text:
\section{Introduction}

\ \ \ \ Given (always bounded, linear)  operators $A$, $B$ on a
Banach space $X$, define $L_A$, $R_B$ on $L(X)$ (the space of
bounded linear operators on $X$) by $L_AT=AT$, $R_BT=TB$.
Operators of the form $L_A R_B$ on $L(X)$ are called {\em
multiplication operators}.  The beginning point of this paper is a
problem raised in 1992 by E. Saksman and H.-O. Tylli \cite {st1}
(see also \cite[Problem 2.8]{st2}):

Characterize the multiplication operators on $L(L_p)$, $1<p\not= 2<\infty$, which are
weakly compact.

Here $L_p$ is $L_p(0,1)$ or, equivalently, $L_p(\mu)$ for any purely non-atomic separable
probability  $\mu$.

In Theorem \ref{multiplication}
 we answer the Saksman-Tylli question.  The characterization is
rather simple but gives rise to   questions about operators on
$L_p$, some of which were asked by Tylli. First we set some
terminology. Given an operator $T:X\to Y$ and a Banach space $Z$,
say that $T$ is $Z$-strictly singular provided there is no
subspace $Z_0$ of $X$  which is isomorphic to $Z$ for which
$T_{|{Z_0}}$ is an isomorphism.  An operator $S:Z\to W$ factors
through an operator $T:X\to Y$  provided there are operators
$A:Z\to X$ and $B:Y\to W$ so that $S=BTA$.  If $S$ factors through
the identity operator on $X$, we say that $S$ factors through $X$.

If $T$ is  an operator on $L_p $, $1<p<\infty$, then $T$ is
$\ell_p$-strictly singular (respectively, $\ell_2$-strictly
singular)  if and only if $I_{\ell_p}$ (respectively,
$I_{\ell_2}$) does not factor through $T$.  This is true because
every subspace of $L_p$ which is isomorphic to ${\ell_p}$
(respectively, ${\ell_2}$) has a subspace which is still
isomorphic to ${\ell_p}$ (respectively, ${\ell_2}$)  and is
complemented in $L_p$.   Actually, a stronger fact is true: if
$\{ x_n \}_{n=1}^{\infty}$ is a sequence in $L_p$ which is equivalent to the unit
vector basis for either $\ell_p$ or $\ell_2$, then $\{ x_n \}_{n=1}^{\infty}$ has a
subsequence which spans a complemented subspace of $L_p$. For
$p>2$, an even stronger theorem was proved by Kadec-Pe\lambda czy\'nski
  \cite{kp}.  When $1<p<2$ and $\{ x_n \}_{n=1}^{\infty}$ is a sequence in
$L_p$ which is equivalent to the unit vector basis for $\ell_2$,
one takes $\{ y_n \}_{n=1}^{\infty}$ in $L_{p'}$ which are uniformly bounded and
biorthogonal to $\{ x_n \}_{n=1}^{\infty}$.  By passing to a subsequence    which is
weakly convergent and subtracting the limit from each $y_n$, one
may assume that $y_n \to 0$ weakly and hence, by the Kadec--Pe\lambda
czy\'nski dichotomy \cite{kp}, has a subsequence that is
equivalent to the unit vector basis of $\ell_2$ (since it is
clearly impossible that $\{ y_n \}_{n=1}^{\infty}$ has a subsequence equivalent to the
unit vector basis of $\ell_{p'}$). This implies that that the
corresponding subsequence of $\{ x_n \}_{n=1}^{\infty}$ spans a complemented subspace
of $L_p$. (Pe\lambda czy\'nski showed this argument, or something
similar, to one of the authors many years ago, and a closely related result was proved in \cite{pelros}.) Finally, when
$1<p<2$ and $\{ x_n \}_{n=1}^{\infty}$ is a sequence in $L_p$ which is equivalent to
the unit vector basis for $\ell_p$, see the comments after the
statement of Lemma \ref{ellpsequence}.

Notice that the comments in the preceding paragraph yield that an operator on $L_p $,
$1<p<\infty$, is
$\ell_p$-strictly singular (respectively, $\ell_2$-strictly singular)  if and only if
$T^*$ is $\ell_{p'}$-strictly singular (respectively, $\ell_2$-strictly singular),
where $p'={p \over {p-1}}$ is the conjugate index to $p$. Better known is that
an operator on $L_p$, $1<p<\infty$, is strictly singular if it is both $\ell_p$-strictly singular and $\ell_2$-strictly singular (and hence $T$ is strictly singular if and only if $T^*$ is strictly singular).  For $p>2$ this is immediate from  \cite{kp}, and Lutz Weis \cite{weis} proved the case $p<2$.

Although Saksman and Tylli did not know a complete
characterization of the weakly compact multiplication operators on
$L(L_p)$, they realized that a classification  must involve
$\ell_p$ and $\ell_2$-strictly singular operators on $L_p$.  This
led Tylli to ask us about possible classifications of the $\ell_p$
and $\ell_2$-strictly singular operators on $L_p$. The $\ell_2$
case is known. It is enough to consider the case $2<p<\infty$.  If
$T$ is an operator on $L_p$, $2<p<\infty$, and $T$ is
$\ell_2$-strictly singular, then it is an easy consequence of the
Kadec-Pe\lambda czy\'nski dichotomy that $I_{p,2}T$ is compact, where
$I_{p,r}$ is the identity mapping from $L_p$ into $L_r$. But then
by \cite{j}, $T$ factors through $\ell_p$. Tylli then asked
whether the following conjecture is true:

{\bf Tylli Conjecture.} If $T$ is an $\ell_p$-strictly singular
operator on $L_p$, $1<p<\infty$, then   $T$ is in the closure (in
the operator norm) of the operators on $L_p$ that factor through
$\ell_2$. (It is clear that the closure is needed because not all
compact operators on $L_p$, $p\not= 2$, factor through $\ell_2$.)

We then formulated a weaker conjecture:

{\bf Weak Tylli Conjecture.} If $T$ is an $\ell_p$-strictly
singular operator on $L_p$, $1<p<\infty$, and $J:L_p\to
\ell_\infty$ is an isometric embedding, then   $JT$ is in the
closure of the operators from $L_p$ into $\ell_\infty$ that factor
through $\ell_2$.

It is of course evident that an operator on $L_p$, $p\not= 2$,
that satisfies the conclusion of the Weak Tylli Conjecture must be
$\ell_p$-strictly singular. There is a slight subtlety in these
conjectures: while the Tylli Conjecture for $p$ is equivalent to
the Tylli Conjecture for $p'$, it is not at all clear and   is
even false that the Weak Tylli Conjecture for $p$ is equivalent to
the Weak Tylli Conjecture for $p'$.  In fact, we observe in Lemma
\ref{weak} (it is simple) that for $p>2$ the Weak Tylli Conjecture
is true, while the example in Section \ref{example} yields that
the Tylli Conjecture is false for all $p\not= 2$ and the Weak
Tylli Conjecture is false for $p<2$.

There are however some interesting consequences of the Weak Tylli
Conjecture  that are true when $p<2$.  In Theorem
\ref{maintheorem} we prove that if $T$ is an $\ell_p$-strictly
singular operator on $L_p$, $1<p<2$, then $T$ is $\ell_r$-strictly
singular for all $p<r<2$. In view of theorems of Aldous \cite{ald}
(see also \cite{km}) and Rosenthal \cite{ros},  this proves that
if such a $T$ is an isomorphism on a subspace $Z$ of $L_p$, then
$Z$ embeds into $L_r$ for all $r<2$.  The Weak Tylli Conjecture
would imply   that $Z$ is isomorphic to $\ell_2$, but the example
in Section \ref{example} shows that this need not be true.  When
we discovered Theorem \ref{maintheorem}, we thought its  proof
bizarre and assumed that a more straightforward argument would
yield a stronger theorem. The example suggests that in fact the
proof may be ``natural".

In Section  \ref{biased} we discuss   convolution by a biased coin on
$L_p$ of the Cantor group, $1\le p<2$.  We do not know whether such an operator $T$ on $L_p$, $1<p<2$, must satisfy the Tylli Conjecture or the weak Tylli conjecture.  We do prove, however, that if  $T_{|X}$ is an
isomorphism for some reflexive subspace $X$ of $L_p$, $1\le p<2$, then $X$ is
isomorphic to a Hilbert space. This answers an old question of
H.~P.~Rosenthal \cite{rosAltgeld}.

The standard Banach space theory terminology and background we use
can be found in \cite{lt}.

\section{ Weakly compact multiplication operators on $L(L_p)$}

\ \ \ \ We use freely the result \cite[Proposition 2.5]{st2} that
if $A$, $B$ are in $L(X)$ where  $X$ is a  reflexive Banach space
with the approximation property, then the multiplication operator
$L_AR_B$ on $L(X)$ is weakly compact if and only if for every $T$
in $L(X)$, the operator $ATB$ is compact. For completenes, in section \ref{appendix}  we give another proof of this under the weaker assumption that $X$ is reflexive and has the compact approximation property.
This theorem implies that for
such an $X$, $L_AR_B$ is weakly compact on $L(X)$  if and only if
$L_{B^*}R_{A^*}$ is a weakly compact operator on $L(X^*)$.
Consequently, to classify weakly compact multiplication operators
on $L(L_p)$, $1<p<\infty$, it is enough to consider the case
$p>2$. For $p\le r$ we denote the identity operator from $\ell_p$
into $\ell_r$ by $i_{p,r}$. It is immediate from \cite{kp} that an operator $T$  on $L_p$, $2<p<\infty$, is compact if and only if $i_{2,p}$ does not factor through $T$.

\begin{thm}\label{multiplication} Let  $2<p<\infty$ and let $A$, $B$ be bounded linear operators on
$L_p$. Then the multiplication operator $L_AR_B$ on $L(L_p)$ is weakly compact if and only if
one of the following (mutually exclusive) conditions hold.
\begin{itemize}
\item[(a)] $i_{2,p}$ does not factor through $A$ (i.e., $A$ is compact)
\item[(b)] $i_{2,p}$   factors through $A$ but $i_{p,p}$ does not factor through $A$
(i.e., $A$ is $\ell_p$-strictly singular) and $i_{2,2}$ does not factor through $B$
(i.e., $B$ is $\ell_2$-strictly singular)
\item[(c)] $i_{p,p}$ factors through $A$ but $i_{2,p}$ does not factor through $B$
(i.e., $B$ is compact)
\end{itemize}
\end{thm}

 \noindent {\bf Proof: \  } The proof is a straightforward application of the Kadec-Pe\lambda
czy\'nski dichotomy principle \cite{kp}: if $\{ x_n \}_{n=1}^{\infty}$ is a
semi-normalized (i.e., bounded and bounded away from zero) weakly
null sequence in $L_p$, $2<p<\infty$, then there is a subsequence
which is equivalent to either the unit vector basis of $\ell_p$
or of $\ell_2$ and spans a complemented subspace of $L_p$. Notice
that this immediately implies the ``i.e.'s" in the statement of
the theorem so that (a) and (c) imply that $L_AR_B$ is weakly
compact.  Now assume that (b) holds and let $T$ be in $L(L_P)$.
If $ATB$ is not compact, then there is a normalized weakly null
sequence $\{ x_n \}_{n=1}^{\infty}$ in $L_p$ so that $ATBx_n$ is bounded away from
zero.  By passing to a subsequence, we may assume that $\{ x_n \}_{n=1}^{\infty}$ is
equivalent to  either the unit vector basis of $\ell_p$  or of
$\ell_2$.  If $\{ x_n \}_{n=1}^{\infty}$ is equivalent to  the unit vector basis of
$\ell_p$, then since $TBx_n$ is bounded away from zero, we can
assume by passing to another subsequence that also $TBx_n$ is
equivalent to   the unit vector basis of $\ell_p$ and similarly
for $ATBx_n$, which contradicts the assumption that $A$ is
$\ell_p$-strictly singular. On the other hand, if $\{ x_n \}_{n=1}^{\infty}$ is
equivalent to  the unit vector basis of $\ell_2$, then since $B$
is $\ell_2$-strictly singular we can assume by passing to a
subsequence that $Bx_n$ is equivalent to   the unit vector basis
of $\ell_p$ and continue as in the previous case to get a
contradiction.

