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 \centerline {\bf O.I.~Reinov (Saint Petersburg)
  }
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 { \bf
   Properties $ \mathrm{AP}(l_p)$ and \\
  $ l_p$-factorizations of operators%
\footnote%text
{This work was done with partial support of
FCP ``Integracija", by a grant of "Scientific schools",
and by the Swedish Royal Academy
of Sciences.
This research is supported by a grant from Higher Education Commission,
Pakistan.
 }
 %}
}
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Is it true that if the adjoint to $ T$ operator $ T^*$
compactly factors through the space $ l_p,$
then the operator $ T$ itself can be (strictly) factored
through the space $ l_{p'}?$

The answer to this question is negative if $ p=1$ or $ p=\infty$
(see [2], [3], [4]).
The answer is still unknown for the other values of parameter
 $ p,$ i.e. in the case where $ 1<p\neq2<\infty.$

Let us stop for a while and look at the situation somewhat clearing
the difficulties in solving this problem.
Definitions and details connecting with the  considerations to be
given now can be found below.


Let us consider the Banach tensor product
$ X^*\widetilde{\widetilde{\otimes}}_p Y$ (the completion in
the norm (1) of $ X^*{\otimes} Y$),
associated with the ideal
$ \mathrm{K}_p$ of the compactly
$ l_p$-factored operators in Banach spaces
(in particular, every tensor in the considering situation
generates an operator which can be factored through the space
$ l_p$ compactly).
Each tensor $ z\in X^*\widetilde{\widetilde{\otimes}}_p Y,$
as well as every operator $ S\in \mathrm{ K}_p(X,Y),$
admits a representation in the form (3), and, moreover,
for a given number
$ \varepsilon>0,$ this representation for $z$ can be taken
in such a way that the relation  (4) if fulfilled,
where  $ k_p^0(z)$ is the norm of the element $ z$
in its tensor product.
Every operator $ S,$ as was said just before, also admit
an expansion  of such a kind into a series of 1-dimensional operators,
but with the estimation at the right
by its  $ \mathrm{ K}_p$-norm:
$ \dots\le k_p(S)+\varepsilon.$
Besides, the norms $ k_p^0$ and $ k_p$ are defined for the finite
rank operators of the kind (2) very similarly
(see below).

At the first sight, it is not seen any difference between the tensor
$ z\in  X^*\widetilde{\widetilde\otimes}_p Y$
and the corresponding to it operator $ S=\widetilde z,$
especially, if to look at their expansions into the series of the kind
of (3).
However, an essential difference in their definitions can be
seen if we consider the finite dimensional operators (or tensors):
for the finite dimensional operator $ S$ from (2)
(let $z$ be a tensor which corresponds to it),
its norm in the space $ \mathrm{K}_p(X,Y)$ is computed
as the infimum of the left part of the relation (4)
{\it over all}\,
corresponding representations of the operator $S$ in the form (3) ,
while the norm of the associated tensor element $ z$ ---
as the infimum  of the left part of the relation (4)
over all expansions of the operator  $ S$
{\it into a finite sum of one dimensional mappings.}
This difference appears
to be very essential, and we know that at least for the cases where
$ p=1, p=\infty,$
the considered norms, generally speaking, are not equal each to other
even on the finite dimensional operators
(about the case $ p=\infty$ see [2, Corollary 3.4 and
the proofs before the corollary] and also [4];
the case  $ p=1$ was considered in  [3],
but the corresponding assertion was appeared, without any proof,
already in the paper  [2, Corollary 4.2]; cf. also [4]).

If  $ 1\le p\le\infty$ and
 $ S$ is a finite dimensional operator from $ X$ to $ Y,$
then its
 $ \mathrm{K}_p^0$-norm is defines as the quantity
$$   k_p^0(S):=
  \inf \left\{ (\sup_{1\leq i\leq} |a_i|)\,
        \varepsilon_p(x'_j)\,\varepsilon_{p'}(y_j)
       \right\}, \eqno{(1)}
$$
where the sup is taken over all  {\it finite} representations of the
operator $ S$ in the form
$$ S=\sum_{j=1}^n a_j\,x'_j\otimes y_j.  \eqno{(2)}
$$
The completion of the algebraic tensor product $ X^*\otimes Y$ (
which we consider as a linear space of all finite dimensional
operators) in this norm  $ k_p^0$ will be denoted by
$ X^*\widetilde{\widetilde\otimes}_p Y,$
and for the norm in this completion we take the same notation $
k_p^0.$
Every tensor element $ z\in  X^*\widetilde{\widetilde\otimes}_p Y$
can be represented in the form
of a convergent series
$$ \sum_{j=1}^{\infty} a_j\,x'_j\otimes y_j,\qquad  \alpha_j\to 0, \eqno{(3)}
$$
and, for a given number
$ \varepsilon>0,$
this representation can be chosen in such a way that
$$ (\sup_{1\leq i<\infty}  |a_i|)\,
\varepsilon_p(x'_j)\,\varepsilon_{p'}(y_j)
                                    \leq k_p^0(z)+\varepsilon    \eqno{(4)}
$$

