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%%%%           APsAPPL II  18.05.2017 8:24:10
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   %
\def\ove#1{\overline{#1}}    %
\def\ovs#1#2{\overset{#1}\to{#2}}
     \def\({\big(}       \def\al{\alpha}           \def\lee{\leqslant}
     \def\){\big)}      \def\e{\varepsilon}    \def\gee{\geqslant}
     \def\[{\left[}       \def\la{\lambda}
     \def\]{\right]}      \def\ffi{\varphi}
                          \def\be{\beta}
     %\def\{{\left\{}
                                      \def\ot{\otimes}
     \def\<{\langle}                 \def\wh{\widehat}
     \def\>{\rangle}                 \def\wt{\widetilde}
                 \def\sbs{\subset}
\def\tr{\operatorname{trace}\,}
                              \def\id{\operatorname{id}\,}
                   \def\det{\operatorname{det}\,}

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\def\Gr{\operatorname{Gr}}
\def\AP{\operatorname{AP}}
\def\BAP{\operatorname{BAP}}
\def\N{\operatorname{N}}
\def\I{\operatorname{I}}
\def\id{\operatorname{id}}
\def\L{\operatorname{L}}
\def\QN{\operatorname{QN}}
\def\J{\operatorname{J}}
\def\R{\operatorname{R}}
\def\dim{\operatorname{dim}}
\def\dual{\operatorname{dual}}
                     \def\sbs{\subset}

            \def\co{\operatorname{co}\,}
\def\span{\operatorname{span}}
        \def\op{\operatorname}
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\documentclass[12pt,oneside]{amsproc}
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%\usepackage[russian]{babel}
\usepackage{amssymb}
\usepackage{amsthm}
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\textheight=22cm \textwidth=15cm \oddsidemargin=5mm
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\title[Approximation properties associated with operator ideals]
{Approximation properties associated with quasi-normed operator ideals of
$(r,p,q)$-nuclear operators}
%{}
\author{Oleg Reinov}
\address{ St. Petersburg State University,
Saint Petersburg, RUSSIA.}
\email{orein51@mail.ru}

\thanks{%${ }^\maltese$
AMS Subject Classification 2010: 46B28 Spaces of operators; tensor products; approximation properties.
}
\thanks{${ }$ Key words:  nuclear operator; tensor product; approximation property; eigenvalue.}

\begin{document}

  \maketitle

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

\begin{abstract}
We consider quasi-normed tensor products lying between
Laprest\'e tensor products and spaces of $(r,p,q)$-nuclear operators.
We define and investigate the corresponding approximation properties for Banach spaces.
An intermediate aim is to answer a question of Sten Kaijser.
 \end{abstract}


 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \medskip

   %%%%%%%%%%%%%%%%%%%%%% BEGINNING
  % {\bf Introduction.}\,                                             %!!!   Begin text with this
    %%%
   \begin{comment}
In 1955, A. Grothendieck \cite{Gr} has introduced the  notion of the              %!!!ref
approximation property (the AP) and has shown that {\it there exists
a Banach space without the approximation property
iff there exists an operator $U: c_0\to c_0$
such that the (nuclear) trace of $U$ equals 1 but $U^2=0$ identically}\,
\cite[Chap. I, "Proposition" 37. $(a')\Leftrightarrow (f''),$ pp. 170-171]{Gr}.        %!!!ref
Let us recall a proof of the "only if" part of this result.

As is known \cite{Gr}, the AP for $X$ is equivalent to the injectivity               %!!!ref
of the canonical map from the projective tensor product $X^*\wh\ot X$ into
the space $L(X)$ of all linear operators in $X.$
 Suppose $X$ does not  have the AP and let $z$ be an element of the projective
 tensor  product $X^*\wh\ot X$ such that the associated operator $\wt z$
 is identically zero. Take a representation $z=\sum \mu_j x'_j\ot x_j$
 in $X^*\wh\ot X$ with $1\ge ||x'_j||\to 0,\, ||x_j||=1,$ $(\mu_j)\in l_1$ and
 $\tr z=1$ (note that $||z||_{\land} \le \sum \mu_j ||x'_j||).$
Consider a diagram
$$
 X\overset{A}\to c_0 \overset{\Delta}\to l_1 \overset{B}\to X
  \overset{A}\to c_0 \overset{\Delta}\to l_1 \overset{B}\to X
   \overset{A}\to c_0,
$$
where $Ax= (\<x'_j, x_j\>)\in c_0,$
$\Delta (\al_j)= (\mu_j \al_j),$
$B(\al_j)= \sum_j \beta_j x_j$ and $B\Delta A= \wt z=0.$
Putting $T:= AB\Delta$ we get a nuclear operator $T$ with $T^2=0$ and
$T(\al_j)= \sum_k \mu_k \al_k ABe_k,$ where $(e_k)$ is a sequence of unit vectors
i $c_0$ (and also in $l_1).$ Thus, $T= \sum \mu_k e_k\ot Ax_k \in l_1\wh\ot c_0$
and $\tr T= \sum \mu_j \<e_k, Ax_k= \sum \mu_k \<A^*e_k, x_k\>\>=
\sum \mu_k \<x'_r, x_k\>= \tr z =1.$ Note that $||T||_\land\le \sum \mu_j$
while $||z||_\land\le \sum \mu_j) ||x'_j||$  (where $x'_j\to0).$
       \end{comment}
        
        
        %%%
                                                                    %!!! locref
        
        
$$...$$

This is, essentially, a continuation of the author's paper \cite{Trend}                       %!!!ref
and the first part of the work on the approximation properties
connected with the quasi-normed ideals of so-called $(r,p,q)$-nuclear
operators.
             \bigskip
             
   %%
\centerline{\bf 0. Notation, preliminaries}\,    
                                          \bigskip
                                          
Throughout, we denote by $X,Y,\dots, G,F,W\dots$ Banach spaces over a field $\mathbb K$
(which is either $\mathbb R$ or $\mathbb C);$ $X^*, Y^*,\dots$ are
Banach dual to $X, Y, \dots.$ By $x, y,\dots, x',\dots$ (maybe with indices) we denote elements of
$X, Y, \dots, Y^*\dots$ respectively. $\pi_Y: Y\to Y^{**}$ is a natural isometric imbedding.

 Notations
 $l_p,$ $l_p^n$ $(0<p\le\infty, n=1,2,\dots),$ $c_0$ are standard;   $e_k$ 
 $(k=1,2,\dots)$ is the k-th unit vector in $l_p$ or $c_0$
 (when we consider the unit vectors as the linear functionals, we use 
 notation $e'_k$).  
We use  $\id_X$ for the identity map in $X.$
                                 %  $BAP, MAP, AP, C-MAP$     %!!!
       
It is denoted by $F(X,Y)$ a vector space of all linear continuous mappings
from $X$ to $Y.$ By $X\ot Y$ we denote the algebraic tensor product of the spaces
$X$ and $Y.$ 
  $X\ot Y$ can be considered as a subspace of the vector space $F(X^*, Y)$
(namely, as a vector space of all linear weak${^*}$-to-weak continuous
finite rank operators). We can identify also the tensor product (in a natural way)
with a corresponding subspace of $F(Y^*,X).$ If $X=W^*,$ then 
$W^*\ot Y$ is identified with $F(X,Y^{**})$ (or with $F(Y^*,X^*).$
If $z\in X\ot Y,$ then $\wt z$ is the corresponding finite rank operator.
If $z\in X^*\ot X$ and e.g. $z=\sum_{k=1}^n x'_k\ot x_k,$ then
   $\tr z:= \sum_{k=1}^n \<x'_k, x_k\>$ does not depend on representation 
   of $z$ in $X^*\ot X.$     
 $L(X,Y)$ is a Banach space of all linear continuous mappings ("operators")
 from $X$ to $Y$ equipped with the usual operator norm. 
 %We have:
%$X\ot Y\subset F(X^*, Y)\subset L(X^*,Y);$ a completion of the algebraic
%tensor product $X\ot Y$ with respect to the operator norm is denoted by
%$X\wh\ot Y$ ("")
   
  
  If $A\in L(X,W),$ $B\in L(Y,G)$ and $z\in X\ot Y,$ then  a linear map
  $A\ot B: X\ot Y\to W\ot G$ is defined by 
  $A\ot B((x\ot y):= Ax\ot By$ (and then extended by linearity). Since
 $\wt{A\ot B(z)}= B\wt z A^*$ for $z\in X\ot Y,$ we will use notation   
$B\circ z \circ A^*\in W\ot G$ for $A\ot B(z).$
In the case where $X$ is a dual space, say $F^*,$ and $T\in L(W,F)$
(so, $A=T^*: F^*\to W^*),$ 
one considers a composition $B\wt z T;$ in this case $T^*\ot B$ maps
$F^*\ot Y$ into $W^*\ot Y$ and
we use notation
$B\circ z\circ T$ for $T^*\ot B (z).$

If $\nu$ is a tensor quasi-norm (see \cite[0.5]{166}), %on $X\ot Y$ and $W\ot G$,       %!!!ref
then $\nu(A\ot B (z))\le ||A||\, ||B||\, \nu(z)$  and we can extend the map
$A\ot B$ to the completions of the tensor products with respect to the quasi-norm
$\nu,$ having the same inequality. The natural map $(X\ot Y, \nu)\to L(X^*, Y)$
is continuous and can be extended to the completion $\wh{X\ot_\nu Y};$ for
a tensor element $z\in \wh{X\ot_\nu Y},$ we still denote by $\wt z$ the corresponding
operator. The natural mapping
  $\wh{X\ot_\nu Y}\to L(X^*, Y)$  need not to be injective; {\it if it is injective
  for a fixed $Y$ and for all $X,$ then we say that $Y$ has the $\nu$-approximation property}.                                                        
   
A projective tensor product $X\wh\ot Y$ of Banach spaces $X$ and $Y$ is defined
as a completion of $X\ot Y$ with respect to the norm $||\cdot||_{\land}:$
if $z\in X\ot Y,$ then
$$
 ||z||_\land:= \inf \sum_{k=1}^n ||x_k||\, ||y_k||,
$$
where infimum is taken over all representation of $z$ as $\sum_{k=1}^n x_k\ot y_k.$
We can try to consider
    $X\wh\ot Y$ also as operators $X^*\to Y$ or $Y^*\to X,$ but this
    correspondence is, in general, not one-to-one.
Note that $X\wh\ot Y= Y\wh\ot X$ in a sence.
If $z\in X\wh\ot Y, \e>0,$ then one can represent $z$ as
$z=\sum_{k=1}^\infty x_k\ot y_k$ with $\sum_{k=1}^\infty ||x_k||\, ||y_k||<||z||_\land+\e.$
For $z\in X^*\wh\ot X$ with a "projective representation" $z=\sum_{k=1}^\infty x'_k\ot x_k,$
trace of $z, \tr z:=z=\sum_{k=1}^\infty \<x_k, y_k\>,$ does not depend of representation of $z.$
The Banach dual $(X\wh\ot Y)^*= L(Y,X^*)$ by $\<T, z\>=\tr T\circ z.$   