Now suppose that (a), (b), and (c) are all false. If $i_{p,p}$
factors through $A$ and $i_{2,p}$  factors through $B$ then there
is sequence $\{ x_n \}_{n=1}^{\infty}$ equivalent to the unit vector basis of $\ell_2$ or of  $\ell_p$
so that $Bx_n$ is  equivalent to the unit vector basis of $\ell_2$ or of  $\ell_p$ (of course, only three of the four cases are possible)
and $Bx_n$ spans a complemented subspace of $L_p$.  Moreover,
there is a sequence $\{ y_n \}_{n=1}^{\infty}$ in $L_p$ so that both $y_n$ and $Ay_n$
are equivalent to the unit vector basis of $\ell_p$. Since $Bx_n$
spans a complemented subspace of $L_p$, the mapping $Bx_n\mapsto
y_n$ extends to a bounded linear operator $T$ on $L_p$ and $ATB$
is not compact.  Finally, suppose that $i_{2,p}$   factors through
$A$ but $i_{p,p}$ does not factor through $A$ and $i_{2,2}$
factors through $B$. Then there is a sequence $\{ x_n \}_{n=1}^{\infty}$ so that $x_n$
and $Bx_n$ are both equivalent to the unit vector basis of
$\ell_2$ and $Bx_n$  spans a complemented subspace of $L_p$. There
is also a sequence $\{ y_n \}_{n=1}^{\infty}$  equivalent to the unit vector basis of
$\ell_2$ so that $Ay_n$ is  equivalent to the unit vector basis of $\ell_2$ or of
$\ell_p$.  The mapping $Bx_n\mapsto y_n$ extends to a bounded
linear operator $T$ on $L_p$ and $ATB$ is not compact.
\qed \medskip

It is perhaps worthwhile to restate Theorem \ref{multiplication} in a way that the cases where $L_AR_B$ is weakly compact  are not mutually exclusive.

\begin{thm}\label{multiplicationtwo} Let  $2<p<\infty$ and let $A$, $B$ be bounded linear operators on
$L_p$. Then the multiplication operator $L_AR_B$ on $L(L_p)$ is weakly compact if and only if
one of the following   conditions hold.
\begin{itemize}
\item[(a)]  $A$ is compact
\item[(b)] $A$ is $\ell_p$-strictly singular  and  $B$ is $\ell_2$-strictly singular
\item[(c)]  $B$ is compact
\end{itemize}
\end{thm}

\section{ $\ell_p$-strictly singular operators on $L_p$}

\ \ \ \ We recall the well known

\begin{lm}\label{ellpsequence} Let $W$ be a bounded convex symmetric subset of $L_p$, $1\le
p\not=2<\infty$. The following are equivalent:
\begin{enumerate}
\item[1.] No sequence  in $W$   equivalent to the unit vector basis for $\ell_p$ spans
a complemented subspace of $L_p$.
\item[2.] For every $C$ there exists $n$ so that no length $n$ sequence in $W$ is
$C$-equivalent to the unit vector basis of $\ell_p^n$.
\item[3.] For each $\varepsilon>0$ there is $M_\varepsilon$ so that
$W\subset \varepsilon B_{L_p} + M_\varepsilon B_{L_\infty}$.
\item[4.] $|W|^p$ is  uniformly integrable; i.e.,
$\lim_{t\downarrow 0} \sup_{x\in W} \sup_{\mu(E)< t} \| \mathbf{1} _E x\|_p
=0$.
\end{enumerate}
\end{lm}

When $p=1$, the assumptions that $W$ is convex and and $W$
symmetric are not needed, and the conditions in Lemma
\ref{ellpsequence} are equivalent to the non weak compactness of
the weak closure of $ {W}$. This case is essentially proved in
\cite{kp} and proofs  can also be found in books; see, e.g.,
\cite[Theorem 3.C.12]{woj}). (Condition (3) does not appear in
\cite{woj}, but it is easy to check the equivalence of (3) and
(4).  Also, in the proof in \cite[Theorem 3.C.12]{woj}) that not
(4) implies not (1), Wojtaszczyk only constructs a basic sequence
in $W$ that is equivalent to the unit vector basis for $\ell_1$;
however, it is clear that the constructed basic sequence spans a
complemented subspace.)

 For $p>2$, Lemma
\ref{ellpsequence} and stronger versions of condition (1) can be
deduced from \cite{kp}. For $1<p<2$, one needs to modify slightly
the proof in \cite{woj} for the case $p=1$.  The only essential
modification comes in the proof that not (4) implies not (1), and
this is where it is needed that $W$ is convex and symmetric.  Just
as in \cite{woj}, one shows that not (4) implies that there is a
sequence $\{ x_n \}_{n=1}^{\infty}$ in $W$ and a sequence $\{ E_n \}_{n=1}^{\infty}$ of disjoint measurable
sets so that $\inf \|1_{E_n} x_n\|_p >0$.  By passing to a
subsequence, we can assume that $\{ x_n \}_{n=1}^{\infty}$ converges weakly to, say,
$x$.  Suppose first that $x=0$. Then by passing to a further
subsequence, we may assume that $\{ x_n \}_{n=1}^{\infty}$ is a small perturbation of a
block basis of the Haar basis for $L_p$ and hence is an
unconditionally basic sequence. Since $L_p$ has type $p$, this
implies that there is a constant $C$ so that for all sequences
$\{ a_n \}_{n=1}^{\infty}$ of scalars, $\|\sum a_n x_n\|_p\le C(\sum |a_n|^p)^{1/p}$.
Let $P$ be the norm one projection from $L_p$ onto the closed
linear span $Y$ of the disjoint sequence $ \{ \1_{E_n} x_n \}_{n=1}^{\infty}$. Then $Px_n$ is
weakly null in a space isometric to $\ell_p$ and $\|Px_n\|_p$ is
bounded away from zero, so there is a subsequence
$\{Px_{n(k)}\}_{k=1}^\infty$ which is equivalent to the unit
vector basis for $\ell_p$ and whose closed span  is the range of a
projection $Q$ from $Y$.  The projection $QP$ from $L_p$ onto the
the closed span of $\{Px_{n(k)}\}_{k=1}^\infty$ maps $x_{n(k)}$ to
$Px_{n(k)}$ and, because of the upper $p$ estimate on $\{ x_{n(k)} \}_{k=1}^{\infty}$, maps
the closed span of $\{ x_{n(k)} \}_{k=1}^{\infty}$ isomorphically onto the closed span of
   $\{Px_{n(k)}\}_{k=1}^\infty$.  This yields that $\{ x_{n(k)} \}_{k=1}^{\infty}$ is
equivalent to the unit vector basis for $\ell_p$ and spans a
complemented subspace.  Suppose now that the weak limit $x$ of
$\{ x_n \}_{n=1}^{\infty}$ is not zero. Choose a subsequence $\{ x_{n(k)} \}_{k=1}^{\infty}$ so that $\inf
\|1_{E_{n(2k+1)}}(x_{n(2k)}-x_{n(2k+1)})\|_p>0$ and replace $\{ x_n \}_{n=1}^{\infty}$
with $\{{{x_{n(2k)}-x_{n(2k+1)}}\over 2}\}_{k=1}^\infty$ in the
argument above.

Notice that the argument outlined above gives that if $\{ x_n \}_{n=1}^{\infty}$ is a
sequence in $L_p$, $1< p\not=2<\infty$, which is equivalent to the
unit vector basis of $\ell_p$, then there is a subsequence $\{ y_n \}_{n=1}^{\infty}$
 whose closed linear span in $L_p$ is complemented.  This
is how one proves that the identity on $\ell_p$ factors through
any operator on $L_p$ which is not $\ell_p$-strictly singular.

The Weak Tylli Conjecture for $p>2$ is an easy consequence of the
following lemma.

\begin{lm}\label{weak} Let $T$ be an  operator from a
$\mathcal{L}_{1}$ space $V$ into $L_p$, $1<p<2$,  so that $W:=TB_{V}$ satisfies
condition (1) in Lemma \ref{ellpsequence}. Then for each
$\varepsilon>0$  there is an operator $S:V\to L_2$ so that   $\|T-I_{2,p}S\| <\varepsilon$.
 \end{lm}
 \noindent {\bf Proof: \  } Let $\varepsilon>0$.  By condition (3) in Lemma \ref{ellpsequence},
for each norm one vector $x$ in $V$ there is a vector $Ux$ in
$L_2$ with $\|Ux\|_2\le \|Ux\|_\infty\le  M_\varepsilon$ and
$\|Tx-Ux\|_p\le \varepsilon$.  By the definition of $\mathcal{L}_{1}$
space, we can write $V$ as a directed union $\cup_\alpha E_\alpha$
of finite dimensional spaces that are uniformly isomorphic to
$\ell_1^{n_\alpha}$, $n_\alpha=\dim E_\alpha$,   and let
$(x^\alpha_i)_{i=1}^{n_\alpha}$ be norm one vectors in $E_\alpha$
which are, say, $\lambda$-equivalent to the unit vector basis for
$\ell_1^{n_\alpha}$ with $\lambda$ independent of $\alpha$.  Let
$U_\alpha$ be the linear extension to $E_\alpha$ of the mapping
$x^\alpha_i\mapsto Ux^\alpha_i$, considered as an operator into
$L_2$. Then $\|T_{|E_\alpha}- I_{2,p}U_\alpha\|\le \lambda \varepsilon$
and $\|U_\alpha\|\le \lambda M_\varepsilon$.  A standard Lindenstrauss
compactness argument produces an operator $S:V\to L_2$ so that
$\|S\|\le \lambda M_\varepsilon$ and $\|T-I_{2,p}S\|\le \lambda \varepsilon$.
Indeed, extend $U_\alpha$ to all of $V$ by letting $U_\alpha x=0 $
if $x\not\in E_\alpha$. The net $T_\alpha$ has a subnet $S_\beta$
so that for each $x$ in $V$, $S_\beta x$ converges weakly in
$L_2$; call the limit $Sx$. It is easy to check that $S$ has the
properties claimed.
\qed \medskip

\begin{thm}\label{weaktyllithm} Let $T$ be an $\ell_p$-strictly singular operator on
$L_p$, $2<p<\infty$, and let $J$ be an isometric embedding of $L_p$ into an injective $Z$.
  Then for each $\varepsilon>0$ there is an operator $S:L_p\to Z$ so that $S$ factors
through $\ell_2$ and $\|JT-S\| <\varepsilon$.
\end{thm}
 \noindent {\bf Proof: \  } Lemma \ref{weak} gives the conclusion when $J$ is the adjoint
of a quotient mapping from $\ell_1$ or $L_1$ onto $L_{p'}$. The
general case then follows from the injectivity of $Z$.
\qed \medskip

The next proposition, when souped-up via ``abstract nonsense" and
known results,
 gives our main
result about
$\ell_p$-strictly singular operators on $L_p$.  Note that it shows that an
$\ell_p$-strictly singular operator on $L_p$, $1<p<2$, cannot be the identity on
the span
of a sequence of
$r$-stable independent random variables for any $p<r<2$.  We do not know another way of
proving even this special case of our main result.

\begin{pr}\label{proposition1} Let $T$ be an $\ell_p$-strictly singular operator on
$L_p$, $1<p<2$. If $X$ is a subspace of $L_p$ and $T_{|X}=aI_X$ with $a\not= 0$, then
$X$ embeds into $L_s$ for all $s<2$.
\end{pr}
 \noindent {\bf Proof: \  } By making  a change of density, we can by \cite{jj} assume
that $T$ is also a bounded linear operator on $L_2$, so assume,
without loss of generality, that $\|T\|_p\vee \|T\|_2 = 1$, so
that, in particular, $a\le 1$.  Lemma \ref{ellpsequence} gives for
each $\epsilon>0$ a constant $M_\epsilon$ so that
\begin{equation}\label{eqprop1}
T B_{L_p}\subset \epsilon B_{L_p} +  M_\epsilon B_{L_2}.
\end{equation}

Indeed, otherwise condition (1) in Lemma \ref{ellpsequence} gives
a bounded sequence $\{ x_n \}_{n=1}^{\infty}$ in $L_p$ so that $\{ Tx_n \}_{n=1}^{\infty}$   is equivalent
to the unit vector basis of $\ell_p$.  By passing to a subsequence
of differences of $\{ x_n \}_{n=1}^{\infty}$, we can assume, without loss of
generality, that $\{ x_n \}_{n=1}^{\infty}$ is a small perturbation of a block basis of
the Haar basis for $L_p$ and hence is an unconditionally basic
sequence. Since $L_p$ has type $p$, the sequence $\{ x_n \}_{n=1}^{\infty}$ has an
upper $p$ estimate, which means that there is a constant $C$ so
that for all sequences $\{ a_n \}_{n=1}^{\infty}$ of scalars, $\| \sum a_n x_n\| \le
C\| (\sum |a_n|^p)^{1/p}\|$. Since $\{ Tx_n \}_{n=1}^{\infty}$ is equivalent to the
unit vector basis of $\ell_p$, $\{ x_n \}_{n=1}^{\infty}$ also has a lower $p$ estimate
and hence $\{ x_n \}_{n=1}^{\infty}$ is equivalent to the unit vector basis of
$\ell_p$. This contradicts the $\ell_p$ strict singularity of $T$.

