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\journal{Journal of Mathematical Analysis and Applications}

\begin{document}

\begin{frontmatter}

\title{On linear operators with $s$-nuclear adjoints, $0< s\le 1$}


\author{O.I. Reinov}

\address{Saint Petersburg  State University\\
 St. Petersburg, Petrodvorets,
28 Universitetskii pr., 198504 Russia}

\begin{abstract}
We prove that if $ s\in (0,1]$ and $ T$ is a linear operator with $ s$-nuclear
adjoint from a Banach space $ X$ to a Banach space $ Y$ and if one
of the spaces $ X^*$ or $ Y^{***}$ has the approximation property of order $s,$ then
the operator $ T$ is nuclear.
The result is in a sense exact. For example, it is shown that for each $r\in (2/3, 1]$
there exist a Banach space $Z_0$ and a non-nuclear operator
$ T: Z_0^{**}\to Z_0$ so that  $ Z_0^{**}$ has a Schauder basis,  $ Z_0^{***}$
has the $AP_s$ for every $s\in (0,r)$ and $T^*$ is $r$-nuclear.

\end{abstract}

\begin{keyword}
$s$-nuclear operator, Schauder basis,
approximation property, tensor product.

%% MSC codes here, in the form: \MSC code \sep code
%% or \MSC[2008] code \sep code (2000 is the default)
\MSC[2010]  47B10. Operators belonging to operator ideals (nuclear,
p-summing, in the Schatten-von Neumann classes, etc.)
\end{keyword}

\end{frontmatter}

%%
%% Start line numbering here if you want
%%
% \linenumbers

%% main text
\section{Introduction}
\label{Introduction}




 %%%%%%%%%%%%%
    %%%%%%%%%%%%%%%%%%%%%%


\smallskip

We will be interested in the following question from [6, Problem 10.1]:        
Suppose $T$ is a (bounded linear) operator acting between Banach spaces $X$ and $Y,$
and let $s\in(0,1).$ Is it true that if $T^*$ is $s$-nuclear then $T$ is $s$-nuclear too?

As is well known, for $s=1,$ a negative answer was obtained already
by T. Figiel and W.B. Johnson in [4].                 
For $s\in (2/3,1]$ the negative answer can be found, e.g., in [14].     
Here we are going to give some (partially) positive results in this direction as well as to show the sharpness of them.

It is not difficult to see that if 
$T^*$ is $s$-nuclear, then $T$ is $p$-nuclear with 
%$T^*\in N_s(Y^*,X^*)$ then $T\in N_p(X,Y)$ with 
$1/s=1/p+1/2$
%(what is, surely, must be known; 
(see e.g., [13], [14]). This is the best possible general result 
one can obtain without imposing any conditions on the Banach spaces involved.
The sharpness of the assertion $1/s=1/p+1/2,$ 
%in the scale
%of $q$-nuclear operators:                
%the sharpness of the last assertion, 
for $s\in (2/3, 1],$
can be seen, for instance, in [14].                  

Below we consider a slightly %little bit 
different question:
Under which conditions  on the Banach spaces involved %$X$ and $Y$
is it valid that

$(*)$\  an operator $T\in L(X,Y)$ is nuclear if its adjoint $T^*$ is $s$-nuclear?

One of the possibilities for getting some positive answers to $(*)$
is to apply the so-called approximation properties %(the $AP_s)$
of order $s,$\, $s\in(2/3,1]$
(we assume that $s>2/3$ since for $s\le 2/3$ the answer is positive for any Banach spaces; see e.g., [5] or [14]).

We will prove that $(*)$ is true if either $X^*$ or $Y^{***}$ has the $AP_s$ (Theorem 1).
Some examples, given after Theorem 1, will show that these assumptions are, in a sense,
necessary (for example, it is not enough to assume that  $X$ or $Y^{**}$ or even both 
enjoy the $AP_s,$ not even for $s=1).$ 
Let us note that the case where $s=1$ was first investigated
in the paper  [8] by Eve Oja and the author.

%%%%%%%%%%%%%%%%%%%%%%%%  
\vskip0.5cm

%{\bf Notations, preliminaries.}\
\section{Notations and  preliminaries}
%\centerline{\bf \S2. \, Notations and  preliminaries}
\smallskip

Our main reference is  [9].  
All the spaces under considerations, $(X,Y,\dots)$
are real or complex Banach spaces, all linear mappings (operators) are continuous; as usual, $X^*, X^{**}, \dots$
are Banach duals (of $X$), and $x', x'', \dots$ (or $y', \dots)$ are the functionals acting on $X, X^*,\dots$
(or on $Y,\dots).$ By $\pi_Y$ we denote the natural isometric injection of $Y$ into its second dual.
If $x\in X, x'\in X^*$ then $\langle x,x'\rangle=\langle x',x\rangle=x'(x).$
$L(X,Y)$ as usual stands for the Banach space of all linear bounded operators from $X$ to $Y.$

