TITLE: How to understand oriented intersection number in the non-transversal case? (including in the definition of Euler characteristic)
QUESTION [0 upvotes]: Background
I am reading through Guiellman and Pollack's Differential Topology. Ultimately I would like to understand the Euler characteristic. On page 116, they define the Euler characteristic of a compact oriented manifold $Y$ as the oriented self-intersection number of the diagonal $\Delta$ in $Y\times Y$. This definition is based on the definition of oriented intersection number of a pair of oriented submanifolds $X\subset Y$, $Z\subset Y$. When $X$ is transversal to $Z$ (and we have dimensional complementarity so that $Y=X\oplus Z$), I understand the concept of the oriented intersection number: At each intersection point of $X,Y$, just concatenate an oriented basis of $X$ with an oriented basis of $Z$ and see if you get an oriented basis of $Y$. If you do, then the local intersection number at that point is positive, otherwise it's negative.  Then add up the local intersection numbers from each point of intersection. (Note: if you are familiar with the intersection number $I(f,Z)$ as defined for a function $f$ mapping into $Y$ instead of $I(X,Z)$ for a submanifold $X$, the latter notion can be derived from the former by setting $f$ to be the inclusion map $X\hookrightarrow Y$.)
Main Question
Where I am struggling to understand is when $X$ is not transversal to $Y$ (and I think this will basically always be the case when $Z=X$ as in the Euler characteristic definition). I gather that the thing to do is to slightly deform $X$ so it becomes transversal to $Z$ and then look at the intersection number of the deformed version of $X$ with $Z$ as defined above. But we would need to know that the resulting number doesn't depend on how we deform $X$ (within some neighborhood). This is where I am confused.
It may be useful to consider an example where $X=Z$ is the x-axis in $\mathbb{R^2}$ such that the positive direction  (i.e. moving from left to right) is considered positively orientated. It would appear to me that the intersection number we get after deforming $X$ (leaving $Z$ alone) would depend on whether we rotate $X$ clockwise or counterclockwise. If clockwise, I think we would get a positively orientated basis for $\mathbb{R^2}$ (so an intersection number of 1) and if counterclockwise we would get a negatively oriented basis of $\mathbb{R^2}$ (so an intersection number of -1). I think there is some mistake in my understanding here? Note that I am thinking of the tangent space of $X$ as the first element of the basis and of the tangent space of $Z$ as the second element of the basis (of $\mathbb{R^2}$).
Possibly relevant: I have read through the proof that the intersection number (also apparently called degree) is homotopy invariant (from Guilleman and Pollack and from Milnor) and I think I understand this proof (which uses the classication of 1-manifolds) but I don't see how the non-transversal case is accounted for.
As a secondary question, I am also not sure how to define the oriented intersection number when $X$ is transveral to $Z$ but their tangent spaces overlap as in $\mathbb{R^3}$. A third question is how to define oriented intersection number when $X$ is not transveral to $Z$ but where their tangent spaces do not overlap as with two perpendicular lines intersecting at a point in $\mathbb{R^3}$.
Note: I have not had the time to properly work through Guilleman and Pollack line-by-line from the beginning. I have skipped over several parts so it is possible that I have missed some important concept that is typically presumed when presenting this material.

REPLY [1 votes]: Yes, you need to read the earlier parts of  the book. The crucial thing you’re missing is that $X$ must be compact (and $Z$ closed). Indeed, in your example, you can push the $x$-axis up and have no intersections at all.


TITLE: This job selection problem finds the best days to work for a job where **if you work one day, you cant work the day before or after.**
QUESTION [0 upvotes]: This job selection problem finds the best days to work for a job where if you work one day, you cant work the day before or after. The self reduction for this problem is given below:
$ JS(P[1,2,........,n])= \begin{bmatrix}  0 ,  \ \ \ if \ \ n<0 \\ P[1] , \ \ \  if \ n=1 \\ max\{P[1],P[2] \} , \ if \ \ \  n=2 \\ max\{ JS (P[1,2,......,n-2] )+P[n],JS (P[1,2,.., n-1 ]) \}, \ \ \  if \ \ n>2  \end{bmatrix} $
Then how would the self reduction change if the problem changed to $ \to \ $  you can work 2 days in a row but you cant work the day before or after ?
Answer:
The given self -reduction was for the work of  $ \  1 \ \ day  \ $.
We have to find same the formula  for $ 2 \ \ days \ $.
According to me the self-reduction for $ \ 2 \ \ days \ $ work becomes
$ JS(P[1,2,........,n])= \begin{bmatrix}  0 ,  \ \ \ if \ \ n<0 \\ P[1] , \ \ \  if \ n=1 \\ P[1]+P[2]  , \ if \ \ \  n=2 \\ max\{P[1]+P[2], \ P[1]+P[3], \ P[2]+P[3]  \}, \ \ if \ \ n=3 \\  max\{ JS (P[1,2,......,n-2] )+P[n],JS (P[1,2,.., n-1 ]) \}, \ \ \  if \ \ n>3  \end{bmatrix} $
Am I right ?  Is there any help ?

REPLY [0 votes]: Not quite: the last term in the self-reduction still forces you to alternate days on and off. There should be three possible cases - count backwards from the last day you don't work.


text:
\begin{document}

\title[Inclusion-Exclusion Polynomials]
{On Ternary Inclusion-Exclusion Polynomials}
\author{Gennady Bachman}
\address{University of Nevada, Las Vegas\\
 Department of Mathematical Sciences\\
 4505 Maryland Parkway\\
 Las Vegas, Nevada 89154-4020}
\email{bachman@unlv.nevada.edu}

\subjclass{11B83, 11C08}

\keywords{Cylcotomic polynomials, inclusion-exclusion polynomials}

\begin{abstract}
Taking a combinatorial point of view on cyclotomic polynomials
leads to a larger class of polynomials we shall call the
inclusion-exclusion polynomials. This gives a more appropriate
setting for certain types of questions about the coefficients
of these polynomials. After establishing some basic
properties of inclusion-exclusion polynomials we turn to a
detailed study of the structure of ternary inclusion-exclusion
polynomials. The latter subclass is exemplified by cyclotomic
polynomials $\Phi_{pqr}$, where $p<q<r$ are odd primes. Our main
result is that the set of coefficients of $\Phi_{pqr}$ is simply
a string of consecutive integers which depends only on the residue
class of $r$ modulo $pq$.
\end{abstract}
\maketitle

\section{Introduction}\label{S:1}

A ternary cyclotomic polynomial  is a cyclotomic polynomial $\tcp$
where $t$ is a product of three distinct odd primes. More precisely,
\[
\tcp(z)=\prod_{\substack{0<a<t\\(a,t)=1}}\lp z-e^{2\pi ia/t}\rp,
\]
where $t=pqr$ and $p$, $q$, and  $r$ are distinct odd primes.
Following the usual conventions we assume that $p$ is the smallest
of the three primes, let $a_m=a_m(t)$ denote the coefficients of
$\tcp$, and set
\[
A(t)=\max_m|a_m(t)|.
\]
There has been much progress recently in our understanding of
coefficients of $\tcp$ and, especially, of the function $A(t)$.
A long-standing conjecture of M. Beiter \cite{Be} asserted that
the bound
\be{1.1}
A(t)=A(pqr)\le\ff{p+1}2
\end{equation}
holds for all $t$. But in a recent work of Y. Gallot and P. Moree
\cite{GM} this conjecture was disproved in a rather dramatic
fashion and a number of prescriptions of integers $t$ for which
\refE{1.1} fails to hold were given. In particular, it was shown
that if $\gre>0$ is fixed then for every sufficiently large prime
$p$ there exist $q$ and $r$ such that
\be{1.D}
A(t)>\Blp\ff23-\gre\Brp p.
\end{equation}
They conjectured that, in fact,
\be{1.A}
A(t)\le\ff23p.
\end{equation}
Following this B. Lawrence \cite{La} announced that he proved the
validity of \refE{1.A} for $p>10^6$.

Another interesting question that was resolved recently is whether
it is possible to arrange it so that
\be{1.2}
A(t)=1,
\end{equation}
even for arbitrary large $p$. We say that polynomial $\tcp$ is flat
in this case. An old folklore conjecture asserts that there are
flat cyclotomic polynomials of all orders -- the order of $\Phi_n$
is the number of distinct odd prime divisors of $n$ if $n$ is not
a power of 2 and is 1 otherwise. The case of ternary
cyclotomic polynomials has now been settled in the affirmative and
the validity of \refE{1.2} was first established by
this author in \cite{B3}. This result was later extended by
T. Flanagan \cite{Fl} who showed that \refE{1.2} holds for a
larger family of integers $t$. But the best known result in
this direction is due to N. Kaplan \cite{Ka} who showed that
\refE{1.2} holds for every $t$ with $r\equiv\pm1\pmod{pq}$.

As part of his work on \refE{1.2} Kaplan showed that the value of
$A(t)$ is completely determined by the residue class of $r$ modulo
$pq$, where $r>q>p$. More precisely, he showed that if $s>q$ is
another prime and if $r\equiv\pm s\pmod{pq}$ then
\be{1.B}
A(pqr)=A(pqs).
\end{equation}
Moreover, he also obtained the following partial analogue of
\refE{1.B} for the set of coefficients $\ca_t=\{a_m(t)\}$ of $\tcp$.
Set identities
\be{1.C}
\ca_{pqr}=\begin{cases}
\ca_{pqs},  &\text{if $r\equiv s\pmod{pq}$,}\\
-\ca_{pqs}, &\text{if $r\equiv -s\pmod{pq}$,}
\end{cases}\end{equation}
are certainly valid if $r,s>pq$. The first of these identities
was also proved by Flanagan \cite{Fl}. Actually \refE{1.C} is only
implicit in \cite{Ka}. It follows that each residue class
$r_0$ modulo $pq$ determines at most two different sets of
coefficients $\ca_{pqr}$ with $r\equiv r_0\pmod{pq}$.

Residue class of $r$ modulo $pq$ imposes certain structure on
$\tcp$ and its set of coefficients $\ca_t$ and our object here is
to further investigate this structure. The main result of this
paper is that $\ca_t$ is completely determined by the residue class
of $r$ modulo $pq$, that is, that \refE{1.C} holds for $r,s>q$.
We are also interested in an analogue of \refE{1.C} for the case
when $r\equiv s\pmod{pq}$ but $s<q<r$. Kaplan's result on
$r\equiv\pm1\pmod{pq}$ falls into this case and is seen to be a
special case of this general principle (see \refS{3}). In pursuing
this development we shall work in a more general setting of what
we shall call inclusion-exclusion polynomials. Accordingly we begin
with a brief discussion of this class of polynomials and their
relation to cyclotomic polynomials, this is the subject of
\refS{2}. We then concentrate on the ternary case (of the
inclusion-exclusion polynomials) in \refS{3}.

\section{Inclusion-Exclusion Polynomials}\label{S:2}

Let $\rho=\{\,r_1,r_2,\dots,r_s\,\}$ be a set of natural numbers
satisfying $r_i>1$ and $(r_i,r_j)=1$ for $i\ne j$, and put
\[
n_0=\prod_ir_i,\q n_i=\ff{n_0}{r_i},\q
n_{ij}=\ff{n_0}{r_ir_j}\ [i\ne j],\q \dots
\]
For each such $\rho$ we define a function $\qr$ by
\be{2.1}
\qr(z)=\ff{\lp z^{n_0}-1\rp\cdot\prod_{i<j}
\lp z^{n_{ij}}-1\rp\cdot\dots}
{\prod_i\lp z^{n_i}-1\rp\cdot
\prod_{i<j<k}\lp z^{n_{ijk}}-1\rp\cdot\dots} .
\end{equation}
Our first observation is that $\qr$ is, in fact, a polynomial.

\begin{theorem}\label{T:1}
We have
\be{2.A}
\qr(z)=\prod_\gw(z-\gw),
\end{equation}
where the product is taken over all the roots of unity $\gw$
satisfying the condition
\[
\gw^{n_0}=1\q\text{but}\q\gw^{n_i}\ne1\q[1\le i\le s].
\]
Moreover, the degree of $\qr$ is given by
\[
\gf(\rho)=\prod_i(r_i-1).
\]
\end{theorem}

\begin{proof}
The claim follows by routine applications of the
inclusion-exclusion principle and we omit the straight-forward
details.
\end{proof}

We shall refer to polynomials $\qr$ as inclusion-exclusion
polynomials, the term suggested by the construction \refE{2.1}.
Our interest in this class of polynomials is motivated by the study
of coefficients of cyclotomic polynomials, as will become plain
below. From the algebraic point of view the most interesting case
is when parameters $r_i$ are assumed to be distinct prime numbers.
In this case $n_0=\prod_ir_i$ is a canonical factorization of $n_0$
into primes, $n_0$ is a square free integer, and the product in
\refE{2.A} is taken over all primitive $n_0$th roots of unity $\gw$.
In other words, in this case polynomial $\qr$ is better known as
cyclotomic polynomial $\Phi_{n_0}$. In general, as we show presently,
$\qr$ is a certain product of cyclotomic polynomials.

\begin{theorem}\label{T:2}
Given $\rho$ let
\be{2.2}
D=D_{\rho}=
\{\, d:d\mid n_0\text{ and }(d,r_i)>1\text{ for all }i\, \}.
\end{equation}
Then we have
\be{2.3}
\qr(z)=\prod_{d\in D}\Phi_d(z).
\end{equation}
\end{theorem}

\begin{proof}
Since both sides of \refE{2.3} are monic polynomials with roots
of multiplicity 1 it suffices to show that the roots are, in
fact, the same. So consider any root $\gw$ of $\qr$ and let $d$
be the smallest integer such that $\gw^d=1$. Then $\gw$ is a
root of $\Phi_d$. Moreover, by \refT{1}, $d\in D$ since
$d\mid n_0$ and $(d,r_i)>1$.

