
\begin{align*}
\mu (W_{n,j})
& = \sum_{l\ge n \,:\, k\in H_{l,n,j}}\; \mu (R_{l,k}) \\
& \le \bigg( \sum_{l \ge n \,:\, k\in H_{l,n,j}}\; a^{-l\beta} \bigg)^{1/\beta}
\bigg( \sum_{l \ge n \,:\, k\in H_{l,n,j}}\; a^{l\alpha}
\bigl(\mu (R_{l,k})\bigr)^\alpha \bigg)^{1/\alpha}
\end{align*}
Using that $|H_{l,n,j}|=2^{l-n}$, we get:
\begin{displaymath}
\bigg[\sum_{l\ge n \,:\, k\in H_{l,n,j}}\; a^{-l\beta}\bigg]^{1/\beta} =
\bigg[\sum_{l \ge n} 2^{l - n} a^{- l\beta}\bigg]^{1/\beta} =
\biggl( 2^{-n} \frac{(2 a^{-\beta})^n}{1 - 2 a^{-\beta}}\biggr)^{1/\beta}= C_\beta a^{-n} .
\end{displaymath}
Therefore we have:
\begin{align*}
\sum_{n=1}^\infty \sum_{j=0}^{2^n - 1} 2^{n\alpha} \bigl(\mu(W_{n,j})\bigr)^\alpha
&\le \sum_{n=1}^\infty \sum_{j=0}^{2^n - 1} 2^{n\alpha} C_\beta^\alpha a^{-n\alpha}
\sum_{l \ge n \,:\, k\in H_{l,n,j}}\; a^{l\alpha} \bigl(\mu (R_{l,k})\bigr)^\alpha \\
& = C_\beta^\alpha \sum_{l =1}^\infty \sum_{k=0}^{2^l - 1} \;
\bigl(\mu(R_{l,k})\bigr)^\alpha a^{l \alpha} \sum_{{(n,j) \,:\, n\le l\,, k\in H_{l,n,j}}} (2/ a) ^{n\alpha}  \cr
& \lesssim \sum_{l =1}^\infty \sum_{k=0}^{2^l - 1} \;
\bigl(\mu(R_{l,k})\bigr)^\alpha a^{l \alpha} (2/a)^{l \alpha} \\
& = \sum_{l=1}^\infty \sum_{k=0}^{2^l - 1} 2^{l\alpha} \bigl(\mu (R_{l,k})\bigr)^\alpha\,.
\end{align*}
We have hence proved that $(a)$ implies $(b)$ for $\alpha > 1$ and therefore
Proposition~\ref{Luecking vs Carleson} follows. \qed
\medskip

As a corollary we prove a necessary condition that $\rho_\phi$ must satisfy when $C_\phi$ is in the Schatten
class $S_p$.

\begin{proposition}\label{condition necessaire Schatten}
Let $\phi \colon\D\to \D$ be an analytic self-map. If the composition operator
$C_\phi \colon H^2 \to H^2$ is in the Schatten class $S_p$ for some $p> 0$, then, as $h$ goes to $0$:
\begin{equation}\label{limite condition necessaire Schatten}
\rho_\phi (h) = o\, \bigg( h \Big( \log \frac{1}{h} \Big)^{-2/p} \bigg).

\end{equation}
\end{proposition}

\noindent{\bf Proof.} Thanks to Luecking's characterization and the equivalence in
Proposition~\ref{Luecking vs Carleson}, we have, for the pullback mesure $m_\phi$:
\begin{equation}\label{eq 1 condition necessaire}
\sum_{n=1}^\infty \sum_{j=0}^{2^n - 1} 2^{n p/2} \big(m_\phi (W_{n,j}) \big)^{p/2} < +\infty.
\end{equation}
Observe that, for $h=2^{-n}$ each window $W (\xi, h)$ is contained in the union of at most three of the
$W_{n, j}$'s; hence:
\begin{displaymath}
\big(\rho_\phi (2^{-n})\big)^{p/2} \leq
\Big(3 \max_{0\le j\le 2^n - 1} m_\phi (W_{n,j}) \Big)^{p/2}
\le 3^{p/2} \sum_{j=0}^{2^n - 1} \big(m_\phi (W_{n,j}) \big)^{p/2}  \,,
\end{displaymath}
and \eqref{eq 1 condition necessaire} yields:
\begin{displaymath}
\sum_{n=1}^\infty \bigl(\rho_\phi (2^{-n})\bigr)^{p/2} 2^{n p/2} < +\infty\,.
\end{displaymath}
Hence, setting:
\begin{equation}
\gamma_n = \sum_{n/2\le k \le n} \bigl(\rho_\phi (2^{-k})\bigr)^{p/2} 2^{kp/2}\,,
\end{equation}
we have:
\begin{equation}\label{eq 2 condition necessaire}
\lim_{n\to\infty} \gamma_n = 0\,.
\end{equation}
\medskip

Now, by using \cite{CompOrli}, Theorem~4.19, we get a constant $C > 0$ such that, for $k\le n$:
\begin{displaymath}
C \rho_\phi (2^{-k}) \ge 2^{n - k} \rho_\phi (2^{-n})\,,
\end{displaymath}
and so:
\begin{equation}\label{eq 3 condition necessaire}
C^{p/2} \gamma_n \ge (n/2) \Big(\frac{\rho_\phi(2^{-n})}{2^{-n}}\Big)^{p/2} \,.
\end{equation}
\par

To finish the proof, it remains to consider, for every $h\in (0,1/2)$, the integer $n$ such that
$2^{-n-1} < h \le 2^{- n}$; then \eqref{eq 2 condition necessaire} and \eqref{eq 3 condition necessaire} give:
\begin{displaymath}
\lim_{h\to 0^+} \Big( \frac{\rho_\phi (h)}{h}\Big)^{p/2} \log(1/h) = 0,
\end{displaymath}
as announced. \qed
\medskip

\noindent{\bf Remark.} We can also deduce Corollary~\ref{coro Luecking} from the following result.

\begin{proposition}
If $\mu$ is a $\beta$-Carleson probability measure on $\D$, with $\beta > 2$, then the Poisson integral
${\cal P} \colon L^2 (\T) \to L^2 (\mu)$ is in the Schatten class $S_p$ for any $p > 2/ (\beta - 1)$.
\end{proposition}

\noindent{\bf Proof.} We may assume that $p \leq 2$ since $S_{p_1} \subseteq S_{p_2}$ when
$p_1 \leq p_2$. To have ${\cal P} \in S_p$, it suffices then to have
$\sum_{n\in \Z} \| {\cal P}(e_n) \|_{L^2 (\mu)}^p < +\infty$, where $e_n (\e^{it}) = \e^{int}$
(see \cite{Koenig}, Proposition~1.b.16, page 40, for example).\par
But $({\cal P} e_n ) (z) = z^n$ for $n \geq 0$ and $({\cal P} e_n ) (z) = \bar{z}^{|n|}$ for $n \leq - 1$. Hence:
\begin{displaymath}
\sum_{n\in \Z} \| {\cal P}(e_n) \|_{L^2 (\mu)}^p
\leq 2\sum_{n = 0}^\infty \Big( \int_{\D} |z|^{2n}\,d\mu \Big)^{p/2} .
\end{displaymath}
But
\begin{align*}
\int_{\D} |z|^{2n}\,d\mu
& = \int_0^1 2n t^{2n -1} \mu (|z | \geq t) \,dt \\
& = 2 n \int_0^1 (1 - x)^{2n - 1} \mu (|z | \geq 1 - x) \,dx \\
& \lesssim n \int_0^1 (1 - x)^{2n - 1} \frac{1}{x}\, x^\beta \,dx \\
& \quad \text{since $\{ | z | \geq 1 - x \}$ can be split in $O (1/x)$ windows $W (a, x)$} \\
& = n \int_0^1 (1 - x)^{2n - 1} x^{\beta - 1}\,dx  \lesssim n.n^{-\beta}.
\end{align*}
Hence
\begin{displaymath}
\sum_{n\in \Z} \| {\cal P}(e_n) \|_{L^2 (\mu)}^p \lesssim \sum_{n \geq 1} \frac{1}{n^{(\beta - 1)p/2}}
\raise 1,5pt \hbox{,}
\end{displaymath}
which is finite since $(\beta - 1)p/2 > 1$. \qed

\bigskip

\subsection{A general contruction}\label{general construction}

In this subsection, we are going to describe a general way to construct symbols with some prescribed conditions.
A particular case of this construction has been used in \cite{CompOrli}, Theorem~4.1. We also shall use it in
\cite{Orlicz-Hardy}.
\medskip

Let
\begin{displaymath}
f (t) = \sum_{k=0}^\infty a_k \cos (kt)
\end{displaymath}
be an even, non-negative, $2\pi$-periodic continuous function, vanishing at the origin: $f (0) = 0$, and such
that:
\begin{equation}\label{deuxieme condition}
\text{$f$ is \emph{strictly} increasing on $[0,\pi]$.}
\end{equation}

The Hilbert transform (or conjugate function) ${\cal H}f$ of $f$ is:
\begin{displaymath}
{\cal H}f (t) =\sum_{k=1}^\infty a_k \sin (kt).
\end{displaymath}
We shall assume moreover that, as $t$ tends to zero:
\begin{equation}\label{derivee de Hf}
\big({\cal H}f\big)' (t) = o\, ( 1/t^2).
\end{equation}
\par

Let now $F \colon \D \to \Pi^+ = \{ \Re z >0\}$ be the analytic function whose boundary values are
\begin{equation}\label{def grand F}
F^\ast (\e^{it}) = f (t) + i {\cal H}f (t).
\end{equation}
One has:
\begin{displaymath}
\qquad F (z) = \sum_{k=0}^\infty a_k z^k, \qquad |z| < 1.
\end{displaymath}
\par
We define:
\begin{equation}\label{def grand Phi}
\qquad \Phi (z) = \exp \big( - F(z)\big), \qquad z\in \D.
\end{equation}
Since $f$ is non-negative, one has
\begin{displaymath}
\Re F (z) = \frac{1}{2\pi} \int_{-\pi}^{\pi} f (t) P_z (t)\,dt > 0,
\end{displaymath}
so that $| \Phi (z) | < 1$ for every $z \in \D$: $\Phi$ is an analytic self-map of $\D$, and
$|\Phi^\ast | = \exp (- f ) < 1$ \emph{a.e.} . Note that the assumption $f (0) = 0$ means that
$\Phi^\ast (1) = 1$; we then have $\| \Phi \|_\infty = 1$, which is necessary for the non-compactness of
$C_\Phi$.\par
\smallskip

An example is $\Phi (z ) = \exp \big( - (1 - z)/2\big)$, for which $f (t) = \sin^2 (t/2)$.\par
\smallskip

\begin{lemme}\label{premier lemme}
Assume that $f$ and ${\cal H}f$ are ${\cal C}^1$ functions.
Then, the Carleson function $\rho_\Phi$ of $\Phi$ is not $o\,(h)$ when $h$ goes to $0$, and so the
composition operator $C_\Phi \colon H^2 \to H^2$ is not compact.
\end{lemme}

Note that the hypothesis of the lemma holds, for example, when:
\begin{equation}\label{premiere condition}
\sum_{k=0}^\infty k\,|a_k| < +\infty.
\end{equation}

\noindent{\bf Proof.} The non-compactness of $C_\Phi$ follows immediately from the ``angular
derivative'' condition (\cite{Shap-livre}, Theorem~3.5: see \eqref{angular derivative} in the remark at the
end of the next section). But we shall give a proof using the Carleson function, in order to illustrate the
methods to be used later on.\par
It will be enough to minorize $\mu_\Phi [W (1,h)]$. Since $f$ and ${\cal H}f$ are
${\cal C}^1$, we have $| f (t) | \leq C |t|$ and $| {\cal H}f (t) | \leq C |t|$ for some positive constant $C$.
Now, if $| t | \leq h/ C$, we see that
\begin{displaymath}
| \Phi^\ast ( \e^{it}) | = \e^{ - f (t) } \geq \e^{- C |t|} \geq \e^{-h} \geq 1 - h,
\end{displaymath}
and $| \arg \Phi^\ast (\e^{it}) | = | {\cal H} f (t) | \leq C |t| \leq h$; hence $\Phi^\ast (\e^{it}) \in W (1,h)$.
This means that
\begin{displaymath}
m_\Phi [W (1, h)] \geq m (\{ \e^{it}\,;\ | t | \leq h / C \}) \geq \frac{1}{\pi C}\, h,
\end{displaymath}
and the lemma follows. \qed
\bigskip

We shall now perturb $\Phi$ by considering
\begin{equation}\label{M}
\qquad M (z) = \exp\Big( - \frac{1+ z }{ 1 -z }\Big)\,, \qquad |z| <1,
\end{equation}
and
\begin{equation}\label{def general phi}
\qquad \phi ( z) = M (z) \Phi (z), \qquad | z | <1.
\end{equation}
One has:
\begin{displaymath}
\phi^\ast (\e^{it})  = \e^{ - f(t)}\,\e^{- i ({\cal H} f (t) + \cot \frac{t}{2} )}.
\end{displaymath}