An operator $T:X\to Y$ is $s$-nuclear $(0<s\le1)$ if it is of the form
$$
 Tx=\sum_{k=1}^\infty \langle x'_k,x\rangle y_k
$$
for all $x\in X,$ where $(x'_k)\subset X^*, (y_k)\subset Y,\, \sum_k ||x'_k||^s\,||y_k||^s<\infty.$ We use
the notation $N_s(X,Y)$ for the quasi-Banach space of all such operators, equipped with the quasi-norm
$$
 ||T||_{N_s}:=\inf (\sum_k ||x'_k||^s\,||y_k||^s)^{1/s},
$$
where the infimum is taken over all representations of $T$ in the above form (see [9, 6.1 and 18.1]). 
If $s=1,$ then $N_1(X,Y)$ is a Banach space, which is usually denoted by $N(X,Y).$ The operators
from $N(X,Y)$ are called also "nuclear operators". It is clear that for $0<s_1<s_2\le1$ one has
the natural continuous injection $N_{s_1}(X,Y)\subset N_{s_2}(X,Y).$

Every $s$-nuclear operator is a canonical image of an element of a projective tensor product. Namely,
denote by $X^*\widehat\otimes_s Y$ the $s$-projective tensor product of $X^*$ and $Y$ 
%that is
%a subspace of the Grothendieck projective tensor product
%$X^*\widehat\otimes Y$ $(= X^*\widehat\otimes_1 Y)$\, 
consisting of all tensor elements $z$ which admit
a representation of the kind
$$
 z=\sum_{k=1}^\infty x'_k\otimes y_k  \ \mathrm{ with }\ \sum_{k=1}^\infty ||x'_k||^s\,||y_k||^s<\infty.
$$
Then every $s$-nuclear operator from $X$ to $Y$ is an image of an element of $X^*\widehat\otimes_s Y$
via the canonical mappings
$$
 X^*\widehat\otimes_s Y \stackrel{j_s}{\rightarrow} X^*\widehat\otimes_1 Y\stackrel{j}{\rightarrow} L(X,Y).
$$
Note that $X^*\widehat\otimes_1 Y$ is exactly the projective tensor product $X^*\widehat\otimes Y$
of A. Grothendieck [5].
If $z\in X^*\widehat\otimes Y$ then we denote the corresponding operator (from $X$ to $Y)$
by $\tilde z.$

 We say, following, e.g., [13] or  [14],        
 that a Banach space $Y$ has the $AP_s$ (the approximation property of order $s),$
 if for every Banach space $X$ 
 the natural map $jj_s$ (from above) is one-to-one
 (note that $AP_1=AP$ of A. Grothendieck).
 It can be seen that, just like for the classical AP, $Y$ has the $AP_s$ if the natural map
 $Y^*\widehat\otimes_s Y \to L(Y,Y)$ is one-to-one.
       %It is the same as to say that the natural map $Y^*\widehat\otimes_s Y \to L(Y,Y)$ is one-to-one.
 Therefore, if $Y$ has the $AP_s,$ we have $X^*\widehat\otimes_s Y = N_s(X,Y)$
 whatever Banach space $X$ we consider.
    Clearly, $AP_s\Rightarrow AP_t$ for $0<t<s\le1.$ Every Banach space has the $AP_{2/3}$
  (Grothendieck's Theorem [5], see also [11], [13] or [14]).         
\smallskip

\noindent
{\it Remark 1}:\
Let $s\in (0,1].$ If the space $Y$ is a dual space, say $Y=Z^*,$
then the tensor product $X^*\widehat\otimes_s Y=X^*\widehat\otimes_s Z^*$
can be naturally (isometrically) identified with the tensor product
$Z^*\widehat\otimes_s X^*$ by definition (e.g., we can consider an element $x'\otimes z'$
also as the element $z'\otimes x').$
It follows that the dual space $Z^*$ has the $AP_s$ iff for every Banach space $X$
the natural map 
$Z^*\widehat\otimes_s X^* \to L(Z,X^*)$ is one-to-one. In this case if $Z^*$ has the $AP_s,$
then for any $X$ we have $N_s(Z,X^*)=Z^*\widehat\otimes_s X^*$ and
$N_s(X,Z^*)=X^*\widehat\otimes_s Z^*.$
We will use this remark in the proof of Theorem 1 (Section  3).

\smallskip

 If $w\in Y^*\widehat\otimes X$ and $U\in L(X,Y),$ then $w\circ U$ denotes the image of $w$
    in $X^*\widehat\otimes X$ under the map $U^*\otimes \mathrm{id}_X$ (see e.g., [5]).
    So if $w=\sum \varphi'_m\otimes \psi_m$ is a representation of $w$ in $Y^*\widehat\otimes X,$ then
    $w\circ U=\sum U^*\varphi'_m\otimes \psi_m$ is a representation of $w\circ U$ in $X^*\widehat\otimes X.$
    Then $\mathrm{trace}\, w\circ U= \sum \langle U^*\varphi'_m,  \psi_m\rangle= 
    \sum \langle \varphi'_m, U\psi_m\rangle= \mathrm{trace}\, U\circ w,$
    where $U\circ w$ is the image of $w$ in $Y^*\widehat\otimes Y$ under the map $\mathrm{id}_{Y^*}\otimes U.$
If $U\in L(X, Y^{**}),$ then the tensor element $U\circ w\in Y^*\widehat\otimes Y^{**}$ is defined by the 
same way.

  Recall that the dual space of $X^*\widehat\otimes Y$ is $L(X,Y^{**})$ and the duality is defined by  the "trace":
  If $z\in X^*\widehat\otimes Y$ and $U\in L(Y, X^{**}),$ then
  $$
   \langle U,z\rangle := \mathrm{trace}\, U\circ z
  $$
  $(= \sum_k \langle x'_k, Uy_k\rangle$ for a projective representation $z= \sum_k x'_k\otimes y_k).$
  So, the element $z\in X^*\widehat\otimes_s Y$ is zero iff it it zero in the projective tensor product $X^*\widehat\otimes Y$
  iff for every $U\in L(Y, X^{**})$ \, $\mathrm{trace}\, U\circ z=0.$ If $z\in X^*\widehat\otimes Y,$ then the corresponding
  operator $\tilde z$ is zero iff for every $R\in Y^*\otimes X$ we have $\mathrm{trace}\, R\circ z=0$ (evidently).