Some more notations:
If $\frak A$ is an operator ideal then
$\frak A^{reg}(X,Y):= \{T\in L(X,Y):\ \pi_YT\in \frak A(X,Y^{**})\},$
$\frak A^{dual}(X,Y):= \{T\in L(X,Y):\ T^*\in \frak A(Y^*,X^*)\}.$

Finally,  
      $$l_p(X):=\{(x_i)\subset X:\ ||(x_i)||_p:=\(\sum ||x_i||^p)^{1/p}<\infty\},$$
    $$l_\infty(X):=\{(x_i)\subset X:\ ||(x_i)||_\infty:=\sup_i ||x_i||<\infty\},$$
   $$l^w_p(X):=\{(x_i)\subset X:\ ||(x_i)||_{w,p}:= \sup_{||x'||\le1}\(\sum |\<x', x_i\>|^p)^{1/p}<\infty\},$$
   $$l^w_\infty(X):=\{(x_i)\subset X:\ ||(x_i)||_{w,\infty}:=\sup_i ||x_i||<\infty\}.$$
 Note that if $p\le q,$ then $||\cdot||_q\le  ||\cdot||_p$ and $||\cdot||_{w,q}\le  ||\cdot||_{w,p}.$
If $0<p\le\infty,$ then $p'$ is a conjugate exponent: 
$1/p+1/p'=1$ if $p\ge1$ and $p'=\infty$ if $p\in (0,1].$   
  
   {\bf Below $0<r, s\le1,$ $0< p,q \le\infty$ and $1/r+1/p+1/q=1/\beta\ge1.$}
   
   %%%
      
%{\bf !: on r,p,q-nuclear: we will use fixed natation r,s,p,q throughout,
%but it'll be ness.-ly --- (t,p,q), (h,p,q) for t,h in [o, infty)...
%t.g. see...}

Let us note that "Remarks" in the paper can contain sometimes  quite important
information comparable to the information presented in Theorems and Propositions. %!!! maybe below?-       
      \bigskip
   %%

       \centerline{\bf 1. The tensor products $X\wh\ot_{r,p,q} Y$}
                            \bigskip
                            
       %$AP_{r,p,q}.$}\,
 We use partially notations from \cite{166}. %!!!
 %    [166] Lapreste, J. T.: Op'erateurs sommants et factorisations `a travers les espaces LP;
%Studia Math. 57(1976)47-83     
%$\wh{X\ot_{r,p,q} Y}$                      $X\wh\ot_{r,p,q} Y$
 % $X\wh\ot Y$           $N_{r,p,q}(X, Y)$ 
  For $z\in X\ot Y$ we put 
  $$\mu_{r,p,q}(z):= \inf\{||(\al_k)||_r ||(x_k)||_{w,p} ||(y_k)||_{w,q}:\ 
  z=\sum_{k=1}^n \al_k x_k\ot y_k\};$$
$X\ot_{r,p,q} Y$ is the tensor product, equipped with this quasi-norm $\mu_{r,p,q}.$  
Note that $\mu_{1,\infty,\infty}$ is the projective tensor norm of A. Grothendieck \cite{Gr}. %!!!
  
Let us denote by $\wh{X\ot_{r,p,q} Y}$ the completion of $X\ot Y$ with respect to
this quasi-norm $\mu_{r,p,q}$ (in \cite{166} --- $X\underset{r,p,q}{\wh\ot} Y).$               %!!!ref
Every tensor element $z\in \wh{X\ot_{r,p,q} Y}$
admits a representation of type
$z= \sum_{k=1}^\infty \al_k x_k\ot y_k,$
where $||(\al_k)||_r ||(x_k)||_{w,p} ||(y_k)||_{w,q}<\infty,$ and
$$\mu_{r,p,q}(z):= \inf ||(\al_k)||_r ||(x_k)||_{w,p} ||(y_k)||_{w,q}$$
(infimum is taken over all such finite or infinite representations) \cite[Proposition 1.3, p. 52]{166}.        %!!!ref
Note that $\wh{X\ot_{1,\infty,\infty} Y}= X\wh\ot Y.$
                    \medskip
                    
{\bf Lemma 1.1}\,        % Prop.?
Let
1)\,
$0<r_1\le r_2\le1,$ $p_1\le p_2$ and $q_1\le q_2$
or
2)\,
 $0<r_1< r_2\le1,$ $p_1\ge p_2,$ $q_1\ge q_2$ and
 $1/r_2+1/p_2+1/q_2\le 1/r_1+1/p_1+1/q_1.$
If $z\in X\ot Y,$ then $\mu_{r_2,p_2,q_2}(z)\le \mu_{r_1,p_1,q_1}(z).$     
In particular,  $\mu_{1,\infty,\infty}(z)\le \mu_{r_1,p_1,q_1}(z).$ 
Consequently, a natural mappings
$X\ot_{r_1,p_1,q_1} Y\to X\ot_{r_2,p_2,q_2} Y\to X\wh\ot Y$ can be extended
to the (natural) continuos maps
  $$\wh{X\ot_{r_1,p_1,q_1} Y}\to \wh{X\ot_{r_2,p_2,q_2} Y}\to X\wh\ot Y.$$
                            \smallskip
                            
 {Proof}.\,
Case 1): If $z=\sum_{k=1}^n \al_k x_k\ot y_k,$ then 
$||(\al_k)||_{r_2} ||(x_k)||_{w,p_2} ||(y_k)||_{w,q_2}\le
||(\al_k)||_{r_1} ||(x_k)||_{w,p_1} ||(y_k)||_{w,q_1}.$

Case 2:
The proof is standard (cf. \cite[18.1.5, p. 246-247]{PiOP}).              %!!!ref
Take $r$ such that $1/r_1=1/r+1/p+1/q,$ where $1/p:= 1/p_2-1/p_1$ and $1/q:=1/q_2-1/q_1.$
Then $r\le r_2$ and $r_1/r+r_1/p+r_1/q=1.$ If $z=\sum_{k=1}^n \al_k x_k\ot y_k,$
then $z=\sum_{k=1}^n \al_k^{r_1/r} (\al^{r_1/p}_k x_k)\ot (\al^{r_1/q}_k y_k)$ and
$$
  ||(\al_k^{r_1/r})||_{r_2} \le ||(\al_k^{r_1/r})||_r= ||(\al_k)||^{r_1/r}_{r_1}.
$$
Since $p_2\le p_1$ and $1-p_2/p_1=p_2/p\le1,$ we can apply Golder inequality to get      %!!!
$$
  ||(\al^{r_1/p}_k x_k)||_{w,p_2} \le
   \(\sum_{k=1}^n |\al_k|^{\frac{p}{p_2}\cdot \frac{r_1p_2}{p}}\)^{\frac{1}{p_2}\cdot \frac{p_2}{p}}\, ||(x_k)||_{w,p_1}=
 $$ 
$$  
   ||(\al_k^{r_1/p})||_{p}\, ||(x_k)||_{w,p_1}= ||(\al_k)||^{r_1/p}_{w,r_1}\,  ||(x_k)||_{w,p_1}.
$$
By the same reason,
 $$
   ||(\al^{r_1/q}_k y_k)||_{w,q_2} \le
   ||(\al_k^{r_1/q})||_{q}\, ||(y_k)||_{w,q_1}= ||(\al_k)||^{r_1/q}_{w,r_1}\,  ||(y_k)||_{w,q_1}.
 $$  
 Hence,
 $$
   ||(\al_k^{r_1/r})||_{r_2}\, ||(\al^{r_1/p}_k x_k)||_{w,p_2}\, ||(\al^{r_1/q}_k y_k)||_{w,q_2}\le
 $$
   $$  
     ||(\al_k)||^{r_1/r}_{r_1}\, ||(\al_k)||^{r_1/p}_{w,r_1}\,  ||(x_k)||_{w,p_1}\,
     ||(\al_k)||^{r_1/q}_{w,r_1}\,  ||(y_k)||_{w,q_1}=
 $$
 $$
  ||(\al_k)||_{r_1}\, ||(x_k)||_{w,p_1}\, ||(y_k)||_{w,q_1}.
 $$
 It follows that $\mu_{r_2,p_2,q_2}(z)\le \mu_{r_1,p_1,q_1}(z).$
\medskip

Let us recall the following useful fact (see Section 0). %!!! Sec 0
If $A\in L(X,W),$ $B\in L(Y,G)$ and $z\in \wh{X\ot_{r,p,q} Y},$ then  
$B\circ z \circ A^*\in \wh{W\ot_{r,p,q} G}$ and 
$\mu_{r,p,q}(B\circ z \circ A^*)\le ||B||\, ||A||\, \mu_{r,p,q}(z).$
Particular cases: $X=W$ and $A=\id_X$ or $Y=G$ and $B=\id_Y.$ 
 
 
\small
The topological dual to $(\wh{X\ot_{r,p,q} Y}, \mu_{r,p,q})$ is the space
$\Pi_{\infty,p,q}(X,Y^*)$ of absolutely $(\infty,p,q)$-summing operators 
from $X$ to $Y^*$ \cite[Theorem 1.3, p. 57]{166}
(recall that $0<r\le1):$ If $\tau\in (\wh{X\ot_{r,p,q} Y})^*$ and 
$x\ot y\in X\ot Y,$ then the corresponding operator $T$
is defined by $\<\tau, x\ot y\>= \<Tx, y\>$ \cite[pp. 56-57]{166}.                %!!!ref
Recall that, by definition, an operator $T: X\to F$ is
absolutely $(\infty,p,q)$-summing if for any finite sequences
$(x_k)$ and $(f'_k)$ (from $X$ and $F^*$ respectively) one has
$$
 \sup_k |\<Tx_k, f'_k\>|\le C\, ||(x_k)||_{w,p} ||(f'_k)||_{w,q}.
$$
With a norm $\pi_{\infty,p,q}(T):= \inf C,$ the space 
 $\Pi_{\infty,p,q}(X, F)$ is a Banach space and in duality above (for $F=Y^*)$
 $\pi_{\infty,p,q}(T)= ||\tau||$ (on the right is the norm of the functional
 $\tau$ in $(\wh{X\ot_{r,p,q} Y})^*$).
 