We will now, according to the various choices of $f$, study the behaviour of $\phi$ with respect to the
Carleson windows.\par
\smallskip

We set:
\begin{equation}
\gamma (t ) =  {\cal H} f (t) + \cot \frac{t}{2} = \frac{2}{t} + r (t),
\end{equation}
where the derivative of the odd function $r$ satisfies $r ' (t) = o\, (1/t^2)$ .\par
Now, we have:

\begin{lemme}\label{deuxieme lemme}
When $h > 0$ goes to $0$, one has:
\begin{equation}
h f^{-1} (h) \lesssim \rho_\phi (h) \lesssim h f^{-1} (2h).
\end{equation}
\end{lemme}

\noindent{\bf Proof.} Let $a = \e^{i\theta} \in \T$, $|\theta | \leq \pi$. We may assume that
$0 < h \leq h_0 \leq 1/2$, for some $h_0$ small enough. \par
We have to analyze the set of $t$'s such that
$\phi^\ast (\e^{-it}) = \e^{- f (t)} \e^{i \gamma (t)} \in W (a, h)$, which
imposes two constraints. Without loss of generality, we may analyze only the set of positive $t$'s, \emph{i.e.}
$0 < t \leq \pi$.\par
\smallskip

\emph{Modulus constraint.} We must have $| \phi^\ast (\e^{-it}) | \geq 1 - h$, \emph{i.e.}
$\e^{- f (t)} \geq 1 - h$, or $f (t) \leq \log \frac{1}{1 - h}$, which is $\leq 2h$ since $h \leq 1/2$. Hence we
must have:
\begin{equation}\label{contrainte module}
0 < t \leq f^{-1} (2h).
\end{equation}

\emph{Argument constraint.} We must have $| \gamma (t) - \theta | \leq h \mod 2\pi$, \emph{i.e.}, since
we have assumed that $t >0$:
\begin{equation}\label{intervalle}
\gamma (t) \in \bigcup_{n \geq 0} [ \theta - h + 2n \pi, \theta + h + 2n \pi] = \bigcup_{n \geq 0} J_n (h).
\end{equation}
Since $\gamma (t) \to +\infty$ as $t \mathop{\to}\limits^{\scriptscriptstyle >} 0$, and since we
already have $t \leq f^{-1}(2h)$ by \eqref{contrainte module}, we know that $\gamma (t) > 2 \pi$ for $h$
small enough; hence we have $\gamma (t) \in J_n (h)$ only for $n \geq N_h$, where the integer $N_h$
goes to infinity when $h$ goes to $0$; in particular, we may assume that, for $h$ small enough, we have
$\gamma (t) \in J_n (h)$ for $n \geq 1$ only.\par
Let $I_n (h) = \gamma^{-1} \big( J_n (h) \big)$. Since $\gamma (t) = \frac{2}{t} + r (t)$, and
$r' (t) = o\, (1/t^2)$, $\gamma (t)$ is decreasing for $0 < t \leq h_0$, for $h_0$ small enough. Hence, for
$h$ small enough:
\begin{displaymath}
I_n (h)  = \big[ \gamma^{-1} (\theta +h + 2n \pi), \gamma^{-1} (\theta - h + 2n \pi) ].
\end{displaymath}
Since $\gamma (t) = \frac{2}{t} + o\, (1/t)$ and $|\gamma ' (t) | = \frac{2}{t^2} + o\, (1/t^2)$, one has:
\begin{equation}\label{encadrement intervalle}
\frac{c_1}{n} \leq \min I_n (h) \leq \max I_n (h) \leq \frac{c_2}{n}\,\raise 1,5pt \hbox{,}
\end{equation}
where $c_1$ and $c_2$ are two universal positive constants. By the mean-value theorem, we get that:
\begin{align*}
2 \pi m \big( I_n (h)\big)
& = \gamma^{-1} (\theta - h + 2n \pi) - \gamma^{-1} (\theta +h + 2n \pi) \\
& = 2h | (\gamma^{-1})' (\xi_n)| = \frac{2h }{|\gamma ' (t_n)|}\,\raise 1,5pt \hbox{,}
\end{align*}
for some $\xi_n \in J_n (h)$ and $t_n \in I_n (h)$. But, \eqref{encadrement intervalle} ensures that
$\frac{c_1}{n} \leq t_n \leq \frac{c_2}{n}$ and, since $|\gamma ' (t) | = \frac{2}{t^2} + o (1/t^2)$, we get
that:
\begin{equation}\label{mesure intervalle}
m\big( I_n (h)\big) \approx \frac{h}{n^2}\cdot
\end{equation}
\smallskip

Now:\par

1) Assume that $\phi^\ast (\e^{-it}) \in W (a, h)$. By \eqref{intervalle}, \eqref{encadrement intervalle} and
\eqref{contrainte module}, we must have $t \in I_n (h)$ with $\frac{c_1}{n} \leq t \leq f^{-1}(2h)$. Hence,
if $n_0$ is the integer part of $\frac{c_1}{f^{-1}(2h)}$, we must have $n \geq n_0$. Now,
\eqref{mesure intervalle} shows that:
\begin{align*}
m\big( \{ t\in ]0,\pi]\,;\ \phi^\ast (\e^{- it}) \in W (a,h)\}\big)
& \leq \sum_{n \geq n_0} m\big( I_n (h)\big) \\
& \lesssim \sum_{n \geq n_0} \frac{h}{n^2} \lesssim \frac{h}{n_0} \lesssim h f^{-1}(2h).
\end{align*}
\par

2) We want to minorize, as it suffices, $m\big( \{ t\in ]0,\pi]\,;\ \phi^\ast (\e^{it}) \in W (1,h)\}\big)$. Let
$n_1$ be the integer part of $\frac{c_2}{f^{-1}(h)} +1$; we have:
\begin{equation}\label{implication fenetre}
t \in \bigcup_{n \geq n_1} I_n (h) \qquad \Longrightarrow \qquad  \phi^\ast (\e^{it}) \in W (1,h)
\end{equation}
because $t \in I_n (h)$ for $n \geq n_1$ implies $t \leq \frac{c_2}{n_1} \leq f^{-1} (h)$, so that
$| \phi^\ast (\e^{it}) | = \e^{ - f(t)} \geq \e^{-h} \geq 1 - h$, and the modulus constraint for
$\phi^\ast (\e^{it}) $ is automatically satisfied. Since the argument constraint is satisfied by construction, as
$t$ belongs to some $I_n(h)$, this proves \eqref{implication fenetre}.\par
As a consequence, we have, using \eqref{mesure intervalle}:
\begin{displaymath}
\rho_\phi (h) \geq m_\phi \big( W (1, h) \big) \geq \sum_{n \geq n_1} m\big( I_n (h)\big) \gtrsim
\sum_{n \geq n_1} \frac{h}{n^2} \gtrsim \frac{h}{n_1} \gtrsim h f^{-1} (h),
\end{displaymath}
and this ends the proof of Lemma~\ref{deuxieme lemme}. \qed
\bigskip

\section{Composition operators with symbol of same modulus}\label{section same modulus}

J. Shapiro and P. Taylor (\cite{Shap-Tay}, Theorem~3.1) characterized Hilbert-Schmidt composition
operators $C_\phi$ on $H^2$ (\emph{i.e.} $C_\phi\in S_2$), and this characterization depends only on
the modulus of $\phi^\ast$ on $\partial \D$. It follows that if $\phi_1$ and $\phi_2$ are two symbols such
that $|\phi_1^\ast | = |\phi_2^\ast|$, and $C_{\phi_2}$ is Hilbert-Schmidt, then $C_{\phi_1}$ is also
Hilbert-Schmidt, and in particular, compact.\par
It appears that this is a limiting case, as we shall see in Theorem~\ref{meme module Schatten}.
\medskip

Actually, in view of Corollary~\ref{coro Luecking}, we may formulate this problem in the following way.
Assume that the composition operator $C_{\phi_1}$ is not compact on $H^2$, and that
$|\phi_1^\ast | = |\phi_2^\ast|$; for which values of $\alpha$, can $m_{\phi_2}$ be an
$\alpha$-Carleson measure?\par
Note that if $m_{\phi_2}$ is an $\alpha$-Carleson measure, we necessarily must have
$\alpha \leq 2$. Indeed, if $\alpha >  2$, Corollary~\ref{coro Luecking} implies that $C_{\phi_2}\in S_2$,
and hence $C_{\phi_1} \in S_2$ as well.\par
\medskip

When $\alpha < 2$, the situation is very different, and one has:

\begin{theoreme}\label{meme module puissance}
For every $\alpha$ with $1 < \alpha < 2$, there exist two symbols $\phi_1$ and $\phi_2$ having the
same modulus on $\partial \D$ and such that $\rho_{\phi_1}(h) \approx h$, but
$\rho_{\phi_2}(h) \approx h^\alpha$.
\end{theoreme}

It follows from Corollary~\ref{coro Luecking} that:

\begin{theoreme}\label{meme module Schatten}
For every $p > 2$, there exist two symbols $\phi_1$ and $\phi_2$ having the same modulus on $\partial \D$
and such that $C_{\phi_1}$ is not compact on $H^2$, but $C_{\phi_2}$ is in the Schatten class $S_p$.
\end{theoreme}

\noindent{\bf Proof of Theorem~\ref{meme module puissance}.}  In \cite{CompOrli}, Theorem~4.1, we proved
a particular case of this result, corresponding to $\alpha =3/2$. We took there
$\phi_1 (z) =\frac{1+z}{2}\,$. This function behaves as
$\exp\big( - [\sin^2 (t/2)] + i {\cal H} [\sin^2 (t/2)]\big)$,
and to prove Theorem~\ref{meme module puissance}, we shall just change the power $2$ of $\sin (t/2)$.
\par\smallskip

We shall use the following lemma, whose proof will be postponed.

\begin{lemme}\label{lemme coeff}
For $0 < \beta < 2$, let $f (t) = |\sin \frac{t}{2}|^\beta$. Then:
\begin{displaymath}
f (t) = c_0 + \sum_{k=1}^\infty c_k \cos kt\,,
\end{displaymath}
with:
\begin{equation}\label{signe coeff}
c_k < 0 \quad \text{for all } k\geq 1;
\end{equation}
and
\begin{equation}\label{estim coeff}
c_k = O\,\Big(\frac{1}{k^{\beta +1}}\Big)\,\cdot
\end{equation}
In particular, for $\beta >1$, the series $\sum_{k\geq 1} k c_k$ is convergent, and
$\sum_{k\geq 1} k c_k  < 0$.
\end{lemme}
\par\medskip

Taking $\beta = \frac{1}{\alpha - 1}$\,, which is $>1$, and $f (t) = | \sin (t/2)|^\beta$ as in
Lemma~\ref{lemme coeff}, observe that $f$ satisfies the assumptions \eqref{premiere condition} and
\eqref{deuxieme condition} of the general construction of the Subsection~\ref{general construction} (note that
for $\beta \geq 2$, $f$ is a ${\cal C}^2$ function, and hence, we have \eqref{premiere condition} directly,
without using Lemma~\ref{lemme coeff}). With the notation of that subsection (see \eqref{def grand F},
\eqref{def grand Phi}, and \eqref{M}), set:

\begin{displaymath}
\phi_1 =  \Phi \qquad \text{and} \qquad \phi_2 = M \Phi.
\end{displaymath}
One has:

\begin{displaymath}
|\phi_1^\ast| = | \phi_2^\ast | \quad a.e.
\end{displaymath}
\par

Lemma~\ref{premier lemme} and MacCluer's theorem show that $C_{\phi_1}$ is not compact on $H^2$.
On the other hand, since $f^{-1}(h) \approx h^{1/\beta}$, Lemma~\ref{deuxieme lemme} shows that
$\rho_{\phi_2} (h) \approx h^\alpha$, with $\alpha = 1 + \frac{1}{\beta} \in ]1,2[$.\par
\smallskip

That ends the proof of Theorem~\ref{meme module puissance}. \qed
\bigskip

\noindent{\bf Proof of Lemma~\ref{lemme coeff}.} Before beginning the proof, it should be remarked that for
$\beta =2$ (\emph{i.e.} $\alpha =3/2$, which is the case processed in \cite{CompOrli}, Theorem~4.1), we
have a trivial situation: $\sin^2 \frac{t}{2} = \frac{1}{2} - \frac{1}{2} \cos t$. \par
\emph{Proof of \eqref{signe coeff}}. For $0 < p < 1$, we have the well-known binomial expansion, for
$-1 \leq x \leq 1$:
\begin{displaymath}
(1 -x)^p = 1 - \sum_{k=1}^{\infty} \alpha_k x^k,
\end{displaymath}
with:
\begin{displaymath}
\alpha_k = \frac{p (1 - p) \cdots (k - 1 -p)}{k!} >0.
\end{displaymath}
Taking $x = \cos t$ and $p= \beta /2$, we get:
\begin{equation}\label{etoile}
2^{\beta/2} \Big| \sin {\frac{t}{2}} \Big|^\beta = 1 - \sum_{k=1}^\infty \alpha_k (\cos t)^k.
\end{equation}
Now, we know that:
\begin{displaymath}
(\cos t)^k = \sum_{j=0}^k b_{j,k} \cos (k - 2j) t \,\raise 1,5 pt \hbox{,}
\end{displaymath}
with $b_{j,k} > 0$. Substituting this in \eqref{etoile}, grouping terms, and dividing by $2^{\beta/2}$, we
get \eqref{signe coeff}.\par