In the opposite direction, fix $d\in D$ and let $\gw$ be any
root of $\Phi_d$. Then $\gw^{n_0}=1$. Moreover, $(d,n_i)<d$ and
we conclude that $\gw^{n_i}\ne1$. Thus, by \refT{1}, $\gw$
is a root of $\qr$.
\end{proof}

We now turn our attention to the question of coefficients of these
polynomials. We begin with a few remarks of general nature on
coefficients of inclusion-exclusion polynomials versus coefficients
of cyclotomic polynomials. As is well known, to study coefficients
of cyclotomic polynomials it suffices to consider only polynomials
$\Phi_n$ with $n$ square free. We have seen that in this case
$\Phi_n=\qr$, where $r_i$ are prime factors of $n$. It is thus
natural to consider properties of coefficients of cyclotomic
polynomials in the larger context of coefficients of
inclusion-exclusion polynomials. In fact, the literature on this
topic contains many results on cyclotomic polynomials that are
actually theorems about inclusion-exclusion polynomials. This
characterization certainly applies to every result which was
obtained by an argument that (i) used identity \refE{2.1} as a
point of departure and (ii) did not crucially depend on parameters
$r_i$ to be prime but actually only required the condition
$(r_i,r_j)=1$, for $i\ne j$. For instance, most of the work on
coefficients of cyclotomic polynomials of low order falls into this
category. Even when it comes to open questions about coefficients of
cyclotomic polynomials, it seems rather clear that some of them are
truly questions about coefficients of inclusion-exclusion
polynomials. Take, for instance, the question of whether there
exist flat cyclotomic polynomials of arbitrary large order. It is
quite evident that this is a question about the structure of
\refE{2.1} and that the algebraic distinction of cyclotomic
polynomials has no bearing on the matter.

Finally, we remark that the setting of inclusion-exclusion
polynomials may offer not only a more appropriate context but it
may actually furnish certain ``practical'' advantages over the
setting of cyclotomic polynomials. This will become plain in the
next section where we take up the case of ternary
inclusion-exclusion polynomials (defined below).

We define the order of $\qr$ to be $s$, if $s=1$ or if $r_i\ge3$
for $1\le i\le s$, and $s-1$ otherwise. This parameter corresponds
to the order of cyclotomic polynomial and plays an important role.
The situation is identical to the more familiar setting of
cyclotomic polynomials and the best known examples of this are
polynomials of orders 1 and 2. Indeed, if $s=1$ and
$\rho=\{\, p\,\}$ ($p\ge2$; not necessarily prime), then
\be{2.B}
Q_{\{p\}}(z)=\ff{z^p-1}{z-1}=\sum_{n=0}^{p-1}z^n.
\end{equation}
Similarly,
\begin{align*}
Q_{\{p,q\}}(z)&=\ff{(z^{pq}-1)(z-1)}{(z^q-1)(z^p-1)}=
(1-z^{pq})(1-z)\sum_{i=0}^\infty z^{iq}\sum_{j=0}^\infty z^{jp}\\
&\equiv(1-z)\sum_{i,j\ge0}z^{iq+jp}\pmod{z^{(p-1)(q-1)+1}},
\end{align*}
by \refT{1}. It follows that if $\chi$ denotes the characteristic
function of integers representable in the form $iq+jp$ with
$i,j\ge0$ then
\be{2.C}
Q_{\{p,q\}}(z)=\sum_{n=0}^{(p-1)(q-1)}\chi(n)z^n-
\sum_{n=0}^{(p-1)(q-1)-1}\chi(n)z^{n+1}.
\end{equation}
Thus, in the sense of \refE{2.B} and \refE{2.C}, the structure of
$\qr$ is determined by the order. In particular, polynomials of
orders 1 and 2 are flat. (For a more detailed discussion of
polynomials of order 2, phrased in terms of cyclotomic polynomials,
see, for example, \cite{Le}.)

The condition $r_i\ge3$ in the definition of order of $\qr$ is
explained by the identity (whose cyclotomic polynomials analogue
is also well-known)
\[
Q_{\{2,r_2,\dots,r_s\}}(z)=Q_{\{r_2,\dots,r_s\}}(-z)\q[s\ge2].
\]
This follows readily from \refT{1} and we omit the details (see,
for example, \cite{Le}). This takes us to polynomials of order
$\ge3$ where the situation is considerably more interesting. As we
already mentioned in the introduction, even the ternary case, that
is $\qr$ of order 3, still presents interesting challenges. This
case is the principal object  of this paper and it will be taken
up in the next section.

The fact that cyclotomic polynomials are reciprocal proved to
be useful in the study of their coefficients. We conclude this
section by observing that the same is true for
inclusion-exclusion polynomials. Indeed, the identity
\[
\qr(z)=z^{\gf(\rho)}\qr(z^{-1})
\]
follows readily from \refE{2.1} and the fact that $\gf(\rho)$
is the degree of $\qr$. From this we infer that if
\[
\qr(z)=\sum_{m=0}^{\gf(\rho)}a_mz^m \q[a_m=a_m(\rho)],
\]
then $a_m=a_{\gf(\rho)-m}$.

\section{The Ternary Case}\label{S:3}

We shall write $\qt$ to denote a ternary inclusion-exclusion
polynomial. For esthetic reasons we normally write
$\gt=\{\, p,q,r\,\}$ rather than $\gt=\{\, r_1,r_2,r_3\,\}$. Thus,
contrary to the conventions of \refS{1}, we now assume only that
parameters $p$, $q$, and $r$ are $\ge3$ and relatively prime in
pairs. At times, however, the use of notation
$\gt=\{\, r_1,r_2,r_3\,\}$ will prove to be the better choice.
Consequently, we consider the two forms to be interchangeable and
shall freely use either one with our choice dictated by convenience.
Adopting other conventions in the introduction we write, by \refT{1},
\be{3.1}
\qt(z)=\sum_{m=0}^{\gf(\gt)}a_mz^m\q[a_m=a_m(\gt)],
\end{equation}
as well as
\be{3.A}
\at=\{\, a_m(\gt)\,\}\q\text{and}\q A(\gt)=\max_m|a_m(\gt)|.
\end{equation}
Moreover, set
\[
A^+(\gt)=\max_ma_m(\gt)\q\text{and}\q A^-(\gt)=\min_ma_m(\gt).
\]
Let us emphasize that we are not assuming any particular order for
the parameters $p$, $q$, and $r$. The structural symmetry of $\qt$
with respect to these parameters is a key aspect of the problem and
it will play an important role in our development. Correspondingly,
we shall explicitly state any additional assumptions on $p$, $q$,
and $r$ when it is appropriate.

Recall from the introduction that we are after the relationship
between polynomials $Q_{\tr}$ and $Q_{\ts}$ with
$r\equiv s\pmod{pq}$. This problem splits into two parts according
to whether
\[
r,\, s>\max(p,q)\q\text{or}\q r>\max(p,q)>s\ge1,
\]
say. The principal focus of this paper is the former condition
and our main result is as follows.

\begin{theorem}\label{T:3}
Set of coefficients $\ca_{\tr}$ is a string of consecutive integers
and, for $r>\max(p,q)$, is completely determined by the residue
class of $r$ modulo $pq$. More precisely, we have
\be{3.B}
\at=[\, A^-(\gt),A^+(\gt)\,]\cap\mathbb Z
\end{equation}
and, for $r,s>\max(p,q)$,
\be{3.C}
\ca_{\tr}=\begin{cases} \ca_{\ts}, &\text{if $r\equiv s\pmod{pq}$,}\\
 -\ca_{\ts}, &\text{if $r\equiv -s\pmod{pq}$.}
 \end{cases}\end{equation}
\end{theorem}

As we pointed out in \refS{2}, much of what is known about
cyclotomic polynomials of low order pertains to corresponding
inclusion-exclusion polynomials. The work of Flanagan and Kaplan
discussed in the introduction is a case in point. In particular,
identity \refE{3.C} was already known to hold under the assumption
$r,s>pq$.

The remark in the preceding paragraph applies to all of our
references in what follows. Thus, for the sake of simplicity, we
shall henceforth ignore the distinction between cyclotomic and
inclusion-exclusion polynomials, when appropriate.

We deduce \refE{3.B} from \refL{2} below. Both of these facts were
discovered independently by Gallot and Moree \cite{GM2}. It is
worth noting that the approach in the works of \cite{Ka} and
\cite{GM2} is rather different from ours.

We derive \refT{3} by a sequence of lemmas some of which are
of independent interest and shed additional light on the
structure of $\qt$. As we remarked earlier, our development
preserves symmetry in the parameters $p$, $q$, and $r$ whenever
it is appropriate. This, in particular, will be handy when
considering the second alternative, namely
\be{3.D}
r\equiv\pm s\pmod{pq}\q\text{and}\q r>\max(p,q)>s\ge1.
\end{equation}

The situation in this case is more complicated and $A(p,q,r)$
need not equal to $A(p,q,s)$ -- in a slight abuse of notation
we shall write $A(p,q,r)$ in place of $A(\{\, p,q,r\,\})$, and
use the same conventions for the functions $A^+$ and $A^-$.
Instead we have the following result. Recall from \refS{2} that
polynomials of order less than 3 are flat, so that $A(p,q,2)=1$.
Moreover, it is convenient to extend the definition of $A$ by
setting $A(p,q,1)=0$. With these conventions we state our result
for \refE{3.D}.

\begin{theorem}\label{T:4}
If $r$ and $s$ satisfy \refE{3.D}, then
\be{3.E}
A(p,q,s)\le A(p,q,r)\le A(p,q,s)+1.
\end{equation}
\end{theorem}

The proof of \refT{4} will require further development and will be
carried out elsewhere. We shall limit ourselves here to a few brief
remarks. Note that, by \refT{3}, we have
\[
A(p,q,r)=A(p,q,pq\pm s),
\]
under \refE{3.D}. In this light \refE{3.E} is seen as a recursive
estimate. Of course, using an absolute upper bound for $A(p,q,s)$
on the right of \refE{3.E} yields the corresponding upper bound for
$A(p,q,r)$. For instance, a simple and convenient estimate
\be{3.F}
A(p,q,r)=A(p,q,pq\pm s)\le s,
\end{equation}
valid for all $s\ge1$, is obtained on combining \refE{3.E} with
the bound (see \cite{B1})
\[
A(p,q,s)\le s-\lc s/4\rc.
\]
Estimate \refE{3.F} sacrifices precision for convenience and is
certainly weaker than \refE{3.E} for $s\ge5$. On the other hand,
\refE{3.F} is sharp for $1\le s\le3$. We do not know if the
equation $A(p,q,pq+4)=4$ has any solutions. Note also that Kaplan's
result on flat cyclotomic polynomials corresponds to \refE{3.F}
with $s=1$.

When iteration of \refE{3.E} is possible it leads to a very rapid
reduction technique. For example, two applications of \refE{3.E}
give
\[
A(p,pq+1,p^2q+p+q)\le A(p,pq+1,q)+1\le2.
\]

Our first step is to observe that, by \refE{2.1}, polynomial $\qt$
has a representation
\be{3.2}\begin{aligned}
\qt(z)&=\ff{(1-z^{pqr})(1-z^r)(1-z^q)(1-z^p)}
{(1-z^{qr})(1-z^{rp})(1-z^{pq})(1-z)}\\
&\equiv(1-z^r)(1-z^q)(1+z+\dots+z^{p-1}) \\
&\phantom{(1-z^r)(1-z^q)}\times \sum_{i=0}^\infty z^{iqr}
\sum_{j=0}^\infty z^{jrp}\sum_{k=0}^\infty z^{kpq}
\pmod{z^{pqr}}.
\end{aligned}\end{equation}
Evidently, of key importance are integers $n$ of the form
\be{3.3}
n=iqr+jrp+kpq,\q i,j,k\ge0,
\end{equation}
and we let $\chi=\chi_{\gt}$ be the characteristic function of
such integers, that is,
\be{3.4}
\chi(n)=\chi_{\gt}(n)=\begin{cases}
1, &\text{if $n$ has representation \refE{3.3},}\\
0, &\text{otherwise.}
\end{cases}\end{equation}
Note that if $n<pqr$ then either representation \refE{3.3} is
not possible or it is unique. Therefore, by \refE{3.1}, \refE{3.2}
and \refE{3.4}, the identity
\be{3.5}
a_m=\sum_{m-p<n\le m}\bigl(\chi(n)-\chi(n-q)-\chi(n-r)
+\chi(n-q-r)\bigr)
\end{equation}
holds for all $m<pqr$. Let us clarify the meaning of this statement.
Recall that $\qt$ is a polynomial of degree $\gf(\gt)$. But in
\refE{3.5} we take $a_m=0$ for $m<0$ and for $\gf(\gt)<m<pqr$ and
then the identity  remains valid in the range $m<pqr$. We shall
find this extension useful for technical reasons.