Iterating this we get for every $n$ and $0<\epsilon<1/2$
\[
a^n B_X \subset T^n B_{L_p}\subset \epsilon^n B_{L_p} +  2M_\epsilon B_{L_2}
\]
or, setting $A:=1/a$,
\[
B_X \subset  A^n\epsilon^n B_{L_p} +  2 A^n M_\epsilon B_{L_2}.
\]

For $f$ a unit vector in $X$ write $f=f_n+g_n$ with $\|f_n\|_2\le  2 A^n M_\epsilon$
and $\|g_n\|_p\le (A\epsilon)^n$. Then $f_{n+1}-f_n=g_n-g_{n+1}$, and since
evidently $f_n$ can be chosen to be of the form $(f\vee -k_n)\wedge k_n$ (with
appropriate interpretation when the set $[f_n=\pm k_n]$ has  positive measure), the
choice of
$f_n$, $g_n$ can be made so that
\[
\|f_{n+1}-f_n\|_2\le \|f_{n+1}\|_2\le 2M_\epsilon A^{n+1}
\]
\[
\|g_n-g_{n+1}\|_p\le \|g_n\|_p\le (A\epsilon)^n.
\]

For $p<s<2$ write ${1\over s}={\theta\over 2} +{{1-\theta}\over p}$.  Then
\[
\|f_{n+1}-f_n\|_s  \le \|f_{n+1}-f_n\|_2^\theta \|g_n-g_{n+1}\|_p^{1-\theta}
\le(2M_\epsilon A)^\theta (A\epsilon^{1-\theta})^n
\]
which is summable if $\epsilon^{1-\theta}<1/A$.   But $\|f-f_n\|_p\to 0$ so
$f=f_1 + \sum_{n=1}^\infty f_{n+1} - f_n$ in $L_p$ and hence also in $L_s$ if
$\epsilon^{1-\theta}<1/A$. So for some constant $C_s$ we get for all $f\in X$ that
$\|f\|_p\le \|f\|_s\le C_s\|f\|_p$. \qed

\medskip
We can now prove our main theorem. For background on ultrapowers
of Banach spaces, see \cite[Chapter 8]{djt}.

\begin{thm}\label{maintheorem} Let $T$ be an $\ell_p$-strictly singular operator on
$L_p$, $1<p<2$. If $X$ is a subspace of $L_p$ and $T_{|X}$ is an isomorphism, then $X$ embeds
into $L_r$ for all $r<2$.
\end{thm}
 \noindent {\bf Proof: \  } In view of Rosenthal's theorem \cite{ros}, it is enough to
prove that $X$ has type $s$ for all $s<2$.  By virtue of of the
Krivine-Maurey-Pisier theorem, \cite{kr} and \cite{mp} (or,
alternatively, Aldous' theorem, \cite{ald} or \cite{km}), we only
need to check that for $p<s<2$, $X$ does not contain almost
isometric copies of $\ell_s^n$ for all $n$. (To apply the
Krivine-Maurey-Pisier theorem we use that the second condition in
Lemma \ref{ellpsequence}, applied to the unit ball of $X$, yields
that $X$ has type $s$ for some $p<s\le 2$). So suppose that for
some $p<s<2$, $X$   contains almost isometric copies of $\ell_s^n$
for all $n$. By applying Krivine's theorem \cite{kr} we get for
each $n$ a sequence $(f^n_i)_{i=1}^n$ of unit vectors in $X$ which
is $1+\epsilon$-equivalent to the unit vector basis for $\ell_s^n$
and, for some constant $C$ (which we can take independently of
$n$), the sequence $(CTf^n_i)_{i=1}^n$ is also
$1+\epsilon$-equivalent to the unit vector basis for $\ell_s^n$.
By replacing $T$ by $CT$, we might as well assume that $C=1$.  Now
consider an ultrapower $T_{\cal U}$, where ${\cal U}$ is a free
ultrafilter on the natural numbers.  The domain and codomain of
$T_{\cal U}$ is the (abstract) $L_p$ space $(L_p)_{\cal U}$, and
$T_{\cal U}$ is defined by $T_{\cal U}(f_1,f_2,\dots) = (Tf_1,
Tf_2,\dots)$ for any (equivalence class of a) bounded sequence
$(f_1,f_2,\dots)$.  It is evident that $T_{\cal U}$ is an isometry on
the ultraproduct of span $(f^n_i)_{i=1}^n$; $n=1,2,\dots$,  and
hence $T_{\cal U}$ is an isometry on a subspace of $(L_p)_{\cal U}$
which is isometric to $\ell_s$. Since condition 2 in Lemma
\ref{multiplication} is obviously preserved when taking an
ultrapower of a set, we see that $T_{\cal U}$ is $\ell_p$-strictly
singular.  Finally, by restricting $T_{\cal U}$ to a suitable
subspace, we get an $\ell_p$-strictly singular operator $S$ on
$L_p$ and a subspace $Y$ of $L_p$ so that $Y$ is isometric to
$\ell_s$ and $S_{|Y}$ is an isometry.  By restricting the domain
of $S$, we can assume that $Y$ has full support and the functions
in $Y$ generate the Borel sets.  It then follows from the
Plotkin-Rudin theorem \cite{pl}, \cite{ru} (see \cite[Theorem
1]{kk}) that $S_{|Y}$ extends to an isometry $W$ from $L_p$ into
$L_p$. Since any isometric copy of $L_p$ in $L_p$ is norm one
complemented (see \cite[\S 17]{la}), there is a norm one operator
$V:L_p\to L_p$ so that $VW=I_{L_p}$. Then $VS_{|Y}=I_Y$ and $VS$
is $\ell_p$-strictly singular, which contradicts Proposition
\ref{proposition1}.
\qed \medskip

\begin{re} The   $\ell_1$-strictly singular operators on
$L_1$ also form an interesting class.  They are the weakly compact
operators on $L_1$.  In terms of factorization, they are just the
closure in the operator norm of the integral operators on $L_1$
(see, e.g., the proof of Lemma \ref{weak}).
\end{re}

\section{The example}\label{example}

\ \ \ \ Rosenthal \cite{rosxp} proved that if $\{ x_n \}_{n=1}^{\infty}$ is a sequence
of three valued, symmetric, independent random variables, then for
all $1<p<\infty$, the closed span in $L_p$ of $\{ x_n \}_{n=1}^{\infty}$ is
complemented by means of the orthogonal projection $P$, and
$\|P\|_p$ depends only on $p$, not on the specific sequence $\{ x_n \}_{n=1}^{\infty}$.
Moreover, he showed that if $p>2$, then for any sequence $\{ x_n \}_{n=1}^{\infty}$ of
symmetric, independent random variables in $L_p$, $\|\sum x_n\|_p$
is equivalent (with constant depending only on $p$) to $(\sum
\|x_n\|_p^p)^{1/p}\vee (\sum \|x_n\|_2^2)^{1/2}$. Thus if $\{ x_n \}_{n=1}^{\infty}$ is
normalized in $L_p$, $p>2$, and  $w_n:=\|x_n\|_2$, then $\|\sum
a_n x_n\|_p$ is equivalent to $\|\{ a_n \}_{n=1}^{\infty}\|_{p,w}:=(\sum
|a_n|^p)^{1/p}\vee (\sum |a_n|^2w_n^2)^{1/2}$. The completion of
the finitely non zero sequences of scalars under the norm
$\|\cdot\|_{p,w}$ is called $X_{p,w}$. It follows that if $w=\{ w_n \}_{n=1}^{\infty}$
is any sequence of numbers in $[0,1]$, then $X_{p,w}$ is
isomorphic to a complemented subspace of $L_p$. Suppose now that
$w=\{ w_n \}_{n=1}^{\infty}$ and $v=\{ v_n \}_{n=1}^{\infty}$ are two such sequences of weights and $v_n\ge
w_n$, then the diagonal operator $D$ from $X_{p,w}$ to $X_{p,v}$
that sends the $n$th unit vector basis vector $e_n$ to
${{w_n}\over{v_n}} e_n$ is contractive and it is more or less
obvious that $D$  is $\ell_p$-strictly singular if
${{w_n}\over{v_n}}\to 0$ as $n\to \infty$. Since $X_{p,w}$ and
$X_{p,v}$ are isomorphic to complemented subspaces of $L_p$, the
adjoint operator $D^*$ is $\ell_{p'}$ strictly singular and
(identifying $X_{p,w}^*$ and $X_{p,v}^*$ with subspaces of
$L_{p'}$) extends to a $\ell_{p'}$ strictly singular operator on
$L_{p'}$.  Our goal in this section is produce weights $w$ and $v$
so that $D^*$ is an isomorphism on a subspace of $X_{p,v}^*$ which
is not isomorphic to a Hilbert space.

For all $0<r<2$ there is a positive constant $c_r$ such that
\[
|t|^r=c_r\int_0^\infty \frac{1-\cos tx}{x^{r+1}} dx
\]
for all $t\in\mathbb{R}$. It follows that for any closed interval
$[a,b]\subset (0,\infty)$ and for all $\varepsilon>0$ there are
$0<x_1<x_2<\dots<x_{n+1}$ such that $\max_{1\le j\le
n}\Big|\frac{x_{j+1}-x_j}{x_j^{r+1}}\Big|\le\varepsilon$ and
\begin{equation}\label{eq:appr}
\Big|c_r\sum_{j=1}^n\frac{x_{j+1}-x_j}{x_j^{r+1}}(1-\cos
tx_j)-|t|^r\Big|<\varepsilon
\end{equation}
for all $t$ with $|t|\in[a,b]$.