\emph{Proof of \eqref{estim coeff}}. We shall separate the cases $\beta = 1$, $0 < \beta < 1$, and
$1 < \beta < 2$.\par
For $\beta =1$, one has, in an explicit way:

\begin{displaymath}
\Big|\sin \frac{t}{2} \Big| =\frac{2}{\pi} - \frac{4}{\pi} \sum_{k=1}^\infty \frac{\cos kt}{4 k^2 - 1}
=c_0 +\sum_{k=1}^\infty c_k \cos kt \,.
\end{displaymath}
\smallskip

Assume $0 < \beta < 1$. Since $\frac{\sin t/2}{t/2} >0$ on $[0,\pi]$, we can write:

\begin{displaymath}
\Big( \frac{\sin t/2}{t/2}\Big)^\beta = 1  + t^2 u(t)\,,
\end{displaymath}
where $u$ is a ${\cal C}^\infty$ function on  $[0,\pi]$. Then:

\begin{displaymath}
c_k = \frac{2}{2^\beta \pi} \int_0^\pi t^\beta \cos kt \,dt +
\frac{2}{2^\beta \pi} \int_0^\pi t^{\beta +2} u(t) \cos kt \,dt.
\end{displaymath}
The second integral is $O (k^{-2})$, as easily seen by making two integrations by parts. The first one
writes:

\begin{align*}
\int_0^\pi t^\beta \cos kt \,dt
& = -\frac{\beta}{k} \int_0^\pi t^{\beta - 1} \sin kt \,dt \\
& =  -\frac{\beta}{k} \int_0^{k\pi} \Big(\frac{x}{k}\Big)^{\beta - 1} \sin x \,\frac{dx}{k} \\
& \sim -\frac{\beta}{k^{\beta+1}} \int_0^{+\infty} x^{\beta - 1} \sin x\,dx.
\end{align*}
This last integral is convergent and positive. Hence, since $\beta + 1 < 2$:
\begin{displaymath}
c_k \sim - \delta k^{-(\beta + 1)},
\end{displaymath}
where $\delta$ is a positive constant.
\par

Before continuing, let us observe that we have similarly, due to the vanishing of the integrated terms:

\begin{equation}\label{grand O integrale}
\int_0^\pi \Big(\sin \frac{t}{2}\Big)^\sigma \sin (2k+1) \frac{t}{2}\,dt
= O (k^{-(\sigma + 1)}) + O (k^{-2}) = O (k^{-(\sigma + 1)}),
\end{equation}
for $0 < \sigma < 1$.
\smallskip

Assume now  $1 < \beta < 2$. We have:

\begin{align*}
\frac{\pi}{2} c_k
& = \int_0^\pi f (t) \cos kt\,dt = - \frac{1}{k} \int_0^\pi f '(t) \sin kt\,dt \\
& = - \frac{\beta}{2k} \int_0^\pi \big(\sin (t/2)\big)^{\beta - 1}  \cos (t/2) \sin (kt)\,dt \\
& = - \frac{\beta}{4k} \bigg[\int_0^\pi \Big(\sin \frac{t}{2} \Big)^{\beta - 1}\sin (2k+1)\frac{t}{2} \,dt \\
& \qquad \qquad \qquad \qquad \qquad
+ \int_0^\pi \Big(\sin \frac{t}{2} \Big)^{\beta - 1} \sin (2k - 1) \frac{t}{2} \,dt \bigg] \\
& = \frac{1}{k}\, O (k^{-\beta}) = O (k^{-(\beta +1)}),
\end{align*}
in view of \ref{grand O integrale}, applied with $\sigma = \beta - 1 \in ]0,1[$.\par
This ends the proof of Lemma~\ref{lemme coeff}. \qed
\bigskip

\noindent{\bf Remark.} The following question arises naturally: does $C_\phi \in S_p$ for some $p < \infty$
imply that $\mu_\phi$ is $\alpha$-Carleson for some $\alpha >1$? We shall see in the next section
(Remark~2 after the proof of Proposition~\ref{prop Schatten}) that the answer is negative.\par
\smallskip

Another question, related to our work, has been raised by K. Kellay: given a compact composition operator
$C_\phi \colon H^2 \to H^2$, and another symbol $\psi \colon \D \to \D$, is the composition
operator $C_{\phi \psi} \colon H^2 \to H^2$ still compact? This is the case if $\phi \psi$ is univalent.
In fact, the compactness of $C_\phi$ implies (\cite{Shap-livre}, Theorem~3.5) that:
\begin{equation}\label{angular derivative}
\lim_{|z| \to 1} \frac{1 - |\phi (z)|}{1 -|z|} = +\infty;
\end{equation}
hence, since $|\phi (z) \psi (z) | \leq |\phi (z)| $ for every $z \in \D$, we have:
\begin{displaymath}
\lim_{|z| \to 1} \frac{1 - |\phi (z) \psi (z)|}{1 -|z|} = +\infty,
\end{displaymath}
which implies the compactness of $C_{\phi\psi}$, thanks to the univalence of $\phi \psi$
(\cite{Shap-livre}, Theorem~3.2).\par

In \cite{CompOrli}, Proposition 4.2, we proved a related result: let $\phi_1, \phi_2 \colon \D \to \D$ be
univalent analytic self-maps such that $|\phi_1^\ast| \leq |\phi_2^\ast|$ on $\partial\D$; if
$C_{\phi_2} \colon H^2 \to H^2$ is compact, and $\phi_2$ vanishes at some point $a\in \D$, then
$C_{\phi_1} \colon H^2 \to H^2$ is also compact. Note that the vanishing condition for $\phi_2$
in that result is automatic (since $\phi_2$ must have a fixed point $a\in \D$ because the
composition operator $C_{\phi_2}$ is compact, see \cite{Shap-livre}, page~84, \S~5.5, Corollary), but the
univalence condition for $\phi_1$ and $\phi_2$ cannot be dropped. In fact, take
$\phi_1 (z) = \exp \big(- (1 - z)/2\big)$ and $\phi_2 (z) = z\, M (z)\,\exp \big(- (1 - z)/2\big)$ (where $M (z)$
is defined by \eqref{M}). One has $|\phi_1^\ast| = |\phi_2^\ast|$ on $\partial \D$, and $\phi_1$ is univalent
(if $\phi_1 (z) = \phi_1 (w)$, $k = 0$ is the only integer such that $\frac{z}{2} - \frac{w}{2} = 2k\pi i$, since
$|z| + |w| \leq 2$). But $\phi_1$ is the function $\Phi$ defined by \eqref{def grand Phi}, with
$f (t) = \sin^2 (t/2)$. Since $f$ and ${\cal H}f$ are ${\cal C}^1$, Lemma~\ref{premier lemme} says that
$C_{\phi_1}$ is not compact. On the other hand, one has $f^{-1} (h) \approx h^{1/2}$, so
Lemma~\ref{deuxieme lemme} gives the compactness of $C_{\phi_2}$ (and even, $C_{\phi_2} \in S_p$
for every $p >4$, by Corollary~\ref{coro Luecking}).\par
It should be pointed out that, however, $\phi_1$ cannot be written $\phi_1 = \phi_2 \psi$ for some analytic
self-map $\psi \colon \D \to \D$.

\section{Composition operators in Schatten classes}\label{log puissance theta}

In \cite{Shap-Tay}, Theorem~4.2, J. Shapiro and P. Taylor constructed a family of composition operators
$C_{\phi_\theta} \colon H^2 \to H^2$, indexed by a parameter $\theta >0$ such that
$C_{\phi_\theta}$ is always compact, but $C_{\phi_\theta}$ is Hilbert-Schmidt if and only if
$\theta > 2$. In this section, we shall slightly modify the symbol $\phi_\theta$,
and shall study the membership of $C_{\phi_\theta}$ in the Schatten classes $S_p$. In \cite{Orlicz-Hardy},
we study on which Hardy-Orlicz spaces $H^\Psi$ these composition operators $C_{\phi_\theta}$
are compact.\par

\begin{theoreme}\label{theo Schatten}
For every $p_0 > 0$, there exists an analytic self-map $\phi \colon \D \to\D$ such that the composition
operator $C_\phi \colon H^2 \to H^2$ is in the Schatten class $S_p$ for every $p > p_0$, but not in
$S_{p_0}$.
\end{theoreme}

\noindent{\bf Proof.} We shall use the same function as J. Shapiro and P. Taylor in
\cite{Shap-Tay}, \S~4, with slight modifications. This modified function will be easier to analyze.\par
Let $\theta > 0$.\par
For $\Re z > 0$, $\log z$ will be the principal determination of the logarithm. Let, for $\eps >0$:
\begin{equation}\label{V indice epsilon}
V_\eps = \{ z\in \C\,;\ \Re z >0 \ \text{and}\ |z | < \eps\}.
\end{equation}
and consider, for $\eps >0$ small enough:
\begin{equation}\label{f indice theta}
f_\theta (z) = z (- \log z )^\theta, \qquad z \in V_\eps.
\end{equation}

\begin{lemme}\label{definition f indice theta}
For $\eps >0$ small enough, one has $\Re f_\theta (r \e^{i \alpha}) > 0$ for $0 < r < \eps$ and
$| \alpha | < \pi/2$. Moreover, one has $\Re f_\theta^\ast ( z) > 0$ for all
$z \in \partial V_\eps \setminus\{0\}$.
\end{lemme}

\noindent{\bf Proof.} Actually, $f_\theta$ can be defined on $\overline{V_\eps} \setminus \{0\}$, and we
shall do that.\par
Let, for $| \alpha | \leq \pi/2$ and $0 < r \leq \eps$:
\begin{displaymath}
Z_\alpha = \big(- \log (r \e^{i\alpha}) \big)^\theta = \Big( \log \frac{1}{r} - i \alpha \Big)^\theta.
\end{displaymath}
One has $f_\theta (r \e^{i\alpha}) = r \e^{i\alpha} |Z_\alpha| \e^{i \arg Z_\alpha}$, so that:
\begin{displaymath}
\Re f_\theta (r \e^{i\alpha}) = r\, |Z_\alpha| \cos (\alpha + \arg Z_\alpha).
\end{displaymath}
On the other hand,
\begin{displaymath}
\arg Z_\alpha = - \theta  \arctan \frac{\alpha}{\log 1/r}\, \cdot
\end{displaymath}
Since $\arctan x \geq x/2$ for $ 0 \leq x \leq 1$, we get, for $0 < r \leq \eps \leq \e^{-\pi/2}$:
\begin{align*}
| \alpha + \arg Z_\alpha |
& \leq |\alpha| \Big( 1 -  \frac{\theta}{\alpha}  \arctan \frac{\alpha}{\log 1/r} \Big)  \\
& \leq \frac{\pi}{2} \Big( 1 - \frac{\theta}{2\log 1/r} \Big) = \Upsilon_r.
\end{align*}
Therefore, for $0 < r \leq \eps$ and $|\alpha | \leq \pi/2$:
\begin{displaymath}
\Re f_\theta (r \e^{i\alpha}) \geq (\cos \Upsilon_r) \,r \Big(\log \frac{1}{r}\Big)^\theta > 0,
\end{displaymath}
as announced in Lemma~\ref{definition f indice theta}. \qed
\bigskip

Let now $g_\theta$ be the conformal mapping from $\D$ onto $V_\eps$, which maps $\T = \partial \D$
onto $\partial V_\eps$, and with $g_\theta (1) = 0$ and $g'_\theta (1) = -\eps/4$. Explicitly, $g_\theta$ is
the composition of the following maps: a) $\sigma \colon z \mapsto -z$ from $\D$ onto itself; b)
$\gamma \colon z \mapsto \frac{z + i}{1 + iz} = \frac{ z + \bar{z} + i ( 1 - |z |^2)}{|1 + i z |^2}$ from $\D$ onto
$P=\{ \Im z > 0\}$; c) $s \colon z \mapsto \sqrt z$ from $P$ onto $Q =\{\Re z > 0\,, \Im z >0\}$; d)
$\gamma^{-1} \colon z \mapsto \frac{ z -i }{ 1 -iz}$ from $Q$ onto $V = \{ |z | < 1\,, \Re z >0\}$, and
e) $h_\eps \colon z \mapsto \eps z$ from $V$ onto $V_\eps$.
\par\smallskip

Let:
\begin{equation}\label{definition phi indice theta}
\phi_\theta = \exp ( - f_\theta \circ g_\theta).
\end{equation}

By Lemma~\ref{definition f indice theta}, the analytic function $\phi_\theta$ maps $\D$ into
$\D$. Moreover, one has $| \phi_\theta^\ast | < 1$ on $\partial \D \setminus \{1\}$.
\par\medskip