  Let us mention some examples of Banach spaces with $AP_s:$ For $s\in[2/3,1]$ and $1/p+1/2=1/s,$ every quotient of any subspace
  of any $L_p$-space (and every subspace of any quotient of any $L_{p'}$-space) has the $AP_s$
  (as well as all their duals; see, e.g., [13] or [14]; for a more general fact, see Lemma 3 below).
  Here $1/p+1/p'=1.$
  
  All  Banach spaces have $AP_s$ for $s\in (0,2/3],$ however if $2/3\le s_1< s_2\le1$
  then $AP_{s_2}\Rightarrow AP_{s_1}$ but $AP_{s_1}$ does not imply $AP_{s_2}.$
  It is known that for every $p\neq 2,$ $p\in [1,\infty],$
  there exists a subspace (a lot of them) of $l_p$ without the Grothendieck approximation property;
  thus, for example, $AP_1\neq AP_s$ if $s\in (0,1).$ Indeed, firstly, for $s\in (0,2/3]$ any Banach space has the $AP_s,$
  but not every Banach space has the $AP_1.$ Secondly, for $s\in (2/3,1)$ and $p$ with $1/p+1/2=1/s$
  every subspace of $l_p$ (as was said) has the $AP_s,$ but there is a subspace of $l_p$ without the $AP_1 (= AP).$
  Some more information about the statement 
  $AP_{s_2}\neq AP_{s_1}$ can be deduced 
   from our Examples 1 and 2, as well as  Theorems 2 and 3.

We will use later the following facts (surely well known, but maybe not mentioned in the literature):
\smallskip

\noindent
{\bf Lemma 1.}\
If $T\in L(X, Y^{**})$ then
$||T^*|_{\pi_{Y^*}(Y^*)}||=||T||$ and $(T^*|_{\pi_{Y^*}(Y^*)})^*|_X=T.$
So, we may  write $L(X,Y^{**})=L(Y^*,X^*),$ where the equality sign refers to the above
identification.
\smallskip

\noindent
{\bf Lemma 2.}\  
Let $0<s\le1.$
If $T\in L(X, Y)$ then

1)\, $\pi_Y T\in N_s(X,Y^{**}) \, \iff \, T^*\in N_s(Y^*, X^*);$

2)\, $T\in N_s(X,Y) \Rightarrow T^*\in N_s(Y^*, X^*).$
\smallskip

\noindent
{\bf Lemma 3.}\
If $E$ is a Banach space of type 2 (respectively, of cotype 2)
and of cotype $q_0$ (respectively, of type $q'_0)$
then $E$ has the $AP_s,$ where $1/s=3/2-1/q_0.$
\smallskip

The proof of Lemma 3 can be found in [13] or [14].         

\smallskip

\noindent
{\bf Lemma 4.}\
Let $s\in (0,1].$ If $Y^*$ has the $AP_s,$ then $Y$ has the $AP_s$ too.
\medskip

\noindent
{\it Proof}.\,
We use the fact (mentioned above) that $Y$ has the $AP_s$ iff
the natural map 
 $Y^*\widehat\otimes_s Y \to L(Y,Y)$ is one-to-one.
As is known [5], the projective tensor product 
 $Y^*\widehat\otimes Y$  is a Banach subspace of the  tensor product 
  $Y^*\widehat\otimes Y^{**}.$
  The tensor product  $Y^*\widehat\otimes_s Y $ is  a linear subspace of  $Y^*\widehat\otimes Y,$ 
  as well as  $Y^*\widehat\otimes_s Y^{**}$ is a linear subspace of  $Y^*\widehat\otimes Y^{**}.$
  Therefore, the natural map  $Y^*\widehat\otimes_s Y \to Y^*\widehat\otimes_s Y^{**} $
  is one-to-one. Now if $Y^*$ has the $AP_s,$ then the canonical map
   $Y^{**}\widehat\otimes_s Y^* \to L(Y^*,Y^*)$ is one-to-one.
   Since we can identify the tensor product $Y^{**}\widehat\otimes_s Y^*$ with 
   the tensor product $Y^{*}\widehat\otimes_s Y^{**}$ (see Remark 1), it follows that
   the natural map $Y^*\widehat\otimes_s Y \to L(Y,Y)$ is one-to-one.
   Thus, if $Y^*$ has the $AP_s,$ then $Y$ has the $AP_s$ too.
   \smallskip
   
   We will use Lemma 4 in the proof of Theorem 1 in the next section.
   