Let $0<q<r<2$ and define $v_j$ and $a_j$, $j=1,\dots,n$, by
\[
v_j^{\frac{2q}{2-q}}=c_r\frac{x_{j+1}-x_j}{x_j^{r+1}} \ , \ \ \ \
\ \frac {a_j}{v_j^{\frac{2}{2-q}}}=x_j.
\]
 Let $Y_j$, $j=1,\dots,n$, be independent, symmetric, three valued
 random variables such that $|Y_j|=v_j^{\frac{-2}{2-q}} \mathbf{1} _{B_j}$
 with ${\rm Prob}(B_j)=v_j^{\frac{2q}{2-q}}$, so that in particular
 $\|Y_j\|_q=1$ and $v_j=\|Y_j\|_q/\|Y_j\|_2$. Then the characteristic function of $Y_j$ is
 \[
 \varphi_{Y_j}(t)=1-v_j^{\frac{2q}{2-q}}+v_j^{\frac{2q}{2-q}}\cos
 (tv_j^{\frac{-2}{2-q}})=1-v_j^{\frac{2q}{2-q}}(1-\cos
 (tv_j^{\frac{-2}{2-q}}))
 \]
 and
\begin{equation}\label{eq:phi}
\begin{array}{rl}
\varphi_{\sum a_jY_j}(t)= &
\prod_{j=1}^n(1-v_j^{\frac{2q}{2-q}}(1-\cos
 (ta_jv_j^{\frac{-2}{2-q}})))\\
 =&
\prod_{j=1}^n(1-c_r\frac{x_{j+1}-x_j}{x_j^{r+1}}(1-\cos
 (tx_j)))
\end{array}
\end{equation}
 To evaluate this product we use the estimates
 on $\frac{x_{j+1}-x_j}{x_j^{r+1}}$ to deduce that, for each $j$
 \[
 \begin{array}{rl}
 |\log(1-c_r\frac{x_{j+1}-x_j}{x_j^{r+1}}(1-\cos
 (tx_j)))+&c_r\frac{x_{j+1}-x_j}{x_j^{r+1}}(1-\cos
 (tx_j))|\\
 &\le C\varepsilon c_r^2\frac{x_{j+1}-x_j}{x_j^{r+1}}(1-\cos
 (tx_j))
 \end{array}
 \]
 for some absolute $C<\infty$. Then, by (\ref{eq:appr}),
 \[
 \begin{array}{rl}
 |\sum_{j=1}^n\log(1-c_r\frac{x_{j+1}-x_j}{x_j^{r+1}}(1-\cos
 (tx_j)))+&c_r\sum_{j=1}^n\frac{x_{j+1}-x_j}{x_j^{r+1}}(1-\cos
 (tx_j))|\\
 &\le C\varepsilon c_r(\varepsilon+b^r).
 \end{array}
 \]
Using (\ref{eq:appr}) again we get
\[
|\sum_{j=1}^n\log(1-c_r\frac{x_{j+1}-x_j}{x_j^{r+1}}(1-\cos
 (tx_j)))+|t|^r|\le (C+1)\varepsilon(\varepsilon+b^r)
 \]
 (assuming as we may that $b\ge 1$), and from (\ref{eq:phi}) we get
 \[
 \varphi_{\sum a_jY_j}(t)=(1+O(\varepsilon))\exp(-|t|^r)
 \]
for all $|t|\in [a,b]$, where the function hiding under the $O$
notation depends on $r$ and $b$ but on nothing else. It follows
that, given any $\eta>0$, one can find $a,b$ and $\varepsilon$, such that
for the corresponding $\{a_j,Y_j\}$ there is a symmetric
$r$-stable $Y$ (with characteristic function $e^{-|t|^r}$)
satisfying
\[
\|Y-\sum_{j=1}^n a_j Y_j\|_q\le\eta.
\]
This follows from classical translation of various convergence
notions; see e.g.   \cite[p. 154]{rosII}.

 Let now $0<\delta<1$. Put
$w_j=\delta v_j$, $j=1,\dots,n$, and let $Z_j$, $j=1,\dots,n$, be
independent, symmetric, three valued
 random variables such that $|Z_j|=w_j^{\frac{-2}{2-q}} \mathbf{1} _{C_j}$
 with ${\rm Prob}(C_j)=w_j^{\frac{2q}{2-q}}$, so that in particular
 $\|Z_j\|_q=1$ and $w_j=\|Z_j\|_q/\|Z_j\|_2$. In a similar manner
 to the argument above we get that,

 \begin{equation*}
\begin{array}{rl}
\varphi_{\sum \delta a_jZ_j}(t)= &
\prod_{j=1}^n(1-w_j^{\frac{2q}{2-q}}(1-\cos
 (t\delta a_j w_j^{\frac{-2}{2-q}})))\\
 = &
 \prod_{j=1}^n(1-\delta^{\frac{2q}{2-q}}v_j^{\frac{2q}{2-q}}(1-\cos
 (t\delta^{\frac{-q}{2-q}} a_j v_j^{\frac{-2}{2-q}})))\\

 =
 & (1+O(\varepsilon))\exp(-\delta^{\frac{q(2-r)}{2-q}}|t|^r)
\end{array}
\end{equation*}
 for all $|t|\in
[\delta^{\frac{q}{2-q}}a,\delta^{\frac{q}{2-q}}b]$, where the $O$
now depends also on $\delta$.

Assuming $\delta^{\frac{q(2-r)}{2-q}}>1/2$ and for a choice of
$a,b$ and $\varepsilon$, depending on $\delta, r, q$ and $\eta$ we get that
there is a symmetric $r$-stable random variable $Z$ (with
characteristic function $e^{-\delta^\frac{q(2-r)}{2-q}|t|^r}$)
such that
\[
\|Z-\sum_{j=1}^n \delta a_j Z_j\|_q\le\eta.
\]
Note that the ratio between the $L_q$ norms of $Y$ and $Z$ are
bounded away from zero and infinity by universal constants and
each of these norms is also universally bounded away from zero.
Consequently, if $\varepsilon$ is small enough the ratio between the $L_q$
norms of $\sum_{j=1}^n a_j Y_j$ and $\sum_{j=1}^n \delta a_j Z_j$
are bounded away from zero and infinity by universal constants.

Let now $\delta_i$ be any sequence decreasing to zero and $r_i$
any sequence such that $q<r_i\uparrow 2$ and satisfying
$\delta_i^{\frac{q(2-r_i)}{2-q}}>1/2$. Then, for any sequence
$\varepsilon_i\downarrow 0$ we can find two sequences of symmetric,
independent, three valued random variables $\{Y_i\}$ and
$\{W_i\}$, all normalized in $L_q$, with the following additional
properties:

\begin{itemize}
\item put $v_j=\|Y_j\|_q/\|Y_j\|_2$ and $w_j=\|Z_j\|_q/\|Z_j\|_2$.
Then there are disjoint finite subsets of the integers $\sigma_i$,
$i=1,2,\dots$, such that $w_j=\delta_i v_j$ for $j\in \sigma_i$.
\item There are independent random variables $\{\bar Y_i\}$ and $\{\bar Z_i\}$
with $\bar Y_i$ and $\bar Z_i$ $r_i$ stable with bounded, from
zero and infinity, ratio of $L_q$ norms and there are coefficients
$\{a_j\}$ such that
\[
\|\bar Y_i-\sum_{j\in\sigma_i}a_jY_j\|_q<\varepsilon_i \ \ \ \mbox {and}\ \
\ \|\bar Z_i-\sum_{j\in\sigma_i}\delta_ia_jZ_j\|_q<\varepsilon_i.
\]
\end{itemize}

From \cite{rosxp} we know that the spans of $\{Y_j\}$ and
$\{Z_j\}$ are complemented in $L_q$, $1<q<2$, and the dual spaces
are naturally isomorphic to the spaces $X_{p,\{v_j\}}$ and
$X_{p,\{w_j\}}$ respectively; both the isomorphism constants and
the complementation constants depend only on $q$. Here $p=q/(q-1)$
and
\[
\|\{\alpha_j\}\|_{X_{p,\{u_j\}}}=\max\{(\sum|\alpha_j|^p)^{1/p},(\sum
u_j^2\alpha_j^2)^{1/2}\}.
\]
Under this duality the adjoint $D^*$ to the operator $D$ that
sends $Y_j$ to $\delta_i Z_j$ for $j\in\sigma_i$ is formally the
same diagonal operator between $X_{p,\{w_i\}}$ and
$X_{p,\{v_i\}}$. The relation $w_j=\delta_i v_j$ for $j\in
\sigma_i$ easily implies that this is a bounded operator.
$\delta_i\to 0$ implies that this operator is $\ell_q$ strictly
singular.
If $\varepsilon_i\to 0$ fast enough, $D^*$ preserves a copy of ${\rm
span}\{\bar Y_i\}$. Finally, if $r_i$ tend to 2 not too fast this
span is not isomorphic to a Hilbert space. Indeed, let $1\le
s_j\uparrow 2$ be arbitrary and let $\{n_j\}_{j=1}^\infty$ be a
sequence of positive integers with $n_j^{\frac{1}{s_j}-\frac12}\ge
j$, $j=1,2,\dots$, say. For $1\le k\le n_j$, put
$r_{n_1+n_2+\dots+n_{j-1}+k}=s_j$. Then the span of
$\{Y_i\}_{i=n_1+\dots+n_{j-1}+1}^{n_1+\dots+n_j}$ is isomorphic,
with constant independent of $j$, to $\ell_{s_j}^{n_j}$ and this
last space is of distance at least $j$ from a Euclidean space.

It follows that if $J:L_q\to \ell_\infty$ is an isometric
embedding, then $JD^*$
 cannot be arbitrarily
approximated by an operator which factors through a Hilbert space,
and hence the Weak Tylli Conjecture is false in the range $1<q<2$.

\section{Convolution by a biased coin}\label{biased}

\ \ \ \ In this section we regard $L_p$ as $L_p(\Delta)$, where
$\Delta=\{-1,1\}^\mathbb{N}$ is the Cantor group and the measure is the
Haar measure $\mu$ on $\Delta$; i.e., $\mu=\prod_{n=1}^\infty
\mu_n$, where $\mu_n(-1)=\mu_n(1)= 1/2$.  For $0<\varepsilon<1$, let
$\nu_\varepsilon$ be the $\varepsilon$- biased coin tossing measure; i.e.,
$\nu_\varepsilon=\prod_{n=1}^\infty \nu_{\varepsilon,n}$, where
$\nu_{\varepsilon,n}(1)=\frac{1+\varepsilon}{2}$ and
$\nu_{\varepsilon,n}(-1)=\frac{1-\varepsilon}{2}$.  Let $T_\varepsilon$ be convolution by
$\nu_\varepsilon$, so that for a $\mu$-integrable function $f$ on $\Delta$,
$(T_\varepsilon f)(x)=(f*\nu_\varepsilon)(x)=\int_\Delta f(xy)\, d\nu_\varepsilon (y)$.  The
operator $T_\varepsilon$ is a contraction on $L_p$ for all $1\le p\le
\infty$. Let us recall how $T_\varepsilon$ acts on the characters on
$\Delta$.  For $t=\{ t_n \}_{n=1}^{\infty} \in \Delta$, let $r_n(t)=t_n$.  The
characters on $\Delta$ are finite products of these {\it
Rademacher functions} $r_n$ (where the void product is the
constant one function).  For $A$ a finite subset of $\mathbb{N}$, set
$w_A=\prod_{n\in A} r_n$ and let $W_n$ be the linear span of
$\{w_A: |A|=n\}$. Then $T_\varepsilon w_A=\varepsilon^{|A|}w_A$.

 We are interested in studying $T_\varepsilon$ on $L_p$, $1\le p
<2$. The background we mention below is all contained in Bonami's
paper \cite{bon} (or see \cite{rosAltgeld}). On $L_p$, $1<p<2$,
$T_\varepsilon$ is $\ell_p$-strictly singular; in fact, $T_\varepsilon$ even maps
$L_p$ into $L_r$ for some $r=r(p,\varepsilon)>p$. Indeed, by interpolation
it is sufficient to check that $T_\varepsilon $ maps $L_s$ into $L_2$ for
some $s=s(\varepsilon)<2$. But there is a constant $C_s$ which tends to $1$
as $s\uparrow 2$ so that for all $f\in W_n$, $\|f\|_2\le C_s^n
\|f\|_s$ and the orthogonal projection $P_n$ onto (the closure of)
$W_n$ satisfies $\|P_n\|_p\le C_s^n$.  From this it is easy to
check that  if $\varepsilon C_s^2<1$,   then $T_\varepsilon$ maps $L_s$ into $L_2$.
We remark in passing that Bonami \cite{bon} found for each $p$ (including $p\ge 2$) and $\varepsilon$ the largest value of $r=r(p,\varepsilon)$ such that $T_\varepsilon$ maps $L_p$ into $L_r$.

Thus Theorem \ref{maintheorem} yields that if $X$ is a subspace of
$L_p$, $1<p<2$, and $T_\varepsilon$ (considered as an operator from $L_p$
to $L_p$) is an isomorphism on $X$, then $X$ embeds into $L_s$ for
all $s<2$. Since, as we mentioned above, $T_\varepsilon$ maps $L_s$ into
$L_2$ for some $s<2$, it then follows from an argument in
\cite{rosAltgeld} that $X$ must be isomorphic to a Hilbert space. (Actually, as we show after the proof, Lemma \ref{rosLemma} is strong enough that we can prove Theorem \ref{rosProblem} without using Theorem \ref{maintheorem}.)
Since \cite{rosAltgeld} is not generally available, we repeat
Rosenthal's argument in Lemma \ref{rosLemma} below.