The tensor product $ X^*\widetilde{\widetilde\otimes}_p Y$
generates naturally a linear subspace of operators
$ \mathrm{K}_p(X,Y)$ in
$ \mathrm{L}(X,Y)$;
this is a factor space of the considered tensor product
over the kernel of the natural map
$ X^*\widetilde{\widetilde\otimes}_p Y\to \mathrm{L}(X,Y).$
The corresponding norm on this space of operators is denoted
by  $ k_p.$ $ \left[\mathrm{K}_p, k_p\right] $ is a Banach
operator
ideal and it is a particular case of the so-called ideals of
$ (r, p ,q)$-nuclear operators ---
$ \left[\mathrm{N}_{(r,p,q)}, \nu_{(r,p,q)}\right]\!:$\ \,
$ \left[\mathrm{K}_p, k_p\right] =
\left[\mathrm{N}_{(\infty,p,p')}, \nu_{(\infty,p,p')}\right]$
(details can be found in the book [1]).
Every operator  $ S\in  \mathrm{K}_p(X,Y)$
can be represented in the form of a convergent in the space
$ \mathrm{K}_p(X,Y)$ series (3),
and, for a given number $ \varepsilon>0,$
this representation can be chosen in such a way that
$$  (\sup_{1\leq i<\infty} |a_i|)\,
\varepsilon_p(x'_j)\,\varepsilon_{p'}(y_j)
                          \leq k_p(S)+\varepsilon. \eqno{(5)}
$$

Now, let us look at the discussed
at the moment operator ideal from another side.
We shall say that an operator
$ S: X\to Y$ {\it compactly factors through the space
$ l_p,$} \, (in the terminology of A. Pietsch [1], is
{\it $ p$-compact}\,),   if there exist two compact operators
$ A: X\to l_p$ and $ B: l_p\to Y,$ for which $ S=BA.$
As a norm of such compactly $l_p$-factored operator we take a quantity
$ \inf\,\|A\|\,\|B\|,$  where the infimum is taken over all possible
factorizations of the operator  $ S$ of the mentioned kind.

One of the main theorems in the theory of compactly $l_p$-factored
operator is formulated as follows:
$$ k_p(S) = \inf\,\{\|A\|\,\|B\|:\, S=BA: X\to l_p\to Y;\, A,B \,
\mathrm{\ are\ compact} \}
$$
and {\it the ideal $ \mathrm{K}_p$
coincides with the ideal of all operators which are
compactly factored through}\ $ l_p$; in the case $ p=\infty$
the compact factorization through $ l_{\infty}$  is just the same
as the compact factorization through $ c_0$
(see [1], Theorems 18.3.2, 18.1.3).

A little bellow we will consider some of the facts, obtained by us and
connected with the presence or the absence
of the approximation properties with respect to the tensor norms
 $ k_p^0$ in Banach spaces $( 1\leq p\leq \infty).$
Let us denote shortly by  $ \mathrm{AP}(l_p)$
the approximation property with respect to $ k_p^0$
We write $ X\in \mathrm{AP}(l_p),$
if the space $ X$ has the property
$ \mathrm{AP}(l_p.)$
It is worthwhile
to recall that  $ X\in\mathrm{AP}(l_p)$
{\it if for every Banach space  $ Y$
the canonical mapping from $ Y^*\widetilde{\widetilde\otimes}_p X$
to $\mathrm{L}(Y,X)$ is one-to-one}.\,
This is equivalent to the fact that
for every $ Y$ we have a canonical isometry
$ Y^*\widetilde{\widetilde\otimes}_p X= \mathrm{K}_p(Y,X).$

It is for a long time not known {\it if there exist the spaces without
the property $ \mathrm{ AP}(l_p)$ for any $ p$
not equal to one, two or infinity}.
It is not hard to show that every Banach space has the property
$ \mathrm{ AP}(l_2).$
On the other hand, it was shown by the author that {\it
there exist separable Banach spaces without the properties
$ \mathrm{ AP}(l_1)$ and $ \mathrm{ AP}(l_{\infty})$}
(cf. [2], [3], [4]).

Recall that a Banach space is said to be {\it completely separable}
if it and all of its duals are separable.