Now Theorem~\ref{theo Schatten} will follow from the following proposition.

\begin{proposition}\label{prop Schatten}
With the above notation, $C_{\phi_\theta} \colon H^2 \to H^2$ is compact for every $\theta >0$ and,
moreover, is in the Schatten class $S_p$ if and only if $p > \frac{4}{\theta}\cdot$
\end{proposition}

Hence, given $p_0 > 0$, if we choose $\theta_0 = \frac{4}{p_0}$, we get that $C_{\phi_{\theta_0}}$
is in $S_p$ if and only if $p > p_0$. \qed
\par\medskip

\noindent{\bf Proof of Proposition~\ref{prop Schatten}} The compactness was proved in \cite{Shap-Tay},
Theorem~4.2; indeed, setting $h_\theta = f_\theta \circ g_\theta$, J. Shapiro and P. Taylor used the
symbol $\phi_\theta = 1 - h_\theta$ and proved that $\lim_{t \to 0} | \phi_\theta '(\e^{it}) | = + \infty$:
see equation (4.4) in \cite{Shap-Tay} (note that, in order to deduce the compactness from this equality,
J. Shapiro and P. Taylor had to prove a theorem: Theorem~2.4 in \cite{Shap-Tay}, whose proof needs to use
Gabriel's Theorem). Since our symbol is $\phi_\theta = \exp ( - h_\theta)$, and
$\exp\big( h_\theta (\e^{it}) \big) \to 1$
as $t \to 0$, it follows that the derivatives have the same behaviour when $t \to 0$. However, we are
going to recover this result by another method. For convenience, we shall write the boundary values
$f_\theta^\ast, g_\theta^\ast, \ldots$ of the different analytic functions $f_\theta, g_\theta,\ldots$ in the
same way as the analytic functions, without the exponent ${}^\ast$.\par
\smallskip

Note that, by Lemma~\ref{definition f indice theta}, $| \phi_\theta |$ is far from $1$ when
$g_\theta (z)$ belongs to the half-circle $\{\eps \e^{i\alpha}\,;\ |\alpha | \leq \pi/2 \}$. Hence, we only have
to study the case where $g_\theta ( \e^{i t}) = it$, $ -\eps \leq t \leq \eps$. Moreover, since
$f_\theta (it) = it \big(\log \frac{1}{| t |} - i\, {\rm sgn}\,(t) \frac{\pi}{2}\big)^\theta$, one has
\begin{displaymath}
| f_\theta (it) | \approx | t |\Big(\log \frac{1}{| t |}\Big)^\theta,
\end{displaymath}
so that $| \phi_\theta \big( g_\theta^{-1}(it)\big) |$ is far from $1$ when $t$ is away from $0$. Therefore,
it suffices to study what happens when $t$ is in a neighbourhood of $0$.\par
Note also that $g_\theta$ is bi-Lipschitz in a neighbourhood of $1$ (so $g_\theta (\e^{it}) \approx it$
when $|t |\leq \pi/2$), so we may forget it, and only consider the measure of the $t$'s for
which $f_\theta (it)$ belongs to the suitable sets. Moreover, for convenience, we only write the proof for
$t > 0$.\par
\smallskip

Since
\begin{align}
f_\theta (it)
& = it \Big(\log \frac{1}{t} \Big)^\theta
\bigg[\Big(1 - \frac{i\pi/2}{\log 1/t} \Big)^\theta \bigg]  \notag\\
& = it \Big(\log \frac{1}{t} \Big)^\theta \bigg[1 - \frac{i \pi \theta /2}{\log 1/t}
+ o\,\Big(\frac{1}{\log 1/t}\Big)\bigg], \label{developpement f theta de it}
\end{align}
one has:
\begin{align}
& \Re f_\theta (it) \approx t \Big(\log \frac{1}{t}\Big)^{\theta - 1} \label{partie Re de f theta}\\
& \Im f_{\theta} (it) \approx t \Big(\log \frac{1}{t}\Big)^\theta. \label{partie Im de f theta}
\end{align}
\par
Now, we have $\exp\big( - f _\theta (it ) \big) \in W (e^{- i\alpha}, h)$ ($0\leq \alpha < 2\pi$) if and only if
\begin{align}
& \Re f_\theta (it) \lesssim h  \label{numero un} \\
& | \Im f_{\theta} (it) - \alpha | \lesssim h. \label{numero deux}
\end{align}
But, when $e^{- i\alpha} \not \simeq 1$, this cannot happen for $h$ small enough (since $t$ goes to $0$
as $h$ goes to $0$). Essentially, we only have to consider the case $\e^{- i\alpha} = 1$, for which one has:
when $t > 0$ goes to $0$, $|\Re f_\theta (it) | \lesssim |\Im f_{\theta} (it) |$, and hence
$\exp\big( - f _\theta (it ) \big) \in W (1, h)$ if and only if $t (\log 1/t)^\theta \lesssim h$, \emph{i.e.}
$t \lesssim h/ (\log 1/h)^\theta$. Actually, we cannot assume $\alpha = 0$, and we have to be more precise,
and must do the following reasoning. When  \eqref{numero un} is satisfied, one has
$t (\log 1/t)^{\theta - 1} \lesssim h$; then $t \lesssim h/(\log 1/h)^{\theta -1}$ and hence:
\begin{displaymath}
t \Big( \log \frac{1}{t} \Big)^\theta \lesssim
\frac{h}{(\log 1/h)^{\theta - 1}} \Big( \log \frac{1}{h}\Big)^\theta
= h \log \frac{1}{h}\cdot
\end{displaymath}
It follows that the condition~\eqref{numero deux}  implies that:
\begin{equation}\label{encadrement alpha}
0 \leq \alpha \lesssim h \log \frac{1}{h}\, \cdot
\end{equation}
Therefore condition~\eqref{numero deux} implies that $t (\log 1/t)^\theta \lesssim h \log 1/h$, which gives:
\begin{displaymath}
t \lesssim \frac{h \log 1/h}{(\log 1/h)^\theta} = \frac{h}{(\log 1/h)^{\theta - 1}} \raise 1,5 pt \hbox{,}
\end{displaymath}
\emph{i.e.} $t(\log 1/t)^{\theta - 1} \lesssim h$: condition~\eqref{numero un} is satisfied (up to a constant
factor for $h$).\par
Since, by \eqref{encadrement alpha}, condition~\eqref{numero deux} is satisfied when
$\alpha - h \lesssim t (\log 1/t)^\theta \lesssim (\alpha + h)$ and implies that
$ - (\alpha + h) \lesssim t (\log 1/t)^\theta \lesssim (\alpha + h)$, it follows that the set of $t$'s
such that \eqref{numero un} and \eqref{numero deux} are satisfied has, since
$\log (\alpha + h) \approx \log h$, a measure $\approx h/(\log 1/h)^\theta$.
\par\smallskip

We then have proved that:
\begin{equation}\label{rho pour phi indice theta}
\rho_{\phi_\theta} (h) \approx \frac{h}{(\log 1/h )^\theta}\cdot
\end{equation}
Since $\rho_{\phi_\theta} (h) = o\, (h)$, MacCluer's criterion gives the compactness of $C_{\phi_\theta}$.
\medskip

Now, we shall examine when $C_{\phi_\theta}$ is in the Schatten class $S_p$, and for that we shall use
Luecking's theorem (Theorem~\ref{theo Luecking}). We have to analyze the behaviour of $\phi_\theta$ with
respect to the Luecking sets $R_{n,j}$; actually, for convenience, we shall work with
$R'_{n,j} = R_{n, 2^n - 1 - j}$. \par
\smallskip

We have to consider the modulus and the argument constraints.
\par\smallskip

\emph{Modulus constraint.} The condition $\frac{h}{2} \leq \Re f_\theta (it) < h$ writes:
\begin{equation}\label{encadrement module f indice theta}
\frac{h}{2} \lesssim t \Big(\log \frac{1}{t}\Big)^{\theta - 1} \lesssim h,
\end{equation}
and reads as:
\begin{equation}\label{encadrement module avec h}
\frac{h/2}{(\log 1/h)^{\theta - 1}} \lesssim t \lesssim \frac{h}{(\log 1/h)^{\theta - 1}}\,\raise 1,5 pt \hbox{,}
\end{equation}
or, when $h= h_n = 2^{-n}$:
\begin{equation}\label{contrainte module Luecking}
\frac{2^{-(n+1)}}{(n+1)^{\theta -1}} \lesssim t \lesssim \frac{2^{-n}}{n^{\theta - 1}}\cdot
\end{equation}
More precisely, one must have $t \in I_n = [a_n, b_n]$, with
$a_n \approx \frac{2^{-(n+1)}}{(n+1)^{\theta -1}}$\raise 1,5 pt \hbox{,}
$b_n \approx \frac{2^{-n}}{n^{\theta - 1}}$\raise 1,5 pt \hbox{,}
and $|I_n| \approx \frac{2^{-n}}{n^{\theta - 1}}\cdot$
\smallskip

\emph{Argument constraint.} We are looking for the set $J_n$ of the indices $j = 0,1, \ldots, 2^n - 1$ for
which we have $\exp\big( - f_\theta (it) \big) \in R'_{n,j}$. We have must have both the modulus constraint
$t \in I_n$ and:
\begin{equation}\label{contrainte argument theta}
j h_n \lesssim t \Big(\log \frac{1}{t} \Big)^\theta \lesssim (j+1) h_n,
\end{equation}
which implies, for $j \geq 1$, since one has \eqref{encadrement module f indice theta}:
\begin{displaymath}
j \lesssim \log \frac{1}{t} \lesssim 2 (j + 1),
\end{displaymath}
and hence, by \eqref{encadrement module avec h} and \eqref{contrainte module Luecking}:
\begin{displaymath}
\frac{n}{2} \log 2 \lesssim j \lesssim n \log 2,
\end{displaymath}
\emph{i.e.} $j \approx n$. The constant coefficients are not relevant here, and hence this estimation means
that $j$ can only take $O (n)$ values.\par
On the other hand, when the modulus constraint \eqref{encadrement module f indice theta} is satisfied,  one has
$\log (1/t)  \approx \log (1/h_n)$, and the argument constraint~\eqref{contrainte argument theta} is equivalent,
with $h=h_n$, to:
\begin{displaymath}
\frac{j h}{ (\log 1/h)^\theta} \lesssim t \lesssim \frac{(j+1)h}{(\log 1/h)^\theta} \cdot
\end{displaymath}
The length of the corresponding interval is $\approx h/ \big(\log (1/h)\big)^\theta $, which is equal, when
$h = h_n = 2^{-n}$, to $2^{-n} / n^\theta$.\par
It follows hence that $m_{\phi_\theta} (R'_{n,j}) \approx 2^{-n} / n^\theta$ for exactly $O(n)$ values of $j$,
and otherwise $m_{\phi_\theta} (R'_{n,j}) =0$; therefore:
\begin{displaymath}
\sum_{n=0}^\infty 2^{np/2} \sum_{j=0}^{2^n - 1} \big[m_{\phi_\theta} (R'_{n,j}) \big]^{p/2}
\approx \sum_{n=1}^\infty 2^{np/2} n \Big( \frac{2^{-n}}{n^\theta} \Big)^{p/2}
= \sum_{n=1}^\infty \frac{1}{n^{\theta \frac{p}{2} - 1}} \raise 1,5pt \hbox{,}
\end{displaymath}
which is finite if and only if $\frac{\theta p}{2} - 1 >1$, \emph{i.e.} $p > 4/ \theta$. \qed
\par
\bigskip

\noindent{\bf Remark 1.} Proposition~\ref{prop Schatten} shows that, for these symbols $\phi_\theta$, the
necessary condition~\eqref{limite condition necessaire Schatten} of
Proposition~\ref{condition necessaire Schatten} is not sharp; in fact, we have proved in
\eqref{rho pour phi indice theta} that $\rho_{\phi_\theta} (h) \approx h /(\log 1/h)^\theta$; hence
\eqref{limite condition necessaire Schatten} gives $\theta > 2/p$ when $C_{\phi_\theta} \in S_p$, even
though we must have $\theta > 4/p$.
\bigskip

\noindent{\bf Remark 2.} These composition operators answer negatively the question asked at the end
of Section~\ref{section same modulus}, since, by \eqref{rho pour phi indice theta}, the measure
$m_{\phi_\theta}$ is $\alpha$-Carleson for no $\alpha > 1$, though $C_{\phi_\theta}$ is
in $S_p$ for every $p >  4/\theta$.\par
\bigskip