Now $T_\varepsilon$ is not $\ell_1$ strictly singular on $L_1$.
Nevertheless, we still get that if $X$ is a reflexive subspace of
$L_1$ and $T_\varepsilon$ (considered as an operator from $L_1$ to $L_1$)
is an isomorphism on $X$, then $X$ is isomorphic to a Hilbert
space. Indeed, Rosenthal showed (see Lemma \ref{rosLemma}) that
then there is another subspace $X_0$ of $L_1$ which is isomorphic
to $X$ so that $X_0$ is contained in $L_p$ for some $1<p<2$, the
$L_p$ and $L_1$ norms are equivalent on $X_0$, and $T_\varepsilon$ is an
isomorphism on $X_0$.  This implies that as an operator on $L_p$,
$T_\varepsilon$ is an isomorphism on $X_0$ and hence  $X_0$ is isomorphic
to a Hilbert space. (To apply Lemma \ref{rosLemma}, use the fact
\cite{ros} that if $X$ is a relexive subspace of $L_1$, then $X$
embeds into $L_{p}$ for some $1<p<2$.)

We summarize this discussion in the first sentence of Theorem \ref{rosProblem}.
The case $p=1$ solves Problem B from Rosenthal's 1976 paper
\cite{rosAltgeld}.

\begin{thm}\label{rosProblem} Let $1\le p < 2$, let $0<\varepsilon<1$, and
let $T_\varepsilon$ be considered as an operator on $L_p$.  If $X$ is a
reflexive subspace of $L_p$ and the restriction of $T_\varepsilon$ to $X$
is an isomorphism, then $X$ is isomorphic to a Hilbert space. Moreover, if $p>1$, then $X$ is complemented in $L_p$.
\end{thm}

We now prove Rosenthal's lemma \cite[proof of Theorem 5]{rosAltgeld} and defer the proof of the ``moreover" statement in
Theorem \ref{rosProblem} until after the proof of the lemma.
.

\begin{lm}\label{rosLemma} Suppose that $T$ is an operator on
$L_p$, $1\le p<r<s <2$, $X$ is a subspace of $L_p$ which is
isomorphic to a subspace of $L_s$, and $T_{|X}$ is an isomorphism.
Then there is another subspace $X_0$ of $L_p$ which is isomorphic
to $X$ so that $X_0$ is contained in $L_r$, the $L_r$ and $L_p$
norms are equivalent on $X_0$, and $T$ is an isomorphism on $X_0$.
\end{lm}

 \noindent {\bf Proof: \  } We want to find a measurable set $E$ so that
\begin{itemize}
\item[(1)] $X_0:= \{ \mathbf{1} _E x : x\in X\}$ is isomorphic to $X$,
\item[(2)] $X_0\subset L_r$,
\item[(3)] $T_{|X_0}$ is an isomorphism.
\end{itemize}
(We did not say that   $\|\cdot\|_p$ and $\|\cdot\|_r$, are
equivalent on $X_0$ since that follows formally from the closed
graph theorem. The isomorphism $X\to X_0$ guaranteed by (a) is of
course the mapping $x\mapsto  \mathbf{1} _E x$.)

Assume, without loss of generality, that $\|T\|=1$. Take $a>0$ so
that $\|Tx\|_p \ge {a} \| x\|_p$ for all $x$ in $X$. Since
$\ell_p$ does not embed into $L_s$ we get from (4) in Lemma
\ref{ellpsequence} that there is $\eta>0$ so that if $E$ has
measure larger than $1-\eta$, then $\| \mathbf{1} _{\sim E}x\|_p \le
{\frac{a}{2}}\|x\|_p$ for all $x$ in $x$. Obviously (1) and (3)
are satisfied for any such $E$.  It is proved in \cite{ros} that
    there is strictly positive   $g$ with $\|g\|_1=1$ so that
$\frac {x}{g}$ is in $L_r$ for all $x$ in $X$.  Now simply choose
$t<\infty$ so that $E:=[g<t]$ has measure at least $1-\eta$; then
$E$ satisfies (1), (2), and (3).
\qed \medskip

Next we remark how to avoid using Theorem \ref{maintheorem} in proving Theorem \ref{rosProblem}.  Suppose that  $T_\varepsilon$  is an isomorphism on a reflexive subspace $X$ of $ L_p$, $1\le p<2$.  Let $s$ be the supremum of those $r\le 2$ such that $X$ is isomorphic to a subspace of $L_r$, so $1<s\le 2$.  It is sufficient to show that $s=2$. But if $s<2$, we get from the interpolation formula that if $r<s$ is sufficiently close to $s$, then $T_\varepsilon$ maps $L_r$ into $L_t$ for some $t>s$ and hence, by Lemma  \ref{rosLemma}, $X$ embeds into $L_t$.

Finally we prove the ``moreover" statement in Theorem \ref{rosProblem}. We now know that $X$ is isomorphic to a Hilbert space.  In the proof of Lemma \ref{rosLemma}, instead of using Rosenthal's result from  \cite{ros}, use Grothendieck's theorem \cite[Theorem 3.5]{djt}, which implies that  there is strictly positive   $g$ with $\|g\|_1=1$ so that
$\frac {x}{g}$ is in $L_2$ for all $x$ in $X$.  Choosing $E$ the same way as in the proof of Lemma \ref{rosLemma} with $T:=T_\varepsilon$, we get that (1), (2), and (3) are true with $r=2$. Now the $L_2$ and $L_p$ norms are equivalent on both $X_0$ and on $T_\varepsilon X_0$. But it is clear that the only way that $T_\varepsilon$ can be an isomorphism on a subspace $X_0$ of $L_2$ is for the orthogonal projection $P_n$ onto the closed span of $W_k$, $0\le k\le n$, to be an isomorphism on $X_0$ for some finite $n$. But then also in the $L_p$ norm the restriction of $P_n$ to $X_0$ is an isomorphism because the $L_p$ norm and the $L_2$ norm are equivalent on the span of $W_k$, $0\le k\le n$ and $P_n$ is bounded on $L_p$ (since $p>1$).  It follows that the operator $S:= P_n \circ  \mathbf{1} _E$ on $L_p$ maps $X_0$ isomorphically onto a complemented subspace of $L_p$, which implies that $X_0$ is also complemented in $L_p$.

We conclude this section with the open problem that started us thinking about $\ell_p$-strictly singular operators.

\begin{problem}\label{problem} Let $1<p<2$ and $0<\varepsilon<1$.  On $L_p(\Delta)$, does
$T_\varepsilon$ satisfy the conclusion of the Tylli Conjecture or the Weak
Tylli Conjecture?
\end{problem}

Of course, the answer to Problem \ref{problem} is ``yes" when
$\varepsilon=\varepsilon(p)$ is sufficiently small, since then $T_\varepsilon$ maps $L_p$
into $L_2$.

\section{Appendix}\label{appendix}

\ \ \  In this appendix we prove a theorem that is essentially due to Saksman and Tylli.  The only novelty is that we assume the compact approximation property rather than the approximation property.

\begin{thm}\label{SaksTylli} Let $X$ be a reflexive Banach spaceand let $A$, $B$ be in $L(X)$.  Then
\begin{itemize}
\item[(a)] If $ATB$ is a compact operator on $X$ for every $T$ in $L(X)$, then $L_A R_B$ is a weakly compact operator on $L(X)$.
\item[(b)] If $X$ has the compact approximation property and $L_A R_B$ is a weakly compact operator on $L(X)$, then $ATB$ is a compact operator on $X$ for every $T$ in $L(X)$.
\end{itemize}
\end{thm}

 \noindent {\bf Proof: \  }  To prove (a), it is enough to recall \cite{kalton} that for a reflexive space $X$, on bounded subsets of $K(X)$ the weak topology is the same as the weak operator topology (the operator $T\mapsto f_T\in C((B_X, \mbox{ weak}) \times (B_{X^*}, \mbox{weak}))$, where $f_T(x,x^*):=\langle x^*,Tx\rangle$,  is an isometric isomorphism from $K(X)$ into a space of continuous functions on a compact Hausdorff space).

To prove (b), suppose that we have a $T\in L(X)$ with $ATB$   not compact. Then there is a weakly null normalized sequence $\{ x_n \}_{n=1}^{\infty}$ in $X$ and $\delta>0$ so that for all $n$, $\|ATB x_n\|>\delta$. Since a reflexive space with the compact approximation property also has the compact metric approximation property \cite{chojohnson}, there are $C_n\in K(X)$ with $\|C_n\|< 1+1/n$, $C_n Bx_i=Bx_i$ for $i\le n$.  Since the $C_n$ are compact, for each $n$, $\|C_n B x_m\| \to 0$ as $m\to \infty$.  Thus $A(TC_n)B x_i=ATB x_i$ for $i\le n$ and $\|A(TC_n)B x_m\|\to 0$ as $m\to \infty$.  This implies that no convex combination of $\{A(TC_n)B\}_{n=1}^\infty$ can converge in the norm of $L(X)$ and hence $\{A(TC_n)B\}_{n=1}^\infty$ has no weakly convergent subsequence.  This contradicts the weak compactness of $L_A R_B$ and completes the proof.
\qed \medskip

\vfill\eject

redpajama_set_name: RedPajamaArXiv


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text:
Published on Sunday Examiner (http://sundayex.catholic.org.hk)
Home > Finding Jesus in stillness and silence
Finding Jesus in stillness and silence
Print Version [1] Email to Friend [2]
WCCM.png [3]
HONG KONG (SE): Speaking to an audience of about 300 people at St. Francis of Assisi College in Fanling on January 27, Benedictine Father Laurence Freeman, director of the World Community for Christian Meditation (WCCM), asked people to picture a roomful of kindergarten children, sitting in a perfect lotus position, meditating in stillness and silence.
He recently witnessed this incredible scene in Myanmar.
The children were meditating naturally, he said, immersed as they were in the contemplative culture of Buddhism.
Father Freeman pointed out that Christianity, too, has a long, long tradition of meditation. The practice originated with the Desert Fathers and Mothers, a monastic movement from the 3rd century that had contemplative prayer at the centre of their lives.
Down the centuries, however, this dimension of prayer had been forgotten, and Father Freeman and WCCM—with 3,000 groups around the world—want to restore Christian meditation as part of our prayer life.
"We have become shallow, we've lost this depth dimension in prayer," he lamented.
"We need meditation to become better disciples of Christ and we need it for an increasingly troubled and divided world," he said.
Father Freeman described meditation as a journey of prayer from the mind to the heart and through one Christian sacred word, "the mantra," repeated faithfully and lovingly, we let go of thoughts, words and imagination and enter into an ever deepening journey into stillness and silence.
The suggested mantra is the word Maranatha, used by St. Paul at the end of his first letter to the Corinthians, which means, "Come, Lord Jesus!"
"Don't think of it as a technique that you have to master and become expert at and you'll get an A grade as a meditator," he warned. "Think of it as an art."
Meditation can also be understood as the work of a relationship, Father Freeman continued, first of all being in relationship with yourself.
"We can't be makers of peace if we are not at peace with ourselves, unless we overcome our interior personal conflicts and division," he said.
As with every relationship, it takes work. Meditation is "simple, but not easy." It is simple because it involves concentrating only on a word, but is not easy because it takes patience and practice. Ideally through daily practice, this prayer of silence will eventually take us to Christ within us.
In his talk, Father Freeman recalled the Apostles, the first disciples of Jesus. They were poor and simple, renounced their possessions, had poverty of spirit but carried themselves with dignity and above all, they loved Jesus.
So how does meditation help to become better disciples? "In the process of meditating and coming to the core of our being, we are also stripped—leaving everything behind—a spiritual renouncing if you will, until there is purity of heart in union with Jesus," he said.
Jesus modelled contemplation for us, spending time praying alone, regularly, seeking solitude and silence for his prayer to the Father. The pope, in a recent letter, said it was absolutely necessary if we are to be holy in our daily lives that we take time off every day for stillness and silence, in other words, for contemplation.
Expanding to the modern world, Father Freeman said human dignity is essential in the face of threats from totalitarianism and state authoritarianism. Meditation is one way to counter this, as we allow our spirit expand and grow, we become more open to love and compassion and feel a common humanity with others.
Meditation will help us become "big, generous, inclusive, not petty, not prejudiced, not sectarian," he said.
Along with the spiritual benefits, Father Freeman said medical and scientific studies have shown that meditation improves the cardiovascular system, the immune system, enhances the quality of sleep and helps one deal with anxiety, depression and anger.
"A doctor said to me recently this is the only area of medical research where we see no negative side effects," said Father Freeman. "So meditation is pretty good for you."
He differentiated meditation from mindfulness. Mindfulness as a relaxation technique has helped many and deserves credit. But mindfulness is basically an observation of the self in the present moment and is taken out of context as an end in itself.
"The American military is teaching mindfulness to its snipers. You can be a mindful sniper!
"I don't think that's what the Buddha had in mind or what Jesus had in mind; to go into your inner room and pray."
Mindfulness has its place and serves as a good prelude or preparation for meditation. "Without meditation, mindfulness has limited value," he said.
This is because the orientation between mindfulness and meditation is vastly different. Mindfulness leads to self-fixation, whereas in meditation, the attention is off oneself and into selflessness, working towards the fruits of the spirit—love, joy, peace, patience, kindness, goodness…
More than ever, "we need this recovery of our own interiority, of our own contemplative dimension."
Sat, 02/23/2019 - Fri, 03/01/2019
More from this section [4]
Previous: Monsignor Jozic bids farewell to Hong Kong [5]
Next: World Youth Day delegate thrilled to walk with the pope [6]
Father Laurence Freeman [7]
World Community for Christian Meditation [8]
Source URL: http://sundayex.catholic.org.hk/node/6352
[1] http://sundayex.catholic.org.hk/print/6352
[2] http://sundayex.catholic.org.hk/printmail/6352
[3] http://sundayex.catholic.org.hk/sites/sundayex.catholic.org.hk/files/WCCM.png
[4] http://sundayex.catholic.org.hk/hongkong
[5] http://sundayex.catholic.org.hk/node/6336
[7] http://sundayex.catholic.org.hk/taxonomy/term/498
[8] http://sundayex.catholic.org.hk/taxonomy/term/1086