 Futhermore, taking a sequence in $X\times F^*,$ consisting of one nonzero element
 $(x,f'),$ we obtain: If $T\in \Pi_{\infty,p,q}(X, F),$ then
 $|\<Tx, f'\>|\le \pi_{\infty,p,q}(T)\, ||x||\, ||f'||;$
 thus, $||T||\le \pi_{\infty,p,q}(T).$
  On the other hand, if $T\in L(X,F),$ then
for any finite sequences $(x_k)$ and $(f'_k)$ we have:
$$\sup_k |\<Tx_k, f'_k\>|\le ||T||\, \sup_k ||x_k||\, \sup_i ||f'_i||\le
 ||T||\, ||(x_k)||_{w,p}\, ||(f'_k)||_{w,q}.$$ 
 Therefore,
 $\Pi_{\infty,p,q}(X, F)= L(X, F).$

 I do not know whether the dual space $\Pi_{\infty,p,q}(X, Y^*)$ separates
 points of $\wh{X\ot_{r,p,q} Y}.$ If so, then the natural map
   $\wh{X\ot_{r,p,q} Y}\to X\wh\ot Y$ is one-to-one. As a matter of fact,
   it follows from the above considerations, that
 {\it the space $\Pi_{\infty,p,q}(X, Y^*)$ separates
 points of $\wh{X\ot_{r,p,q} Y}$ iff the natural map
   $j_{r,p,q}:\, \wh{X\ot_{r,p,q} Y}\to X\wh\ot Y$ is one-to-one.}
       \smallskip
       
 {\bf Definition 1.1.}\,  
 We define a tensor product $X\wh\ot_{r,p,q} Y$ as a linear subspace of
 the projective tensor product $X\wh\ot Y,$ consisting of all tensor elements $z,$
 which admit representations of type
 $$                 
  z=\sum_{k=1}^\infty \al_k x_k\ot y_k,\
  (\al_k)\in l_r,\, (x_k)\in l_{w,p},\, (y_k)\in l_{w,q}
 $$
 and equipped with the quasi-norm $||z||_{\land\!; r,p,q}:= \inf ||(\al_k)||_r\,
  ||(x_k)||_{w,p}\, ||(y_k)||_{w,q},$ where the infimum is taken over all
  representations of $z$ in the above form.
       \smallskip
       
 Note that this tensor product is $\beta$-normed (see \cite{166, PiOP}).  %!!! ref  Check!!!    
           \smallskip
           
 {\it Remark 1.1}.\,
  We can define $X\wh\ot_{r,p,q} Y$ also as a quotient of the space $\wh{X\ot_{r,p,q} Y}$
 by the kernel of the map $j_{r,p,q}$ (i.e. by the annihilator $L(X,Y^*)_{\perp}$ of
$L(X,Y^*)$ in the space $\wh{X\ot_{r,p,q} Y}).$ Therefore:
 
 (i)\,
 The tensor product $X\wh\ot_{r,p,q} Y$ is complete, i.e. a quasi-Banach space.
This, with the injectivity of the natural map $X\wh\ot_{r,p,q} Y\to X\wh\ot Y$
answers a corresponding question of Sten Kaijser ("Why the last map is one-to-one
for the "completion" $X\wh\ot_{r,p,q} Y$?").
 
 (ii)\,
If the dual of $\wh{X\ot_{r,p,q} Y}$ separates points of this space,
 then we can write $\wh{X\ot_{r,p,q} Y}= X\wh\ot_{r,p,q} Y.$ In this case
 "finite nuclear" quasi-norm $\mu_{r,p,q}$ coincides with the tensor quasi-norm
$||z||_{\land\!; r,p,q}$ (compare with \cite[18.1.10.]{PiOP}).   %!!!   ref
              
(iii)\,
 The dual space to $X\wh\ot_{r,p,q} Y$ is still 
 $\Pi_{\infty,p,q}(X,Y^*)$ of absolutely $(\infty,p,q)$-summing operators 
from $X$ to $Y^*$ with its natural quasi-norm.               
                        \medskip
                        
It follows from Lemma 1.1      %!!! L.1--> Prop.
(or, if one wishes, can be proved by the same method)
                      \medskip
                      
{\bf Proposition 1.1}\,
Let
1)\,
$0<r_1\le r_2\le1,$ $p_1\le p_2$ and $q_1\le q_2$
or                  
2)\,
 $0<r_1< r_2\le1,$ $p_1\ge p_2,$ $q_1\ge q_2$ and
 $1/r_2+1/p_2+1/q_2\le 1/r_1+1/p_1+1/q_1.$
If $z\in X\ot Y,$ then 
 $||z||_{\land\!; r_2,p_2,q_2} \le ||z||_{\land\!; r_1,p_1,q_1}.$
     %$\mu_{r_2,p_2,q_2}(z)\le \mu_{r_1,p_1,q_1}(z).$
In particular,  
$||z||_{\land\!; 1,\infty,\infty} \le ||z||_{\land\!; r_1,p_1,q_1}.$
   %$\mu_{1,\infty,\infty}(z)\le \mu_{r_1,p_1,q_1}(z).$ 
Consequently, a natural mappings
$X\wh\ot_{r_1,p_1,q_1} Y\to X\wh\ot_{r_2,p_2,q_2} Y\to X\wh\ot Y$ are continuos
injections of quasi-norms 1.                        
                       \medskip 
                        
                        
  {\bf Proposition 1.2.}\,  
 If $X$ or $Y$ has the bounded approximation property, then 
 $\mu_{r,p,q}= ||\cdot||_{\land\!; r,p,q}$ on $X\ot Y.$ Hence, in this case
 the dual of $\wh{X\ot_{r,p,q} Y}$ separates points, $j_{r,p,q}$ is injective and
 $\wh{X\ot_{r,p,q} Y}= X\wh\ot_{r,p,q} Y$ (and equals to the corresponding space of 
 $(r,p,q)$-nuclear operators; see below Corollary 2.1).  %!!! find: c1
                              \smallskip
                              
 {\it Proof}.\, It is enough to show that the map $j_{r,p,q}$ is injective.
Since $\wh{X\ot_{r,p,q} Y}= \wh{Y\ot_{r,q,p} X},$ it is enough to consider the case,
where $Y\in C$-MAP,\, $C\in [1,\infty).$ Let $z\in X\ot Y$ and let $\wt z: X^*\to Y$ be
an operator, associated with $z$ (note that this is one-to-one correspondence). 
There exists a finite rank operator $R: Y\to Y$ such
that $||R||\le C+1$ and $R\wt z\, (:= (\wt{\id_X\ot R})(z) ) = \wt z$ (see \cite[10.2.5, p. 131]{PiOP}).       %!!!ref
 Fix $\delta>0$ and choose a representation for $z,$ $z=\sum_{k=1}^\infty \al_k\, x_k\ot y_k,$
 with $||(\al_k)||_r\, ||(x_k)||_{w,p}\, ||(y_k)||_{w,q}\le ||z||_{\land\!; r,p,q}\, (1+ \delta).$ 
 Let
 $E:= R(Y)\subset Y,$ $M:= \dim E$ and $\e=\e(M)\in(0,\delta]$ (to be chosen later). Then 
 $\wt z= \sum_{k=1}^\infty \al_k\, \<x_k, \cdot\> \ffi_k,$ where $\ffi_k=Ry_k\in E.$
 Let $N$ be such that 
$||(\al_k)_N^\infty||_r\, ||(x_k)_N^\infty||_{w,p}\, ||\ffi_N^\infty||_{w,q}\le \e\,  ||z||^\beta_{\land\!; r,p,q}.$
                  
Now, since $E$ is finite dimensional, $\id_E$ admit a representation in
$\wh{E^*\ot_{r,p,q} E}$ which give us an estimation from above 
for $\mu_{r,p,q}(\id_E)$ by a constant $C=C(M)$ depending only on $M.$    %!!! \mu -> to ||.||_.. and around
 Indeed, take an isomorphism $A: E\to l_2^M$
with $||A||=1, ||A^{-1}||\le \sqrt M$ (see e.g. \cite[Corollary 3.9]{Pis}).   %!!! ref       
  %Pisier, Gilles, 
  % (Cambridge tracts in mathematics; 94), Cambridge University Press 1989, The volume of convex bodies and Banach space geometry
Since $\id_{l_2^M}=\sum_{k=1}^M e'_k\ot e_k,$ \, $\mu_{r,p,q}(\id_{l_2^M})\le M^{1/\beta}.$
Therefore, $\id_E=A^{-1} \id_{l_2^M} A = \sum_{k=1}^M A^*e'_k\ot A^{-1}e_k$ and
$\mu_{r,p,q}(\id_E)\le M^{1/\beta+ 1/2}.$          
  So, for  any $v\in X\ot E,$ considering 
  $\id_E\circ v$ $(= \id_E \wt v)$ we obtain an inequality
  $\mu_{r,p,q}(\id_E\circ v)\le M^{1/\beta+ 1/2}\, ||\wt v||.$
Since $||\wt v||\le ||v||_{\land\!; r,p,q}$  we get
$$
    \mu_{r,p,q}(\id_E\circ v)\le C(M)\, ||v||_{\land\!; r,p,q}.
$$
Hence, for our $z$ we get
$$
  \mu_{r,p,q}^\beta(z)\le \mu_{r,p,q}^\beta\(\sum_{k=1}^N \al_k\, x_k\ot \ffi_k\) +
 \mu_{r,p,q}^\beta\(\sum_{k=N+1}^\infty \al_k\, x_k\ot \ffi_k\) \le
$$
$$
  %(1+\delta)^\beta\, 
  ||R||^\beta\, 
  \mu_{r,p,q}^\beta\(\sum_{k=1}^N \al_k\, x_k\ot y_k\) +
  %  \e^\beta 
    C(M)^\beta\, ||\sum_{k=N+1}^\infty \al_k\, x_k\ot \ffi_k||^\beta_{\land\!; r,p,q}\le
$$                                                                                              
$$
  % (1+\delta)^\beta\, 
  ||R||^\beta\, (||(\al_k)||_r\, ||(x_k)||_{w,p}\, ||(y_k)||_{w,q})^\beta + C(M)^\beta\, 
   (||(\al_k)_N^\infty||_r\, ||(x_k)_N^\infty||_{w,p}\, ||\ffi_N^\infty||_{w,q})^\beta \le
$$
$$
  (1+\delta)^\beta\, 
    ||R||^\beta\, ||z||^\beta_{\land\!; r,p,q}+  \e^\beta 
    C(M)^\beta\, ||z||^\beta_{\land\!; r,p,q}.
$$
                                             