     \smallskip
   
   \noindent
   {\it Remark 2}:\
  For any $s\in(2/3,1]$ there exists a Banach space, possessing the Grothendieck 
  approximation property, whose dual does not have the $AP_s$ (it is well known for the case
  where $s=1).$
  Moreover,  if $s\in(2/3,1],$ then we can find a Banach space $W$ such that
  $W$ has a Schauder basis and $W^*$ does not have the $AP_s.$
  Indeed, let $E$ be a separable reflexive Banach space without the $AP_s$ (see [13] or [14]).
    Let $ Z$ be a separable space such that $ Z^{**}$ has a basis
  and there exists a linear homomorphism $ \varphi$ from $ Z^{**}$
  onto $ E^*$ 
  %with the kernel
  %$ Z\subset Z^{**}$ 
  so that the subspace $ \varphi^*(E)$ is complemented
  in $ Z^{***}$ and, moreover,
  $Z^{***}\cong \varphi^*(E)\oplus Z^*$
  (see [7, Proof of Corollary 1]). 
  Put $W:= Z^{**}.$ This (second dual) space  $W$ has a Schauder basis and its dual  $W^*$ does not have the $AP_s.$

  \smallskip

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  New for s<1 ...
%{\bf A theorem.}\
\smallskip

\section{A positive result}
%\centerline{\bf \S3.\, A positive result}
\smallskip



\noindent
{\bf Theorem 1.}\
Let $\, s\in (0,1],$ $ T\in L(X,Y)$
and assume that either
$\, X^*\in \,AP_s\ $ or $\, Y^{***}\in \,AP_s.$
If $ T\in N_s(X, Y^{**}),$
then $T\in N_1(X,Y).$ In other words, under these conditions,
from the $ s$-nuclearity of the conjugate operator $ T^*,$ it follows
that the operator $ T$ is nuclear.
%\endproclaim
\medskip

\noindent
{\it Proof}.\,
Suppose there exists  an operator  
    $ T\in L(X,Y)$
 such that
$ T\notin N_1(X,Y),$ but $ \pi_Y\,T\in N_s(X,Y^{**}).$
Since either $ X^*$ or $ Y^{**}$  has the $ AP_s,$
 $ N_s(X,Y^{**})=X^*\widehat\otimes_s Y^{**}$ (see Remark 1 in Section 2).
Therefore the operator $ \pi_Y\,T$ can be identified with the
tensor element
$ t\in X^*\widehat\otimes_s Y^{**}\subset X^*\widehat\otimes_1 Y^{**}.$
In addition, by the choice of $ T,$ \
$  t\notin X^*\widehat\otimes_1 Y$ \ (the space
$  X^*\widehat\otimes_1 Y$
is considered as a closed subspace of the space
$  X^*\widehat\otimes_1 Y^{**}$).
Hence there is an operator
$ U\in L(Y^{**},X^{**})=\left( X^*\widehat\otimes_1 Y^{**}\right)^*$
with the properties that
$ \mathrm{trace}\, U\circ t=\mathrm{trace}\, \left(t^*\circ \left( U^*|_{X^*}\right) \right)=1$ and
$ \mathrm{trace}\, U\circ \pi_Y\circ z=0$ for each
$ z\in X^*\widehat\otimes_1 Y.$
From the last observation it  follows that, in particular, $ U\pi_Y=0$ and
$ \pi_Y^*\,U^*|_{X^*}=0.$
In fact, if
$ x'\in X^*$ and $ y\in Y,$ then
$$ \langle U\pi_Y\,y,x'\rangle  = \langle y, \pi_Y^*\,U^*|_{X^*}x'\rangle 
       = \mathrm{trace}\, \,U\circ (x'\otimes \pi_Y(y))=0.
$$
Evidently, the tensor element
$ U\circ t\in X^*\widehat\otimes_s X^{**}$  induces the operator
$ U\pi_Y T,$ which is equal  to the 0-operator.

If $ X^*\in AP_s$ then
 $ X^*\widehat\otimes_s X^{**}= N_s(X,X^{**})$
and, therefore, this tensor element is zero, which  contradicts
the equality
$ \mathrm{trace}\,\, U\circ t=1.$

Now let  $ Y^{***}\in AP_s.$ In this case
$$ V:= \left( U^*|_{X^*}\right)\circ T^*\circ \pi_Y^*: \
       Y^{***}\to Y^* \to X^*\to Y^{***}
$$
uniquely determines a tensor element
$ t_0$ from the $s$-projective tensor product
$ Y^{****}\widehat{\otimes}_s Y^{***}.$
Let us take any representation $ t=\sum x'_n\otimes y''_n$ for $ t$
as an element of the space $ X^*\widehat{\otimes}_s Y^{**}.$
Denoting  the operator $ U^*|_{X^*}$ by $ U_*,$
we obtain

 $$  Vy'''=U_*\, \left( T^*\pi_Y^*\,y'''\right) =
      U_*\, \left( (\sum y''_n\otimes x'_n) \,\pi_Y^*\, y'''\right) $$%\\
   $$ =U_*\, \left( \sum \langle y''_n, \pi_Y^*\, y'''\rangle  \,x'_n \right) = \\ %= %
      \sum \langle \pi_Y^{**}y''_n,  y'''\rangle  \,U_* x'_n.$$



So, the operator $ V$ (or the element $ t_0$) has in the space
$ Y^{****}\widehat\otimes_s Y^{***}$ the representation
$$ V= \sum \pi_Y^{**}(y''_n)\otimes U_* (x'_n).
$$
Therefore,
$$  \mathrm{trace}\, t_0=\mathrm{trace}\, V= \sum \langle \pi_Y^{**}(y''_n), U_* (x'_n)\rangle  =
        \sum \langle y''_n, \pi_Y^*\,U_* x'_n\rangle  =  \sum 0=0
$$
(since $\pi_Y^*\,U_*=0,$ see above).