\noindent{\bf Remark 3.} Our proof of Theorem~\ref{meme module puissance} fails when
$\alpha =2$ (\emph{i.e.} $\beta =1$). However, we are going to see that in this case, the composition
operator $C_{\phi_1}$ of this Theorem~\ref{meme module puissance} is in $S_p$ for every $p>4$.\par
Indeed, for $\beta =1$, one has, in an explicit way:
\begin{displaymath}
\Big|\sin \frac{t}{2} \Big| =\frac{2}{\pi} - \frac{4}{\pi} \sum_{k=1}^\infty \frac{\cos kt}{4 k^2 - 1}
=c_0 +\sum_{k=1}^\infty c_k \cos kt \,,
\end{displaymath}
with $c_k = O (k^{-2})$. Then:

\begin{displaymath}
{\cal H} f (t) =\sum_{k=1}^\infty \frac{\sin kt}{4 k^2 - 1} = \frac{1}{4} u (t) + v (t),
\end{displaymath}
where:
\begin{displaymath}
u (t) = \sum_{k=1}^\infty \frac{\sin kt}{k^2}\,\raise 1,5pt \hbox{,}
\end{displaymath}
and $v$ is ${\cal C}^2$, since $\hat v (k) = O (k^{-4})$. We have:
\begin{displaymath}
\qquad \qquad u ' (t) = \sum_{k=1}^\infty \frac{\cos kt}{k}
= - \log \Big(2 \sin \frac{t}{2}\Big)\,\raise 1,5pt \hbox{,}
\qquad 0 < t < \pi,
\end{displaymath}
(note that $u \notin {\rm Lip}\,1$, but $u$ is in the Zygmund class). It follows, when $t > 0$ goes to zero, that:
\begin{displaymath}
{\cal H}f (t) \approx - t \log \Big( 2\sin \frac{t}{2} \Big) \approx t \log \frac{1}{t} \cdot
\end{displaymath}
Since the modulus of continuity of ${\cal H}f$ at $0$ is $\approx t \log (1/t)$ (see also \cite{Zyg}, Chapter III,
Theorem~(13.30) for a general result), it follows that the argument condition
on $\phi_1$ is $|t| \lesssim \frac{h}{\log (1/h)}$ and that
$\rho_{\phi_1}(h) =O\big(\frac{h}{\log (1/h)}\big)$. Then $C_{\phi_1}$ is compact. It is actually in $S_p$
for every $p>4$. In fact, when $t > 0$ goes to zero, $\sin (t/2) \sim t/2$, and hence, with the notation of the
Subsection~\ref{general construction}, one has $| \Phi^\ast (t) | \sim t$, whereas
$\arg \big( \Phi^\ast (t) \big) \approx t/ \log (1/ t)$. We have then the same conditions as in the proof of
Proposition~\ref{prop Schatten}, when $\theta =1$. Hence $C_{\phi_1} \in S_p$ for every $p>4$. \qed
\bigskip

\noindent{\bf Remark 4.} We shall characterize in \cite{Orlicz-Hardy} the Orlicz functions $\Psi$ for which
these composition operators $C_{\phi_\theta}$ are compact.
\bigskip

In Theorem~\ref{theo Schatten}, we get, for any $p_0 > 0$, a composition operator which is in $S_p$ if and
only if $p > p_0$. We can modify slightly this operator so as to belong in $S_p$ if and only if $p \geq p_0$.

\begin{theoreme}\label{theo Schatten bis}
For every $p_0 >0$, there exists an analytic self-map $\phi \colon \D \to\D$ such that the composition
operator $C_\phi \colon H^2 \to H^2$ is in the Schatten class $S_{p_0}$, but not in $S_p$, for $p < p_0$.
\end{theoreme}

\noindent{\bf Proof.} We shall use the same method as in Theorem~\ref{theo Schatten}, but by replacing
the function $f_\theta$ by this modified function:
\begin{equation}
\tilde f_\theta (z) = z (- \log z )^\theta \big[ \log ( - \log z) \big]^q,
\end{equation}
where $q > \theta /2$ ($q =\theta$, for example).\par
Call $\tilde \phi_\theta$ the corresponding self-map of $\D$.\par
\smallskip

We shall not give all the details since they are mostly the same as in the proof of
Theorem~\ref{theo Schatten}.\par
We have first to check that:
\begin{lemme}\label{petit lemme}
For $\eps >0$ small enough, one has  $\Re \tilde f_\theta (z) > 0$ for every $z \in V_\eps$.
Moreover $|\tilde \phi_\theta^\ast | <  1$ a.e. on $\partial \D$.
\end{lemme}

\noindent{\bf Proof.} Write $z = r \e^{i\alpha}$ with $0 < r < \eps$ and $|\alpha | \leq \pi/2$. One has:
\begin{displaymath}
\tilde f_\theta (z) = r \e^{i\alpha} \Big( \log \frac{1}{r} - i \alpha\Big)^\theta
\bigg( \log \Big( \log \frac{1}{r} - i \alpha \Big)\bigg)^q.
\end{displaymath}
But:
\begin{displaymath}
\Big( \log \frac{1}{r} - i \alpha\Big)^\theta = \Big( \log \frac{1}{r}\Big)^\theta
\bigg( 1 - \frac{i \alpha \theta}{\log 1/r} + o\, \Big( \frac{1}{( \log 1/r) }\Big) \bigg);
\end{displaymath}
and, on the other hand:
\begin{align}
\log \Big( \log \frac{1}{r} - i \alpha \Big)
& = \log \sqrt{\Big( \log \frac{1}{r} \Big)^2 + \alpha^2 }
+ i \arg \Big(  \log \frac{1}{r} - i \alpha \Big)  \label{log log}\\
& = \Big(\log \log \frac{1}{r} \Big) \big( 1 + o\,(1/ \log (1/r)) \big)
- i \arctan \frac{\alpha}{\log 1/r} \notag \\
& = \Big( \log \log \frac{1}{r} \Big) \big( 1 + o\,(1/ \log (1/r) ) \big)
- i \frac{\alpha}{\log 1/r} \big( 1 + o\,(1) \big)\,\raise 1,5 pt \hbox{,} \notag
\end{align}
and
\begin{align}\label{puissance q}
\bigg[& \log \Big( \log \frac{1}{r}  - i \alpha \Big) \bigg]^q \\
& = \Big(\log\log \frac{1}{r} \Big)^q \bigg[ \Big[ 1 + o\,\Big(\frac{1}{\log 1/r}\Big)\Big]
- i \frac{1}{\log 1/r \log\log 1/r} \big( \alpha q + o\,(1)\big) \bigg]  \notag \\
& =  \Big(\log\log \frac{1}{r} \Big)^q \bigg[ 1 + o\,\Big(\frac{1}{\log 1/r}\Big) \bigg]. \notag
\end{align}
Hence, for $\eps >0$ small enough:
\begin{align*}
\Re \tilde f_\theta (z)
& = r (\log 1/r )^\theta ( \log \log 1/r)^q
\bigg[ \cos \alpha +\frac{\theta \alpha \sin \alpha}{\log 1/r} + o\,\Big( \frac{1}{\log 1/r} \Big) \bigg] \\
& \geq r (\log 1/r )^\theta ( \log \log 1/r)^q \bigg[ \cos \alpha
+ \frac{\theta (\alpha \sin \alpha - 1/4 ) }{\log 1/r} \bigg].
\end{align*}
That gives the result since, on the one hand, $|\theta (\alpha \sin \alpha - 1/4 ) / \log 1/r |$ is $\leq \sqrt 2/4$
for $\eps > 0$ small enough and $\cos \alpha \geq \sqrt 2/2$ when $|\alpha | \leq \pi/4$, and, on the
other hand, when $|\alpha | \geq \pi/4$, one has
\begin{displaymath}
\cos \alpha + \theta (\alpha \sin \alpha - 1/4 ) \geq \theta \big( \pi \sqrt 2/8 - 1/4) > 0.
\end{displaymath}
 That ends the proof of the lemma. \qed
\medskip

Now, by \eqref{log log}, for $t > 0$ going to zero:
\begin{displaymath}
\log \Big( \log \frac{1}{t} - i \frac{\pi}{2} \Big)
= \bigg[ \log \log \frac{1}{t} + O \Big( \frac{1}{(\log t)^2}\Big) \bigg] + i\, O \Big(\frac{1}{\log 1/t}\Big)\cdot
\end{displaymath}

Therefore:
\begin{align*}
& \Re \bigg[ \log \Big( \log \frac{1}{t} - i \frac{\pi}{2} \Big) \bigg]^q
\approx \Big( \log \log \frac{1}{t} \Big)^q \\
& \Im  \bigg[ \log \Big( \log \frac{1}{t} - i \frac{\pi}{2} \Big) \bigg]^q
\approx \frac{(\log \log 1/t)^{q - 1}}{\log 1/t}\, \raise 1,5pt \hbox{,}
\end{align*}
and, using \eqref{developpement f theta de it} (or \eqref{puissance q}, with
$\alpha = \pi/2$), we get:
\begin{align*}
& \Re \tilde f_\theta (it) \approx t \Big( \log \frac{1}{t} \Big)^\theta
\frac{(\log \log 1/t)^q}{\log 1/t}
= t \Big( \log \frac{1}{t} \Big)^{\theta - 1} \Big(\log \log \frac{1}{t} \Big)^q  \\
& \Im \tilde f_\theta (it) \approx t \Big( \log \frac{1}{t} \Big)^\theta
\Big( \log \log \frac{1}{t} \Big)^q.
\end{align*}
Hence the modulus constraint gives, with $h = h_n = 2^{-n}$:
\begin{equation}\label{blabla}
t \approx \frac{h}{\big( \log 1/h \big)^{\theta - 1} \big( \log \log 1/h\big)^{q - 1}}\,\raise 1,5 pt \hbox{,}
\end{equation}
and the argument constraint:
\begin{displaymath}
jh \lesssim t \Big( \log \frac{1}{t}\Big)^\theta \Big( \log \log \frac{1}{t} \Big)^q \lesssim (j+1)h.
\end{displaymath}
One gets:
\begin{displaymath}
j \approx \log\frac{1}{t} \approx \log \frac{1}{h} \approx n,
\end{displaymath}
and:
\begin{displaymath}
m_{\tilde \phi_\theta} (R'_{n,j}) \approx \frac{h}{\big( \log 1/h)^\theta \big( \log \log 1/h\big)^q}
\approx \frac{2^{-n}}{n^\theta (\log n)^q}\cdot
\end{displaymath}
It follows that Luecking's criterion becomes:
\begin{displaymath}
\sum_{n=1}^\infty 2^{np/2} n \frac{2^{-np/2}}{n^{\theta p/2} (\log n)^{q p/2}}
= \sum_{n=1}^\infty \frac{1}{n^{\theta p/2 - 1} (\log n)^{q p/2}}\cdot
\end{displaymath}
\par

This series converges if and only if $p > 4/\theta$ or else $p= 4/\theta$, since then
$q p/2 = q (4/\theta) /2 = q/ (\theta /2) > 1$. Hence $C_{\tilde \phi_\theta} \in S_p$ if and only if
$p \geq 4/\theta$, and that proves Theorem~\ref{theo Schatten bis}. \qed

\bigskip

In \cite{Carroll-Cowen}, T. Carroll and C. Cowen showed that there exist compact composition operators on
$H^2$ which are in no Schatten class $S_p$ for $p < \infty$ (see also \cite{Zhu} and \cite{Jones}). We shall
give an explicit example of such an operator.

\begin{theoreme}\label{no Schatten}
There exist compact composition operators $C_\phi \colon H^2 \to H^2$ which are in no Schatten class
$S_p$ with $p < \infty$.
\end{theoreme}

\noindent{\bf Proof.} The proof follows the lines of those of Theorem~\ref{theo Schatten} and
Theorem~\ref{theo Schatten bis}, but with, instead of $f_\theta$ or $\tilde f_\theta$:
\begin{equation}
f ( z) = z \log (- \log z).
\end{equation}
For $\eps >0$ small enough, we have $\Re f (z) > 0$, so that the corresponding function $\phi$ sends
$\D$ into itself; moreover, one has $|\phi^\ast | < 1 $ \emph{a.e.} on $\partial \D$. Indeed, if
$z = r \e^{i\alpha} \in V_\eps$, then $- \log z = R\,\e^{i\beta}$, with
\begin{align*}
R & = \sqrt{(\log 1/r)^2 +\alpha^2} = \log (1/r) + o\,(\log 1/r) \\
\beta & = - \arctan \frac{\alpha}{\log 1/r} = - \frac{\alpha}{\log 1/r} + o\, \big(1/ \log (1/r) \big).
\end{align*}
Then $\log (- \log z) = \log R + i\beta = (\log R) \big( 1 + \frac{i \beta}{\log R}\big)$ and
\begin{displaymath}
\Re f (z) = r (\log R) \Big( \cos \alpha - \frac{\beta \sin \alpha}{\log R} \Big).
\end{displaymath}
But
\begin{displaymath}
\cos \alpha - \frac{\beta \sin \alpha}{\log R} = \cos \alpha +
\frac{\alpha \sin \alpha}{(\log 1/r ) (\log \log 1/r) } + o\, \big(1/ (\log 1/r) (\log \log 1/r) \big)
\end{displaymath}
and we see, as in the proof of Lemma~\ref{petit lemme} that this quantity is, for $\eps >0$ small enough,
greater than a positive constant, uniformly for $|\alpha | \leq \pi/2$.
\par
\smallskip