redpajama_set_name: RedPajamaCommonCrawl


This is a quick kind of off-topic blog about an amazing organization I came across during a recent trip to Costa Rica. I donate a percentage of my profits each year to various organizations and this year I'll be sending some love to the Monkey Farm.
There are three new pieces for you to find on the site.
Hi, all. Welcome to the first blog from my new website. Seems like a good opportunity to talk about the new site!


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The other major banks implemented eftpos systems during 1984, initially with petrol stations.These cards are accepted by merchants who accept credit cards and are processed through the eftpos terminal in the same way as credit cards.Refund and replacement policy : Cards are not refundable for cash.14 In Australia, this facility (known as debit card cashback in many other countries) is known as "cash out".All debit cards now have a magnetic stripe citation needed on which is encoded coles insurance promotional code 2016 the card 's service codes, consisting of three-digit values.If so, we may either provide a refund or a replacement Card of equivalent value unless we reasonably suspect fraud in relation to the Card.Gift, card redeemable at all stores in centre that accept eftpos.In the case of an ATM, the card will not be returned, and the cardholder will need to visit the branch to retrieve the card, or request a new card to be issued.Bank debit cards and other credit cards do not currently offer a contactless payment facility.The second is operated by eftpos New Zealand which is fully owned by VeriFone Systems, following its sale by ANZ New Zealand in December 2012.
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Subsequently, networks facilitating the process of money transfer and payment settlement between the consumer and the merchant grew from a small number of nationwide systems to the majority of payment processing transactions.In 2011, these figures increased to 750,000 terminals, with 325,000 individual businesses, processing over 2 billion transactions with combined value of approximately 131 billion for the year."Mobile drive in Asia-Pacific".The New Zealand Herald.31 A merchant who wishes to accept eftpos payments must enter an agreement with one of the seven merchant service providers, which rent the terminal to the merchant.Cash out is not available in association with credit card sales because on credit card transactions the merchant is charged a percentage commission based on the transaction value, and also because cash withdrawals are treated differently from purchase transactions by the credit card company.Consequently, anyone in possession of the."Commonwealth Bank rolls out contactless terminals and cards".
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text:
Justice League of America (DC Universe Rebirth) Volume 3: Panic in the Microverse
Ryan Choi leads the JLA on a rescue mission into the Microverse to rescue his long-lost mentor, Ray Palmer, the original Atom, and unravel the mystery of his disappearance in DC UNIVERSE: REBIRTH #1. But when disaster strikes and the JLA's shrink ship crashes on a strange world, the team encounters a mysterious being who claims a connection to the missing Ray Palmer. But nobody's going anywhere without the help of Shahn-Zi, a sentient planet with the power to find Palmer.
Collects JUSTICE LEAGUE OF AMERICA #12-17.


text:
Investigate Midwest (https://investigatemidwest.org/tag/salmonella/page/2/)
Late-November recalls include 80 tons of food with 'undeclared allergens'
By Robert Holly | December 3, 2013
The U.S. Food and Drug Administration and the Food Safety and Inspection Service announced recalls for more than 80 tons of food in late November. The recalled products included frozen chicken fettuccine alfredo, dried seaweed and blue cheese. Officials announced the recalls because of dangerous contaminations and undisclosed ingredients.
Food safety officials tweak Salmonella testing process
By Sam Robinson/Midwest Center for Investigative Reporting | September 19, 2013
The Food Safety and Inspection Service announced changes to the process it uses to detect Salmonella in ground beef. Among the changes, beef samples that inspectors examine will increase from 25 grams to 325 grams.
Cracks in the System: The five most common strains of Salmonella
By Sam Robinson/Midwest Center for Investigative Reporting | August 2, 2013
An estimated 1.2 million Salmonella-related illnesses occur each year in the United States. Approximately 400 people die.
While Salmonella is most often associated with poultry products, outbreaks are linked to a wide variety of sources including contaminated ground beef, fruits and vegetables, dog food, turtles and hedgehogs. Here's a look at the five most common
Cracks in the System: Salmonella proves to be a problem in beef too
Salmonella tainted ground beef could be the biggest challenge facing the industry, said a leading beef researcher. Scientists have realized they may have misidentified the source of Salmonella in beef cattle. They now realize it may be in the lymphatic system of cattle, making it harder to prevent than E. coli. As recently as March of 2013, Salmonella Typhimurium in ground beef was linked to more than 20 human illnesses in six states. In September 2012 nearly 50 people in nine states became ill from eating ground beef tainted with Salmonella Enteritidis, according to the Centers for Disease Control and Prevention
"It was always our working assumption that E. coli interventions should be controlling Salmonella," said James Marsden, professor of animal science at Kansas State University.
New egg safety plan shows cracks in system
By Sam Robinson/Midwest Center for Investigative Reporting | July 31, 2013
It's been almost three years since more than 500 million eggs were recalled in 2010 because of an outbreak of Salmonella that caused nearly 2,000 illnesses - the largest outbreak of its kind on record.
Yet under a new egg safety plan approved shortly before the recall, all egg production facilities are still not inspected as required by the plan. In fact, it could be years before they are.


text:
BELOW THE FOLD
By Interactive Strategies
Dead Technologies: Thank You For Being a Friendster
The dawn of "social media" is something that my generation has had the dubious honor of participating in, and indeed pioneering, since before the term "social media" even meant anything (..still not entirely sure what it means). Over the years, my personal level of participation has been undertaken sometimes with trepidation, sometimes with reckless abandon, but always with willing curiosity.
My family "got" Internet in 1995. Try as I might, My 13 year-old brain could not fully wrap itself around the idea of the web, much less the concept of "e-mail". How did it work? I resolved that somehow it must be a lot like the movie Lawnmower Man.
Having briefly experienced the youthful glee of chat rooms at a friend's house (thank you AOL), my formative years were spent trolling message-boards lying in wait for opportunities to flame nitwitted users to my heart's content. As these things go, the novelty soon wore off and throughout high school I largely relegated my online persona to AIM chats and email. Admittedly, I did flirt with blogging for a bit. First with Xanga, followed by then-ubiquitous LiveJournal, both of which remain surprisingly not defunct. However, my first true foray into the "social" sphere was without a doubt, Friendster.
For a lot of 20-somethings in the early 2000's, Friendster is the original "social network." Long before Twitter and Google+, Friendster represented a paradigm shift in the way we shared information about ourselves and yes, "socialized" on the web. For me, it was the first experience I had with things like user profiles, comments ("testimonials") and friend requests. Nowadays, these features are de rigeur components of any web-community worth its salt but in the heyday of Friendster, it was unprecedented. Gone were the salad days of anonymity and cryptic chat-room handles. Connecting with friends and taking full advantage of Friendster meant the laborious yet self-aggrandizing task of composing one's profile, down to the minutest detail. In college, "are you on Friendster?" became a coded IRL means of asking "is it alright if I stalk you (online)?"
Of course, the evolutionary wheel of social media continued to turn. The friendliness of Friendster soon gave way to the creepiness of MySpace, the upstart start-up which took social voyeurism and exhibitionism to new heights. The cacophony of Myspace eventually paved the way for the non-threatening, corn-flower blue blandness of TheFacebook.com which slowly spread from university to university, and the rest as they say is history.
In May 2011, Friendster was acquired by MOL Global, an international media company based in Asia. Friendster has since repositioned itself as some kind of game-based social platform targeting mostly asian markets. A few months prior, I received an email from Friendster offering me the option to export my profile page in anticipation of an "exciting new Friendster." Years since my last login, it seems I'd forgotten to tell Friendster that I'd already moved on. Like many of my former college buddies, Friendster and I simply have nothing in common anymore. So, Friendster if you're listening… I'd like to tell you in no uncertain terms-- we're through. Don't write any testimonials, don't message me. Don't make it weird. It was great while it lasted! But let's face it, we've both changed.
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text:
2006 World Chess Championship
Compiled by Penguincw
★★★ INTRO ★★★
The 2006 World Chess Championship was between 31 year old Vladimir Kramnik of Russia who was the "classical" champion and 31 year old Veselin Topalov of Bulgaria who was the FIDE champion.
★★★ BACKGROUND ★★★
Prior to this matchup, there were 2 chess championships: the "Classical" championship (formed by the PSA) and FIDE championship. During the 2005 FIDE championship (see Game Collection: 2005 FIDE World Chess Championship), Kramnik got invited to participate, but since he was the "classical" champion, he refused to participate. However, he did agree to play the winner. Topalov won the tournament with a round to spare, and soon enough played Kramnik to determine one world champion.
★★★ CONDITIONS ★★★
The match was played in Elista, Russia. The match was a best of 12, meaning the first to 6 1/2 points would be declared the winner. If it was tied after 12 games, 4 rapid games would follow. If it was still tied, then 2 blitz games would follow followed by one armageddon game if necessary. The winner would be declared world chess championship and compete in the 2007 World Championship (see Game Collection: 2007 World Chess Championship) and earn the right for a rematch in 2008 if they lost. The loser would not compete in the world championship next year, but would be directly entered into the final match to decide the challenger for the 2010 match. The first game was played on September 23rd, 2006.
★★★ NOTABLE GAMES ★★★
Game 2: Topalov vs Kramnik, 2006
Game 8: Kramnik vs Topalov, 2006
Game 10: Kramnik vs Topalov, 2006
★★★ FINAL ★★★
Entering Game 12, Kramnik had the white pieces with the match tied up 5 1/2-5 1/2 (Kramnik vs Topalov, 2006 for game). That game was drawn, forcing a tiebreak. Game 13 was drawn (Topalov vs Kramnik, 2006), Kramnik won Game 14 (Kramnik vs Topalov, 2006), but Topalov won Game 15 (Topalov vs Kramnik, 2006), setting up an indentical situation as before, where Kramnik had the white pieces, and the winner of the game winning it all.
With the win, Kramnik had become the one and only world champion by a final score of 8 1/2-7 1/2.
★★★ MINI TABLE ★★★
Kramnik 1 1 1/2 1/2 0 1/2 1/2 0 0 1 1/2 1/2 1/2 1 0 1 (5 wins, 4 losses and 7 draws) <56.7% score>
Topalov 0 0 1/2 1/2 1 1/2 1/2 1 1 0 1/2 1/2 1/2 0 1 0 (4 wins, 5 losses and 7 draws) <43.3% score>
★★★ OTHER LINKS ★★★
See also: Kramnik - Topalov World Championship Match (2006)
Head to Head Record Prior to 2006: http://www.chessgames.com/perl/ches... (19 wins, 9 draws and 34 draws in favour of Kramnik)
Previous "Classical" Championship: Game Collection: 2004 FIDE World Chess Championship
Previous FIDE Championship: Game Collection: 2005 FIDE World Chess Championship
Next World Championship: Game Collection: 2007 World Chess Championship
Game 1, Kramnik leads 1-0
Kramnik vs Topalov, 2006
(E04) Catalan, Open, 5.Nf3, 75 moves, 1-0
Topalov vs Kramnik, 2006
(D18) Queen's Gambit Declined Slav, Dutch, 63 moves, 0-1
Game 3, Kramnik leads 2 1/2-1/2
(E04) Catalan, Open, 5.Nf3, 38 moves, 1/2-1/2
(D47) Queen's Gambit Declined Semi-Slav, 54 moves, 1/2-1/2
Game 5, Kramnik 3-2
(A00) Uncommon Opening, 0 moves, 0-1
Game 6, Kramnik leads 3 1/2-2 1/2
(D17) Queen's Gambit Declined Slav, 31 moves, 1/2-1/2
(D27) Queen's Gambit Accepted, Classical, 60 moves, 1/2-1/2
Game 8, Match tied 4-4
(D47) Queen's Gambit Declined Semi-Slav, 52 moves, 0-1
Game 9, Topalov leads 5-4
(D12) Queen's Gambit Declined Slav, 39 moves, 1-0
Game 10, Match tied 5-5
(E10) Queen's Pawn Game, 43 moves, 1-0
Game 11, Match tied 5 1/2-5 1/2
(D18) Queen's Gambit Declined Slav, Dutch, 47 moves, 1/2-1/2
Game 14, Kramnik leads 7 1/2-6 1/2
Game 16, Kramnik wins 8 1/2-7 1/2
» View all game collections by Penguincw
» Search entire game collection library
» Clone this game collection (copy it to your account)
» FAQ: Help with Game Collections