In considering integers representable in the form \refE{3.3} it
is helpful to observe that every integer $n$ has a unique
representation in the form
\be{3.6}
n=x_nqr+y_nrp+z_npq+\gd_npqr,
\end{equation}
with $0\le x_n<p,\ 0\le y_n<q,\ 0\le z_n<r,$ and
$\gd_n\in\mathbb{Z}$. It follows that $n$ is representable in the
form \refE{3.3} if and only if $\gd_n\ge0$. But if $n<pqr$, as we
shall assume henceforth, then $\gd_n\le0$ and we obtain the
characterization
\be{3.7}
\chi(n)=1\q\text{if and only if}\q \gd_n=0 \q[n<pqr].
\end{equation}

We shall deduce \refE{3.B} from \refL{1} below. Recall our
convention of using $\{\, p,q,r\,\}$ and
$\{\, r_1,r_2,r_3\,\}$ interchangeably.

\begin{lemma}\label{L:1}
We have
\[
\bigl|\chi(n)-\sum_i\chi(n-r_i)+\sum_{i<j}\chi(n-r_i-r_j)
-\chi(n-r_1-r_2-r_3)\bigr|\le1.
\]
\end{lemma}

\begin{proof}
Let $\gt'=\{\,\pm r_1,\pm r_2,\pm r_3\,\}$ and for every
pair of $u,v\in\gt'$ with $|u|\ne|v|$ set
\be{3.G}
\psi_{uv}(n)=\chi(n)-\chi(n-u)-\chi(n-v)+\chi(n-u-v).
\end{equation}
Observe that if $w$ is another element of $\gt'$ and $|u|$, $|v|$,
and $|w|$ are all distinct then
\[
\bigl|\chi(n)-\sum_i\chi(n-r_i)+\sum_{i<j}\chi(n-r_i-r_j)
-\chi(n-{\textstyle\sum_ir_i})\bigr|= \bigl|\psi_{uv}(n')-\psi_{uv}(n'-w)\bigr|,
\]
where $n'=n+(u-|u|)/2+(v-|v|)/2+(w-|w|)/2$. Therefore to prove the
lemma it suffices to show that the inequality
\be{3.H}
\psi_{uv}(n)-\psi_{uv}(n-w)\le1
\end{equation}
holds for all $n$ (in an appropriate range depending on parameters
$u$, $v$, and $w$).

In \cite[Lemma 2]{B1} it was shown that
\be{l11}
|\psi_{uv}(n)|\le1.
\end{equation}
Actually, this estimate was given explicitly only for $\psi_{qr}$,
with $q,r>p$, but the argument applies for every choice of
$u$ and $v$ in $\gt'$. The rest of this proof is essentially an
extension of \cite[Proof of Lemma 2]{B1} and, in particular,
\refE{l11} will serve as a convenient reduction tool. In the first
place, by \refE{3.H} and \refE{l11}, it suffices to
show that there is no $n$ for which
\be{l13}
\psi_{uv}(n)=1\q\text{and}\q\psi_{uv}(n-w)=-1.
\end{equation}
To reach a contradiction let us assume that \refE{l13} holds for
some $n$. Noting that, by \refE{3.G}, we have
\[
\psi_{uv}(n)=\psi_{(-u)(-v)}(n-u-v),
\]
shows that, in addition to $\psi_{uv}(n)=1$, there is no loss in
generality in assuming that $\chi(n)=1$. Similarly, by \refE{3.G}
and symmetry, it follows that in addition to $\psi_{uv}(n-w)=-1$
we may assume that $\chi(n-w-u)=1$. It now follows from \refE{l11}
with $v$ replaced by $w$ that we may also assume that $\chi(n-w)=1$,
say. But then, since $\psi_{uv}(n-w)=-1$, we must also have
$\chi(n-w-v)=1$ and $\chi(n-w-u-v)=0$. Finally, in view of
$\psi_{uv}(n)=1$, we may further assume that $\chi(n-u)=0$, say.
To summarize, to show that \refE{l13} is not
possible it suffices to show that there is no $n$ for which
\[
\chi(n)=\chi(n-w)=\chi(n-w-u)=\chi(n-w-v)=1
\ \text{and}\ \chi(n-u)=\chi(n-w-u-v)=0.
\]
But this readily follows from \refE{3.6} and \refE{3.7} by
chasing the coefficients in representations \refE{3.6} of all
the relevant integers (see \cite[Proof of Lemma 2]{B1}).
\end{proof}

\begin{lemma}\label{L:2}
We have $\bigl|a_m(\gt)-a_{m-1}(\gt)\bigr|\le1$.
\end{lemma}

\begin{proof}
By \refE{3.5} and \refE{3.G},
$a_m-a_{m-1}=\psi_{qr}(m)-\psi_{qr}(m-p)$,
and the claim follows from \refL{1}.
\end{proof}

The following simple observation is rather useful. By \refE{3.6} and
\refE{3.7}, we have
\begin{gather}
\chi(n)=\chi(n-pq),\q\text{unless $\gd_n=z_n=0$},\label{E:3.12}\\
\chi(n)=1\ \text{and}\  \chi(n-pq)=0,\q
 \text{if $\gd_n=z_n=0$},\label{E:3.13}
\end{gather}
as well as the analogues of \refE{3.12} and \refE{3.13} with $pq$
and $z_n$ replaced by $qr$ and $x_n$ or by $rp$ and $y_n$,
respectively.

\begin{lemma}\label{L:3}
Let $R_m$ be the set of all integers appearing as an argument of
$\chi$ in the summation \refE{3.5}. Then we have
\be{l3}
a_m=a_{m-pq},\q\text{unless there is $n\in R_m$ with $\gd_n=z_n=0$,}
\end{equation}
as well as the analogues of \refE{l3} with $pq$ and $z_n$ replaced
by $qr$ and $x_n$ or by $rp$ and $y_n$, respectively.
\end{lemma}

\begin{proof}
This is an immediate consequence of \refE{3.5} and \refE{3.12}.
\end{proof}

\begin{lemma}\label{L:3B}
The estimate
\be{3.I}
|a_m-a_{m-r_ir_j}|\le2
\end{equation}
holds unconditionally for every pair of parameters $r_i\ne r_j$.
Moreover, if $r\ge p+q$ then we have
\be{3.J}
|a_m-a_{m-pq}|\le1.
\end{equation}
\end{lemma}

\begin{proof}
Consider representation \refE{3.5} with parameter $p$ given by
$p=\min(r_1,r_2,r_3)$. We shall now prove \refE{3.I} with
$r_ir_j=qr$, the remaining case follows in the same way. Put
$I_n=(n-p,n]\cap\mathbb Z$, so that
\be{3.K}
R_m=I_m\cup I_{m-q}\cup I_{m-r}\cup I_{m-q-r}.
\end{equation}
Note that $x_n=0$ if and only if $n$ is a multiple of $p$. But
each of the four intervals on the right of \refE{3.K} contains
exactly one multiple of $p$, say $\ga_ip$. Therefore, by
\refE{3.5}, \refE{3.12} and \refE{3.13}, we get
\[
|a_m-a_{m-qr}|\le
|\chi(\ga_1p)-\chi(\ga_2p)-\chi(\ga_3p)+\chi(\ga_4p)|,
\]
and \refE{3.I} follows.

To prove \refE{3.J} we argue in the same way but take advantage
of the condition $r\ge p+q$. By \refL{3}, we may assume that $R_m$
contains multiples of $r$ -- these are integers $n$ with $z_n=0$.
But, in the present case, the range $I_m\cup I_{m-q}$ contains
at most one such multiple, say $\ga r$. Therefore,
by \refE{3.5}, \refE{3.12} and \refE{3.13}, we have
\[
|a_m-a_{m-pq}|=|\chi(\ga r)-\chi(\ga r-r)|\le1,
\]
as claimed.
\end{proof}

Lemmas 3 and 4 have a number of interesting consequences. Observe
that if we take $r$ to be the largest of the three parameters then,
by \refE{l3}, it suffices to consider coefficients $a_m$ with
$m=\ga r+\gb$ and $0\le\gb<\min(r,p+q)$. We will use this fact
below. The parallel between \refE{3.J} and \refL{2} is immediate.
Unlike \refL{2}, however, \refE{3.J} has a hole in the form of the
range $\max(p,q)<r<p+q$. Estimate \refE{3.I} may be used to give a
simple upper bound for coefficients of $\qt$ as follows. Let
$p<\min(q,r)$ and take $r_ir_j=qr$ in \refE{3.I}. One then readily
verifies that iterating \refE{3.I} yields the bound
\be{3.L}
|a_m|\le2\lc m/(qr)\rc+1.
\end{equation}
Recall that we may assume that $m\le\gf(\gt)/2<pqr/2$. Thus
\refE{3.L} suggests that coefficients of largest size are to
occur near the middle of the range of the index $m$ and gives a
nontrivial bound for ``small'' $m$. Note also that replacing 2 by
1 on the right of \refE{3.I} would have the same effect on the
right of \refE{3.L}. But this would imply the bound
$A(\gt)\le\lc p/2\rc+1$, contradicting \refE{1.D}. It follows
that, in general, \refE{3.I} is sharp.

Another immediate consequence of \refE{3.6} is that $\gd_n\ge0$
if and only if $x_nqr+y_nrp\le n$. Therefore, if
\be{3.8}
f(n)=f_{\gt,r}(n)=x_nq+y_np,
\end{equation}
then, by \refE{3.7},
\be{3.9}
\chi(n)=1\q\text{if and only if}\q f(n)\le\lf n/r\rf.
\end{equation}
This observation, first made in \cite{B2}, plays a key role in
our analysis. Note that $f(n)\equiv nr^*\pmod{pq}$, where $r^*$ is
the multiplicative inverse of $r$ modulo $pq$. Now let $[N]_{pq}$
denote the least nonnegative residue of $N$ modulo $pq$ and let
$\rep$ be the set of integers representable as a nonnegative linear
combination of $p$ and $q$, that is,
\[
\rep=\{\, N\mid N=xq+yp,\ x,y\ge0\,\}.
\]
Then, by \refE{3.6} and \refE{3.8}, we have
\be{3.10}
f(n)=\begin{cases} [nr^*]_{pq}, &\text{if $[nr^*]_{pq}\in\rep$}, \\
 [nr^*]_{pq}+pq, &\text{otherwise.}
\end{cases}\end{equation}
It is now evident that $f$ is determined by the residue class of
$r$ modulo $pq$. Before stating this formally, let us introduce the
convention of writing $f_r$ in place of $f_{\gt,r}$, as long as it
is understood that the parameters $p$ and $q$ are fixed. Similarly,
we will find it convenient to write $\chi_r$ in place of
$\chi_{\gt}$ under the same circumstances.

\begin{lemma}\label{L:4}
If $r\equiv s\pmod{pq}$ and $n_1\equiv n_2\pmod{pq}$ then
$f_r(n_1)=f_s(n_2)$.
\end{lemma}

\begin{proof}
Since $[n_1r^*]_{pq}=[n_2s^*]_{pq}$ the conclusion follows from
\refE{3.10}.
\end{proof}

In addition to assumptions of \refL{4} we need to impose certain
further restrictions in order to guarantee that
$\chi_r(n_1)=\chi_s(n_2)$.

\begin{lemma}\label{L:5}
Suppose that $\max(p,q)<r<s$ and that $r\equiv s\pmod{pq}$. Then
\be{l51}
\chi_r(kr+j)=\chi_s(ks+j),
\end{equation}
for all $k<pq$ and $|j|<r$. Moreover, if $|j|<\min(r,pq)$ then
we also have
\be{l52}
\chi_r(kr+j-r)=\chi_s(ks+j-r).
\end{equation}
\end{lemma}

\begin{proof}
The first claim follows from \refE{3.9} and \refL{4}. The second
claim with $j\ge0$ follows in exactly the same way (\refE{l51}
contains \refE{l52} for $j>0$). In the remaining case we have
\be{l53}
\Big\lfloor\ff{kr+j-r}r\Big\rfloor=k-2\q\text{and}\q
\Big\lfloor\ff{ks+j-r}s\Big\rfloor=k-1.
\end{equation}
Moreover, if $j\not\equiv0\pmod{pq}$ then
\be{l54}
f_r(kr+j-r)\ne k-1,
\end{equation}
since $f_r(kr+j-r)\equiv k-1+jr^*\pmod{pq}$. Combining
\refE{l53} and \refE{l54} with \refE{3.9} and \refL{4}
completes the proof of the lemma.
\end{proof}