Taking $\e<\delta$ to have $\e^\beta C(M)^\beta<\delta,$  we obtain 
 $\mu_{r,p,q}^\beta(z) \le [(1+ \delta)^\beta\, (C +1)^\beta +\delta]\, 
 ||z||^\beta_{\land\!; r,p,q}.$                
 
This inequality means, in particular, that the quasi-norms 
 $\mu_{r,p,q}$ and $||\cdot||_{\land\!; r,p,q}$ are equivalent
 on the space $X\ot Y,$ thus giving the same completions.
 Since the space $X\wh\ot_{r,p,q} Y$ is complete (Remark (i) above),
 we obtain that the natural quotient map
$\wh{X\ot_{r,p,q} Y}\to X\wh\ot_{r,p,q} Y$ is injective (recall that
the last space is a subspace of the projective product $X\wh\ot Y).$
                        \smallskip
                        
{\it Remark 1.2.}\,
For an "operator" situation, see Corollary 2.1 below % c1 to find
and (for $1\le p,q,\le \infty)$ \cite[pp. 249-251]{PiOP}.                  %!!!ref
\smallskip

      \begin{comment}
??? {\bf Proposition 2. --- NONCLEAR!!!}\,
  The natural map $Y\widehat\otimes_{r,p,q} X\to Y\widehat\otimes_{r,p,q} X^{**}$
  is an isometric imbedding (and, in particular, one-to-one).
  
  %\noindent
  {\it Proof}.\, 
As was mentioned in Section 1, Remark 1, (iii), % previous!!!
the dual space to $Y\widehat\otimes_{r,p,q} X$ is the space
$\Pi_{\infty,p,q}(X,Y^*).$ ... ... ... 
   \end{comment}
          \bigskip
          
          
\centerline{\bf Approximation properties}
\bigskip

We begin with the main definition.
%\smallskip
 %{\bf General information.}
\small

 {\bf Definition 2.1.}\,
 A Banach space $X$ has the approximation property $AP_{r,p,q}$ if
 for every Banach space $Y$ the canonical mapping $Y\wh\ot_{r,p,q} X\to L(Y^*,X)$
 is one to one.
 \medskip
 
 {\bf Proposition 2.1.}\,
  The following conditions are equivalent:
  
  1)\,
$X$ has the $AP_{r,p,q}.$                        

2)\,
For every Banach space $W$
the natural map $W^*\wh\ot_{r,p,q} X\to L(W,X)$ is one-to-one.

3)\,
The natural map $X^*\wh\ot_{r,p,q} X\to L(X):=L(X,X)$ is one-to-one.
        \small
        
{\it Proof.}\,
Implications $1)\implies 2) \implies 3)$ are evident.

$3)\implies 1).$\,
%The proof repeats word for word the proof of Proposition 6.1 from \cite{Trend}      %!!!
%(instead of $\wh\ot_{\alpha}$ and $Y^*$ there we have to put now the tensor products 
%$\wh\ot_{r,p,q}$ and $Y$ respectively).
            %\centerline                          {\bf part 1 FOR ARXIVE}
%For the sake of completeness, we give the proof here.
      Suppose that the canonical map $X^*\widehat\otimes_{r,p,q} X\to L(X)$ is one-to-one,
but there exists a Banach space $Y$ such that the natural map
$Y\widehat\otimes_{r,p,q} X\to L(Y^*, X)$ is not injective. Let $z\in Y\widehat\otimes_{r,p,q} X$
be such that $z\neq0$ and the associated operator $\widetilde z$ is a zero operator.
Then we can find an operator $V$ from $L(Y,X^*)$ (the dual space to the
projective tensor product $Y\widehat\otimes X)$ so that $\operatorname{trace}\, V\circ z^t=1,$
where, as usual, $z^t$ is the transposed tensor element, $z^t\in X\widehat\otimes Y.$
Since $V\circ z^t\in X\widehat\otimes X^*$ and   $\operatorname{trace}\, V\circ z^t=1,$ the tensor element
$(V\circ z^t)^t$ (which, evidently, belongs to $X^*\widehat\otimes_{r,p,q} X)$
is not zero. Contradiction.
%\centerline                          {\bf end part 1 FOR ARXIVE}
 \medskip
                                         
 {\it Remark 2.1}.\
 One can introduce also  (in a similar way) some new notions of the approximation properties by using
 the Laprest\'e tensor products. We do not consider these properties here because
 we do not know how to work with the tensor products if their Banach duals
 do not separate points.
               \medskip

 The following assertion is an analogue of \cite[Prop. 6.2]{Trend}. %Again,                %!!!Ref
Its proof is contained in the corresponding proof of Proposition 6.2 from \cite{Trend}.           %!!!
But since a situation now is a little bit different from the one there
(quasi-norms are here not "selfadjoint"), we present a proof here.              %"self..."?
%(essential different from the one in [rei Holl]).  %!!!                       

\medskip
 
  {\bf Proposition 2.2.}\,
 If $X^*$ has the $AP_{r,p,q},$ then $X$ has the $AP_{r,q,p}.$ 
             \small
             
  %\noindent
  {\it Proof}.\,
  We will use Proposition 1.
  As it is known \cite{Gr}, the projective tensor product                         %!!!ref
   $X^*\widehat\otimes X$  is a Banach subspace of the  tensor product
    $X^*\widehat\otimes X^{**}.$
    The tensor product  $X^*\widehat\otimes_{r,q,p} X $ is  a linear subspace of  $X^*\widehat\otimes X,$
    as well as  $X^*\widehat\otimes_{r,q,p} X^{**}$ is a linear subspace of  $X^*\widehat\otimes X^{**}.$
    Therefore, the natural map  $X^*\widehat\otimes_{r,q,p} X \to X^*\widehat\otimes_{r,q,p} X^{**} $
    is one-to-one. Now if $X^*$ has the $AP_{r.p.q},$ then the canonical map
     $X^{**}\widehat\otimes_{r,p,q} X^* \to L(X^*,X^*)$ is one-to-one.
     Since we can identify the tensor product $X^{**}\widehat\otimes_{r,p,q} X^*$ with
     the tensor product $X^{*}\widehat\otimes_{r,q,p} X^{**},$ it follows that
     the natural map $X^*\widehat\otimes_{r,q,p} X \to L(X,X)$ is one-to-one.
     Thus, if $X^*$ has the $AP_{r,p,q},$ then $X$ has the $AP_{r,q,p}.$ 
     \medskip

   %  We will use Lemma 4 in the proof of Theorem 1 in the next section.

      % \smallskip

     %%%%%%%%   Zamechanie o tochnosti - perepravit' !!!
  %   \noindent
     {\it Remark 2.2.}\,
        The inverse statement is not true. Some examples
     are given in \cite[Remark 6.1]{Trend}.    %!!!Ref
    % See also  ... below.                            %!!! below - corrol-s for some r,p,q: [r,p]...
             \medskip
             
 {\it Remark 2.3.}\, 
(i)\,
From the proof it follows that: {\it For any $X$ and $Y$
the natural map  $X\widehat\otimes_{r,q,p} Y \to X\widehat\otimes_{r,q,p} Y^{**}$
 is one-to-one.}
 
 (ii)\,
On the other hand: {\it For any $X$ and $Y$
the natural map  $\widehat{X\otimes_{r,q,p} Y} \to 
\widehat{X\otimes_{r,q,p} Y^{**}}$} is an isometric embedding. To prove this,
it is enough to apply Principle of Local Reflexivity (see e.g. \cite[E.3.1]{PiOP}) 
as it was done in \cite[18.1.12]{PiOP} for the case $1\le p,q\le\infty.$
             \medskip
             
Recall that a linear map $T: X\to Y$ is called $(r,p,q)$-nuclear
if it has a representation $T= \sum_{k=1}^\infty \al_k\, \<x'_k, \cdot\> y_k,$
where $(\al_k)\in l_r,$ $(x'_k)\in l_{w,p}(X^*)$ and $(y_k)\in l_{w,q}(Y).$
Every such a map is continuous. The space $N_{r,p,q}(X,Y)$ of all
$(r,p,q)$-nuclear operators from $X$ to $Y$ can be considered as a quotient
of the tensor product $X^*\wh\ot_{r,p,q} Y$ (as well as a quotient of
$\wh{X^*\ot_{r,p,q} Y})$ by the kernel of the natural map
$X^*\wh\ot_{r,p,q} Y \to L(X,Y).$ We equip this space with the induced
quasi-norm $(\beta$-norm)
denoted by $\nu_{r,p,q}.$ If the corresponding quotient map has 
a trivial kernel, then we write $N_{r,p,q}(X,Y)= X^*\wh\ot_{r,p,q} Y$
(respectively, $N_{r,p,q}(X,Y)= \wh{X^*\ot_{r,p,q} Y}).$
Thus, $X$ has the $AP_{r,p,q}$ iff for every Banach space $Y$ the equality
 $N_{r,p,q}(Y,X)= Y^*\wh\ot_{r,p,q} X$ holds.
        \small
        
  {\it Remark 2.4.}\,
  1)\, 
 If $t\in (0, +\infty],$ then $N_{t,p,q}(X, Y)$ (the space of  $(t,p,q)$-nuclear operators)
 can be defined by the analogues way:  an operator $T: X\to Y$ is $(t,p,q)$-nuclear,
 if it can be written in the form
 $T = \sum_{k=1}^\infty \al_k\, \<x'_k, \cdot\> y_k,$ where 
$(\al_k)\in l_t,$ $(x'_k)\in l_{w,p}(X^*)$ and $(y_k)\in l_{w,q}.$ The quasi-norm
$||\cdot||_{N_{t,p,q}}$ is defined for $T$ as
$\inf ||(\al_k)||_t ||(x'_k)||_{w,p} ||(y_k)||_{w,q},$
where infimum is taken over all appropriate representations of $T.$ If $1/t + 1/p + 1/q=1,$
then it is a norm. %We will use these spaces in Section ...      %!!! where - write

  2)\,
  In notation, we follow J.-T. Laprest\'e \cite{166}, only changing a triple         %!!! ref
  $(p,r,s)$ there to $(r,p,q)$  here
(see also \cite{PiOP} ; nota bene: A. Pietsch \cite[18.1]{PiOP} uses different notations        %!!! ref
for this class of operators and considers the cases where $1\le p,q\le\infty.)$      
\medskip