Now, one has, as above, for $t $ going to zero:
\begin{align*}
& \Re f (it ) \approx \frac{|t|}{\log 1/|t|} \\
& \Im f (it) \approx t \log \log \frac{1}{|t|}\cdot
\end{align*}
It follows that $C_\phi$ is compact since $\rho_\phi (h) \approx h/\log\log 1/h$: indeed, we have to control
the two conditions:
\begin{align}
0 & < \frac{|t|}{\log 1/|t|} \lesssim h \label{un}\\
\alpha - h & \lesssim t \log \log \frac{1}{|t|}  \lesssim \alpha + h
\,\raise 1,5pt \hbox{,} \label{deux}
\end{align}
for $0 \leq \alpha < 2\pi$. When $\alpha \not\simeq 0$, condition~\eqref{deux} cannot happen
for $h$ small enough, because of condition~\eqref{un}; more precisely, \eqref{un} and \eqref{deux}
imply that
\begin{equation}\label{trois}
0 \leq \alpha \lesssim h + h \log \frac{1}{h} \log \log \frac{1}{h}
\lesssim  h \log \frac{1}{h} \log \log \frac{1}{h} \raise 1,5 pt \hbox{,}
\end{equation}
and then \eqref{deux} implies:
\begin{displaymath}
- h \lesssim t \log \log \frac{1}{|t|} \lesssim h \log \frac{1}{h} \log \log \frac{1}{h} \raise 1,5 pt \hbox{;}
\end{displaymath}
\emph{ a fortiori}:
\begin{displaymath}
|t|  \log \log \frac{1}{|t|} \lesssim h \log \frac{1}{h} \log \log \frac{1}{h} \raise 1,5 pt \hbox{,}
\end{displaymath}
and then:
\begin{displaymath}
|t| \lesssim h \log \frac{1}{h} \cdot
\end{displaymath}
Therefore condition~\eqref{deux} implies condition~\eqref{un}. Since, \eqref{trois} implies that
\begin{displaymath}
\log (\alpha + h) \approx \log h ,
\end{displaymath}
we get that the measure of the $t$'s satisfying \eqref{deux} is $\approx h\log \log 1/h$: \eqref{deux}  is
implied by $-h \leq t \log \log 1/|t| \leq \alpha +h$ and implies that
$- (\alpha +h) \leq t \log \log 1/|t| \leq (\alpha +h)$.\par

Hence, we have got that $\rho_\phi (h) \approx h / \log \log 1/h$, showing that $C_\phi$ is compact.
\par
By Proposition~\ref{condition necessaire Schatten}, $C_\phi$ is in no Schatten class $S_p$, $p > 0$.\par
In order to make more transparent the behaviour of the measure $m_\phi$, we shall give a direct
proof, using Luecking's criterion.
\smallskip

For testing Luecking's criterion, we must have, assuming $t > 0$:
\begin{displaymath}
t \approx h\log \frac{1}{h}
\end{displaymath}
and:
\begin{displaymath}
jh \lesssim t \log \log \frac{1}{t} \lesssim (j + 1) h,
\end{displaymath}
so $j \approx \log 1/h \log \log 1/h \approx n \log n$, and:
\begin{displaymath}
m_\phi (R'_{n,j}) \approx \frac{h}{\log \log 1/h} \approx \frac{2^{-n}}{\log n}\cdot
\end{displaymath}
Therefore Luecking's condition becomes:
\begin{displaymath}
\sum_{n=1}^\infty 2^{np/2} \bigg[n\log n \bigg(\frac{2^{-np/2}}{(\log n)^{p/2}} \bigg) \bigg]=
\sum_{n=1}^\infty \frac{n}{(\log n)^{p/2 - 1}} = +\infty.
\end{displaymath}
It follows that $C_\phi \notin S_p$. \qed
\bigskip

As a corollary, we get:

\begin{theoreme}\label{no Schatten meme module}
There exist two symbols $\phi$ and $\psi$ of same modulus $|\phi^\ast| = |\psi^\ast|$ on $\T$ such that
$C_\phi$ is compact on $H^2$, but in no Schatten class $S_p$ for $p < \infty$, whereas $C_\psi$ is in
$S_p$ for every $p > 2$.
\end{theoreme}

\noindent{\bf Proof.} Set
\begin{displaymath}
\psi = \phi M,
\end{displaymath}
where $\phi$ is the function used in Theorem~\ref{no Schatten}, and $M$ is the singular inner function given
by \eqref{M}. Since $\Re f (it ) \approx t/ \log (1/t)$, Lemma~\ref{deuxieme lemme} gives, if one sets
$\tilde f (t) = t/ \log (1/t)$:
\begin{displaymath}
\rho_\psi (h) \approx h \tilde f^{-1} (h) \approx h^2 \log (1/h) .
\end{displaymath}
It follows that $m_\psi$ is an $\alpha$-Carleson measure for any $\alpha <  2$. Hence, given any $p >2$,
and choosing an $\alpha < 2$ such that $p > 2/(\alpha - 1)$, it follows from Corollary~\ref{coro Luecking}
that $C_\psi \in S_p$. \qed
\bigskip

A funny question about Schatten classes is: for which Orlicz functions $\Psi$ is the above composition operator
$C_\phi$ in the Orlicz-Schatten class $S_\Psi$?

\section{Questions}

We shall end this paper with some comments and questions.\par
\smallskip

{\bf 1.} G. Pisier recently suggested that one should study more carefully the approximation numbers of
composition operators. Since an operator $T$ on a Hilbert space is in the Schatten class $S_p$ if and
only if $\sum_{n=1}^\infty a_n^p < +\infty$, where $a_n$  is the $n^{th}$ approximation number of $T$,
our present work may be seen as a, very partial, contribution to that study. \par
Note that Luecking's proof of its trace-class theorem does not make explicit mention of singular numbers,
but relies instead on an interpolation argument.
\par\smallskip

{\bf 2.} It is clear, by our results, that the membership in $S_p$ for the composition operator $C_\phi$ cannot
be characterized in terms of the growth of the Carleson function $\rho_\phi$. But we have given a sufficient
condition in Corollary~\ref{coro Luecking}, and a necessary one in
Proposition~\ref{condition necessaire Schatten}. Are these two conditions sharp? Can they be improved?\par
\smallskip

{\bf 3.} Do there exist two symbols $\phi_1$ and $\phi_2$ having the same modulus on $\partial \D$ such
that $C_{\phi_1}$ is not compact on $H^2$, but $C_{\phi_2}$ is in $S_p$ for every $p > 2$?\par
\smallskip

{\bf 4.} (K. Kellay) If $C_\phi \colon H^2 \to H^2$ is compact (or even in $S_p$, with $p >2$) and
$\psi \colon \D \to \D$ is analytic, is $C_{\phi\psi} \colon H^2 \to H^2$ compact?\par
\smallskip

config: arxiv


TITLE: Suggestions for high school?
QUESTION [11 upvotes]: I am currently a sophomore in high school who is very interested in mathematics and (theoretical) physics, and was wondering if the diverse set of mathematicians at MSE had any suggestions as to any self study texts I could read in order to enrich my knowledge of pure mathematics. First, I suppose, you should have some quick background as to my current level and interest in the field:
I have completed the majority of Tom M. Apostol's Calculus of a single variable (I more than likely would have finished if school had not started back up before I could), and really enjoyed the emphasis of set theory and proofs in the first chapter or so. I also enjoyed the treatment of integration (definite), before differentiation (I had knowledge of both through MIT open course ware). I especially enjoyed the proofs of basic properties using the axioms of the real number system, which I attempted mostly using first order logic (although I must admit a general lack of rigor to it all).
I have also perused some linear algebra here and here, although I can in no way be said to be 'proficient' in the techniques of linear algebra. In addition I have looked at some differencial equations, but am (again) relatively clueless with respect to them. Of the two subjects, however, I found differential equations to be the most interesting, although I do not have an awfully good measure of either subject.
Through my studies of quantum field theory I have come across stochastic calculus and statistical mechanics, which (in my naive understanding) seem rather interesting, but not exactly my favorite. In my study of General Relativity I became acquainted with tensors (to some extent), but more interestingly introduced into topology. Topology is intensely interesting and is both highly abstracted (as opposed to statistics), and is conceptually (somewhat) comprehensible. Another subject I have come across, although not through physics, is analysis (I know that is very broad), and really enjoyed the strict definitions of functions and the proofs of elementary properties from axioms. Though I have mostly seen real analysis, I think the little bit of complex I have seen is also rather interesting, but I have such a shallow understanding of it that I know I can not fully appreciate it.
Lastly are my encounters with algebra, these have come in many forms from many places, but most notably in programming, number theory, and physics. I have seen (but not fully understood) rings, fields (e.g. ordered fields), et cetera, and have found it to be extremely interesting. Every time I see a new algebra concept I see the logical connections between things I never would have seen before, but a lot of it is over my head. I feel that is true of the majority of the aforementioned mathematics, I can start to approximate an understanding but I know I can not truly appreciate the mathematics. I think a good example of this is exponents: I have known and used exponent rules since sixth grade, but I had never really appreciated their power and relationships until I discovered the proof of them (while working through Apostol).
With that I ask you: What texts will help my mathematical reasoning (more than my ability to calculate) at my level? What introductory text would you suggest from the various fields of mathematics that interest you? What text would you suggest to any high school students? What would you suggest outside of math texts (i.e online resources, coding projects, et cetera)?

REPLY [4 votes]: If you've worked through the majority of Apostol, then I highly recommend Hubbard and Hubbard's Vector Calculus, Linear Algebra, and Differential Forms. (Get it straight from the publisher; much cheaper than elsewhere). It is written very well, with the reader in mind. You'll learn linear algebra and some other requisites to start with. From there, you learn a bit of very basic topology and you end up doing some differential geometry-esque stuff later. There is SO much in this book... you could study it for a couple of years (especially if you dig into the appendix). Moreover, despite its rigor, there are many applications (that are actually very interesting).
As well, there is some courseware from Harvard that uses this book (look for Math 23a,b and Math 25a,b). And there is a (partial) solutions manual floating around.
You might also like to read:
-Set Theory and Metric Spaces by Kaplansky
-anything from the New Math Library (from the MAA)
-Particularly Basic Inequalities, Geometric Transformations, or Mathematics of Choice.
-the linear algebra in Apostol (be sure to get Vol 2 also!)
-the whole second volume of Apostol!
-Discrete Mathematics by Biggs (get the first edition!)
-Finite Dimensional Vector Spaces by Halmos,
-Principles of Mathematical Analysis by Rudin with this and this and these awesome lectures.
-Algebra by Artin is amazing, but hard! Enjoy these lectures. Vinberg's algebra text is supposed to be amazing and in a similar flavor to Artin (but a bit more gentle).
And you might also like these (great) reading lists:
-PROMYS
-Chicago Undergraduate Mathematics Bibliography


TITLE: Solving $2^y = 615+x^2$ for $y$
QUESTION [0 upvotes]: How can I transform the following equation:
$$2^y = 615+x^2$$
to something of the form
$$y = \cdots$$
SideQuest:
What would the values of $x$ and $y$ that make the equality True, and how would you get them?
Thanks!

REPLY [1 votes]: For sidequest : hint
Assuming that you want a positive integer solution in $x,y$.
$2^y$ can not be congruent to $0$ mod 3.
$x^2$ can not be congruent to $2$ mod 3.
$615$ congruent to $0$ mod 3.
Therefore, $2^y$ and $x^2$ must both be congruent to $1$ mod 3.
Therefore, $y$ must be an even.
This greatly reduces the problem, because now,
$2^y - x^2$ can be readily factored.


TITLE: Free group with 3 generators and free group with 2 generators
QUESTION [1 upvotes]: Let $F=<a,b>$ the free group with 2 generators and $G=<u,v,w>$ the free group with 3 generators. How can I show that the map $f:G\rightarrow F$ that sends $u\rightarrow ab,v\rightarrow a^2b^2, w\rightarrow a^3b^3$ and extends to $G$ with the obvious way  is a monomorphism?

I can show that $f$ is a homomorphism but I can't show that it is $1-1$.
  Is there a way to maybe show it has an inverse?

REPLY [2 votes]: Note that the subgroup $H$ generated by $ab$, $a^2b^2$, $a^3b^3$ is also generated by $x = ab$, $y = a^2b^2(ab)^{-1} = a^2ba^{-1}$, and $z = a^3(ab)^{-2}= a^3ba^{-2}$. It is sufficient to prove that $H$ is freely generated by $x,y,z$ because if it is then it is also freely generated by $ab$, $a^2b^2$, and $a^3b^3$.
The point about doing that replacement is that the set $x,y,z$ is Nielsen reduced, but you do not need to know what that means to prove the result.
Note that in any non-cancelling product of two of $x,y,z,x^{-1},y^{-1},z^{-1}$, the terms $b$ or $b^{-1}$ in the generators do not cancel. For example $yx = a^2b^2$, or $yz^{-1}= a^2bab^{-1}a^{-3}$.
So in any reduced word of length $n$ in the generators $x,y,z$, there will be exactly $n$ letters equal to $b$ or $b^{-1}$. Hence no such word reduces to the identity and it follows that $H$ is freely generated by $x,y,z$.