We are now ready to consider functions $A^+(\gt)$ and $A^-(\gt)$.
In view of \refE{3.B}, these functions capture all the information
about the coefficients of $\qt$ as a set. It is plain from our
introductory discussion of the function $A(\gt)$ (in the form of
$A(t)$) that there are basic gaps in our understanding of these
functions. It is interesting to note that in contrast to this the
quantity $A^+(\gt)-A^-(\gt)$ is more transparent. Indeed,
it is known \cite{B1, B2} that the bound
\[
A^+(\gt)-A^-(\gt)\le p
\]
is valid for all $\gt$ and that it is sharp. Our present aim is
the identity \refE{3.C} for which we need to show that if $p$ and
$q$ are fixed then for $r>\max(p,q)$ functions $A^{\pm}(p,q,r)$
depend only on the residue class of $r$ modulo $pq$.

\begin{lemma}\label{L:6}
If $r,s>\max(p,q)$ and $r\equiv s\pmod{pq}$ then
\[
A^+(p,q,r)=A^+(p,q,s)\q\text{and}\q A^-(p,q,r)=A^-(p,q,s).
\]
\end{lemma}

\begin{proof}
Consider the function $A^+$. With $p$ and $q$ fixed, let us write
$A^+(r)$ for $A^+(\gt)$. It suffices to show that
if $r>\max(p,q)$ and $s=r+pq$ then
\be{l61}
A^+(r)=A^+(s).
\end{equation}

Throughout this argument we adopt the convention that $a_m$ and
$b_l$ denote coefficients of $Q_{\tr}$ and $Q_{\ts}$,
respectively. We show first that if $0\le\gb<r$ then
\be{l62}
a_{\ga r+\gb}=b_{\ga s+\gb}.
\end{equation}
Set $m=\ga r+\gb$ and $l=\ga s+\gb$. Observe that, by \refL{5},
we have
\be{l63}
\chi_r(m-j)=\chi_s(l-j),
\end{equation}
for all $0\le j<p+q$. Moreover, \refE{l63} also holds with $m$
and $l$ replaced by $m-r$ and $l-s$, respectively. Combining
this with \refE{3.5} establishes \refE{l62}.

Of course, the inequality $A^+(r)\le A^+(s)$ is an immediate
consequence of \refE{l62}. In fact, \refE{l62} implies \refE{l61}
for we will show that for some $\ga$ and $0\le\gb<r$ we have
\be{l64}
A^+(s)=b_{\ga s+\gb}.
\end{equation}
Thus it only remains to prove \refE{l64}.

Let $l_0$ be the smallest index for which $b_{l_0}=A^+(s)$.
Note that, by \refE{3.5}, $b_0=1$ and $b_q=-1$, so $l_0$ (as
well as the corresponding quantity for the function $A^-$) are
well defined -- see remarks following \refE{3.5}. Applying
\refE{l3} to $b_{l_0}$ shows that we must have
$l_0=\ga_0s+\gb_0$, with $0\le\gb_0<q+p$, and that $\ga_0s$
and $\ga_0s-s$ are in $R_{l_0}$ (given by \refE{3.K} with $r$
replaced by $s$). If, in fact, $\gb_0<r$ then we are done. So
suppose that $r\le\gb_0<q+p$. We claim that in this case \refE{l64}
holds with $\ga s+\gb=l_0+pq$. In the first place, we have
\[
l_0+pq=(\ga_0+1)s+\gb_0-r,\q 0\le\gb_0-r<r.
\]
Therefore, to prove \refE{l64} it remains to show that
\be{l65}
b_{l_0+pq}=b_{l_0}.
\end{equation}

Recall that $\ga_0s, \ga_0s-s\in R_{l_0}$ and observe that they
are the only multiples of $s$ in $R_{l_0}$. Moreover our
assumption on $\gb_0$ implies that
\begin{gather}
\ga_0s\in I_{l_0-q},\q\ga_0s-s\in I_{l_0-q-s},  \label{E:3.M} \\
\ga_0s+r\in I_{l_0},\q\text{and}\q\ga_0s-s+r\in I_{l_0-s}. \label{E:3.N}
\end{gather}
Therefore, by \refE{3.5}, \refE{3.12}, \refE{3.13}, and \refE{3.M},
we get
\be{l66}
b_{l_0-pq}=b_{l_0}-\chi_s(\ga_0s-s)+\chi_s(\ga_0s).
\end{equation}
Whence $\chi_s(\ga_0s)=0$ (and $\chi_s(\ga_0s-s)=1$, but we
will not need this). Also, by \refE{3.N}, $\ga_0s+s\in I_{l_0+pq}$
and $\ga_0s\in I_{l_0+pq-s}$, and they are the only multiples of
$s$ in $R_{l_0+pq}$. Therefore, reasoning as in \refE{l66},
we now get
\[\begin{aligned}
b_{l_0}&=b_{l_0+pq}+\chi_s(\ga_0s)-\chi_s(\ga_0s+s) \\
&=b_{l_0+pq}-\chi_s(\ga_0s+s).
\end{aligned}\]
This implies \refE{l65} and completes the proof in the case
of the function $A^+$.

Essentially identical argument works for the function $A^-$
and we omit the details.
\end{proof}

\begin{proof}[Proof of \refT{3}]
\refE{3.B} follows from \refL{2}.

The first conclusion in \refE{3.C} follows from \refL{6} and
\refE{3.B}.

Recall that Kaplan \cite{Ka} has proved \refE{3.C} for $r,s>pq$.
Using this we deduce the second conclusion in \refE{3.C} from
the first.

Since our method is different from that of \cite{Ka} it is of
interest to give a self-contained treatment for the case
$r\equiv-s\pmod{pq}$. We thus conclude this paper with a sketch of
our argument. We will show that if $a_m$ is an arbitrary coefficient
of $Q_{\tr}$ then there is a coefficient $b_l$ of $Q_{\ts}$
such that $b_l=-a_m$. To do this we make an additional assumption
that $r,s\ge p+q$; this is permissible in view of what we already
proved. By \refE{l3}, we may take $m=\ga r+\gb_1$, with
$0\le\gb_1<p+q$. Following Kaplan, we claim that
\[
a_{\ga r+\gb_1}=-b_{\ga s+\gb_2},\q\text{with}\ \gb_2=p+q+1-\gb_1.
\]
To this end we observe that $[(kr+j)r^*]_{pq}=[(ks-j)s^*]_{pq}$,
so that
\[
f_r(kr+j)=f_s(ks-j).
\]
This is the present case equivalent of \refL{4}. From this we
deduce that if $|j|<p+q$ then
\be{3.O}
\chi_r(kr+j)=\chi_s(ks-j),
\end{equation}
the equivalent of \refE{l51} in \refL{5}. Now apply \refE{3.O} to
the representations of $a_{\ga r+\gb_1}$ and $b_{\ga s+\gb_2}$
given by \refE{3.5}. The proof is completed on observing that the
left and the right sides of \refE{3.O} contribute with the opposite
signs to the values of $a_{\ga r+\gb_1}$ and $b_{\ga s+\gb_2}$,
respectively.
\end{proof}

config: arxiv


text:
\begin{document}

\maketitle

\noindent{\bf Abstract.} \emph{We compare the compactness of composition
operators on $H^2$ and on Orlicz-Hardy spaces $H^\Psi$. We show in particular that exists an Orlicz function
$\Psi$ such that $H^{3+\eps} \subseteq H^\Psi \subseteq H^3$ for every $\eps >0$, and a composition
operator $C_\phi$ which is compact on $H^3$ and on $H^{3+\eps}$, but not compact on $H^\Psi$.}
\medskip

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

\noindent{\bf Key-words.}  Carleson function -- Carleson measure -- composition operator -- Hardy spaces --
Hardy-Orlicz spaces -- Schatten classes

\section{Introduction}

Composition operators on the Hardy space $H^2$ have given rise to a lot of papers since the beginning of
the seventies (see \cite{Shap-livre} and \cite{Co-McC}, and references therein, for an overview until the middle
of the eighties). In particular, criteria of compactness (\cite{Shap}, \cite{McCluer}) or of membership in
Schatten classes (\cite{Lue} \cite{Lue-Zhu}) were found. But only few concrete examples of such operators
were known (see \cite{Shap-Tay}, \cite{Carroll-Cowen}, \cite{Jones}, \cite{Zhu}). In \cite{JFA}, we constructed
explicit examples of compact composition operators $\C_\phi \colon H^2 \to H^2$ which are, or not, in some
Schatten classes $S_p$. On the other hand, we study in \cite{CompOrli} composition operators on Hardy-Orlicz
spaces $H^\Psi$, and characterize their compactness. In this paper, we shall continue our investigation of
composition operators on Hardy-Orlicz spaces, and study for which Orlicz functions $\Psi$ the explicit
examples of \cite{JFA} are compact.
\medskip

Let us describe more precisely the content of this paper.\par
In Section~\ref{Delta indice deux} and Section~\ref{Delta super deux}, we consider the Hardy-Orlicz
space $H^\Psi$ besides the Hardy space $H^2$, and we prove that for every Orlicz function $\Psi$ which does
not satisfy the condition $\Delta_2$, there exists a symbol $\phi$ such that $C_\phi$ is compact on $H^2$,
but not compact on $H^\Psi$ (Theorem~\ref{non psi-compact}). When the Orlicz function $\Psi$ grows
fast enough, we can ensure, beyond the non compactness of $C_\phi$ on $H^\Psi$, that, for every $p > 2$,
$C_\phi \colon H^2 \to H^2$ is moreover in the Schatten
class $S_p$ (Proposition~\ref{croissance Psi}). But if $\Psi$ does not grow too fast, this is not possible: if
$C_\phi \in S_p$ for some $p > 0$, we must have the compactness of $C_\phi$ on $H^\Psi$
(Proposition~\ref{Proposition Psi pas tres croissante}). When $\Psi$ grows very fast, \emph{i.e.} satisfies the
condition $\Delta^2$, we show that $C_\phi$ may be in $S_p$ for every $p > 0$, although $C_\phi$ is
not compact on $H^\Psi$ (Proposition~\ref{proposition Schatten et Delta super deux}); conversely, if
$C_\phi$ is compact on $H^\Psi$, then $C_\phi$ must be in $S_p$ for every $p > 0$
(Proposition~\ref{Delta super deux et Carleson}). \par
In Section~\ref{phi indice theta dans H Psi}, we then characterize (Theorem~\ref{Psi delta 1}) the Orlicz
functions $\Psi$ for which the composition operator $C_{\phi_\theta}$, which appeared in
\cite{JFA}, Section~5, is compact on $H^\Psi$. When $\Psi$ grows fast enough (in
particular if $\Psi$ satisfies the so-called condition $\Delta^1$), then $C_{\phi_\theta}$ is not compact on
$H^\Psi$ (Theorem~\ref{meilleure condition sur Psi}); this gives, in particular, an improvement of
Proposition~\ref{croissance Psi}. This characterization allows us to get the following striking
result (Theorem~\ref{pas d'interpolation}): there exists an Orlicz function $\Psi$ such that
$H^{3+\eps} \subseteq H^\Psi \subseteq H^3$ for every $\eps >0$, and a composition operator $C_\phi$
which is compact on $H^3$ and on $H^{3+\eps}$, but not compact on $H^\Psi$.\par
\bigskip

\section{Notation}

We shall denote by $\D$ the open unit disk 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 continuously every Hardy space $H^p$ ($p >0$) into itself, as
well as (\cite{CompOrli}, Proposition~3.12) the Hardy-Orlicz spaces $H^\Psi$ (see below for their
definition).\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 pullback
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
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
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 $H^2$). Recall (\cite{CompOrli}, Theorem~4.3) that, for every Orlicz
function $\Psi$, as defined below,  $C_\phi \colon H^2 \to H^2$ is compact whenever
$C_\phi \colon H^\Psi \to H^\Psi$ is.\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 $0$.
\bigskip

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

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
$W (\xi, h)$, 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.
\medskip

Let us now recall the definition of the Hardy-Orlicz spaces.\par
An Orlicz function is a non-decreasing convex function $\Psi\colon [0, \infty] \to [0,\infty]$ such that
$\Psi(0) = 0$ and $\Psi(\infty) =\infty$. To avoid pathologies, we assume that the Orlicz function $\Psi$
has the following additional properties: $\Psi$ is continuous at $0$, strictly convex (hence increasing), and
such that
\begin{displaymath}
\frac{\Psi(x)}{x}\mathop{\longrightarrow}_{x\to \infty} \infty.
\end{displaymath}
This is essentially to exclude the case of $\Psi(x) = ax$.\par
The Orlicz space $L^\Psi (\T)$ is the space of all (equivalence classes of) measurable functions
$f\colon \T\to\C$ for which there is a constant $C > 0$ such that:
\begin{displaymath}
\int_\T \Psi\bigg(\frac{\vert f (t)\vert} {C}\bigg)\,dm(t) < +\infty
\end{displaymath}
and then $\Vert f\Vert_\Psi$, the \emph{Luxemburg norm}, is the infimum of all possible
constants $C$ such that this integral is $\leq 1$. The Hardy-Orlicz space $H^\Psi$ is the space of all
$f \in H^1$ such that the boundary values function $f^\ast$ of $f$ is in $L^\Psi (\T)$. We refer to
\cite{CompOrli}, Section~3  for more details.\par