It follows from Proposition 1.1:
\medskip

{\bf Proposition 2.3}\,
%In conditions of Proposition 0, 
Let
1)\,
$0<r_1\le r_2\le1,$ $p_1\le p_2$ and $q_1\le q_2$
or                  
2)\,
 $0<r_1< r_2\le1,$ $p_1\ge p_2,$ $q_1\ge q_2$ and
 $1/r_2+1/p_2+1/q_2\le 1/r_1+1/p_1+1/q_1.$
If 
$X$ has the $AP_{r_2,p_2,q_2},$ then $X$ has the $AP_{r_1,p_2,q_3}.$
In particular, the $AP$ of A. Grothendieck implies any $AP_{r,p,q}.$
\medskip

{\bf Corollary 2.1.}\,     %c1 for find
(i)\,
If $X$ has the bounded approximation property, then for all $r,p,q$ and $Y$
the equalities
$N_{r,p,q}(Y,X)= Y^*\wh\ot_{r,p,q} X = \wh{Y^*\ot_{r,p,q} X}$
hold (with the same quasi-norms).
  (ii)\,
If $Y^*$ has the bounded approximation property, then for all $r,p,q$ and $X$
the equalities
$N_{r,p,q}(Y,X)= Y^*\wh\ot_{r,p,q} X = \wh{Y^*\ot_{r,p,q} X}$
hold (with the same quasi-norms).
                        \small
                        
{\it Proof}.\,
Apply Propositions 1.2 and 2.3.
              \medskip

{\bf Lemma 2.1.%to3.
}\,                          %!!! change X <-> Y as above and below   DO NOT WANT!
%1)\,
The tensor product $\wh{\cdot\ot_{r,p,2}\cdot}$ is injective, i.e.
if $i: Y\to W$ is an isometric imbedding and $z\in \wh{X\ot_{r,p,2}Y},$
then $\mu_{r,p,2}(z)=\mu_{r,p,2}(i\circ z).$

%2)\,
%The tensor product $\wh{\cdot\ot_{r,2,q}\cdot}$ is surjective, i.e.         %!!! �����, ���� ��� ��������?
%if $Q: V\to X$ is a quotient map and $z\in \wh{X\ot_{r,2,q}Y},$
%then $\mu_{r,2,q}(z)=\mu_{r,2,q}(i\circ z).$
                       \small
                       
{\it Proof}.\,
%1)\,
It is clear that $\mu_{r,p,2}(z)\ge\mu_{r,p,2}(i\circ z).$
Let $\e>0$ and $\sum_{k=1}^N \al_k x_k\ot \ffi_k$ be a finite representation of $i\circ z$
   %in $\wh{X\ot_{r,p,2}W}$ such that 
in $X\ot W$ such that
$||(\al_k)||_r ||(x_k)||_{w,p} ||(\ffi_k)||_{w,q}\le (1+\e) \mu_{r,p,2}(i\circ z).$
Define an operator $S\in L(l^N_2,W)$ and a tensor element $z_0\in \wh{X\ot_{r,p,2}l^N_2}$ by
$S = \sum e'_k\ot \ffi_k$ and $z_0=\sum \al_k x_k\ot e_k.$ Let $E:=\ove{\wt{z_0}(X^*)}\subset l_2^N$ 
and $P: l_2^N\to l_2^N$ be an orthogonal projector from $l_2^n$ onto $E.$
Then $S\circ z_0=i\circ z,$ $\wt{SP\circ z_0}=\wt{i\circ z}$ and $SP(l_2^N)\subset i(Y).$
It follows that $z= (i^{-1}|_{i(Y)}SP)\circ z_0= \sum \al_k x_k\ot i^{-1}|_{i(Y)}SPe_k$
(as the elements of $X\ot Y)$
and 
$$\mu_{r,p,2}(z)\le ||S||\, ||(\al_k)||_r ||(x_k)||_{w,p}=
||(\al_k)||_r ||(x_k)||_{w,p} ||(\ffi_k)||_{w,q}\le (1+\e) \mu_{r,p,2}(i\circ z).$$ 
 Therefore, the natural map $\id_X\ot i: X\ot Y\to X\ot W$ is an isometric imbedding
 and it extends to the isometry 
 $\id_X\ove{\ot} i: \wh{X\ot_{r,p,2} Y}\to \wh{X\ot_{r,p,2} W}.$
                  \medskip
                  
% 2)\,
{\it Remark 2.5.% to3
}.\,                                                                               %!!!  to3 and below
It follows from Lemma 2.1 %to3 
and from definitions of $\wh\ot_{r,p,2}$ (and $N_{r,p,2})$
that $\wh\ot_{r,p,2}$ (and $N_{r,p,2})$ are injective (consider the quotient maps
$\wh{\ot_{r,p,2}}\to \wh\ot_{r,p,2}\to N_{r,p,2}$). On the other hand, the injectivity
of $\wh\ot_{r,p,2}$ (and $N_{r,p,2})$ can be proved in the same way as above 
by consideration the infinite representations of $z, \wt z$ (instead of finite ones) 
in the given proof.         
Also, we see from the proof that if
$\sum_{k=1}^\infty \al_k x_k\ot \ffi_k$ is a representation of $i\circ z$
   in $\wh{X\ot_{r,p,2}W}$ (see \cite[Proposition 1.3, p. 52]{166}, then
   the corresponding representation for $z$ in $\wh{X\ot_{r,p,2} Y}$ 
   can be taken of the type
   $\sum_{k=1}^\infty \al_k x_k\ot y_k$ with $||(y_k)||_{w,2}\le ||(\ffi_k)||_{w,2}.$
The same is true for $\wh\ot_{r,p,2}$ and $N_{r,p,2}.$ 
                 \medskip
                 
         % repr. Lapr-Th on ..-nucl. [166, Proposition 1.3, p. 52]. 

The first part of the following fact is partially known 
(cf. \cite[18.1.15-18.1.16]{PiOP} for $1\le p,q\le\infty).$
\medskip

 {\bf Proposition 2.4.}\,
 For any Banach spaces $X,Y$ the equalities
 $$
    N_{r,p,2}(Y,X)= Y^*\wh\ot_{r,p,2} X = \wh{Y^*\ot_{r,p,2} X} \ \mathrm{ and } \
    N_{r,2,q}(Y,X)= Y^*\wh\ot_{r,2,q} X = \wh{Y^*\ot_{r,2,q} X}
 $$
 hold (with the same quasi-norms). 
 In particular, every Banach space has the $AP_{r,p,2}$ and the $AP_{r,2,p}.$
  \small
  
  % This fact is essentially known (see [PiOP, 18.1.15-18.1.16] for $1\le p,q\le\infty).
 
 %\centerline                          {\bf part 2 FOR ARXIVE}
          \medskip
          
 {\it Proof}.\,                                                              
 As is known, the operator ideal $N_{r,p,2}$ is injective, (see \cite[18.1.8]{PiOP}           %%!!!Ref
 for the case $1\le p,q\le\infty);$ apply (factorization) Theorem 2.5 \cite{166}         %!!!ref
 in other cases).         %!!!ref
 I.e., if $X\subset G,$ $T\in L(Y,X)$ and $T\in N_{r,p,2}(Y,G),$ then $T\in N_{r,p,2}(Y,X)$
 (with the same quasi-norm). Also, the tensor product $\wh{\cdot\ot_{r,p,2} \cdot}$
 is injective too (Lemma 2.1)%to3) %apply the same factorization theorem to "finite" tensors $z\in E\ot F).$      %!!! to3
 Now, let $z\in \wh{Y^*\ot_{r,p,2} X}$ and $i: X\to L_\infty$ be an isometric 
 embedding of $X$ into an $L_\infty$-space. Since $L_\infty$ has the MAP,
 $\nu_{r,p,2}(\wt{i\circ z})= \mu_{r,p,2}(i\circ z)$ (see Corollary 2.1).   %!!! find: c1  - CHECK IT
 Hence, $\mu_{r,p,2}(z)=\mu_{r,p,2}(i\circ z)=\nu_{r,p,2}(\wt{i\circ z})=
 \nu_{r,p,2}(i \wt z)\le \nu_{r,p,2}(\wt z)\le \mu_{r,p,2}(z).$ 
 
  To get the last two equalities it is enough to apply the surjectivity of     %!!! details! or else...
 the operator ideal $N_{r,2,q}$ and Corollary 2.1 (second part), by using
 the same idea as above, or just to apply Lemma 2.1   %to3 
 and Remark 2.5    %to3
  (second part):    %!!! Rem to3 - ������?
Take $z\in \wh{Y^*\ot_{r,2,q} X}$ and a quotient map $Q: L_1\to Y.$ Considering
$(z\circ Q)^t$ as an element of $\wh{X\ot_{r,q,2} L_\infty},$
we get:
$$
  \nu_{r,q,2}(\wt{Q^*\circ z^t})= \mu_{r,q,2}(Q^*\circ z^t) =\mu_{r,q,2}(z^t)=
  \mu_{r,2,q}(z).
$$
But $\nu_{r,q,2}(\wt{Q^*\circ z^t})\le \nu_{r,2,q}(\wt{z\circ Q})\le \nu_{r,2,q}(z);$
thus, $ \mu_{r,2,q}(z)\le \nu_{r,2,q}(z).$
\medskip

%{\bf ...}
                                                                                                               , 
%\centerline                          {\bf end part 2 FOR ARXIVE} 
        %  \small
  
  % [Pi consid. only (1,2,\infty) 27.4.10 ---later on Gr-Lids     !!!
                       
{\it Remark 2.6.}\,
The fact that every $X$ has the $AP_{1,2,\infty}$ is essentially contained in
\cite[27.4.10, Proposition]{PiOP}. It is strange, but it seems that
a corresponding fact for $AP_{1,\infty,2}$ appears here for the first time.
Note that this fact follows also from the preceding by virtue of Proposition 2.2:
if every $X$ has the $AP_{1,2,\infty},$ then $X^*$ possesses this property,
and by Proposition 2.2 $X$ has the $AP_{1,\infty,2}.$

  \medskip

Many of the above approximation properties were considered earlier, e.g. in the
papers \cite{RQ, Rs, Trend} etc:   %!!!refs    -  below 

(i)\,
For $p=q=\infty,$ we get the $AP_r$ from \cite{Rs, Trend}.              %!!!