On the other hand,
$$  Vy'''= U_* \left( \pi_Y T\right)^* y'''= U_*\circ t^* (y''')=
     U_*\, \left( \sum \langle y''_n, y'''\rangle  \, x'_n\right)=
      \sum \langle y''_n, y'''\rangle  \, U_* x'_n,
$$
whence $ V=\sum y''_n\otimes U_*(x'_n).$   Therefore
$$ \mathrm{trace}\, t_0=\mathrm{trace}\, V= \sum \langle y''_n, U_* x'_n\rangle  = \sum \langle Uy''_n, x'_n\rangle 
= \mathrm{trace}\, U\circ t=1.$$
The obtained contradiction completes the proof of the theorem.

%\enddemo
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  End Th1 for s<1...

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vskip1.1cm
%{\bf Examples.}\

\smallskip

\section{Examples}
%\centerline{\bf \S4.\, Examples}
\smallskip




We need two examples to show that all conditions, imposed on $X$ and $Y$ in Theorem 1,
are essential. Ideas of such examples are taken from the author's work [13] (not translated,
as far as we know, from Russian).


\smallskip


%\vskip0.1cm


\noindent
{\bf Example 1.}\
Let $r\in(2/3,1], q\in[2,\infty), 1/r=3/2-1/q.$
There exist a separable reflexive Banach space $Y_0$ and a tensor element
$w\in Y_0^*\widehat\otimes_r Y_0$ so that
$w\neq0, \tilde w=0,$ the space $Y_0$ (as well as $Y_0^*)$
has the $AP_s$ for every $s<r$
(but, evidently, does not have the $AP_r).$
Moreover, $Y_0$ is of type 2 and of cotype $q_0$ for any $q_0>q.$
\vskip0.3cm

   %\demo{Proof}
   
\noindent
{\it Proof}.\
%%%%%%%
We will use a variant of Per Enflo's  example [3] of a Banach space without the approximation property
given in the book [9, 10.4.5]. Namely, it follows from the constructions in [9] that           
there exist a Banach space $X$ and a tensor element $z\in X^*\widehat\otimes X$ so that
$\mathrm{trace}\, z\neq0,$ the operator $\tilde z,$ generated by $z,$ is identically zero and $z$ can be represented
in the following form:
$$(1) \quad  z=\sum_{N=1}^{\infty} \sum_{n=1}^{3\cdot 2^N}N^{1/2}2^{-3N/2}
  x'_{nN}\otimes x_{nN},
$$
where the sequences $(x'_{nN})$ and $ (x_{nN})$ are norm bounded by 1
(see [2] or [9, 10.4.5]).              

Fix $r\in(2/3,1]$ and
put  $1/q=3/2-1/r$ (thus, $q\in [2,\infty)$).
Let  $\{\varepsilon_N\}_1^{\infty}$ be a  sequence of numbers such that
 $\sum N^{-1-\varepsilon_N}<+\infty,$ put
$\gamma_N=2+3\varepsilon_N/2$ and let $q_N$ be a number such that
$1/q-1/q_N=N^{-1}\log_2 N^{\gamma_N}$
(therefore, $q_N>q$).
Set $Y=\left(\sum_N l_{q_N}^{3\cdot 2^N}\right)_{l_q}.$


%%%%%%%%%

Denote by $e_{nN}$ (and $e'_{nN}$) the unit vectors in $Y$ and in $Y^*$ respectively
($N=1,2,\dots; n=1,2,\dots, 3\cdot 2^N$), and put
$$
   z_1=\sum_N\sum_n 2^{-N/r}\,N^{-(1+\varepsilon_N)/r}\, e'_{nN}\otimes x_{nN};
   $$
   $$
   T=\sum_N\sum_n 2^{-(3/2-1/r)N}\, N^{(1/2+1/r+\varepsilon_N/r)}\, x'_{nN}\otimes e_{nN}.
$$

Let us show that $z_1\in Y^*\widehat\otimes_r X$ and $T\in \L(X,Y).$
Then $z=z_1\circ T.$ %!!!!
%\footnote{
  %If $w\in Y^*\widehat\otimes X$ and $U\in L(X,Y)$ then $w\circ U$ denotes the image of $w$
  %in $X^*\widehat\otimes X$ under the map $U^*\otimes \mathrm{id}_X.$
 % So, if $w=\sum \varphi'_m\otimes \psi_m$ is a representation of $w$ in $Y^*\widehat\otimes X$ then
 % $w\circ U=\sum U^*\varphi'_m\otimes \psi_m$ is a representation of $w\circ U$ in $X^*\widehat\otimes X.$
 % Then $\mathrm{trace}\, w\circ U= \sum \langle U^*\varphi'_m,  \psi_m\rangle= \sum \langle \varphi'_m, U\psi_m\rangle= \mathrm{trace}\, U\circ w,$
 % where $U\circ w$ is the image of $w$ in $Y^*\widehat\otimes Y$ under the map $\mathrm{id}_{Y^*}\otimes U.$
  %}).
The first inclusion is evident (because of the choice of $\varepsilon_N).$

If $||x||\leqslant1$ then
$$
||Tx||\leqslant 3\left(\sum_N \left(N^{1/2+1/r+\varepsilon_N/r}/2^{(3/2-1/r-1/q_N)N}\right)^q\right)^{1/q}.
$$
Since $3/2-1/r-1/q_N=1/q-1/q_N=N^{-1}\log_2 N^{\gamma_N},$
we get from the last inequality:
$$
||T||\leqslant 3\left(\sum N^{(1/2+1/r+\varepsilon_N/r-\gamma_N)q}\right)^{1/q}=
       3 \left(\sum N^{-1-\varepsilon_N}\right)^{1/q}<\infty.
$$