TITLE: Newton raphson convergence order
QUESTION [0 upvotes]: As you know convergence order of newton raphson method is:
$
E_{t,i+1}=-\frac {f''(x_r)}{2f'(x_r)}E_{t,i}^2
$
What's newton raphson method convergence order if $f'(x_r)=0$.

REPLY [0 votes]: In the hypothesis of the Newton's method there is one about $f'$ calculated on the sequence $\{x_r\}$, i.e.
$$
f'(x_r) \neq 0 \quad \forall \, x_r \in \{x_r\}
$$
From a geometrical point of view if for an $r$ happen $f'(x_r) = 0$ you find a point such that has got the derivate parallel to $x$ axis, so you can't intersect it.
Instead if your case is
$$
f(\alpha) = 0 \quad f'(\alpha) = 0
$$
i.e $\alpha$ is the root (I change the notation because $x_r$ is simular to a sequence) and the derivate calculate in $\alpha$ is zero ($\alpha$ is a root with multiciplity $>1$) see this answer. Dominique's answer explains the convergence argument in general for multiple roots.


TITLE: How many move paths on a 2-d grid?
QUESTION [0 upvotes]: On a 2-d grid how many different move paths can be made that begin at $(0,0)$ and end at $(x,y)$ with $x\gt 0$,$y\gt 0$. Restriction : left,down and diagonally moves aren't allowed .

REPLY [0 votes]: At each position except the boundaries we can go two different directions: "up" and "right".
At point ($x_0,y_0$) this is $f(x_0,y_0) =  f(x_0,y_0-1)+f(x_0-1,y_0)$. We also have the boundary conditions above that there is only one path along each boundary $f(k,1) = f(1,k) = 1$. We can now write a linear equation system to express $f(a,b)$ which we can solve with our favourite method.

However we have a symmetry of our problem which would allow us to simplify it a great deal if you have studied some more algebra. For simplifying using this symmetry we can take a look at this question which uses matrix exponentiation to find the numbers. This can in some sense be compared to using the numerical conjugate-gradient scheme on the above equation system - which results in investigating the Krylov Subspace $\{Iv,Av,A^2v,\cdots,A^nv\}$ which is precisely what the exponentiation above does but in a more compact form.
In particular, we have the
$$A = \left[\begin{array}{ccc}1&0&0\\1&1&0\\0&1&1\end{array}\right]$$
$$(A^{13})_{3,1} = 78 \hspace{0.5cm} \text{and} \hspace{0.5cm} nCr(13,2) = 78$$
Where nCr is the same function as Alex mentions in his answer. If we want to find a solution for larger grids we just need to increase the size of $A$ accordingly.


TITLE: Dijkstra’s algorithm / path is this done correctly?
QUESTION [1 upvotes]: im doing this assignment and it seems as if my teacher has made a mistake.
according to me in order to find the minimum spanning treee from a-z , you start from a and then go to :
a,f,d,c,b,e,z,g
however my teacher wrote in the facit that it is
a,f,d,c,b,e,g,z
i just cant see how that is possible. to go from e,g,z ?
My version:

original

REPLY [1 votes]: This notation probably is the sequence, in which nodes are visited. Not the resulting minimal path. Always the node with the least distance gets picked next. In the state, where e, g and z are not yet picked, their respective distance is 8, 10 and 13. So e is picked, z updated to 12 and g remains. Since 10 is lower than 12, g needs to be considered before z.


TITLE: Largest zero of the difference of Brownian motion and its running maximum
QUESTION [1 upvotes]: In what follows $B$ denotes Brownian motion and $M$ its running maximum.
I have two random variables, $\eta_t$ and $\xi_t$, defined as
\begin{align}
\eta_t &= \sup\{s \leq t: B_s = 0\} \\
\xi_t &= \sup\{s \leq t: M_s - B_s = 0\}
\end{align}
I am trying to prove that $\eta_t \stackrel{d}{=} \xi_t$.
I know that $M_s - B_s \stackrel{d}{=} \lvert B_s\rvert$. If their fdds were also the same, then I would be done but I doubt that is the case. On the other hand, I don't have proof of this either.
I tried applying the time reversal property of $B$, the Markov property of both $B$ and $M-B$ and many other things but I haven't been able to arrive at a proof.

REPLY [1 votes]: It is the case that $\{M_s-B_s: s\ge 0\}$ has the same distribution as $\{|B_s|: s\ge 0\}$. This result is due to P. Lévy. See p. 97 of the second edition of Brownian Motion and Stochastic Calculus by Karatzas and Shreve.


TITLE: Can the law of the acceleration be applicable to a man falling from plane?
QUESTION [1 upvotes]: Is acceleration law
$a = \Delta v/\Delta t$
applicable on somebody who is falling from an air plane? Does the acceleration increase when some body is at some altitude and does his speed increases by  $9.81 m/s$   every second (because of the gravity of the average earth) ?

REPLY [3 votes]: If I understand what you asked, yes that is the situation for which acceleration due to gravity is applied.
However, there is a terminal velocity due to air resistance, and the Acceleration and resistance will reach equilibrium. The wikipedia page discusses skydiving first, and that is what the activity of purposefully jumping out of a plane is called.

REPLY [2 votes]: Yes speed will increase $9.8$ m/s every second provided that there is no other force acting as resistance to the fall.
In reality air resistance depends on speed and the faster someone falls the more the resistance of the air around them. Eventually any object falling through the atmosphere will reach 'terminal velocity' the speed at which the force downwards of gravity is balanced exactly by a force upwards of wind resistance.
One final point is that at high altitude the acceleration due to gravity will be slightly less as the acceleration due to gravity depends on the distance from the centre of the earth. - but then the air is less dense so the air resistance will be lower and the terminal veocity will be higher.


TITLE: If you partition a generating set of $(\mathbb{R},+)$ into finitely many pieces, are one of the parts a generating set?
QUESTION [1 upvotes]: If you partition a generating set $S \subseteq \mathbb{R}$ (as a group) of $(\mathbb{R},+)$ into finitely many sets, $S_1,...,S_n$, are one of the parts a generating set? It is enough to study the case of $n=2$, since if it is true for 2, you will know that one of $S_1, S_2 \cup...\cup S_n$  is a generating set, if it isn't $S_1$ you know it is the other, so you can split that into $S_2, S_3 \cup...\cup S_n$. Now that I think about it, I would also be interested in the same question for the rationals.
I have been playing around/thinking about whether or not studying some (classes of)groups using some sort of infinite combinatorial method would be interesting, and figure I need an example group, like the reals, to generate good questions. In general I am interested in groups where every generating set partitioned into finitely many pieces has one of the pieces a generating set. Similarly I am interested in groups where you replace "generating set" with "normally generating set" in the previous sentence. (This is a "Ramseyan" condition for groups)
If these sorts of things have been studied, I would like some direction (like papers) of where to look. The idea to look into this is some set theoretic topology which studies combinatorics of open covers.

REPLY [2 votes]: Since $\mathbb{R}$ is an infinite dimensional vector space over $\mathbb{Q}$, it can be expressed (assuming the axiom of choice) as a direct sum $V_1\oplus V_2$ of two proper subspaces. But then $V_1\cup V_2$ is a generating set, but neither $V_1$ nor $V_2$ is.
As for the rationals, you could take $S_1$ to be the set of rationals with denominator a power of $2$, and $S_2$ the set of rationals with odd denominator.


TITLE: Lie derivative of the product of a function and a form
QUESTION [3 upvotes]: If $V$ is a vector field and $\alpha$ is a linear form,
$$ L_V\alpha: X\mapsto V(\alpha(X))-\alpha([V,X]) $$
for every field $X$, is $\mathbb R$-linear, which is trivial since $V$ is a field and $\alpha$ is linear, and $[V,\cdot]$ is also linear. (Is this right?) Also, $L_V:\mathfrak X^*\to\mathfrak X^*$,$ \alpha\mapsto L_V\alpha$ is $\mathbb R$-linear, but I would like to prove that
$$ L_V(f\alpha) = V(f)\alpha+ fL_V\alpha, $$
which is like a Leibniz property.
Any help?

REPLY [0 votes]: I find $\alpha$ a little clumsy so instead of using Greek letters for covectors I will use capital letters with bars, just as vectors are capital letters.
Indeed, vector fields are derivations mapping scalar fields to scalar fields; and given two such fields there is a function-composition $U \circ V$ which works out to also be a derivation for the commutator/Lie bracket $[U, V] = U \circ V - V \circ U.$ With type-overloading we can define $L_U V = [U, V]$ for vectors and $L_U f = U f$ for scalars, with $L_U (V f) = U (V f)$ straightforwardly, but the reverse (scalar multiplication, not function application) is going to be $L_U(f\cdot V) x = [U, f\cdot V] x$ which by the Leibniz property of $U$ is going to be $U f \cdot V x + f \cdot U (V x) - f \cdot V(U x),$ so $L_U(f \cdot V) = L_U f \cdot V + f \cdot L_U V.$
Now you are introducing another type overload, onto covectors: $$L_U \bar V = U \circ \bar V - \bar V \circ L_U.$$(1) Yes, you are correct to infer that $L_U \bar V$ is a covector, and it is for the linearity reason that you gave.
(2) If you want to know what this does to the one-form $f\cdot\bar V$ the best way to go is to apply the resulting one-form to an actual vector. For, you know that this is $$L_U (f\cdot \bar V)~X = U ((f \cdot \bar V) X) - (f \cdot \bar V) L_UX.$$The former term is $U(f\cdot \bar V X) = U f \cdot \bar V X + f\cdot U(\bar V X)$ while the latter term is $-f\cdot \bar V(L_U X),$ so we get straightforwardly $$L_U(f \cdot \bar V) X = U f \cdot \bar V X + f \cdot \big(U(\bar V X) - \bar V (L_U X)\big) = Uf \cdot \bar V X + f\cdot (L_U \bar V~ X).$$


TITLE: An inequality for Shannon's entropies
QUESTION [0 upvotes]: This question concerns Shannon's entropy of discrete random variables.
I want to show $H(X,Y)+H(Y,Z)-H(Y) \geq H(X,Z)$
On the left, we have $-\sum_{x,y}p(x,y) \log p(x,y) -\sum_{y,z}p(y,z) \log p(y,z)+\sum_{y}p(y) \log p(y) \\ =-\sum_{x,y,z}p(x,y,z) \log (\frac{p(x,y)p(y,z)}{p(y)})=-\sum_{x,y,z}p(x,y,z) \log [p(x|y)p(y,z)]$
(where $p(x,y,z)$ is the joint probability distribution of $X,Y,Z$)
On the right $\sum_{x,y,z}p(x,y,z) \log p(x,z)$

REPLY [1 votes]: A comment on your approach: for these type of inequalities, it is rarely, if ever, useful to expand the entropies in terms of probabilities. Instead, the typical tools are the chain rule of entropy, and some basic positivity statements (more generally, some kind of submodularity). These are called Shannon-type inequalities, and R. Yeung has some nice work on this that you might want to read. In addition, a standard text like Cover and Thomas will have enough problems to give you some intuition about how to go about proving such things.

For the question itself, by subtracting $H(Z)$ from both sides of the inequality, and using the chain rule $H(A,B) = H(A) + H(B|A),$ the stated inequality is equivalent to \begin{align} H(X|Y) + H(Y|Z) \overset{?}\ge H(X|Z).\end{align}
Since conditioning reduces entropy, it holds that $$ H(X|Y) + H(Y|Z) \ge H(X|Y,Z) + H(Y|Z) = H(X,Y|Z).$$
But $$H(X,Y|Z) = H(X|Z) + H(Y|X,Z) \ge H(X|Z)$$ since entropies are non-negative, and we're done.
This also tells us when equality will hold: If and only if $H(X|Y) = H(X|Y,Z)$ and $H(Y|X,Z) = 0$. The first says that $X$ and $Z$ are conditionally independent given $Y$, and the second says that $Y$ is a deterministic function of $X$ and $Z$.


TITLE: Composition of continuous and nonmeasurable function is measurable.
QUESTION [0 upvotes]: I'm stuck on part (b) of the question below. This is another question from a practice preliminary exam.  Thanks in advance!
Problem

a) Let $g$ be a monotone function on $\mathbb{R}$.  Prove that for every measurable function $f$ on a measurable set $E$ the composition $g \circ f$ is measurable.
b) Show that for every continuous not strictly monotone function $g$ on $\mathbb{R}$ there exists a non-measurable function $f$ such that $g \circ f$ is measurable.

My question relates to part (b).  I've solved part (a).  I'm not sure if they're trying to say, "every continuous [qualifiers removed] function," or if "not strictly monotone" is trying to state that $g$ is monotone but perhaps not strictly so.  Just wondering if anyone can solve and/or provide comment or corrections to the second part of the problem above.
Edited part (b)

Show that for every continuous function $g$ on $\mathbb{R}$ which is not strictly monotone there exists a nonmeasurable function $f$ such that $g \circ f$ is measurable.