(ii)\,
For $p=\infty,$ we get the $AP_{[r,q]}$ from \cite{RQ, Trend}.              %!!!        
                       
(iii)\,
For $q=\infty,$ we get the $AP^{[r,p]}$ from \cite{RQ, Trend}.              %!!!
                       
%[RQ] O. I. Reinov, Q. Latif, Distribution of eigenvalues of nuclear operators and
%Grothendieck-Lidski type formulas, Journal of Mathematical Sciences, Springer,
%Vol. 193, No. 2, August, 2013, 312-329.                       
%
%[Rs] O.I. Reinov, On linear operators with s-nuclear adjoints, $0 < s \le 1,$ J. Math.
%Anal. Appl. 415 (2014) 816-824.
%                       
%[Rrp] Oleg Reinov, Some Remarks on Approximation Properties, with Applications, 
%Ordered Structures and Applications: Positivity VII Trends in Mathematics,
%371-394. 2016 Springer International Publishing      
\smallskip

Following notations from \cite{Trend} (see also \cite{RQ}), we denote              %!!!ref

$N_{r,\infty,\infty}$ by $N_r,$ 

$N_{r,\infty,q}$ by $N_{[r,q]},$

$N_{r,p,\infty}$ by $N^{[r,p]},$ 

$\wh\ot_{r,\infty,\infty}$ by $\wh\ot_{r},$

$\wh\ot_{r,\infty,q}$ by $\wh\ot_{[r,q]},$

$\wh\ot_{r,p,\infty,}$ by $\wh\ot^{[r,p]}.$ 

The corresponding notations are used also for the $AP_{r,p,q}$
(see above (i)--(iii)).

Almost all the information about Banach spaces without (or with) the properties 
 $AP_r,$ $AP_{[r,q]}$ and  $AP^{[r,p]}$ which is known to us by now,
 can be found in \cite{RQ, Rs, Trend}.
 Other results in this direction are the subject of the forthcoming paper of the author. 
                \bigskip
                
               \begin{comment}
{\bf In $L_p$-spaces.}%Approximation properties in spaces of type and cotype.}
\medskip                 



                
      \bigskip
      
%%                     SECTION BIG NEXT

\centerline{SECTION II.}
\bigskip                       
                       
                       
                       $$- - - $$
                       
{\bf On $T^*$ and $T$ for (r,p,q)-nuclerity.}                       
                       
 It is clear that we have:
 \small
 
 {\bf Proposotion D.}\,
 If  $T\in N_{r,p,q},$ then $T^*\in N_{r,q,p}.$                      
                       \small
                       
--- [(r,p,2) ]...                       
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
                                         \end{comment}

 %%%%%%%%%%%%%%%%%%%%%%
  \begin{thebibliography}{00}
  
  \bibitem{Gr}
  A.~Grothendieck: \textit{Produits tensoriels topologiques et \'espaces nucl\'eaires},
 {Mem. Amer. Math. Soc.} \textbf{16}(1955). 
  
     \bibitem{RQ}
  O. I. Reinov, Q. Latif, \textit{Distribution of eigenvalues of nuclear operators and
Grothendieck-Lidski type formulas}, Journal of Mathematical Sciences, Springer,
Vol. 193, No. 2, August, 2013, 312-329.                       

\bibitem{Rs}
 O.I. Reinov, \textit{On linear operators with s-nuclear adjoints, $0 < s \le 1,$} J. Math.
Anal. Appl. 415 (2014) 816-824.
                       
  
  
   \bibitem{Trend}
  Oleg Reinov,
\textit{Some Remarks on Approximation Properties with Applications},
Ordered Structures and Applications: Positivity VII
Trends in Mathematics, 371-394, 2016

\bibitem{166}
Lapreste, J. T.: \textit{Op\'erateurs sommants et factorisations \`a travers les espaces $L_p$},
Studia Math. 57(1976)47-83

    \bibitem{PiOP} A. Pietsch:
 \textit{Operator Ideals}, North Holland (1980).
  
   \bibitem{Pis}
 G. Pisier, \textit{The volume of convex bodies and Banach space geometry},
 Cambridge tracts in mathematics {\textbf 94}, Cambridge University Press 1989. 
  
  
 \end{thebibliography}
  
  
 

\end{document}

  %%%%%%%%%%%
  
  {\bf Lemma 2.7.}\,
   % {Lemma 2.}\,  %!!! rename it!
 (i)\,  Let $z\in \wh{Y^*\ot_{r,p,2} X}.$ 
 If $i: X\to G$ be an isomorphic embedding such that $\nu_{r,p,2}(\wt{i\circ z})>0,$
 then $\nu_{r,,p,q}(\wt z)\ge0.$
 ii)\,  Let $z\in \wh{Y^*\ot_{r,2,q} X}.$ 
 If $Q: W\to Y$ be a surjection such that $\nu_{r,2,q}(\wt{i\circ z})>0,$
 then $\nu_{r,,2,q}(\wt z)\ge0.$
  
{\it Proof}.\,
 (i)\, Fix $\e>0$ and 
    consider the operator $i\circ \wt z.$ 
  By Theorem 2.5 [Lapr],   %!!!
$i\circ\wt z=BA,$ where $B\in L(l_2,G), ||B||=1,$ $A\in N_{r,p,2}(Y,l_2)$  and
$0<\nu_{r,p,2}(A)\le \nu_{r,p,2}(\wt{i\circ z})+\e.$  Let $P: l_2\to H:=B^{-1}(\over{BA(Y)}$
be an orthogonal projector. We have $\wt{i\circ z}= B|_{H}PA,$
$B|_{H}PA(Y)\subset i(X)$ and
$\wt z= i^{-1}|_{i(X)} B|_{H}PA.$
Hence, 0<\nu_{r,p,2}(\wt {i\circ z})        
              
   %%%%
   
 ... We apply Lemma 2.7:   ...
The proof is essentially known (cf. \cite[18.1.15 and 18.1.16]{PiOP} for $1\le p,q\le\infty).$
Let $T\in N_{r,p,2}(Y,X)$ and $i: X\to L_\infty(\mu)$ be an isometric embedding.
 By Theorem 2.5 [Lapr],   %!!!
$T=BA,$ where $B\in L(l_2,X)$ and $A\in N_{r,p,2}(Y,l_2).$  
 
 {\bf Remark.}\, ... ...
             
    In the cases where $p=2,$ or $q=2$ Pietsch....    
    
 %%%%
    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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  in the sense of Grothendieck, Math. Nachr. \textbf{283}, No. 2 (2010), 232--261.

                             \bibitem{9}
M. J. Kadec, M. G. Snobar: \textit{Certain functionals on the Minkowski compactum} (Russian),
 {Mat. Zametki} \textbf{10} (1971), 453-458.

    \bibitem{10} H. Konig:
   \textit{Eigenvalue distribution of compact operators},  Birkh\"auser (1986).

    \bibitem{11} H. Konig:
  \textit{On the eigenvalue spectrum of certain operator ideals},
   Coll. Math. \textbf{44} (1981), 15--28.

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 V.\,B.~Lidski\v{\i}:  \textit{Nonselfadjoint operators having a trace} (Russian),
 {Dokl. Akad. Nauk SSSR} \textbf{125} (1959), 485--487.

 \bibitem{13} J. Lindenstrauss: \textit{On James' paper "Separable Conjugate Spaces"},
  Israel J. Math. \textbf{9} (1971), 279--284.

    \bibitem{14} J. Lindenstrauss, L. Tzafriri:
  \textit{Classical Banach spaces, vol.1: Sequence spaces},
  Berlin-Heidelberg-New York (1977).

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  \textit{Eigenwertverteilungen von Operatoren in Banachr\"aumen},
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 \textit{Operator Ideals}, North Holland (1980).

  \bibitem{17} A. Pietsch:
 \textit{Eigenvalues and s-numbers}, Cambridge Univ. Press (1987).

   \bibitem{18} A. Pietsch:
  \textit{History of Banach spaces and linear operators},
  Birkh\"auser (2007).

                \bibitem{19} G. Pisier:
 \textit{Estimations des distances {\`a} un espace euclidien et
des constantes de proj{\'e}ction des espaces de Banach de dimensions finie},
Seminaire d'Analyse Fonctionelle 1978-1979, Centre de Math., Ecole Polytech.,
Paris (1979), expos{\'e} 10, 1--21.

      \bibitem{20} O.I. Reinov:
      \textit{A simple proof of two theorems of A. Grothendieck} (Russian),
        Vestn. Leningr. Univ. \textbf{7} (1983), 115-116.

     \bibitem{21} O.I. Reinov:
\textit{Disappearance of tensor elements in the scale of p-nuclear operators} (Russian),
Theory of Operators and Theory of Functions, Leningrad , LGU, vol. 1 (1983), 145-165.

      \bibitem{22} O.I. Reinov:
      \textit{A survey of some results in connection with Grothendieck approximation property},
   Math. Nachr. \textbf{119} (1984), 257-264.

 \bibitem{23} O.I. Reinov:
 \textit{Approximation properties $AP_s$ and p-nuclear operators
 (the case $0 <s\le 1)$},
 Journal of Mathematical Sciences \textbf{115}, No. 3 (2003), 2243-2250.

\bibitem{24} O.I. Reinov:
\textit{Some more remarks on Grothendieck-Lidski\v{\i}
trace formulas}, Journal of Prime
Research in Mathematics \textbf{8} (2012), 148-154.

\bibitem{25} O. Reinov, Q. Latif:
\textit{Grothendieck-Lidski\v{\i} theorem for subspaces of  $L_p$-spaces},
Math. Nachr. \textbf{286}, No. 2-3 (2013), 279--282.

 \bibitem{26} O. Reinov, Q. Latif:
 \textit{ Distribution of eigenvalues of nuclear operators and Grothendieck-Lidski\v{\i} type
      formulas}, Journal of Mathematical Sciences  \textbf{193}, No. 2 (2013), 312--329.

    \bibitem{27} O.~I. Reinov, \textit{On linear operators with $s$-nuclear adjoints, $0<s\le1,$},
  J. Math. Anal. Appl. \textbf{415} (2014) 816-824.

            \bibitem{28} O.~I. Reinov, Q. Latif:
\textit{Grothendieck-Lidski\v{\i} theorem for subspaces of quotients of  $L_p$-spaces},
 Banach Center Publications (to appear).

    \bibitem{29} R. Schatten:
 \textit{A theory of cross-spaces}, Princeton University Press (1950).