Hence,
$$
 z=z_1\circ T,\ \ \tilde{z}:\ X\stackrel{T}{\rightarrow} Y\stackrel{\tilde{z}_1}{\rightarrow} X.
$$

Now, let $Y_0:=\overline{T(X)}\subset Y, $\, $T_0: X\to Y_0$ be induced by $T$ and
$z_0:= z_1\circ j$ where $j: Y_0\hookrightarrow Y$ is the natural embedding.
Then $T_0\in L(X,Y_0), z_0\in Y_0^*\widehat\otimes_r X,$
$z=z_1\circ T=z_0\circ T_0,$ $\mathrm{trace}\, z_0\circ T_0\neq0$
(so, $z_0\neq0)$  and $\tilde z_0=0.$

Write $z_0$ as $z_0=\sum_{m=1}^\infty f'_m\otimes f_m,$ where
$(f'_m)\subset Y_0^*, (f_m)\subset X$ and $\sum_m ||f'_m||^r ||f_m||^r<\infty.$
We get
$$
\mathrm{trace}\, z_0\circ T_0 = \sum \langle T_0^*f'_m, f_m\rangle = \sum \langle f'_m, T_0f_m\rangle= \mathrm{trace}\, T_0\circ z_0.
$$
Therefore, $w:= T_0\circ z_0\in Y_0^*\widehat\otimes_r Y_0,$ $\mathrm{trace}\, w\neq0$ and $\tilde w=0.$

Since the space $Y$ is of type 2 and of cotype $q_0$ for every $q_0>q$ (and $Y^*$ is
of cotype 2 and of type $q'_0$),
the space $Y_0$ (respectively, $Y_0^*)$ has the $AP_s,$
where $1/s=3/2-1/q_0,$ for every $s<r$ (Lemma 3).

  %\enddemo

%%%%%%%%%%%%%%


 \vskip0.3cm
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\noindent
{\it Remark 3}:\
We have a nice "by-product consequence" of Example 1.
For $q=2$ (that is, $r=1)$, the space $Y_0$ is a subspace of a space of  the type
$\left(\sum_j l_{p_j}^{k_j}\right)_{l_2}$ with $p_j\searrow 2$ and $k_j\nearrow \infty.$
Every such space is an asymptotically Hilbertian space (for definitions and some discussion,
see [1]). So we have obtained
\smallskip

\noindent
{\bf Corollary.}\
There exists an asymptotically Hilbertian space without the Grothendieck approximation property.
 \smallskip

 The first example of such a space was constructed by the author in 1982 [10],         
 where A. Szankowski's results were used
 (let us note that in that time there was not yet such  notion as "asymptotically Hilbertian space").
 Later,  in 2000, by applying Per Enflo's example in a version due to  A.M. Davie [2],     
 P.~G. ~Casazza,   C.~L. ~Garc\'{\i}a and   W.~B. ~Johnson [1]                     
 gave another example of an asymptotically Hilbertian space  which fails the approximation property.
 Here we have got it  (accidentally)
 by using the construction from [9].          



\vskip0.1cm


\noindent
{\bf Example 2.}\
Let $r\in [2/3,1), q\in(2,\infty], 1/r=3/2-1/q.$
There exist a subspace $Y_q$ of the space $l_q$ and a tensor element
$w_q\in Y_q^*\widehat\otimes_1 Y_q$ so that
$w_q\in Y_q^*\widehat\otimes_s Y_q$ for each $s>r,$
$w_q\neq0, \tilde w_q=0$  and the space $Y_q$ (as well as $Y_q^*)$
has the $AP_r$
(but, evidently, does not have the $AP_s$ if $1\ge s>r).$
Clearly, $Y_q$ is of type 2 and of cotype $q$ for $q<\infty.$
\vskip0.5cm

     %\demo{Proof}
     
     \noindent
{\it Proof}.\
We are going to follow the way  indicated in the proof of  Example 1.
Let $X$ and $z$ be as in that proof, so that $z$ has the form (1).
Fix $r\in [2/3,1).$ Now $1/q=3/2-1/r,$ and we put $\varepsilon_N=0$ and $\gamma_N=2$
for all $N;$ all $q_n$'s are equal to $q.$ Let us fix also an $\alpha=\alpha(q)>0$ (to be specified later).
Consider the space $Y:= \left(\sum_N l_{q}^{3\cdot 2^N}\right)_{l_q}$ (in the case $q=\infty$
"$l_q$" means "$c_0$").