\section{Orlicz functions without condition $\Delta_2$} \label{Delta indice deux}

In \cite{CompOrli}, Corollary~3.26, we showed that for every Orlicz function $\Psi$ which satisfies the
growth condition $\Delta^2$, there exists a symbol $\phi\colon\D \to \D$ which induces a compact
composition on $H^2$, but a non-compact composition operator on $H^\Psi$. We shall generalize this
below. Recall that an Orlicz function $\Psi$ satisfies the $\Delta_2$ condition if there exists a constant
$C > 0$ such that $\Psi (2x) \leq C \Psi (x)$ for $x$ large enough.

\begin{theoreme}\label{non psi-compact}
For every Orlicz function $\Psi$ which does not satisfy the condition $\Delta_2$, there exists an analytic
self-map $\phi \colon \D \to \D$ such that $C_\phi$ is compact on $H^2$ but not compact on $H^\Psi$.
\end{theoreme}

\noindent{\bf Proof.} $\Psi \notin \Delta_2$ means that
$\limsup\limits_{x\to \infty} \frac{\Psi (2x)}{\Psi (x)} =+\infty$. We can hence find a sequence $(x_n)$,
which increases to $+\infty$, such that $\frac{\Psi (2x_n)}{\Psi (x_n)}$ increases to $+\infty$.\par
Set $h_n= \frac{1}{\Psi(x_n)}$ and $c_n = \frac{\Psi (x_n)}{\Psi (2x_n)}\,$. By construction, $(h_n)_n$
and $(c_n)_n$ decrease to $0$. We may assume that $c_n < \pi$, $n\geq 1$.\par
We are going to construct an analytic function $\phi\colon \D \to \D$ such that:
\begin{equation}\label{condition petit o}
\rho_\phi (h) = o\, (h),
\end{equation}
but:
\begin{equation}\label{pas R0}
\hskip 2cm \rho_\phi (h_n) \geq c_n h_n\,,\qquad  \text{for all } n\geq 1.
\end{equation}
Condition \eqref{condition petit o} will ensure that $C_\phi\colon H^2 \to H^2$ is compact, by MacCluer's
theorem, and condition \eqref{pas R0}, which reads:
\begin{displaymath}
\hskip 2cm \rho_\phi (h_n) \geq \frac{1}{\Psi \big( 2 \Psi^{-1} (1/h_n)\big)}\,\raise 1,5pt \hbox{,}
\qquad  \text{for all } n\geq 1,
\end{displaymath}
will ensure (by \cite{CompOrli}, Theorem~4.11), that $C_\phi\colon H^\Psi \to H^\Psi$ is not compact.\par
\medskip

For that purpose, we shall use the ``general construction'' made in \cite{JFA}, \S~3.2. Let us recall this
construction.\par
Let
\begin{equation}\label{forme de f}
f (t) = \sum_{k=0}^\infty a_k \cos (kt)
\end{equation}
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}
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}
and 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}
$P_z$ being the Poisson kernel at $z$,
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 condition $f (0) = 0$ means that $\Phi^\ast (1) = 1$,
so that $\| \Phi \|_\infty =1$.\par
We then 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}\label{lemme 3.6}
We proved in \cite{JFA}, Lemma~3.6, that
\begin{equation}
h f^{-1} (h) \lesssim \rho_\phi (h) \lesssim h f^{-1} (2h)
\end{equation}
when $h$ tends to $0$.
\medskip

We shall now rely on the following simple lemma, whose proof is postponed.

\begin{lemme}\label{lemme construction fonction}
Let $(h_n)$ and $(c_n)$ be two decreasing vanishing sequences of positive numbers, with
$0< h_n< \pi$, $0 < c_n < \pi$.\par
There exists an even $2\pi$-periodic ${\cal C}^2$ function $f \colon \R \to \R$ such that: \par
1) $f (0) = 0$ and $f$ is strictly increasing on $[0,\pi]$;\par
2) $f (c_n ) \leq h_n$, for all $n \geq 1$.
\end{lemme}

We are in the situation of the general construction, and by\eqref{lemme 3.6}, we have:
\begin{displaymath}
\rho_\phi (h) = O \big( h f^{-1} (2h) \big) = o\, (h),
\end{displaymath}
since $f^{-1}(2h) \to 0$ as $h \mathop{\to}\limits^{\scriptscriptstyle >} 0$. MacCluer's criterion implies that
$C_\phi \colon H^2 \to H^2$ is compact. On the other hand, \eqref{lemme 3.6}  again implies that
\begin{displaymath}
\rho_\phi (h_n ) \gtrsim h_n f^{-1} (h_n) \gtrsim c_n h_n,
\end{displaymath}
by Lemma~\ref{lemme construction fonction}.\par
Therefore \eqref{pas R0} holds, and this ends the proof of Theorem~\ref{non psi-compact}. \qed
\medskip

\noindent{\bf Remark.} Actually, we do not use that $f$ is ${\cal C}^2$, but only that it is continous. However,
if $f\in {\cal C}^2$, then $\widehat{f ''} \in \ell_2 (\Z)$ and it follows, by the Cauchy-Schwarz inequality,
since $\widehat{f ''}(k) = (ik)^2 \hat f (k)$, that $\sum_{k=0}^\infty k\,|\hat f (k)| < +\infty$; hence
both $f$ and ${\cal H}f$ are ${\cal C}^1$ functions, and we proved in \cite{JFA}, Lemma~3.5,
that 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, although $|\Phi^\ast| = |\phi^\ast|$
on $\partial \D$ and $C_\phi \colon H^2 \to H^2$ is compact.
\bigskip

\noindent{\bf Proof of Lemma~\ref{lemme construction fonction}.} Let $h_0 = c_0 =\pi$, and
$\varphi \colon [0, \pi] \to \R_+$ be a step function such that:
\begin{equation}\label{step function}
\int_0^{c_n} \varphi (t)\,dt = h_n\,, \qquad n=1,2,\ldots
\end{equation}
We can take $\varphi (t) = \frac{h_j - h_{j+1}}{c_j - c_{j+1}}$ for $ t \in (c_{j+1}, c_j)$, $j= 0,1, \ldots$, since
then:
\begin{align*}
\int_0^{c_n} \varphi (t)\,dt
& = \sum_{j=n}^\infty \int_{c_{j+1}}^{c_j} \varphi (t)\,dt =
 \sum_{j=n}^\infty (c_j - c_{j+1})  \frac{h_j - h_{j+1}}{c_j - c_{j+1}} \\
& = \sum_{j=n}^\infty (h_j - h_{j+1}) = h_n.
\end{align*}
Let now $f \colon \R \to \R$ be the even $2\pi$-periodic function whose values on $[0,\pi]$ are given by:
\begin{displaymath}
f (t) = \frac{1}{\pi^3} \int_0^t (t - u)^3 \varphi (u)\,du.
\end{displaymath}
This function is clearly ${\cal C}^2$, with
\begin{displaymath}
f '' (t) = \frac{6}{\pi^3} \int_0^t (t -u) \varphi (u)\,du,
\end{displaymath}
and $ f( 0) = f ' (0) = f '' (0) =0$; it is strictly increasing on $[0, \pi]$, and
\begin{displaymath}
f (c_n) = \frac{1}{\pi^3} \int_0^{c_n} (c_n - u)^3 \varphi (u)\,du
\leq \frac{c_n^3}{\pi^3} \int_0^{c_n} \varphi (u)\,du \leq \int_0^{c_n} \varphi (u)\,du = h_n,
\end{displaymath}
due to \eqref{step function}. Therefore $f$ fulfills all the requirements of
Lemma~\ref{lemme construction fonction}. \qed
\bigskip

\noindent{\bf Remark.} It is not clear whether one can ensure in Theorem~\ref{non psi-compact} that
the composition operator is moreover in some Schatten class $S_p$ for $p < \infty$. It would be the case, by
\cite{JFA}, Corollary~3.2, if one can construct a symbol $\phi$ such that its pull-back measure $m_\phi$
is $\alpha$-Carleson for some $\alpha > 1$. This can be obtained if we can construct, in
Lemma~\ref{lemme construction fonction}, our function $f$ such that $f^{(N)} (0) \not= 0$ for some
$N > 1$, since then $f (h) \approx h^N$ and $f^{-1}(h) \approx h^{1/N}$, and $\phi$ would be
$\alpha$-Carleson with $\alpha = 1 + \frac{1}{N}$, which would imply that $C_\phi \in S_p$ for
$p > 2N$. However, in general, we cannot obtain that $f^{(N)} (0) \not= 0$ for some $N > 1$. Indeed,
take $\Psi (x) = \e^{(\log (x+1) )^2} - 1$; then
$h_n = \frac{1}{\Psi (x_n)} \approx \exp \big[- (\log x_n)^2\big]$, whereas, since
\begin{displaymath}
\Psi (2x) \approx \exp\big[ (\log x)^2\big]\,\exp (2\log x),
\end{displaymath}
one has $c_n =\frac{\Psi (x_n)}{\Psi (2 x_n)} \approx \exp (- 2 \log x_n)$, so that
$h_n = o\,(c_n^N)$ for every $N < \infty$. Since $f (c_n) \leq h_n$, all the possible derivatives of $f$ at $0$
must be zero.\par
\bigskip

Besides, when $\Psi$ does not grow too fast, one has the following result.

\begin{proposition}\label{Proposition Psi pas tres croissante}
Let $\Psi$ be an Orlicz function such that, for every $A > 1$, one has, as $x$ goes to infinity:
\begin{equation}\label{autre condition sur Psi}
\Psi (Ax) = O \big( \Psi (x) \big(\log \Psi (x) \big)^\eps \big)\,, \qquad \forall \eps >0.
\end{equation}
Then, if  $C_\phi \in S_p$ for some $p > 0$, the composition operator $C_\phi \colon H^\Psi \to H^\Psi$ is
compact.
\end{proposition}

\noindent{\bf Remark.} If the condition~\eqref{autre condition sur Psi} is satisfied for some $A >1$, it is
actually satisfied for every $A > 1$. In fact, $\Psi (Ax) = O \big( \Psi (x) \big(\log \Psi (x) \big)^\eps \big)$
implies that $\Psi (A^2 x) = O \big( \Psi (x) \big(\log \Psi (x) \big)^{2 \eps} \big)$, and we can replace $A$
by $A^2$, $A^4$, $A^8$, \dots Note also that we can replace the \emph{big-oh} assumption by a
\emph{little-oh} one: if the proposition holds with a \emph{little-oh} condition, it also holds with a
\emph{big-oh} one, because, for every $\eps' > 0$, if one has \eqref{autre condition sur Psi}, then, with
$\eps < \eps'$:
\begin{displaymath}
\frac{\Psi (A x)}{\Psi (x) \big(\log \Psi (x) \big)^{\eps'}} \leq K_\eps\,\big( \log \Psi (x) \big)^{\eps - \eps'}
\mathop{\longrightarrow}_{x \to \infty} 0.
\end{displaymath}
\medskip

\noindent{\bf Proof.} If $C_\phi \in S_p$, then, by \cite{JFA}, Proposition~3.4, one has:
\begin{displaymath}
\rho_\phi (h) = o\,\bigg( \frac{h}{[\log (1/h)]^\delta}\bigg)\,,
\end{displaymath}
with $\delta =2/p > 0$. Then, \eqref{autre condition sur Psi} with $x = \Psi^{-1} (1/h)$, gives:
\begin{displaymath}
\rho_\phi (h) = o\,\bigg( \frac{1}{\Psi (x) [\log \Psi (x) ]^\delta}\bigg)
= o\,\bigg( \frac{1}{\Psi (Ax)}\bigg) = o\,\bigg( \frac{1}{\Psi \big( A \Psi^{-1} (1/h)\big) }\bigg)\,,
\end{displaymath}
which implies, by \cite{CompOrli}, Theorem~4.18, that $C_\phi$ is compact on $H^\Psi$. \qed
\bigskip

For example, we have  \eqref{autre condition sur Psi} if, for $x$ large enough:
\begin{displaymath}
\Psi (x) = \exp\big( \log x \,\log \log \log x\big).
\end{displaymath}
In fact, one has:
\begin{displaymath}
\frac{\Psi (Ax)}{\Psi (x)} =
\exp \bigg[ (\log x)\, \bigg( \log \frac{\log (\log x + \log A)}{\log \log x} \bigg) \bigg]  \,
\exp \big( \log A\, \log\log\log (Ax) \big),
\end{displaymath}
and $\exp \big( \log A\, \log\log\log (Ax) \big) = \big( \log \log (Ax) \big)^{\log A}$, whereas
\begin{displaymath}
\log (\log x +\log A)
= \log \log x + \log \Big( 1 + \frac{\log A}{\log x} \Big)
\leq \log \log x + \frac{\log A}{\log x} \, \raise 1,5 pt \hbox{,}
\end{displaymath}
so:
\begin{displaymath}
\log \frac{\log (\log x + \log A)}{\log \log x}
\leq \log \Big( 1 + \frac{\log A}{\log x \log \log x} \Big)
\leq \frac{\log A}{\log x \log \log x} \,\cdot
\end{displaymath}
Hence:
\begin{displaymath}
\frac{\Psi (Ax)}{\Psi (x)} \leq \exp \Big( \frac{\log A}{\log \log x} \Big)\,\big( \log \log (Ax) \big)^{\log A}
\lesssim \big( \log \log (Ax) \big)^{\log A} ,
\end{displaymath}
which gives \eqref{autre condition sur Psi} since, for every $\eps > 0$:
\begin{displaymath}
\big( \log \log (Ax) \big)^{\log A} = o\,\big( (\log x)^\eps \big) = o\,\big( [\log \Psi (x) ]^\eps \big).
\end{displaymath}
\medskip