\vskip 0.3 cm
 {\bf Theorem 1.} {\it
Let $ p\in[1,\infty].$
Suppose  that there exists a Banach space without the property
$ \mathrm{AP}(l_p).$
Then

$ 1)$ there exist  a completely separable Banach space
 $ X$ and a separable reflexive space $ Y,$
an operator $ U\in \mathrm{QN}_{p}^{\mathrm{dual}}\circ \mathrm{QN}_{p'}(X,Y)$
such that $ X\in \mathrm{AP},$
$ U$ lies in the closure of the subspace
 $ X^*\otimes Y$
of the space $ \mathrm{ D}_{p'},$
dual to $ X^*\otimes Y,$
in the topology $ \sigma ( \mathrm{ D}_{p'}(X,Y), Y^*
  \widetilde{\widetilde\otimes}_p X)$
(and, therefore, in the topology of compact convergence), but
$ U$ is not in the closure of the ball of radius $ C$
of the subspace $ X^*\otimes Y$
 of the space $ \mathrm{ D}_{p'}(X,Y)$
in the topology of the compact convergence for any $ C<\infty$;

$ 2)$
there exist   a completely separable Banach space
 $ X$ and a separable reflexive space $ Y,$
an operator $ T\in \mathrm{L}(Y,X)$
such that $ X\in \mathrm{AP},$ $ T\in\mathrm{K}_p(Y,X^{**})$ and
$T\notin \mathrm{K}_p(Y,X);$
moreover, there exists a sequence of  operators
$ \left\{ T_n\right\}$ v $ Y^*\otimes X,$
which is convergent to $ T$ in norm induced from
$ \mathrm{K}_p(Y, X^{**}).$
Thus, in this case  the natural injection
$
 \mathrm{ K}_p(Y,X)\subset
\mathrm{ K}_p(Y,X^{**})
$
is not isometric.}
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{\bf Theorem 2.} {\it
If there exists a Banach space which does not possess the
property $ \mathrm{AP}(l_p)$ then
there exist a separable reflexive Banach space
$ X,$ a completely separable Banach space $ Z$ and an operator
$ T\in \mathrm{L}(X,Z)$ such that $ Z^{**}$ has a basis,
$ X\notin \mathrm{AP}(l_p),$
$T^{**}\in \mathrm{ K}_p(X,Z^{**}),$ but
$ T\notin \mathrm{ K}_p(X,Z).$
Moreover, in this case the natural injection
$
 \mathrm{ K}_p(X,Z)\subset
\mathrm{ K}_p(X,Z^{**})
$
is isometric, but not onto.}
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Now, let us gather some of the facts obtained by us about the spaces
with the properties $ \mathrm{AP}(l_p)$ and
formulate them as one very general theorem.
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{\bf Theorem 3.} {\it
For every $ p\in [1,\infty]$ the following statements
are equivalent.

$ 1)$ Every Banach space possesses the property $\mathrm{AP}(l_p);$

$1')$ Every separable reflexive Banach space possesses the property
$\mathrm{AP}(l_p);$

$ 2)$  Every Banach space possesses the property $\mathrm{AP}(l_{p'});$

$ 2')$  Every separable reflexive Banach space possesses the property
$\mathrm{AP}(l_{p'});$

$ 3)$ There does not exist an operator acting in Banach spaces
which is not a $ \mathrm{ K}_p$-operator, but whose second
adjoint lies in the ideal $ \mathrm{ K}_p;$

$ 3')$  There does not exist an operator acting in completely
separable Banach spaces with bases
which is not a $ \mathrm{ K}_p$-operator, but whose second
adjoint lies in the ideal $ \mathrm{ K}_p;$

$ 4)$ the ideal $ \mathrm{ K}_p$ is regular;

$ 5)$ the ideal $ \mathrm{ K}_{p'}$ is regular;

$ 6)$ the ideal $ \mathrm{ K}_p^{ \mathrm{ reg}}$
is minimal;

$ 7)$ the ideal $ \mathrm{ K}_{p'}^{ \mathrm{ reg}}$
is minimal;

$ 8)$ the ideal of the finite dimensional operators
is dense in the ideal
    $ \mathrm{ K}_{p}^{ \mathrm{ reg}};$

$ 9)$  the ideal of the finite dimensional operators
is dense in the ideal
     $ \mathrm{ K}_{p'}^{ \mathrm{ reg}};$

$10)$ for every separable reflexive Banach space
$ X$ the canonical embedding
$ X^*\widetilde{\widetilde\otimes}_p X\to \mathrm{L}(X,X)$
is one-to-one;

$11)$  for every  Banach space
$ X$ the canonical embedding
$ X^*\widetilde{\widetilde\otimes}_p X\to \mathrm{L}(X,X)$
is one-to-one.}
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\centerline {REFERENCES}
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1. Pietsch~A.
 {\it Operator ideals} \rm --
North-Holland, 1980.

2. Reinov~O.~I.
{\it  Approximation properties of order p and the existence of
  non-p-nuclear operators with p-nuclear second adjoints} //
 Math. Nachr. -- 109 (1982). -- P.~125-134.

3. Reinov~O.~I.
{\it On factorization of operators through the spaces $ l^p$} //
  Vestnik SPb GU Ser.1 -- 2 (2000). -- P.~27-32 (in Russian).

4.  Reinov~O.~I.
{\it  Approximation properties and some classes of operators} //
  Problemy matematicheskogo analiza -- 23 (2001). -- P.~147-205
 (in Russian).

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