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Explain why correcting the legislative drafting mistake, also known as "Scrivener's Error" is a difficult process. How would you change Scrivener's Error to more of a streamlined process? Would you propose statues or laws to identify the error(s) in order to reduce misinterpretation? Provide previous evidence showcasing the strict processes of correcting these so-called errors.


text:
The results has been filter on Tags containing Italy.
Rapid impact
p.p1 {margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px 'Helvetica Neue'}p.p2 {margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px 'Helvetica Neue'; min-height: 14.0px}Making a new product a market leader in one year through machine learning, psychology, and...
Catalogue: Congress 2019: The Global Data & Insights Summit
Authors: Genevieve Hall, Vivek Banerji, Daniel Rayner, Alessandro D'Angelo, Dan Atkins
Companies: Insight Dojo, Shionogi
Healthcare / Pharmaceutical / Wellness
New Product Research
Hall, Banerji, Rayner, D'Angelo and Atkins (2019a, September 08). Rapid impact. ANA - ESOMAR. Retrieved January 28, 2023, from
https://ana.esomar.org/documents/rapid-impact
The case of Alimentare Emiliana
In October 1968, the Chairman and Managing Director of Alimentare Emiliana S.p.A., Mr. Limentani, had considered the advisability of entering into a new sector of pre-packed foods, for which the prospects of development in Italy seemed to be...
Catalogue: Seminar 1968: Operational Research In Marketing
Author: Pietro Gennaro
Packaging Research
Gennaro, P. (1968a, November 01). The case of Alimentare Emiliana. ANA - ESOMAR. Retrieved January 28, 2023, from
https://ana.esomar.org/documents/the-case-of-alimentare-emiliana
Control techniques in a sample survey
The analysis described in this report concerns the control techniques used in sample surveys , and is based on a ten-year experience made by field work in this area and specifically applied to the survey on the Italian Periodical Press Readership by...
Catalogue: ESOMAR Congress 1979: The Challenge Of Eighties
Authors: Angela Amoroso, Costantino Jannacone
Readership Study
Amoroso and Jannacone (1979a, June 15). Control techniques in a sample survey. ANA - ESOMAR. Retrieved January 28, 2023, from
https://ana.esomar.org/documents/control-techniques-in-a-sample-survey
Business requirements and future expectations in competitive bank services
An experimental study on attitudes towards time and on the behavioural implications of such attitudes was conducted on 46 staff members of an Italian bank. The methods used were a Semantic Differential (SD) composed of three stimulus words (the past,...
Catalogue: Seminar 1988: Research For Financial Services
Author: Gabriele Morello
Time Perception
Attitude Research
Semantic Differential Technique
Morello, G. (1988a, June 15). Business requirements and future expectations in competitive bank services. ANA - ESOMAR. Retrieved January 28, 2023, from
https://ana.esomar.org/documents/business-requirements-and-future-expectations-in-competitive-bank-services
Advertising research (French)
This paper is intended to present the point of view of the users of Marketing Research. In this, particular instance, the user is a large oil company operating and marketing in Italy: Esso Standard Italiana. This is the case history of a very...
Catalogue: ESOMAR Conference 1960
Author: Enrico Zanelli
Company: ESSO
Energy / Utilities
Sales-Advertising Relationships
Zanelli, E. (1960a, June 15). Advertising research (French). ANA - ESOMAR. Retrieved January 28, 2023, from
https://ana.esomar.org/documents/advertising-research-french-
In what way does the quantitative "mix" of subject-matter pre-determine sales and readership profiles?
The quantitative mix of a publication's editorial matter undoubtedly plays a basic role in pre-determining the reader's profile and consequently its circulation. I have intentionally inverted the two elements, because circulation is a function of the...
Catalogue: Seminar 1974: Editorial Research In The Publishing Industry
Author: Maria Macchi
Editorial Research
Audience Research
Reader-Magazine-Relationship
Macchi, M. (1974a, June 15). In what way does the quantitative "mix" of subject-matter pre-determine sales and readership profiles?. ANA - ESOMAR. Retrieved January 28, 2023, from
https://ana.esomar.org/documents/in-what-way-does-the-quantitative-mix-of-subject-matter-pre-determine-sales-and-readership-profiles-
Clustering Italian agriculture
This paper outlines the results of an integrated study carried out for I.C.I. Solplant by Burke B.I.S. in 1985 which aimed at identifying differentiated typologies of farmers on the Italian market. The study was divided into two phases: 1) The...
Catalogue: Seminar 1987: Identifying Change in European Agriculture
Authors: Alberto Garbarino, Roberto Luperini
Company: Burke, Inc.
Sampling Designs
Garbarino and Luperini (1987a, June 15). Clustering Italian agriculture. ANA - ESOMAR. Retrieved January 28, 2023, from
https://ana.esomar.org/documents/clustering-italian-agriculture
From number of sales to market share
Grocery chain supermarkets, are subject to local conditions which may greatly influence their performance. The marketing function has the responsiblity for investigating the current conditions and the potential for each store. The companies...
Catalogue: Seminar 1989: Adding Value To Retail Offerings
Authors: Franco Cloni, Massimo Milleri
Cloni and Milleri (1989a, June 15). From number of sales to market share. ANA - ESOMAR. Retrieved January 28, 2023, from
https://ana.esomar.org/documents/from-number-of-sales-to-market-share
New language and communication for social advertising
The goal of this paper Is to investigate social advertising in Italy as a concept (what It is, what it means, how people see it) and as communication through some examples of current advertising. It considers a qualitative research concerning the...
Catalogue: Seminar 1990: Qualitative Research- How Are We Preparing For The Future?
Author: Cristina Nava
Nava, C. (1990a, June 15). New language and communication for social advertising. ANA - ESOMAR. Retrieved January 28, 2023, from
https://ana.esomar.org/documents/new-language-and-communication-for-social-advertising


Such a great awning for a great price!
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text:
News / Palestine
ICRC warns Palestinian prisoner enters medically critical phase after over 85 days of hunger strike
Thursday, 22 October 2020 7:04 PM [ Last Update: Thursday, 22 October 2020 7:17 PM ]
This file photo shows Palestinian prisoner and hunger striker Maher al-Akhras.
Palestinian prisoner Maher al-Akhras is entering a medically "critical phase", the International Committee of the Red Cross (ICRC) warns, as the inmate has entered the 88th consecutive day of his open-ended hunger strike against Israel's so-called policy of administrative detention.
"More than 85 days into the hunger strike, we are concerned about potentially irreversible health consequences," said Yves Giebens, the head of the ICRC's health department in Israel and the occupied Palestinian territories, in a statement on Thursday.
He also stressed that "from a medical perspective, he is entering a critical phase", adding that the ICRC had been "closely monitoring" the situation, including a visit with Akhras earlier in the day.
The 49-year-old Palestinian prisoner, the father of six children, was detained on July 27 and held under the administrative detention order, with no charge. This has led him to start a hunger strike in an attempt to seek justice against the unfair detention.
Physicians have already warned of damage to several organs of the Palestinian prisoner's body, such as the kidneys, liver, and heart, adding that the inmate's senses of hearing and speaking have also been affected negatively.
"The ICRC encourages the patient, his representatives and the competent authorities involved to find a solution that will avoid any loss of life," the statement further said.
Separately on Thursday, the Palestinian Information Center, citing a report by Asra Media Office, said Akhras's health condition has reached a critical stage due to his prolonged hunger strike and refusal to take any vitamins and fluids.
It added that the inmate's health is extremely worrying and deteriorating as he suffers from serious physical weakness, repeated convulsions and loss of consciousness, and constant headaches and stomach aches.
Physicians have warned that Akhras's heart might stop at any moment.
Palestinian prisoner's organs failing as his hunger strike enters 84th consecutive day
Palestinian prisoner Maher al-Akhras's organs are failing as he enters the 84th consecutive day of his open-ended hunger strike.
"I am committed to my decision and will eat food only in my home, and I will not break my will. I am now in Kaplan Hospital, I do not drink anything but water, and I will continue doing so until I return to my house," he said last week.
In 1989, he was arrested for the first time and his detention continued for seven months. Back in 2004, he was arrested for a second time and spent two years behind bars. In 2009, he was rearrested and remained in administrative detention for 16 months. In 2018, Akhras was arrested again and was held in custody for 11 months.
Despite the fact that there is no criminal offense that the Tel Aviv regime is holding Akhras for, the Israeli authorities refuse to release him, even as his health increasingly deteriorates on a daily basis.
The Israeli regime has so far rejected all calls to release Akhras.
Hundreds of Palestinian prisoners are held under administrative detention, in which Israel keeps the detainees for up to six months, a period which can be extended an infinite number of times. Women and minors are among these detainees.
Palestinian detainees have continuously resorted to open-ended hunger strikes in an attempt to express their outrage at the detention. Palestinians hold Israeli authorities fully responsible for any deterioration of the circumstances in jails.
More than 7,000 Palestinians are reportedly held in Israeli jails.
Israel Palestine ICRC administrative detention prison hunger strike Maher al-Akhras
MORE FROM Palestine
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President Abbas decrees first Palestinian elections in 15 yrs
Israeli forces injure Palestinian protesters in occupied West Bank
Palestinians warn about deteriorating condition of hunger striking prisoner
Palestinian hunger-striking inmate may die any moment: Report
Palestinian on hunger strike in Israeli jail in critical condition


Moms And Dads — How Did You Know You Were Done Having Kids?
"You know you're done having babies when you hear a pregnancy announcement and you're relieved it isn't you."
Recently, we asked our readers to share the moment they knew they were ready for another kid, and the responses were totally relatable.
But then it occurred to us that parents who have decided they're finished having kids may have experienced a similar 'aha' moment.
So we want to know the moment you realized you were done having kids.
Maybe it was when you got a taste of what another one — or two — would be like, and you were overwhelmed.
Or perhaps it was your little one's epic tantrum that convinced you to pack up shop.
Or maybe it was when you realized that parenthood is a dirtier job than you thought.
Whatever made you decide you were done having kids, we want to hear about it! Tell us in the Dropbox below, and you could be featured in an upcoming BuzzFeed post!


text:
Q: How to specify the minor version of Microsoft.NetCore.App I have a web app made in .NET Core 3.1, which uses the Microsoft.NetCore.App framework.
My csproj file looks like this:
<PropertyGroup>
    <TargetFramework>netcoreapp3.1</TargetFramework>
    <ApplicationIcon />
    <Win32Resource />
    <NoWin32Manifest>true</NoWin32Manifest>
    <RunPostBuildEvent>OnOutputUpdated</RunPostBuildEvent>
  </PropertyGroup>

When I run the app in my local machine, both are taking by default the version that I have installed: 3.1.29. I know this, because when I run the app I see its taking the dlls from this folder: C:\Program Files\dotnet\shared\Microsoft.NETCore.App\3.1.29
I assume its taking the latest version, by default.
The problem is, in the servers, I have 3.1.28 installed. So when I run the app in the server I get this error:

HTTP Error 500.31 - ANCM Failed to Find Native Dependencies Common
solutions to this issue: The specified version of
Microsoft.NetCore.App or Microsoft.AspNetCore.App was not found.