On the other hand, if we require that the Orlicz function $\Psi$ grows sufficiently, we have:

\begin{proposition}\label{croissance Psi}
Assume that the Orlicz function $\Psi$ satisfies the condition:
\begin{equation}\label{condition croissance Psi}
\limsup_{x \to \infty} \frac{\Psi (Ax)}{[\Psi (x)]^2} > 0,
\end{equation}
for some $A > 1$.\par
Then, for every $p > 2$, there exists an analytic self-map $\phi \colon \D \to \D$ such that
$C_\phi \colon H^2 \to H^2$ is in the Schatten class $S_p$, but $C_\phi \colon H^\Psi \to H^\Psi$ is not
compact.
\end{proposition}

Note that condition~\eqref{condition croissance Psi} if not satisfied when $\Psi \in \Delta_2$, or for the
Orlicz function $\Psi (x) = \e^{(\log (x+1) )^2} - 1$, \emph{i.e.}
$\Psi (x) \approx \exp \big( (\log x)^2 \big)$.\par
\medskip

\noindent{\bf Proof.} Condition~\eqref{condition croissance Psi} implies that there exist some $\delta > 0$ and
a sequence of positive numbers $x_n$ increasing to infinity such that:
\begin{displaymath}
\qquad \qquad \frac{\Psi (A x_n)}{[\Psi (x_n)]^2} \geq \delta, \qquad \text{for all } n \geq 1.
\end{displaymath}
If one sets $h_n = \frac{1}{\Psi (x_n)}\,$\raise 1,5pt \hbox{,} one has:
\begin{equation}\label{majoration par h carre}
h_n^2 \geq \frac{\delta}{\Psi \big[ A \Psi^{-1}( 1/h_n)\big]}\,\cdot
\end{equation}
Now, take $\alpha$ such that $\frac{2}{p} + 1 < \alpha < 2$, and consider the symbol $\phi =\phi_2$
constructed in \cite{JFA}, Theorem~4.1. Its composition operator
$C_\phi \colon H^2 \to H^2$ belongs to $S_p$ (\cite{JFA},  Theorem~4.2). On the
other hand, $C_\phi \colon H^\Psi \to H^\Psi$ is not compact, by \cite{CompOrli}, Theorem~4.18, since
$\rho_\phi (h) \approx h^\alpha$ and hence $\rho_\phi (h_n)$ is not
$\displaystyle o\,\bigg(\frac{1}{\Psi \big[ A \Psi^{-1}( 1/h_n)\big]}\bigg)$,
by \eqref{majoration par h carre}. \hfill \vbox{}\qed
\bigskip

Condition~\eqref{condition croissance Psi} is a weaker growth condition than condition $\Delta^2$. Recall
(see \cite{CompOrli}, \S~2.1) that the Orlicz function $\Psi$ satisfies the condition $\Delta^2$ if, for some
$A > 1$, one has $[\Psi (x)]^2 \leq \Psi (Ax)$ for $x$ large enough. But condition $\Delta^2$ can be re-stated
as:
\begin{equation}\label{Delta super 2 bis}
\liminf_{x \to \infty} \frac{\Psi (Ax)}{[\Psi (x)]^2} > 0.
\end{equation}
\par

Indeed, if  one has $[\Psi (x)]^2 \leq \Psi (Ax)$ for $x$ large enough, one obviously has
$\liminf_{x \to \infty} \frac{\Psi (Ax)}{[\Psi (x)]^2} \geq 1$. Conversely, let
$\delta = \liminf_{x \to \infty} \frac{\Psi (Ax)}{[\Psi (x)]^2}\,$. If $\delta > 1$, one clearly have
$[\Psi (x)]^2 \leq \Psi (Ax)$ for $x$ large enough. If $0 < \delta \leq 1$, then, for $x$ large enough:
\begin{displaymath}
[\Psi (x)]^2 \leq \frac{2}{\delta} \Psi (Ax) \leq \Psi \Big(\frac{2}{\delta} Ax\Big)\,,
\end{displaymath}
by the convexity of $\Psi$, since $2/\delta \geq 1$. Hence $\Psi \in \Delta^2$ (with constant $2A /\delta$).
\par

\section{Orlicz functions with condition $\Delta^2$}\label{Delta super deux}

When $\Psi \in \Delta^2$, we have proved in \cite{CompOrli}, Corollary~3.26 a better result than the one in
Proposition~\ref{croissance Psi}: \emph{Whenever $\Psi \in \Delta^2$, there exists a symbol
$\phi \colon \D \to \D$ such that the composition operator $C_\phi \colon H^2 \to H^2$ is Hilbert-Schmidt,
whereas $C_\phi \colon H^\Psi \to H^\Psi$ is not compact}.
\medskip

We have this improvement:

\begin{proposition}\label{proposition Schatten et Delta super deux}
Let $\Psi$ be an Orlicz function satisfying condition $\Delta^2$. Then, there exists a symbol $\phi$ such
that $C_\phi \colon H^2 \to H^2$ belongs to $S_p$ for every $p > 0$, but such that
$C_\phi \colon H^\Psi \to H^\Psi$ is not compact.
\end{proposition}

\noindent{\bf Proof.} We shall use a \emph{lens map}. One can find a definition of such maps in
\cite{Shap-livre}, page~27, but we shall use a slight variant of it: we shall use the same construction as in
\cite{JFA}, Theorem~5.1 and Theorem~5.6, but with $f (z) = z^{1/2}$ instead of
$f_\theta (z) = z (- \log z)^\theta$ or $f (z) = z \log ( - \log z)$. We shall recall this construction at the
beginning of the next section. Note that
$\Re (r\,\e^{i\alpha})^{1/2} = \sqrt{r}  \,\cos (\alpha /2)$ is positive ($| \alpha | < \pi /2$), so
the associate symbol $\phi$ maps $\D$ into itself. \par
As in the proof of \cite{JFA}, Theorem~5.1, we need only to consider $f (it)$, with $t > 0$. But
$f (it ) = \exp \big(\frac{1}{2} \log t + i \frac{\pi}{4} \big)$, so:
\begin{displaymath}
\Re f (it) = \sqrt{t}\cos (\pi /4) = \sqrt{t/2} \qquad \text{and} \qquad
\Im f (it) = \sqrt{t}\sin (\pi/4) = \sqrt{t/2}.
\end{displaymath}
\par

Let us estimate $\rho_\phi (h)$. The condition $|\Re f (it)| \leq h$ implies $t \in [ -2h^2, 2h^2]$, and,
just using the modulus constraint, we get $\rho_\phi (h) \lesssim h^2$. For the converse, if
$t \in [ -2h^2, 2h^2]$, we have $|\Re f (it)| \leq h$ and $|\Im f (it) | \leq h$; then
$\exp \big( f (it) \big) $ belongs to a window centered at $1$ and with size $Ch$. Hence
$h^2 \lesssim \rho_\phi (h)$, and therefore $\rho_\phi (h) \approx h^2$.\par
It follows that $C_\phi \colon H^\Psi \to H^\Psi$ is not compact if $\Psi \in \Delta^2$. In fact, since
$\Psi \in \Delta^2$, there is some $A_0 > 1$ such that $\Psi (A_0 x) \geq [\Psi (x)]^2$ for
$x$ large enough. We get, with $x = \Psi^{-1} (1/h)$:
\begin{displaymath}
\frac{1}{\Psi \big(A_0 \Psi^{-1} (1/h) \big)}  \leq h^2.
\end{displaymath}
If $C_\phi$ were compact on $H^\Psi$, we should have (\cite{CompOrli}, Theorem~4.11):
\begin{displaymath}
\rho_\phi (h) = o \,\bigg( \frac{1}{\Psi \big(A_0 \Psi^{-1} (1/h) \big)} \bigg)\,,
\end{displaymath}
which is not the case.\par
\smallskip

On the other hand, $C_\phi$ is in $S_p$ for every $p > 0$. Indeed, $\phi (\e^{it})$ is in the
dyadic Carleson window:
\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}
($j=0, 1, \ldots, 2^n - 1$, $n=1,2,\ldots$) if and only if, for $h_n =2^{-n}$, one has, up to constants:
\begin{displaymath}
t \leq 2h_n^2 \qquad \text{and} \qquad j^2 h_n^2 \leq t \leq (j + 1)^2 h_n^2.
\end{displaymath}
This is possible for at most a fixed number, say $N$, of values of $j$; hence:
\begin{align*}
\sum_{n=1}^\infty \sum_{j=0}^{2^n -1} 2^{n p/2} m_\phi ( W_{n,j} )^{p/2}
& \leq N \sum_{n=1}^\infty 2^{n p/2} \big( \rho_\phi (2^{-n}) \big)^{p/2} \\
& \leq N \sum_{n=1}^\infty 2^{n p/2} 2^{-2n p/2} = N \sum_{n=1}^\infty 2^{- n p/2} < +\infty.
\end{align*}
Then Luecking's theorem (\cite{Lue}) and \cite{JFA}, Proposition~3.3 imply that $C_\phi \in S_p$. \qed
\bigskip

Now, in the opposite direction, one has proved in \cite{CompOrli}, Theorem~3.24, that if $C_\phi$ is
compact on $H^\Psi$, with $\Psi \in \Delta^2$, then $C_\phi \in S_p (H^2)$ for every $p > 0$ (though we
only stated that $C_\phi \in S_1$). We have the following improvement.

\begin{proposition}\label{Delta super deux et Carleson}
Let $\Psi$ be an Orlicz function satisfying condition $\Delta^2$. Assume that $C_\phi$ is compact on
$H^\Psi$. Then $m_\phi$ is an $\alpha$-Carleson measure, for every $\alpha \geq 1$, and hence
$C_\phi \in S_p$ for every $p > 0$.
\end{proposition}

\noindent{\bf Proof.} Since $\Psi \in \Delta^2$, there is some $A_0 > 1$ such that
$\Psi (A_0 x) \geq [\Psi (x)]^2$, for $x$ large enough. Hence, for every positive integer $n$, one has
$\Psi (A_0^n x) \geq [\Psi (x)]^{2^n}$. Taking $x = \Psi^{-1} (1/h)$\,, one gets:
\begin{displaymath}
\Psi \big( A_0^n \Psi^{-1} (1/h) \big) \geq \frac{1}{h^{2^n}}\,\cdot
\end{displaymath}
Now, if $C_\phi$ is compact on $H^\Psi$, one has, by \cite{CompOrli}, Theorem~4.11,~1), with $A = A_0^n$:
\begin{displaymath}
\rho_\phi (h) \leq \frac{1}{\Psi \big( A_0^n \Psi^{-1} (1/h) \big)}\,;
\end{displaymath}
therefore $\rho_\phi (h) \leq h^{2^n}$, and $m_\phi$ is a $2^n$-Carleson measure.\par
The last assertion now follows from \cite{JFA}, Corollary~3.2. \qed

\section{Case of the symbols $\phi_\theta$} \label{phi indice theta dans H Psi}

In \cite{JFA}, Section~5, we considered composition operators $C_{\phi_\theta}$ which are, for
every $\theta > 0$, compact on $H^2$, and hence on $H^p$ for every $p < \infty$. We are now going to
examine when these composition operators are compact on $H^\Psi$.\par
\smallskip

Let us recall how the $\phi_\theta$'s are constructed.\par
For $\Re z > 0$, consider the principal determination of the logarithm $\log z$. For $\eps >0$ small enough,
the function $f_\theta (z) = z (- \log z )^\theta$ has strictly positive real part on
$V_\eps = \{ z\in \C\,;\ \Re z >0 \ \text{and}\ |z | < \eps\}$ and
$\Re f_\theta^\ast ( z) > 0$ for all $z \in \partial V_\eps \setminus\{0\}$. Now, if
$g_\theta$ is the conformal mapping from $\D$ onto $V_\eps$, which maps $\T = \partial \D$
onto $\partial V_\eps$, and with $g_\theta (1) = 0$ and $g'_\theta (1) = -\eps/4$, we set:
\begin{displaymath}
\phi_\theta = \exp ( - f_\theta \circ g_\theta).
\end{displaymath}
$\phi_\theta$ maps $\D$ into itself and $|\phi_\theta^\ast| < 1$ on $\partial \D \setminus\{1\}$, and we
proved (\cite{JFA}, Proposition~5.3) that:
\begin{equation}\label{rho pour phi indice theta}
\rho_{\phi_\theta} (h) \approx \frac{h}{(\log 1/h )^\theta}\cdot
\end{equation}
\bigskip