Denote by $e_{nN}$ (and $e'_{nN}$) the unit vectors  in $Y$ and in  $Y^*$ respectively
($N=1,2,\dots; n=1,2,\dots, 3\cdot 2^N$), and set this time

$$
   z_1=\sum_N\sum_n 2^{-N/r}\,N^{\alpha}\, e'_{nN}\otimes x_{nN};
   $$
   $$
   T=\sum_N\sum_n 2^{-(3/2-1/r)N}\, N^{1/2-\alpha}\, x'_{nN}\otimes e_{nN}.
$$

Then $z_1\in Y^*\widehat\otimes_{s} X$ for every $s>r.$
Indeed, if $s\in (r,1]$ then
$$
 \sum_N\sum_n [2^{-N/r}\,N^{\alpha}]^{s}=  \sum_N 3\cdot 2^N\cdot 2^{-Ns/r}\,N^{\alpha\, s}=
 \sum_N 3\cdot 2^{-\varepsilon_0 N}\, N^{\alpha s}<\infty,
$$
where $\varepsilon_0=s/r-1>0.$

We show that  $T\in L(X,Y)$ (clearly, then $z=z_1\circ T$).
Indeed,
if $||x||\leqslant1$ then, for $q<\infty,$
$$
 ||Tx||_{l_q}^q\le
 \sum_N\sum_n \left(2^{-N/q}\, N^{1/2-\alpha}\,  |\langle x'_{nN},x\rangle|\right)^q\le
  \sum_N 3\cdot 2^{N-(N/q)q}\, N^{(1/2-\alpha)q} = \sum_N 3\cdot 1\cdot N^{\alpha_0},
$$
where $\alpha_0=(1/2-\alpha)q.$
Now, take $\alpha>0$ such that  $\alpha_0=-2.$
For $q=\infty$ take $\alpha=\alpha(\infty)=1.$

Therefore,
$$
 z=z_1\circ T,\ \ \tilde{z}:\ X\stackrel{T}{\rightarrow} Y\stackrel{\tilde{z}_1}{\rightarrow} X;
$$

As in the case of the previous proof (in Example 1),
 let $Y_q:=\overline{T(X)}\subset Y, T_q: X\to Y_q$ be induced by $T$ and
$z_q:= z_1\circ j$ where $j: Y_q\hookrightarrow Y$ is the natural embedding.
Then $T_q\in L(X,Y_q), z_q\in Y_q^*\widehat\otimes_s X$ for all $s>r,$
$z=z_1\circ T=z_q\circ T_q,$ $\mathrm{trace}\, z_q\circ T_q\neq0$
(so, $z_q\neq0)$  and $\tilde z_q=0.$


Write $z_q$ as $z_q=\sum_{m=1}^\infty f'_m\otimes f_m,$ where
$(f'_m)\subset Y_q^*, (f_m)\subset X$ and $\sum_m ||f'_m||\, ||f_m||<\infty.$
We get:
$$
\mathrm{trace}\, z_q\circ T_q = \sum \langle T_q^*f'_m, f_m\rangle = \sum \langle f'_m, T_qf_m\rangle=
\mathrm{trace}\, T_q\circ z_q.
$$
Therefore, $w_q:= T_q\circ z_q\in Y_q^*\widehat\otimes_s Y_q$ for every $s>r,$
$\mathrm{trace}\, w_q\neq0$ and $\tilde w_q=0.$
Finally, Lemma 3 says that the space $Y_q$ has the $AP_r$ if $q<\infty.$
If $q=\infty$ then, as we know, any Banach space has the $AP_{2/3}.$
        %\enddemo
\medskip

%%%%%%% end EXample 2

\noindent
{\it Remark}:\
The space $Y_\infty$ from Example 2 not only does not have the $AP_s$
for any $s\in (2/3, 1],$ but also does not have the $AP_p$ (in the sense of the paper [12])    
for any $p\in[1,2)$ (this follows from some facts proved in [13]).         
\smallskip

\vskip0.5cm

\section{Applications of Examples}
%\centerline{\bf  \S4.\, Applications of Examples}
\vskip0.5cm




The next two theorems show that the conditions "$X^*$ has the $AP_s$" and
"$Y^{***}$ has the $AP_s$" are essential in Theorem 1 and can not be replaces 
by the weaker conditions (see Remark 2)
"$X$ has the $AP_s$" (even by "$X$ has the $AP_1$") or "$Y^{**}$ has the $AP_s$";
moreover, even "both $X$ and $Y^{**}$ have the $AP_1$" is not enough for the conclusion
of Theorem 1 to hold.
\smallskip

\noindent
{\bf Theorem 2.}\
Let $r\in (2/3,1]. $ %, q\in [2,\infty), 1/r=3/2-1/q.$
There exist a Banach space $Z_0$ and an operator $T\in L(Z_0^{**}, Z_0)$ so that

(1)\,
$Z_0^{**}$  has a Schauder basis;

(2)\,
all the duals of $Z_0$ are separable;