How can I force my app to run using the 3.1.28 version of this framework? I cannot update the framework in the server, because I got no permissions.
I tried with <TargetFramework>netcoreapp3.1.28</TargetFramework>, but that won't work.
I also tried to specify the version in a global.json file, but it seems is not doing anything:
{
  "frameworks": {
    "netcoreapp3.1": {
      "version": "3.1.28"
    }
  }
}

Edit: This is what I got in the server, in case you are wondering if something is missing:
dotnet --info
.NET SDK (reflecting any global.json):
Version: 6.0.400
Commit: 7771abd614

Runtime Environment:
OS Name: Windows
OS Version: 10.0.14393
OS Platform: Windows
RID: win10-x86
Base Path: D:\Program Files (x86)\dotnet\sdk\6.0.400\

global.json file:
Not found

Host:
Version: 6.0.8
Architecture: x86
Commit: 55fb7ef977

.NET SDKs installed:
1.1.14 [D:\Program Files (x86)\dotnet\sdk]
2.1.526 [D:\Program Files (x86)\dotnet\sdk]
2.2.109 [D:\Program Files (x86)\dotnet\sdk]
3.1.118 [D:\Program Files (x86)\dotnet\sdk]
3.1.420 [D:\Program Files (x86)\dotnet\sdk]
3.1.421 [D:\Program Files (x86)\dotnet\sdk]
3.1.422 [D:\Program Files (x86)\dotnet\sdk]
5.0.406 [D:\Program Files (x86)\dotnet\sdk]
5.0.408 [D:\Program Files (x86)\dotnet\sdk]
6.0.301 [D:\Program Files (x86)\dotnet\sdk]
6.0.302 [D:\Program Files (x86)\dotnet\sdk]
6.0.400 [D:\Program Files (x86)\dotnet\sdk]

.NET runtimes installed:
Microsoft.AspNetCore.All 2.1.30 [D:\Program Files (x86)\dotnet\shared\Microsoft.AspNetCore.All]
Microsoft.AspNetCore.All 2.2.14 [D:\Program Files (x86)\dotnet\shared\Microsoft.AspNetCore.All]
Microsoft.AspNetCore.App 2.1.30 [D:\Program Files (x86)\dotnet\shared\Microsoft.AspNetCore.App]
Microsoft.AspNetCore.App 2.2.14 [D:\Program Files (x86)\dotnet\shared\Microsoft.AspNetCore.App]
Microsoft.AspNetCore.App 3.0.3 [D:\Program Files (x86)\dotnet\shared\Microsoft.AspNetCore.App]
Microsoft.AspNetCore.App 3.1.27 [D:\Program Files (x86)\dotnet\shared\Microsoft.AspNetCore.App]
Microsoft.AspNetCore.App 3.1.28 [D:\Program Files (x86)\dotnet\shared\Microsoft.AspNetCore.App]
Microsoft.AspNetCore.App 5.0.15 [D:\Program Files (x86)\dotnet\shared\Microsoft.AspNetCore.App]
Microsoft.AspNetCore.App 5.0.17 [D:\Program Files (x86)\dotnet\shared\Microsoft.AspNetCore.App]
Microsoft.AspNetCore.App 6.0.7 [D:\Program Files (x86)\dotnet\shared\Microsoft.AspNetCore.App]
Microsoft.AspNetCore.App 6.0.8 [D:\Program Files (x86)\dotnet\shared\Microsoft.AspNetCore.App]
Microsoft.NETCore.App 1.0.16 [D:\Program Files (x86)\dotnet\shared\Microsoft.NETCore.App]
Microsoft.NETCore.App 1.1.13 [D:\Program Files (x86)\dotnet\shared\Microsoft.NETCore.App]
Microsoft.NETCore.App 2.0.9 [D:\Program Files (x86)\dotnet\shared\Microsoft.NETCore.App]
Microsoft.NETCore.App 2.1.30 [D:\Program Files (x86)\dotnet\shared\Microsoft.NETCore.App]
Microsoft.NETCore.App 2.2.14 [D:\Program Files (x86)\dotnet\shared\Microsoft.NETCore.App]
Microsoft.NETCore.App 3.0.3 [D:\Program Files (x86)\dotnet\shared\Microsoft.NETCore.App]
Microsoft.NETCore.App 3.1.27 [D:\Program Files (x86)\dotnet\shared\Microsoft.NETCore.App]
Microsoft.NETCore.App 3.1.28 [D:\Program Files (x86)\dotnet\shared\Microsoft.NETCore.App]
Microsoft.NETCore.App 5.0.15 [D:\Program Files (x86)\dotnet\shared\Microsoft.NETCore.App]
Microsoft.NETCore.App 5.0.17 [D:\Program Files (x86)\dotnet\shared\Microsoft.NETCore.App]
Microsoft.NETCore.App 6.0.7 [D:\Program Files (x86)\dotnet\shared\Microsoft.NETCore.App]
Microsoft.NETCore.App 6.0.8 [D:\Program Files (x86)\dotnet\shared\Microsoft.NETCore.App]

redpajama_set_name: RedPajamaStackExchange


text:
Q: Angular ReactiveForm - form stays valid even after add required (by addValidators) to one formControl In my ReactiveForm form, I add required validator to one formControl by condition.
if(condition) setRequired();
setRequired() { this.form.get(['company','id']).addValidators(Validators.required);}
but form.valid is true, even the field is empty.
Do I need to add something else?
Thanks.

A: After remove/add validator to your form element, you have to invoke function with updateValueAndValidity to update the state.
Like: this.form.updateValueAndValidity();
Or: this.form.get(['company','id']).updateValueAndValidity();


Corn and soybeans prices were firmer again last night as December corn futures made a new high for the move for the 11th consecutive session. We are of course treading through a weather market at the moment as the trade is starting to get worried about weather for the summer. The hot weekend forecast isn't helping calm anyone down and when you see a few drier extended forecasts… the long side of the market is the much more comfortable play.
Remember – this Friday at 11am we will have the next S&D released. Looking at trade estimates, we are already trading reduced carryouts in both corn and soybeans. We will have to wait until Friday to see if the USDA shocks the market one way or the other on that. While this June S&D will be interesting, the June 30th stocks and final acreage numbers tend to bring the most excitement.
Outside markets… crude oil is continuing to march higher, now up over $51 while the US Dollar is weaker – all of that is lending support to our corn/soybean markets. There was also some data out showing that the Chinese economy is not slipping as much as originally thought – which I suppose you can take as friendly.


text:
Upbeat on Singapore economy
WITH 2007 drawing to a close, Singapore's economy looks set to end the year on a high.
The Monetary Authority of Singapore — which is due to release its semi-annual Macroeconomic Review and monetary policy statement today — has said that it expects the economy to grow between 7 and 8 per cent this year.
And yesterday, Prime Minister Lee Hsien Loong told some 1,000 unionists and delegates at the National Trades Union Congress National Delegates' Conference that he was optimistic the economy would grow at "the higher end of this range".
"The outlook is generally upbeat. The economy is doing very well … With good growth, the labour market has tightened. Many reasons for unionists and workers to smile," said Mr Lee, noting the stellar job and wage growth.
But in spite of the rosy conditions, Mr Lee said the Government — whose "job is not just to smile but continue to watch for signs of problem" — is keeping a close eye on two issues: The shaky global financial markets and the red-hot domestic property market.
The stock market jitters in recent months might have started from the sub-prime mortgage woes in the United States, but Mr Lee observed that the financial markets had been "over-confident and overdue for some correction".
Said Mr Lee: "If it's not been the sub-prime mortgage, it could have been something else."
While the jury is still out on whether a recession in the US is on the horizon, Mr Lee was in no doubt that the Singaporean economy would be affected should that happen.
Even so, the Republic's economic links with China and India — which are both growing unimpeded — would "help us weather a US downturn", the Prime Minister added.
At home, Singapore faces an "acute shortage" of prime office space due to the economic boom. The Government is taking steps to increase the supply over the next two to three years, "to not just stabilise the market but … so that lots more businesses can come and set up in Singapore".
Policy-makers are also monitoring the residential property market closely — especially in terms of keeping housing affordable, said the Prime Minister.
Last Friday, the Ministry of National Development withdrew its Deferred Payment Scheme for property purchases — a move that Mr Lee said would "help to dampen excessive speculation and inject some reality into the market".
Other measures will follow if necessary, he assured Singaporeans.
Currently, the Government is releasing more land for Executive Condominiums, targeted at first-time buyers whose household income exceed $8,000 and are thus ineligible for a Central Provident Fund housing grant in buying an HDB flat.
Adding that the move was a response to the "many appeals" on raising the income ceiling, Mr Lee said there "should not be a sandwich group".
"There is enough land in Singapore. There's no need for anybody to get alarmed that this is the last chance and if you don't get on, you will miss the boat," he said.
At a separate event, Trade and Industry Minister Lim Hng Kiang said the economy is not overheating but the rising oil and commodity prices are causing inflationary pressure.
Adding that the Government still expects inflation to average between 1.5 per cent and 2 per cent this year, Mr Lim said: "I don't see a serious problem with overheating (in the economy) … There are some supply side constraints and we're taking steps to address them."
Tharman: S'pore not headed for recession
SM Goh says Singapore can cope even if US goes into recession
Polls show that hiring outlook for Singapore remains strong


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text:
Moulton to drop presidential bid
By Colin A. YoungState House News Service
BOSTON - Congressman Seth Moulton will end his bid for the Democratic Party's presidential nomination in a speech Friday afternoon, his campaign said.
He will also announce he will seek re-election to Congress.
The Salem Democrat, who was the second member of the Massachusetts congressional delegation to wade into to the crowded fray of Democratic candidates for 2020, will make the announcement in a speech at the Democratic National Committee's Summer meeting.
As he ends his presidential campaign, Moulton is expected to talk about the importance of service and how his campaign helped to chip away at the stigma of mental health issues, especially among fellow veterans.
"As a party, we need to spend these next months doing everything we can to determine who our strongest nominee will be. And as we do so, we can't be distracted by the latest outrage on Twitter or the reality show politics of the day," Moulton will say, according to a copy of his prepared remarks. "We need to stay grounded in what it's going to take to win over voters - all kinds of Democrats, independents like the Obama-Trump voters, and even disaffected Republicans - and bring the country together around beating Donald Trump. Because you can be a Democrat, or a Republican, or an independent. But if you're an American, you know that we're better than this."
While Moulton has been running for president, other Democrats have been eyeing his 6th Congressional District seat. In May, the News Service reported that state Rep. Lori Ehrlich of Marblehead, former state Sen. Barbara L'Italien of Andover, Salem State University trustee Jamie Zahlaway Belsito of Topsfield and former State House aide and communications specialist Angus McQuilken were considering campaigns for the seat.