Now, recall (\cite{CompOrli}, Theorem~4.18) that $C_{\phi_\theta}$ is compact on $H^\Psi$ if and
only if, for every $A > 0$:
\begin{displaymath}
\rho_{\phi_\theta} (h) = o\bigg( \frac{1}{\Psi [ A \Psi^{-1} (1/h)]}\bigg)\,, \qquad h \to 0.
\end{displaymath}
By \eqref{rho pour phi indice theta}, that gives, with $u = \Psi^{-1}(1/h)$:

\begin{theoreme}\label{Psi delta 1}
The composition operator $C_{\phi_\theta}$ is compact on $H^\Psi$ if and only if for every $A > 1$, one has:
\begin{equation}\label{compacite phi indice theta sur H Psi}
\qquad \qquad \Psi (Au) = o\,\big( \Psi (u) [\log \Psi (u)]^\theta\big)\,,\qquad \text{as}\ u \to \infty.
\end{equation}
\end{theoreme}

For example, for every $\theta > 0$, $C_{\phi_\theta}$ is compact  on $H^\Psi$ if
$\Psi (x) \approx x^{\log \log \log x} = \e^{(\log x)(\log \log \log x)}\!$, but is not compact on
$H^\Psi$ if $\Psi (x) \approx x^{\log \log x} = \e^{(\log x)(\log \log x)}$.
\par\smallskip

Remark that this condition~\eqref{compacite phi indice theta sur H Psi} is almost the same as
condition~\eqref{autre condition sur Psi}: in \eqref{compacite phi indice theta sur H Psi} one is requiring the
condition to be satisfied for some $\theta > 0$, whereas in \eqref{autre condition sur Psi} one is
requiring it to be satisfied for every $\eps > 0$ (see however the remark at the end of this Section). The
\emph{little-oh} assumption in \eqref{compacite phi indice theta sur H Psi} is actually not very important:
suppose that, for every $ A > 1$, one has $\Psi (Au)  \leq C\,\Psi (u) [\log \Psi (u)]^\theta$ for $u \geq u_A$
(we need a uniform constant $C > 0$), then, for every $ A > 1$, the convexity of $\Psi$ gives, for every
$\eps > 0$,  $\Psi (A u) \leq \Psi (A_\eps u)$, with $A_\eps = A/\eps$; hence we get
$\Psi (Au) \leq C\,\eps\, \Psi (u) [\log \Psi (u)]^\theta$, and the \emph{little-oh} property follows.
\par
\medskip

Note that, since $C_{\phi_\theta}$ is in $S_p$ for $p > 4/\theta$ (\cite{JFA}, Proposition~5.3),
Theorem~\ref{Psi delta 1} is an improvement of Proposition~\ref{croissance Psi}, since
\eqref{compacite phi indice theta sur H Psi} is not satisfied if and only if there exists some $A >1 $ such that
$\limsup_{x \to \infty} \frac{\Psi (Ax)}{\Psi (x) [\log \Psi (x)]^\theta} > 0$, which is of course implied
by condition \eqref{condition croissance Psi}:

\begin{theoreme}\label{meilleure condition sur Psi}
Let $p > 0$ and $\theta > 4/ p$. Assume that the Orlicz function $\Psi$ satisfies, for some $A > 1$, the
condition:
\begin{equation}\label{Delta 1, theta}
\limsup_{x \to \infty} \frac{\Psi (Ax)}{\Psi (x) [\log \Psi (x)]^\theta} > 0.
\end{equation}
Then the composition operator $C_{\phi_\theta}$, which is in $S_p (H^2)$, is not compact on $H^\Psi$.
\par
If the Orlicz function $\Psi$ satisfies condition $\Delta^1$, then condition~\eqref{Delta 1, theta} is
satisfied.
\end{theoreme}

Remark that, for $\theta > 2$, Theorem~\ref{meilleure condition sur Psi} improves
Corollary~3.26 of \cite{CompOrli}. Recall that the Orlicz function $\Psi$ satisfies the condition
$\Delta^1$ if there is a constant $A > 1$ such that $x \Psi (x) \leq \Psi (A x)$ for $x$ large enough
(see \cite{CompOrli}, \S~2.1). \par
\medskip

\noindent{\bf Proof of Theorem~\ref{meilleure condition sur Psi}.} The first part was explained before
the statement. For the second part, assume that, for every $A > 1$,
condition~\eqref{compacite phi indice theta sur H Psi} is satisfied; in particular, we should have, for some
$u_0 > 0$:
\begin{equation}\label{inegalite Psi}
\qquad \qquad \Psi (Au) \leq \Psi (u) \big( \log \Psi (u) \big)^\theta, \qquad u \geq u_0.
\end{equation}
Let
\begin{displaymath}
p_n = \Psi (A^n) \qquad \text{and} \qquad l_n = \log p_n.
\end{displaymath}
One has then, by \eqref{inegalite Psi}, for $n$ large enough:
\begin{displaymath}
p_{n+1} \leq p_n (\log p_n)^\theta \qquad \text{and} \qquad l_{n+1} \leq l_n + \theta \log l_n.
\end{displaymath}
Let
\begin{displaymath}
h (t) =\int_2^t \frac{dx}{\theta \log x}\cdot
\end{displaymath}
One has, since $h'$ decreases:
\begin{displaymath}
0  < h (l_{n+1}) - h (l_n) \leq (l_{n+1} - l_n) h' (l_n) \leq (\theta \log l_n) \frac{1}{\theta \log l_n}
\leq 1.
\end{displaymath}
Hence $h (l_n) \lesssim n$, \emph{i.e.}:
\begin{displaymath}
\int_2^{l_n} \frac{dx}{\theta \log x} \lesssim n.
\end{displaymath}
Since this integral is $\gtrsim \frac{l_n}{\theta \log l_n}$, we get:
\begin{equation}\label{inegalite log Psi}
\frac{\log \Psi (A^n)}{\theta \log \log \Psi (A^n)} \lesssim n.
\end{equation}
Hence:
\begin{displaymath}
\log \log \Psi (A^n) - \log \log \log \Psi (A^n) \lesssim \log n,
\end{displaymath}
and so:
\begin{displaymath}
\limsup_{n \to \infty} \frac{\log \log \Psi (A^n)}{\log n} \leq 1.
\end{displaymath}
\par

On the other hand, $\Psi (A^n) \geq A^n$, \emph{i.e.} $\log \Psi (A^n) \geq n \log A$. \par
Therefore $\log \log \Psi (A^n) \sim \log n$, and hence, using \eqref{inegalite log Psi}:
\begin{displaymath}
\log \Psi (A^n) \lesssim  2\theta n \log n.
\end{displaymath}
But this prevents $\Psi$ from satisfying the condition $\Delta^1$: if $\Psi \in \Delta^1$, there exists
$A > 1$ such that $u \Psi (u) \leq \Psi (A u)$ , for $u$ large enough, and this implies, for $n$ large
enough, that $A^n \Psi (A^n) \leq \Psi (A^{n+1})$, and hence
$A^{n(n+1)/2} \Psi (A) \lesssim \Psi (A^n)$, so that $\log \Psi (A^n) \gtrsim n^2$. \qed
\bigskip

\noindent{\bf Remark.} As a consequence of Theorem~\ref{Psi delta 1}, one has the following striking result.

\begin{corollaire}\label{pas d'interpolation}
There exists an Orlicz function $\Psi$ such that $H^{3+\eps} \subseteq H^\Psi \subseteq H^3$ for every
$\eps >0$, and a composition operator $C_\phi$ which is compact on $H^3$ and on $H^{3+\eps}$, but
not compact on $H^\Psi$.
\end{corollaire}

\noindent{\bf Remark.} It is known (\cite{Shap-Tay}, Theorem 6.1), that the compactness of $C_\phi$ on
$H^{p_0}$ for some $p_0 <\infty$ implies its compactness on $H^p$ for all $p < \infty$; this result follows
from the Riesz factorization Theorem. It follows from Corollary~\ref{pas d'interpolation} that there is no
such factorization for Hardy-Orlicz spaces in general.
\medskip

\noindent{\bf Proof.} Consider the symbol $\phi = \phi_\theta$.
The compactness of  $C_\phi$ from $H^2$ into itself implies its compactness from
$H^p$ into itself for every $p \geq 1$; in particular its compactness on $H^3$ and on $H^{3+\eps}$.\par
Let now $\Psi$ the Orlicz function contructed in \cite{critere}. This function is such that
$x^3/3 \leq \Psi (x)$ for all $x\geq 0$, but $\Psi (n!) \leq (n!)^3$ for all positive integer $n \geq 1$.
Condition~\eqref{compacite phi indice theta sur H Psi} is not satisfied when $\theta < 1$ because
$\Psi (3.k!) \geq k.(k!)^3$ and
\begin{displaymath}
\Psi (k!) [\log \Psi (k!)]^\theta \leq (k!)^3[ 3 \log (k!)]^\theta \lesssim (k!)^3 (k \log k)^\theta;
\end{displaymath}
hence $C_{\phi_\theta}$ is not compact on $H^\Psi$ if $\theta < 1$.\qed
\medskip

\noindent{\bf Remark.} Condition \eqref{compacite phi indice theta sur H Psi} depends on $\theta$, but
if $\Psi$ is \emph{regular}, the compactness of $C_{\phi_\theta}$ on $H^\Psi$ does not actually depend on
$\theta$, because, writing:
\begin{displaymath}
[\log \Psi (u)]^\theta \gtrsim \frac{\Psi (Au)}{\Psi (u)}
= \frac{\Psi (Au)}{\Psi (\sqrt{A} u)} \frac{\Psi (\sqrt{A} u)}{\Psi (u)} \,\raise 1,5pt \hbox{,}
\end{displaymath}
one gets:
\begin{displaymath}
\frac{\Psi (\sqrt{A} u)}{\Psi (u)} \lesssim [\log \Psi (u)]^{\theta/2}
\qquad \text{or else} \qquad
 \frac{\Psi (A u)}{\Psi (\sqrt{A} u)} \lesssim [\log \Psi (u)]^{\theta/2} \cdot
\end{displaymath}
Hence, if $\Psi$ satisfies the following condition of regularity:
\begin{equation}\label{nabla}
\frac{\Psi ( \sqrt{A} u)}{\Psi (u)} \lesssim \frac{\Psi (Au)}{\Psi ( \sqrt{A} u)}\,\raise 1,5pt
\hbox{,} \qquad \forall A > 1, \forall u > 0,
\end{equation}
we get, for every $A >1$ and every $u >0$:
\begin{displaymath}
\frac{\Psi (\sqrt{A} u)}{\Psi (u)} \lesssim [\log \Psi (u)]^{\theta/2} \,\raise 1,5pt \hbox{,}
\end{displaymath}
and condition \eqref{compacite phi indice theta sur H Psi} remains true by replacing
$\theta$ with $\theta/2$. \par
Note that condition~\eqref{nabla} is satisfied whenever the Orlicz function $\Psi$ satisfies the condition
$\nabla_0$ defined in \cite{CompOrli}, Definition~4.5; indeed, by Proposition~4.6 of \cite{CompOrli}, $\Psi$
satisfies $\nabla_0$ if and only if  there exists some $u_0 >0$ such that, for every $\beta > 1$, there exists
$C_\beta \geq 1$ such that:
\begin{displaymath}
\frac{\Psi (\beta u)}{\Psi (u)} \leq \frac{\Psi (\beta C_\beta v)}{\Psi (v)}\,\raise 1,5 pt \hbox{,}
\qquad u_0 \leq u \leq v.
\end{displaymath}
This condition implies \eqref{nabla}, by taking  $\beta = \sqrt A$ and $v = \sqrt{A}u$, since, by convexity,
$\Psi (\beta C_\beta v) \leq C_\beta \Psi (\beta v)$.\par
When we impose in condition~\eqref{nabla} an inequality with factor $1$, for $u$ large enough, one has:
\begin{displaymath}
\big[\Psi ( \sqrt{A} u)\big]^2 \leq \Psi (u)\, \Psi (Au)\,, \qquad \forall A > 1, \forall u \geq u_0,
\end{displaymath}
and if one sets $u = \e^x$ and $Au = \e^y$, we get:
\begin{displaymath}
\log \Psi \Big[ \exp \Big( \frac{x + y}{2} \Big)\Big]
\leq \frac{1}{2} \big[\log \Psi (\e^x) + \log \Psi (\e^y)\big]\,,\qquad x_0 \leq x < y,
\end{displaymath}
which means that the function $\kappa (x) = \log \Psi (\e^x)$ is convex for $x$ large enough. By
\cite{CompOrli}, Proposition~4.7, that means that the Orlicz function $\Psi$ satisfies the condition
$\nabla_0$ with constant $1$. This regularity condition is satisfied, for example if
$\Psi (x) = x^p$, $p\geq 1$, or $\Psi (x) = \e^{(\log (x+1))^\alpha} - 1$, or else
$\Psi (x) = \e^{x^\alpha} -1$, $\alpha \geq 1$.