\bibitem{30} A. Szankowski: \textit{Subspaces without approximation property},
   Isr. J. Math. \textbf{30} (1978), 123-129

\bibitem{31} H. Weyl: \textit{Inequalities between the two kinds of eigenvalues of a linear transformation},
   Proc. Nat. Acad. Sci. U.S.A. \textbf{35} (1949), 408-411.

\bibitem{32} M.C. White:
\textit{Analytic multivalued functions and spectral trace},
Math. Ann. \textbf{304} (1996), 665-683.




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

 \end{thebibliography}


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






%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% INTROD
{\bf Introduction.}\,                                             %!!!   Begin text with this
    %%%
%In 1955 ...
In 1955 A. Grothendieck [7] has introduced the notion of the projective tensor product
of the locally convex vector spaces and developed the corresponding theory.
%which gave
It was a very deep generalization %of the theory
of Schatten--von Neumann  theory of $S_p$-spaces
and the corresponding theory of tensor products of Hilbert spaces [29].
One of the nice properties of Grothendieck's  projective tensor product of type $E\widehat\otimes F$
is that
its topological dual can be described as %is exactly
the space $B(E,F)$ of all continuous bilinear forms on $E\times F.$
In the particular case of the projective product $Y^*\widehat\otimes X$ of Banach spaces
$Y^*$ (dual to $Y)$ and $X,$ one can identify the Banach dual to $Y^*\widehat\otimes X$
with the space $L(X, Y^{**})$  of all linear continuous operators
from $X$ to $Y^{**}$  in a natural way (by using a linear
continuous functional "trace"; see below).
It turned out that the topological (locally convex) dual to the subspace $L(X,Y)$ of $L(X,Y^{**}),$
equipped with the topology of compact convergence $(L_c(X,Y)$ in notations of [7]),
can be identified with a quotient of $Y^*\widehat\otimes X$ ([7], Chap. I, Prop. 22).
As was shown by A. Grothendieck,
it follows from the last statement that
the injectivity of the canonical map
$Y^*\widehat\otimes X\to L(Y,X)$ is equivalent to the density of the set $X^*\otimes Y$
of all finite rank operators from $X$ to $Y$ in $L_c(X,Y).$
  If $X=Y,$ then the last is equivalent to the fact that the identity map is in the closure
  of the set of finite rank operators in the topology of compact convergence.
 This leads to
 the famous Grothendieck's definition of the notion of the
 $AP$ (approximation property) for a Banach space:                                 % $X:$
Following A. Grothendieck, we say that a Banach space $X$ has the $AP,$ if
for every compact subset $K$ of $X$ and for any $\varepsilon>0$ one can find a finite rank operator
$R$ in $X$ such that $\sup_{x\in K}\, ||Rx-x||\le \varepsilon.$
The property is so important that  we can find its applications in a great number  of papers
%in different directions of (e. g.)
devoted to the theory of operators in  Banach spaces.

Let us reformulate (following [7])     %!!!R
the definition of the $AP$ in terms of tensor products: $X$ has the $AP$ iff the natural
mapping $j:\, X^*\widehat\otimes X\to L(X,X)$ is one-to-one
(here by "natural map" we mean the unique extension of the
natural linear inclusion $X^*\otimes X\to L(X,X)$ from the normed space
$(X^*\otimes X, ||\cdot||_{\land})$\, (i.e. with the projective norm) to the completion
$X^*\widehat\otimes X).$
This fact  becomes evident %is immediate
if we note that the map $j$ is one-to-one iff the  closure
of $X^*\otimes X$ in $L_c(X,X)$ is the whole space $L(X,X)$
(or, what is the same, the identity map $\operatorname{id}_X$ is in this closure).
The image of the tensor product $X^*\widehat\otimes X$ in $L(X,X)$ is, by definition,
the space $N(X,X)$
of nuclear operators in $X$ (with the norm induced from $X^*\widehat\otimes X).$ Thus,
$X$ has the $AP$ iff (as we can write) $N(X,X)= X^*\widehat\otimes X.$

On the space $X^*\otimes X,$ the usual linear  functional "trace" is defined which is continuous
on the normed space $(X^*\otimes X, ||\cdot||_{\land}).$ After extension to the projective
tensor product $X^*\widehat\otimes X,$ this linear functional is still bounded and it can be seen that
a Banach space $Y$
has the $AP$ iff every tensor element $z\in Y^*\widehat\otimes Y$ which generates a 0-operator
in $L(Y,Y)$  has the property that $\operatorname{trace}\, U\circ z=0$ for every $U\in L(Y,Y^{**}).$
%is zero itself.

In Chapter II of [7],                            %!!!R
A. Grothendieck has generalized the notion  of nuclear operators and also considered
the more general tensor products: In the terminology of [7],          %!!!R
 an element of $X^*\widehat\otimes Y$ is said to be a "noyau de Fredholm de puissance p.\`eme sommable"
 $(p\in (0,1]),$ if it is of the form $\sum_{i} \lambda_i\, x'_i\otimes y_i,$ where $(\lambda_i)\in l_p$ and
 $(x_i)$ (resp. $y_i)$ is a bounded sequence in $X^*$ (resp. $Y).$
 We will use the notation $X^*\widehat\otimes_p Y$ for the corresponding tensor product
 (in [7], it is denoted by $X^* \overset{(p)}{\otimes} Y).$ This is a linear subspace of the projective     %!!!R
 tensor product $X^*\widehat\otimes Y,$ and with the natural metric (see [7],  Chapter II, \S1) it is             %!!!R
 a complete metric space.

If $z\in X^*\widehat\otimes_p Y,$ then the associated operator $\widetilde z$ from $X$ to $Y$ is called
(by A. Grothendieck) as "une application de puissance p.\`eme sommable".
The natural inclusion
$X^*\widehat\otimes_p Y \hookrightarrow X^*\widehat\otimes Y$ is one-to-one for any pair
of Banach spaces $X$ and $Y$ (a priori, it is not evident; see [7], Chapter II, \S1 for an explanation).    %!!!R

One of the interesting questions  considered in [7], is the connection between    %!!!R
the "order" of a tensor element $z\in X^*\widehat\otimes_p Y$ and the "order"
of the sequence of all eigenvalues
of the corresponding operator $\widetilde z$ (evidently, $\widetilde z$ is compact).
Among the results in this direction, let us mention only the following facts (which were obtained
in [7], Chapter II, \S1, Section 4):                    %!!!R
Let $X$ and $Y$ be Banach spaces and $0<p\le 2/3.$
Then the canonical map $j_p:\, X^*\widehat\otimes_p Y\to L(X,Y)$ is injective.
Moreover, for every $z\in X^*\widehat\otimes_p X$ the sequence $(z_i)$ of all eigenvalues
of the associated operator (counted according to their
multiplicities) %$\widetilde z$
is absolutely summable and
$\operatorname{trace}\, z= \sum_i z_i$ (the assertions are true for any locally convex vector space $X).$

Remembering one of the equivalent definitions of the $AP,$
we see that it was, in fact, shown by A. Grothendieck that every Banach space has some
approximation properties "of type $p$" for all $p\in (0, 2/3].$
We can use, e.g., a notation "$AP_p$". Thus,  A. Grothendieck  considered
the notion of "$p$-approximation property" already in 1955, though implicitly.

Let us mention that A. Grothendieck  (applying deep results of complex analysis and
H. Weyl's [31]                          %!!!R ?? Weil??
theorem on the Schatten -- von Neumann classes $S_p$ of compact operators in Hilbert spaces)    %!!! ??
has proved firstly the "eigenvalue theorem" for the case where $0<p\le 2/3$ and then, as a consequence,
obtained the injectivity of the above maps $j_p$ (surely, the main case is $p=2/3).$
In the paper [20] of the author, the reader can find a more simple proof of these
theorems, where it was shown firstly that the map $j_{2/3}$ is one-to-one and then the
eigenvalue result was obtained (by applying the Lidski\v{\i} theorem for the trace-class                    %!!! ??
operators in Hilbert spaces [12]).

%%    History    !!! !!!

The question about the injectivity of the maps $j_p$ for $p\in(2/3,1)$ was not considered
by A. Grothendieck  in [7] explicitly. He posed the corresponding question only
for the case $p=1.$ This famous approximation problem was solved in negative in 1972
by Per Enflo [6] (for the further information see [3], [14], [16], [30] .          %!!! Ref !!??

 It seems that the notion of  the approximation property "of type $p$" (for $0<p<1)$ was
 (explicitly)  considered firstly by the author in the paper [21], where                               %!!!Ref i nizhe 2
 it appeared as the "approximation property of order $p$". In [21], we used
 the tensor product definition (i. e., the injectivity of the map $j_p).$
 Some simple facts and different (counter)examples were presented in [21].
  Instead of the term "une application de puissance p.\`eme sommable", we used there
  the name "a $p$-nuclear operator". Later $p$-nuclear operators (for $p\in(0,1)$)
  were studied, e.g.,
  in [22], [23], [27], [8].                                               %!!!Refs !??
We refer the reader to these papers for the further information.

%%
Our aim in these notes is to discuss several old and new definitions of different approximation properties
and to formulate (and, partially, to present the proofs of) some results in this direction.
Also, we give applications (in particular, to eigenvalues problems).