REPLY [1 votes]: (b) If $g$ is not strictly monotone, then there are $a,b\in \mathbb R,$ $a\ne b,$ such that $g(a)=g(b).$ Let $E\subset\mathbb R$ be nonmeasurable. Set $f = a \chi_E +b\chi_{\mathbb R \setminus E}.$ Then $f$ is not measurable. However, on $E,$ $g\circ f = g(a),$ and on $\mathbb R\setminus E,$ $g\circ f = g(b).$ Since $g(a)=g(b),$ $g\circ f$ is constant and hence measurable.


TITLE: Existence of a self-complementary graph
QUESTION [8 upvotes]: Show that there exists a self-complementry graph of order $ n $ if and only if $ n = 0$  or $ 1 \ (\mbox{mod } 4) $.
My thoughts so far:
I've proven the 'only if' direction.
For the 'if' direction, I need to show that if $ n = 0$  or $ 1 \ (\mbox{mod } 4) $ then I can construct a self-complementary graph. I've started by focusing on the case $ n = 0 \ (\mbox{mod } 4) $; in particular, I've drawn such a graph in the case n = 4 (a Z shape). But I'm struggling to generalise my method to n = 8 and higher (in fact, I haven't successfully constructed a self-complementary graph of order 8). I'm not really sure how to think about this; I'm new to Graph Theory and so haven't had much practice yet.
EDIT: I've had a further thought. The graph I construct must have exactly $ \frac{n(n-1)}{4} $ edges.
I'd prefer hints to full answers, at least until I can reach the answer myself. Thanks for your help!

REPLY [1 votes]: Recursive constructions are very popular. The construction given in this answer (which I posted in a previous life) seems to be much less popular, but it is also quite simple. To wit:

Take a complete graph with vertex set $V$ and edge set $E={V\choose2}$. Let $\alpha$ be any permutation of $V$ in which the length of each cycle is a multiple of $4$, except for at most one $1$-cycle. (Of course, such permutations exist if and only if $|V|\equiv 0$ or $1\pmod 4$.)

Let $\beta$ be the permutation of $E$ induced by $\alpha$. Observe that $\beta$ contains only cycles of even length. Color the edges in each cycle alternately black and white. The graph $G$ consisting of the black edges is self-complementary. (The permutation $\alpha$ is an anti-automorphism of $G.$)

Example. To construct self-complementary graphs of order $5$, take $V=\{a,b,c,d,e\}$ and let $\alpha=(a\;b\;c\;d)(e)$ so that $\beta=(ab\;bc\;cd\;ad)(ac\;bd)(ae\;be\;ce\;de)\;$. If we choose the edges $ab,cd$ and $ac$ (i.e. "color them black") we get a $4$-point path $P_4$. Now we can choose the edges $be,de$ obtaining the self-complementary graph $C_5$, or else we can choose $ae,ce$ obtaining the other self-complementary graph of order $5$, the one that looks like the letter A.

This construction is completely general, in that every self-complementary graph can be obtained as an instance of this construction.


TITLE: Prove or Disprove that there is an integer $n$ such that $n^4 + 1 < 2n^2$
QUESTION [1 upvotes]: I am a beginner in writing proofs and want to show that an integer $n$ such that the title is true does not exist. I went about it this way and am wondering if this is even considered a valid proof, and if not, some assistance.
First, let $n = 0$
$0^4 + 1< 2(0)^2$
$1 < 0$. False, $1 > 0$, therefore, at $n = 0$, $ n^2 > 2n^2$
Let us simplify $n^4 + 1 < 2n^2$
$= n^4 < 2n^2 -1$
$= (n^2)(n^2) < 2n^2-1$
$= n^2 < \frac{(2n^2-1)}{n^2}$
We know that the $\lim_{n\to\infty} n^2 = \infty$ and $\lim_{n\to\infty}\frac{(2n^2-1)}{n^2} \approx 2$
So we will use candidate values $n = 1$ and $n = 2$
$ (1)^2 < \frac{(2(1)^2-1)}{(1)^2}$
$=  1 < 1/2$
False, $1 > 1/2$, therefore, at $n = 1$, $ n^2 > \frac{(2(n)^2-1)}{(n)^2}$
Then,
$ (2)^2 < \frac{(2(2)^2-1)}{(2)^2}$
$=  4 < 7/4$
False, $4 > 7/4$, therefore, at $n = 2$, $ n^2 > \frac{(2(n)^2-1)}{(n)^2}$
We have already verified that at $n = 0$ and $n = 1$, $ n^2 > \frac{(2(n)^2-1)}{(n)^2}$. Because at  $n = 2$, the value of $n^2$ is already greater than $\lim_{n\to\infty}\frac{(2n^2-1)}{n^2} = 2$, and $\lim_{n\to\infty} n^2 = \infty$, $n^2$ will continue to grow to $\infty$ while $\frac{(2(n)^2-1)}{(n)^2}$ approches $2$. Thus we have shown that there is no integer $n$ such that $n^4 + 1 < 2n^2$

REPLY [2 votes]: IMHO, it's a (mostly) valid proof, but it's a bit sloppy. It does more work than it needs to, and neglects mentioning the case of $n<0$.
Also, good proofs are economical: they achieve the task with the minimal mathematical machinery required. Eg, you really don't need that stuff about limits. (BTW, $\lim_{n\to\infty}\frac{(2n^2-1)}{n^2} = 2$. The limit is exact, it's not an approximation).
The comments show a couple of very succinct ways to prove your statement. However, it's not expected that your proof is the smallest possible proof. The history of mathematics is full of examples of proofs by good mathematicians that were later superceded by new proofs that were more compact, elegant, or more general. The main thing is to write a proof in such a way that your readers will be able to follow your train of thought, and verify that your claims are legitimate. You'll get better at that with practice, and by reading good proofs written by others. In that respect, the art of proof writing is similar to that of writing computer programs.


TITLE: Pigeonhole Principle question
QUESTION [2 upvotes]: There is a row of 35 chairs. Find the minimum number of chairs that must be occupied such that there are some consecutive set of 4 chairs or more occupied.
I would like to have some hints as to approach this problem. This isn't for homework or anything, I'm just curious as to what would be the best strategy for this problem.

REPLY [0 votes]: This is not really an answer ........Fill the pigeonholes in blocks of 3 with 1 separator (shown as ~)
123~456~789~101112~131415~161718~192021~222324~252627
You can see that 27 is the max that can be occupied with 4 in a row. So if 28 or more seats are occupied, you always have 4 or more consecutive occupied seats.
I just made many people's explanation into that...........What is The GENERALISING statement for this type of pigeonhole questions?


TITLE: Computing Expected Value
QUESTION [1 upvotes]: I am completely lost on what the question is saying. I am trying to think of a Probability Distribution to find the expected value, but I cant understand what is going on. I know how to compute the expected value but how do I figure out the probability distribution. Here is the question:
A college offers 10 courses and each student takes only one per semester. Three of the courses have enrollment 100, four courses have enrollment 25 and the rest three courses have enrollment 10.
X is a random variable that tells the enrollment of an randomly chosen class.
Y is a random variable that tells, for a randomly chosen student, what’s the enrollment of his/her class.
a) find E(X) and E(Y).
b) What do you think of these two expectations? If you are a student, which of the above expectations is more important to you?
My main trouble is with understanding what X and Y are I think. I am so lost on how to think about it.

REPLY [1 votes]: When we choose one of the classes at random, there is a probability of $\frac{3}{10}$ that this class has enrollment $100$, since there are $3$ out of the $10$ courses that have this number of students. So $$\mathbb{P}(X=100)=\frac{3}{10}$$
Similary
$$\mathbb{P}(X=25)=\frac{4}{10}\;\;\;\;\;\mathbb{P}(X=10)=\frac{3}{10}$$
And for the students, there are $300$ out of $3*100+4*25+3*10=430$ students that are in a class with $100$ members, so:
$$\mathbb{P}(Y=100)=\frac{300}{430}$$
And similarly
$$\mathbb{P}(Y=25)=\frac{100}{430}\;\;\;\;\; \mathbb{P}(Y=10)=\frac{30}{430}$$
Now you know the distributions of $X$ and $Y$, and I think you can take on from here with what you already know.


TITLE: Is this true :if $x\in [0.2] $ then $f(x)=\frac{2x+3}{x+2} \in [0.2]$?
QUESTION [0 upvotes]: I'm sorry to ask this question mayeb it's a trivial question but i would like to confirme if i have this function $f(x)=\frac{2x+3}{x+2}$ which  $x$ is a real number in $[0.2] $ then $f(x) \in [0.2]$ ?
My question here is :
Is this true :if $x\in [0.2] $ then $f(x) \in [0.2]$ and  why is not in $]0.2]$  ?
Note : The function is increasing and no values in the above area satisfy $f(x)=0$

REPLY [0 votes]: You only need to show that if $0\leq x\leq 2$ then $$0\leq f(x)\leq 2$$ but it's not needed that $f(x)=0$ for some $x\in [0,2]$.
Since $$f(x)=\frac{2x+3}{x+2}=\frac{2x+4-1}{x+2}=2-\frac{1}{x+2}\qquad\qquad\text{for all }x\neq -2$$
It follows that $$0\leq x\leq 2\quad\implies \quad2\leq x+2\leq 4\quad\implies\quad \frac14\leq \frac1{x+2}\leq\frac12$$
Then $$2-\frac12\leq2-\frac1{x+2}\leq2-\frac14\qquad\implies\qquad \frac32\leq f(x)\leq\frac74\quad \forall x\in[0,2]$$
Since $[\frac32,\frac74]$ is a subset of $[0,2]$ the result follows.


TITLE: Show that System $(I)$ is stable iff $X(t)$ is bounded.
QUESTION [4 upvotes]: I have a theorem:
For a linear homogeneous system:
$$\dfrac{dx}{dt}=A(t)x \tag{I}$$
Where $A(t)=(a_{ij}(t))_{n \times n} \in C(\mathbb{R}^+,\mathbb{R}^{n \times n})$
Suppose that $X(t)$ be the fundamental matrix solution of the following reference system $(I)$.
Let $K(t,s)=X(t)X^{-1}(s)$ be the Cauchy matrix of the following reference system $(I)$.
Prove that:

a/ System $(I)$ is stable iff $X(t)$ is bounded, it means $\exists M>0$ such that $$\|X(t)\| \le M, \forall t \ge 0 $$
b/ System $(I)$ is asymptotically stable iff  $$\lim_{t \to +\infty}X(t)=0$$

=================================================================
I have stuck when I try to show this theorem.
I have tried using the definition of stable, asymptotically stable. But I have no solution.
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Ps: (Or) if somebody knows/reads this theorem (book/pdf/djvu...) then you can post it.
Can anyone help me!
Any help will be appreciated! Thanks/

REPLY [2 votes]: Here's my solution for Question a:
a/ A linear homogeneous system $\dot{x}=A(t)x$ is stable iff $X(t)$ is bounded.
Without loss of generality we can assume that $X(t_0)=E$, then $x(t)=X(t)x_0$.

If $X(t)$ is bounded. Whence $$\left \| x(t) \right \|=\left \| X(t)x_0 \right \| \le \left \| X(t) \right \| \left \| x_0 \right \| \le M\left \| x_0 \right \|<\epsilon, \text{for}\left \| x_0 \right \|<\delta:=\frac{\epsilon}{M}$$
Hence, $x \equiv 0$ is stable.
If $x \equiv 0$ is stable. Whence

$$\forall \varepsilon >0, \exists \delta >0 \ \text{such that}\  \left \| x_0 \right \| \le \delta \implies \left \| x(t) \right \| = \left \| X(t)x_0 \right \| < \epsilon, \forall t \ge t_0$$
Thus $$\left \| X(t)\alpha  \right \|=\left \| X(t)\alpha  \cdot \dfrac{\delta}{\left \| \alpha \right \|}\right \|\cdot \dfrac{\left \| \alpha \right \|}{\delta}<\dfrac{\epsilon}{\delta}\left \| \alpha \right \|, \forall \alpha \in \mathbb{R}^n$$
Hence, $\left \| X(t)  \right \| \le M:=\dfrac{\epsilon}{\delta} \blacksquare $.
====================================================
My solution for Question b:

If $\lim_{t \to \infty}X(t)=0$ then since $x(t)=X(t)X^{-1}(t_0)x_0$ we have:
$$\left \| x(t) \right \| \le \left \| X(t) \right \|\cdot  \left \| X^{-1}(t_0)\right \|\cdot \left \| x_0 \right \|$$
Thus,  $\lim_{t \to \infty}\left \| x(t;t_0,x_0) \right \| =0$, forall $x(t)$ of $(I)$.

Hence, $(I)$ is asymptotically stable.

If $(I)$ is asymptotically stable. How we can show that $\lim_{t \to \infty}X(t)=0$?


TITLE: Minkowski sum (Polyhedra)
QUESTION [0 upvotes]: I'd like to understand and visualize the minkowski sum using GeoGebra.
I already managed to create the sum of a triangle and a square - very simple for beginners like me!
Now I wonder how I can visualize the sum of a cube and a ball!
It doesn't have to be GeoGebra - if I'll be able to draw it by hand I'll be satisfied. However, I can imagine that drawing it by hand would be extremely difficult?
And, above all, what does $cube + ball$ look like?

REPLY [1 votes]: It's a cube with rounded edges and corners.