\section{Comparison with $H^\infty$}\label{Other results}

It is well-known, and easy to see, that $C_\phi$ is compact on $H^\infty$ if and only if $\| \phi\|_\infty < 1$.
For composition operators, $H^\infty$ is a very special case; indeed, one has:

\begin{proposition}
\hfill\par
1) Given any analytic self-map $\phi \colon \D \to \D$ such that $\| \phi\|_\infty = 1$, there exists an Orlicz
function $\Psi$ such that $C_\phi$ is not compact on $H^\Psi$.\par
2) Given any Orlicz function $\Psi$, there exists an analytic self-map $\phi \colon \D \to \D$ such that
$\| \phi\|_\infty = 1$ and for which $C_\phi$ is compact on $H^\Psi$.
\end{proposition}

\noindent{\bf Proof.} 1) Since $\| \phi\|_\infty = 1$, one has $\rho_\phi (h) > 0$ for every $0 < h < 1$.\par
We construct piecewise a strictly convex function $\Psi \colon [0,+\infty[ \to \R_+$ such that:
\begin{displaymath}
\Psi (2^{n+1}) \geq \frac{1}{\rho_\phi \big( 1 / \Psi (2^n)\big)} \, \raise 1,5 pt \hbox{,}
\end{displaymath}
Then, if $h_n = 1/ \Psi (2^n)$, one has $h_n \to 0$, and
\begin{displaymath}
\rho_\phi (h_n) \geq \frac{1}{\Psi \big( 2 \Psi^{-1} (1/h_n) \big) }\cdot
\end{displaymath}
It follows from \cite{CompOrli}, Theorem~4.18, that $C_\phi$ is not compact on $H^\Psi$.\par
2) Let $a \in M^\Psi (\T)$, the Morse-Transue space (see \cite{CompOrli}) a positive function such that
$a \notin L^\infty (\T)$ and such that $\Psi \circ a \geq 2$, and let:
\begin{displaymath}
h = 1 -\frac{1}{\Psi \circ a}\cdot
\end{displaymath}
Since $a$ is not essentially bounded, one has $\| h\|_\infty =1$. Moreover, one has $h \geq 1/2$, so
$\log h$ is integrable; we can hence define the outer function:
\begin{displaymath}
\phi (z) =\exp\bigg[ \int_\T \frac{ u + z}{u - z} \log h (u)\,dm (u) \bigg]\,, \qquad | z | < 1.
\end{displaymath}
One has $| \phi^\ast (\e^{i\theta})| = h (\e^{i\theta})$, so $\phi$ is an analytic self-map from $\D$ into
$\D$, and $\| \phi \|_\infty = 1$. Since, for every $A >0$:
\begin{displaymath}
\Psi \bigg[A \Psi^{-1} \Big( \frac{1}{1 -|\phi^\ast|} \Big)\bigg] = \Psi (A a)
\end{displaymath}
is integrable, because $a \in M^\Psi (\T)$, the composition operator $C_\phi$ is $M^\Psi$-order bounded
(see \cite{CompOrli}, Proposition~3.14), and hence is compact on $H^\Psi$. \qed
\bigskip


text:
{\bf Problem.} What is the probability that Fatima gets fewer heads than tails if she flips 10 coins?
{\bf Level.} Level 5
{\bf Type.} Counting & Probability
{\bf Solution.} We could solve this problem using casework, but using a little bit of symmetry and complementary probability gives us a more elegant solution. Since each coin flips heads or tails with equal probability, by the principle of symmetry the probability of getting fewer heads than tails is the equal to the probability of getting fewer tails than heads. Additionally, there are only three possible outcomes: getting fewer heads than tails, getting fewer tails than heads, or getting the same number of both. If we let $x$ represent the probability of the first outcome (which is the same as the probability of the second outcome) and $y$ represent the probability of the third outcome, we get the equation $2x + y = 1 \Rightarrow x=\dfrac{1-y}{2}$. So all we need to do is calculate the probability of getting the same number of heads and tails and we can then easily solve for what we want using the principle of complementary probability. Since there are two equally likely outcomes for each flip, there are a total of $2^{10}$ equally likely possible outcomes for flipping the 10 coins. We will have the same number of both heads and tails if we have exactly 5 of each, which we can count by selecting 5 out of the 10 flips to be heads which can occur in $\binom{10}{5}=252$ ways. So $y=\dfrac{252}{1024}=\dfrac{63}{256}$, and substituting that back into our first equation gives us the probability that we want: $x=\boxed{\dfrac{193}{512}}$.

set_name: MATH


TITLE: Calculate volume of a cone using its angle
QUESTION [0 upvotes]: I'm supposed to compute a way to calculate the volume of a cone using its angle at the top (which is given), using the next formula:
$V=\iiint\rho^2\cos(\theta)\sin(\theta)d\rho d\theta d\phi$
With:

$0<\rho<R$, $R$ being the radius of the base of the cone
$0<\theta<\pi/4$ which is the variable corresponding to the angle at the top of the cone (the angle being $\pi/2$)
$0<\phi<2\pi$

By both using sympy and calculating by hand, I get $V=\dfrac{\pi R^3}{6}$
I know it's not really complicated, but I've looked on internet for a bit but I couldn't find anything about calculating the volume of a cone that way, and couldn't verify if my result is correct. I guess it is, but I'm now curious why there's so little information on the internet. I guess it might not be the most efficient way to do it, but any why barely nobody speaks about that method ?

REPLY [0 votes]: The integral is then,
$V=\iiint_{V}dV=$$\iiint_V \rho^2 cos \theta sin\theta \,d\rho\,d\theta\,d\phi$$ $
$=$$\int_{\rho=0}^{R} \rho^2\int_{\theta=0}^{\frac{\pi}{4}} \int_{\phi=0}^{2\pi} cos \theta sin\theta \,d\rho\,d\theta\,d\phi$$ $
$=$$\int_{\rho=0}^{R} \rho^2 \int_{\phi=0}^{2\pi}\int_{\theta=0}^{\frac{\pi}{4}} cos \theta sin\theta \,d\rho\,d\theta\,d\phi$$ $
$=$$\int_{\rho=0}^{R} \rho^2 d\rho \int_{\phi=0}^{2\pi}d\phi\int_{\theta=0}^{\frac{\pi}{4}} cos \theta sin\theta \,d\theta$$ $
$=$$[\frac{\rho^3}{3}]_{\rho=0}^{R} \int_{\phi=0}^{2\pi}d\phi\int_{\theta=0}^{\frac{\pi}{4}} cos \theta sin\theta \,d\theta$$ $
$=$$\frac{R^3}{3}[\phi]_{\phi=0}^{2\pi}\int_{\theta=0}^{\frac{\pi}{4}} cos \theta sin\theta \,d\theta$$ $
$=$$\frac{R^3}{3}2\pi\frac{1}{2}\int_{\theta=0}^{\frac{\pi}{4}}2 cos \theta sin\theta \,d\theta$$ $
$=$$\frac{R^3}{3}2\pi\frac{1}{2}\int_{\theta=0}^{\frac{\pi}{4}} sin2\theta \,d\theta$$ $
$=$$\frac{R^3}{3}2\pi\frac{1}{2}[\frac{-cos2\theta}{2}]_{\theta=0}^{\frac{\pi}{4}}$$ $
$=$$\frac{R^3}{3}2\pi\frac{1}{2}\frac{1}{2}[0-(-1)]$$ $
$=\frac{\pi R^3}{2.3}$
$=\frac{\pi R^3}{6}$
*You can draw a cone and mention its respective angel as directed then it will be cleared.When we will get individual constant limit of the each  integral then it is easy otherwise need to transfer others formula(Stoke's,Green's if Possible) or change the integral their places.


TITLE: Parallel vectors proof. Is it correct?
QUESTION [1 upvotes]: Theorem: If $\overrightarrow{u} \neq \overrightarrow{0}$, then $\overrightarrow{v} = \alpha \overrightarrow{u}$ for some $\alpha$ if and only if $\overrightarrow{u}$ and $\overrightarrow{v}$ are parallels.
This is my demonstration procedure:
$\overrightarrow{u}=(a_{1},b_{1})$
$\alpha\overrightarrow{u}=\alpha(a_{1},b_{1})$
$\alpha\overrightarrow{u}=(\alpha a_{1},\alpha b_{1})$
Then,
$\overrightarrow{v}=\alpha\overrightarrow{u}=(\alpha a_{1},\alpha b_{1})$
To check if these two vectors -$\overrightarrow{u}$ and $\overrightarrow{v}$- are parallels, we need to find and angle to 0 radians:
$\cos{\theta}=\frac{\overrightarrow{u}\cdot \overrightarrow{v}}{|\overrightarrow{u}||\overrightarrow{v}|}$
Then
$\overrightarrow{u}\cdot \overrightarrow{v}= \alpha a_{1}^{2} + \alpha b_{1}^{2}$
$|\overrightarrow{u}|=\sqrt{a_{1}^{2}+b_{1}^{2}}$
$|\overrightarrow{v}|=\alpha\sqrt{a_{1}^{2}+b_{1}^{2}}$
Therefore,
$\cos{\theta}= \frac{\alpha (a_{1}^{2}+b_{1}^{2})}{\sqrt{a_{1}^{2}+b_{1}^{2}}\alpha\sqrt{a_{1}^{2}+b_{1}^{2}}}$
Once simplified, we have:
$\cos{\theta}=\frac{a_{1}^{2}+b_{1}^{2}} {a_{1}^{2}+b_{1}^{2}}=1$
And finally:
$\theta=\cos^{-1}{1}=0$
Is this proof correct?

REPLY [1 votes]: You have proven that if $v=\alpha u$, then the vectors are paralel. The proof you wrote is correct.
However, the statement says that the vectors are paralel if and only if $v=\alpha u$, which means you need to prove the other way around as well. You must show that if the vectors are paralel, then $v=\alpha u$.


TITLE: Linear ODE Cauchy problem: issue in the distributional solution
QUESTION [0 upvotes]: I am considering the following Initial Value problem:
$$
\begin{aligned}
&u'+\alpha u = \cos \omega t\\
& u(0)=u_0
\end{aligned}
$$
The solution is:
$$u(t) ={\rm e}^{-\alpha t} \left(u_0-{\frac {
\alpha}{{\alpha}^{2}+{\omega}^{2}}} \right)+\frac {\alpha\cos\omega t+\omega\sin\omega t}{\alpha^2+\omega^2}
$$
I now want to solve the equation using a distribution of the form $U(t)=H(t)u(t)$ where $H(t)$ is the Heaviside distribution. Since $U'=u'+u_0\delta$, we have to solve
$$U'+\alpha U=\cos \omega t + u_0\delta$$
in the sense of distributions, which means, using classical convolution formulation
$$(\delta'+\alpha\delta)*U=\cos\omega t + u_0\delta$$
The convolution inverse of $\delta'+\alpha\delta$ is $G(t)=H(t)\exp(-\alpha t)$ and so, if I am correct, the sought solution is $U(t)=G(t)*(\cos\omega t+u_0\delta)$ that is:
$$U(t)=H(t)\Bigl[u_0\exp(-\alpha t)+\frac{\alpha\cos \omega t +\alpha\sin\omega t}{\alpha^2+\omega^2}\Bigr]$$
Obviously there is something wrong in the derivations above, since the two solutions do not match. I cannot find where.
Solution
Based on md2perpe's comments and reply, below is a detailed solution. From $U'+\alpha U=H(t)\cos \omega t + u_0\delta$, we can say that
$$U(t)=G(t)*(H(t)\cos\omega t+u_0\delta)$$
where $G(t)=H(t)\exp(-\alpha t)$. Expanding the convolution yields:
$$U(t)=\int_0^{t}{\rm e}^{-\alpha s}\cos\omega (t-s)\mathrm{d}s+u_0{\rm e}^{-\alpha t}$$
that is
$$U(t)=H(t)\Bigl({\rm e}^{-\alpha t} \left(u_0-{\frac {
\alpha}{{\alpha}^{2}+{\omega}^{2}}} \right)+\frac {\alpha\cos\omega t+\omega\sin\omega t}{\alpha^2+\omega^2}\Bigr)$$
as expected.

REPLY [1 votes]: The solutions to $u' + \alpha u = \cos \omega t$ on all of $\mathbb{R}$ have the form
$$u(t) = A \cos \omega t + B \sin \omega t + C e^{-\alpha t}.$$
Let $v(t) = u(t)$ for $t>0$ and $v(t) \equiv 0$ for $t<0$. This can be written
$v(t) = u(t) \, H(t).$
What happens when we insert this into the differential equation, treating $v$ as a distribution? We have
$$v'(t) = u'(t) \, H(t) + u(t) \, H'(t) = u'(t) \, H(t) + u(0) \, \delta(t)$$
giving
$$\begin{align}
v'(t) + \alpha v(t)
&= \left( u'(t) \, H(t) + u(0) \, \delta(t) \right) + \alpha \left( u(t) \, H(t) \right) \\
&= \left( u'(t) + \alpha u(t) \right) H(t) + u(0) \, \delta(t) \\
&= \cos \omega t \, H(t) + u(0) \, \delta(t).
\end{align}$$
We can see here that you have missed the factor $H(t)$ after $\cos \omega t$.