(3)\,
$Z_0^{***}$ has the $AP_s$ for every $s\in (0,r);$

(4)\,
$\pi_{Z_0} T\in N_r(Z_0^{**}, Z_0^{**});$

(5)\,
$T\notin N_1(Z_0^{**}, Z_0);$

(6)\,
$Z_0^{***}$ does not have the $AP_r.$

\medskip

%%

\noindent
{\it Proof}.\,
Let us fix $r\in (2/3,1], q\in [2,\infty), 1/r=3/2-1/q$\, and take the pair $ (Y_0, w)$
from Example 1.
Let $ Z_0$ be a separable space such that $ Z_0^{**}$ has a basis
and there exists a linear homomorphism $ \varphi$ from $ Z_0^{**}$
onto $ Y_0$ with the kernel
$ Z_0\subset Z_0^{**}$ so that the subspace $ \varphi^*(Y_0^*)$ is complemented
in $ Z_0^{***}$ and, moreover,
$Z_0^{***}\cong \varphi^*(Y_0^*)\oplus Z_0^*$
(see [7, Proof of Corollary 1]). Lift the tensor element
$ w,$ lying in $ Y_0^*\widehat\otimes_r Y_0,$  to an element\footnote{
   If $ w=\sum_{k=1}^\infty \,y'_k\otimes\,y_k $ is any representation of $ w$
in  $ Y_0^*\widehat\otimes_r Y_0$ with $(y_k)\in l_r(Y_0),$  
then we take $ \{ z''_n\}\subset Z_0^{**}$
in such a way that
the last sequence is absolutely $ r$-summing and
$ \varphi(z''_n)=y_n$ for every $ n.$
   }
$ w_0\in Y_0^*\widehat\otimes_r Z_0^{**},$
so that $ \varphi\circ w_0=w,$ and set $ T:= w_0\circ \varphi.$
Since $ \mathrm{trace}\, w_0\circ \varphi=\mathrm{trace}\, \varphi\circ w_0=\mathrm{trace}\, w=1$ and $ Z_0^{**}$ has the
AP, then $ \widetilde {w_0}=w_0\neq 0.$
Besides, the operator $ \widetilde{\varphi\circ w_0}:Y_0\to Z_0^{**}\to Y_0,$
associated with the tensor $ \varphi\circ w_0,$ is equal to zero. Therefore
$ w_0(Y_0)\subset \mathrm{ Ker}\varphi= Z_0\subset Z_0^{**},$
that is, the operator
$ w_0$  acts from $ Y_0$ into $ Z_0.$

Since the subspace $ \varphi^*(Y_0^*)$ is complemented in $ Z_0^{***},$
we have that $ w_0\circ\varphi\in Z_0^{***}\widehat\otimes_r Z_0= N_r(Z_0^{**},Z_0)$ iff
$ w_0\in Y_0^{*}\widehat\otimes_r Z_0= N_r(Y_0,Z_0).$

If $ w_0\in  N_r(Y_0,Z_0),$
then, for its arbitrary (nonzero!) $N_r$-representation
of the form $ w_0=\sum y'_n\otimes z_n,$ the composition $ \varphi\circ w_0$ is
a zero tensor element in
$ Y_0^*\widehat\otimes_r Y_0;$ but this composition
represents the element $ w, $ which, by its choice, can not be zero.
Thus, $ w_0\notin  N_r(Y_0,Z_0)$ and, thereby,
 $ w_0\circ\varphi\notin Z_0^{***}\widehat\otimes_r Z_0= N_r(Z_0^{**},Z_0).$
On the other hand, certainly,
 $ w_0\circ\varphi\in Z_0^{***}\widehat\otimes_r Z_0^{**}= N_r(Z_0^{**},Z_0^{**}).$

 Finally, since $Z_0^{***}\cong \varphi^*(Y_0^*)\oplus Z_0^*,$ one has that
 the space $Z_0^{***}$ has the $AP_s$ for every $s\in (0,r).$
 %$\quad\blacksquare$
%\enddemo

%%   end TH 2 PROOF


   %%%%%%%%%%%%%%%  TH 3
\medskip

\noindent
{\bf Theorem 3.}\
Let $r\in [2/3,1), q\in (2,\infty], 1/r=3/2-1/q.$
There exist a Banach space $Z_q$ and an operator $T\in L(Z_q^{**}, Z_q)$ so that

(1)\,
$Z_q^{**}$  has a Schauder basis;

(2)\, if $q<\infty,$ then
all the duals of $Z_q$ are separable;

(3)\,
$Z_q^{***}$ has the $AP_r;$

(4)\,
$\pi_{Z_q} T\in N_s(Z_q^{**}, Z_q^{**})$ for every $s\in (r,1];$

(5)\,
$T\notin N_1(Z_q^{**}, Z_q);$

(6)\,
$Z_q^{***}$ does not have the $AP_s$ for any $s\in (r,1];$

\medskip

%%

\noindent
{\it Proof}.\,
Let us fix $r\in [2/3,1), q\in (2,\infty], 1/r=3/2-1/q$\, and take the pair $ (Y_q, w_q)$
from Example 2.
Let $ Z_q$ be a separable space such that $ Z_q^{**}$ has a basis
and there exists a linear homomorphism $ \varphi$ from $ Z_q^{**}$
onto $ Y_q$ with the kernel
$ Z_q\subset Z_q^{**}$ so that the subspace $ \varphi^*(Y_q^*)$ is complemented
in $ Z_q^{***}$ and, moreover,
$Z_q^{***}\cong \varphi^*(Y_q^*)\oplus Z_q^*$
(as in the proof of Theorem 2, see [7, Proof of Corollary 1]).

Construct $ w_0\in Y_q^*\widehat\otimes_1 Z_q^{**}$ (following the way of the proof
of Theorem 2)
so that $ w_0\in Y_q^*\widehat\otimes_s Z_q^{**}$ for every $s\in (r,1]$ and
$ \varphi\circ w_0=w_q$
(it is possible to apply a "simultaneous lifting" procedure --- see Footnote 1  ---
since $w$ has a form from the proof of the assertion of Example 2).
Set $ T:= w_0\circ \varphi.$
From this point, the proof repeats the arguments of the proof of Theorem 2,
and we have to mention only:
since $Z_q^{***}\cong \varphi^*(Y_q^*)\oplus Z_q^*,$ one has that
 the space $Z_q^{***}$ has the $AP_r.$



\bigskip

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\end{thebibliography}

\bigskip

Reinov Oleg

St. Petersburg State University, 

198504 St. Petersburg, Petrodvorets, 

28 Universitetskii pr., Russia

E-mail address: \, orein51@mail.ru


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