                             %%%%%%%%%%%%%%%%%%%%%%      Opisanie 1-...
             \medskip

             %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   Predvorilka
{\bf Preliminaries.}\,
All the spaces under considerations $(X,Y,\dots)$
are Banach, all linear mappings (operators) are continuous; as usual, $X^*, X^{**}, \dots$
are Banach duals (to $X$), and $x', x'', \dots$ (or $y', \dots)$ are the functionals on $X, X^*,\dots$
(or on $Y,\dots).$
If $x\in X, x'\in X^*$ then $\langle x,x'\rangle=\langle x',x\rangle=x'(x).$
$L(X,Y)$ stands for the Banach space of all linear bounded operators from $X$ to $Y.$
Every Banach space is considered as a Banach subspace of its second dual. If needed,
by $\pi_Y$ we denote the natural isometric injection of $Y$ into $Y^{**}.$
   % ---

We consider the algebraic tensor product $X^*\otimes Y$ as the linear space of all continuous
finite rank operators from $X$ to $Y.$ The projective tensor product $X^*\widehat\otimes Y$
of the spaces $X^*$ and $Y$ is the completion of $X^*\otimes Y$ with respect to the norm
$||z||_{\land}:= \inf \{\sum |\lambda_k|\},$ where the infimum is taken over all finite
representations of $z\in X^*\otimes Y$ in the form $z= \sum \lambda_k\, x'_k\otimes y_k$
with $||x'_k||=||y_k||=1.$ Every element $z\in X^*\widehat\otimes Y$ admits a representation
$z=\sum_{k=1}^\infty \lambda_k x'_k\otimes y_k$ such that $\sum |\lambda_k|<\infty$ and
$||x'_k||=||y_k||=1.$ If $X=Y,$ then the functional "trace" on the tensor product
$X^*\widehat\otimes X$ is well defined by the formula
$\operatorname{trace}\, z:= \sum \lambda_k\, \langle x'_k, y_k\rangle.$ The Banach dual
to $X^*\widehat\otimes Y$ can be identify with the space $L(Y, X^{**})$ with duality
given by "trace":  for $z\in X^*\widehat\otimes Y$ and $U\in L(Y, X^{**})$ we put
$\langle U, z\rangle := \operatorname{trace}\, U\circ z=\sum \lambda_k\, \langle x_k, Uy_k\rangle.$

 We use standard notations for the classical Banach spaces such as $L_p(\mu),$ $C(K),$ $l_p,$ $c_0$ etc.
   For the theory of (sequence) Lorentz spaces, we refer to [1], [16], [17, Section 2.1]; see also [8, Section 5].     %!!!Ref
%% ---------------------------------
%We will apply mainly the results that can be found, e.g.,  in [1], [6], [8] and [9].
For the definitions of the notions of type and cotype, see any of these references:
[4], [16],    %!!!Ref
[18], [19]
(Rademacher type p = Gauss type p and Rademacher cotype q = Gauss cotype q;
so, we can apply results from G. Pisier's lecture [19], assuming that we are working with
Rademacher notions).

Let us collect some facts we need. Recall that a subspace $E$ of a Banach space $X$
is $b$-complemented $(b>0)$ in $X,$ if there exists a linear continuous projection $P$
from $X$ onto $E$ such that $||P||\le b.$ As usual, if $p\in [1,\infty],$ then $p'$ is
the conjugate exponent: $1/p+1/p'=1.$


%\begin{proposition}{\it
Let $X$ be a Banach space and $1<p\le2,$ $2\le q<\infty.$
$1)$  If $X$ is of type $p$ (cotype $p)$ then every subspace is
of type $p$ (cotype $p);$
{\rm 2)  [4, Proposition 11.11]} If $X$ is of type $p$ then any quotient of $X$      %!!!Ref Diest Jarh.
is of type $p;$
{\rm3)  [4, Proposition 11.10]} If $X$ is of type $p$ then $X^*$ is of cotype $p';$
$4)$ If $X^*$ is of type $p$ then $X$ is of cotype $p';$
$5)$  If $X$ is of type $p$ then any subspace of any quotient (and any quotient of
any subspace) of $X$ is of type $p;$
{\rm 6)  [4, Corollary 11.9]} A Banach space has the same type or cotype as its bidual;
{\rm 7)  [4, Corollary 11.7]} Each $L_r$-space $(1\le r<\infty)$ has type $\min \{r,2\}$
and cotype $\max \{r, 2\};$
{\rm8)  [19, see Theorem 4.1 and its Corollaries]} If $X$ is of type $p$ and of cotype $q$
then there is a constant $D_{p,q}>0$ such that
every finite dimensional subspace $E$ of $X$ is
$D_{p.q}\, ( {\operatorname{dim} } E)^{1/p-1/q}$-complemented in $X.$
 %}
%\end{proposition}


Recall also the well known general fact (due to M. J. Kadec and M. G. Snobar [9]; see also         %!!!Ref i nizhe
[16, 28.2.6. Lemma]): in any Banach space every $n$-dimensional
subspace is $n^{1/2}$-complemented.

%%   Our main refer....               %��� � ������ �����


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   BEGIM.TEX
%\vskip 0.3cm

Our main reference is [16]. All the notions, notations and facts,              %!!!Ref i 4-ja nizhe
given here without any explanation,
%we use without any reference,
can be found in [1, 5, 8, 14, 16, 17, 18].


             %%%%%%%%%%%%%%%%%%%%%%%%%  Content
              \medskip
{\bf Contents.}\,
In Section I we reformulate the definition of Grothendieck approximation property
in terms of 0-sequences that leads us to the consideration of  more general
approximation properties $\widetilde{AP_s}$ for $s\in(0,1].$ We define them in terms of
approximation of the identity maps in Banach spaces by finite rank operators on $l_p$-sequences
$(p$ depends on $s).$ It seems that such properties, for the first time, were considered
by the author in [22] (cf. Lemma 2.1 there). The simplest examples                 %!!!Ref
of Banach spaces possessing the properties of such a kind are subspaces of quotients
of $L_p$-spaces.

In Section II we reformulate the $\widetilde{AP_s}$ in terms of tensor products showing that that
$\widetilde{AP_s}=AP_s,$ where $AP_s$ is the approximation property of order $s$ introduced
in the author paper [21] and investigated also in [2], [23], [27].             %!!!Ref

In Section III we introduce and investigate new notions of the approximation properties
$AP_{t;p,r}$ and $AP_{(r,w)}$ defined by some "Lorentz tensor products" (tensor products
generated by Lorentz sequence spaces). These notions are new and considered here for the
first time. In particular, we obtain a characterization of the $AP_{(r,1)}$ in terms of
approximation of the identity maps by finite rank operators on some (Lorentz)
0-sequences (Theorem 3.3).

In Section IV, with the help of a theorem of M.C. White [32], we prove Theorem 4.1        %!!!Ref i nizhe 1-2 strok
which gives us sufficient conditions for the famous Grothendieck-Lidski{\v{\i}}
trace formula to be valid for certain quasi-Banach operator ideals.
We present an application of the previous results about approximation properties
to some eigenvalue problems. Theorem 4.1 is applied then to the case of $(2/3,1)$-nuclear
operators (related to the Lorentz space $l_{2/3\,1}).$      %!!!?? related?

In Section V the results of Sections I about subspaces of quotients of $L_p$-spaces together with
White's theorem are applied for proving some more theorems concerning the distribution
of eigenvalues of the nuclear operators.
We give new relatively simple proofs of some recent results from the papers [25] and [28].    %!!!Ref   2nd - to appear!!!

In Section VI we prove two statements about the approximation properties considered
in Sections I--V. For instance, it is well known that if $X^*$ has the $AP$ of Grothendieck,
then $X$ has the $AP$ too. We show that the same is true for all the natural approximation properties
considered here, such that $AP_s, AP_{(r,w)}$ etc.

In Section VII we introduce, following [24] and [26], two more notions of the         %!!!Ref i nizhe v abzatce
approximation properties by using the spaces of so-called $(r,p)$-nuclear operators
(a partial case of a class of $(s,q,t)$-nuclear operators from [16, 18.1]). Two theorems
about eigenvalues of the $(r,p)$-nuclear operators are proved. In these theorems, trace formulas
of Grothendieck-Lidski{\v{\i}}type are established for the cases where $1/r-1/p=1/2.$
The first one was proved in [24] and [26] with the help of Fredholm Theory; the second theorem
(Theorem 7.3) was obtained before by the same authors, again by using Fredholm Theory
(but its proof was unpublished). Here, the different (more simple) method is used.
Firstly, we show that every Banach space has the corresponding approximation properties
$AP_{[r,p]}$ and $AP^{[r,p]}.$ After this, we obtain the eigenvalue results by using  simple
 arguments.

Finally, in Section VIII, examples are given (they are taken from the paper [26]).       %!!!Ref
The examples give a possibility to conclude that all the positive results of
Sections I--VII concerning the approximation properties and  trace formulas are sharp.

                %%%%%%%%%%%%%%%%%%%%%%      sections 1-...




%%%%%%%%%%%%%%%%%%%     I-...-VIII         %777 - begin to change refrnc No's
\medskip

{\bf I.}\
It is well known that every compact subset of a Banach space is contained in the closed               % vstavil i dal ssylki 11 ������� 2014 �. 22:48:39
convex hull of a sequence converging to 0 (see, e.g., [7],  p. 112 in Ch.I, Lemme 12, or
[14], p. 30, Proposition 1.e.2). Therefore,
the Grothendieck approximation property for a Banach space $X$ can be defined as follows:
$X$ has the $AP$ iff for every sequence $(x_n)_{n=1}^\infty\subset X$ tending to zero, for any $\varepsilon>0$
there exists a finite rank (continuous) operator $R$ in $X$ such that  for each $n\in\Bbb N$
one has $||Rx_n-x_n||\le\varepsilon.$  Consider a natural question: for which sequences $(x_n)\in c_0(X),$
under some additional assumptions, the identity map $\operatorname{id}_X$  can be approximated
by finite rank operators, as above, and which of those conditions are sharp (or, if one wishes, optimal)?

One of the simplest fact (we think, known for more than 30 years) is that

$(*)$\, if $(x_n)\in l_2(X),$ \,  $X$ is any, then the answer is positive.

Here is a reason of this: Assuming $||x_n||\searrow 0,$ take any $N\in \Bbb N$ and consider
the linear span $E_N:=\operatorname{span}[x_n]_1^N$ as a subspace of $X.$ Define, fixing an $\varepsilon>0,$
a finite rank operator $R$ to be a projection from $X$ onto $E_N$ whose norm $\le \sqrt N.$

Now if $N$ is such that, for every $n\ge N,$ we have $||x_n||\le \frac {\varepsilon}{\sqrt N+1},$ then
$$
  ||Rx_n-x_n||=0 \ \text{  if  } \ n\le N,
$$
and
$$
 ||Rx_n-x_n||\le (||R||+1)\, ||x_n|| \le \varepsilon   \ \text{  if  } \ n\ge N.
$$

Of course, instead of $(*)$ we can consider  the statement

$(**)$\, if $(x_n)\in l_{2,\infty}^0(X) $ \,  (Lorentz space with "o" small ---             %!!! popravil 11 ������� 2014 �. 22:41:11
$l^{min}_{(2, \infty)}(X)$ in notations of [16, 13.9.3 Remark]), $X$ is any, then
the answer is affirmative.

 The idea of the above proof is very simple and can be applied in some more general situations.
 For instance, every subspace of finite dimension $n$ of an $L_p$-space is
 $n^{|1/2-1/p|}$-complemented in that $L_p$-space. So,
  if $p\in [1,\infty],$  $\alpha=|\frac{1}{2}-\frac1p|$ and $X$ is a subspace of an $L_p$-space, then

 $(***)$\,  for every sequence $(x_n)\in l_{q,\infty}^0(X),$ where $1/q=\alpha,$
 the answer is affirmative.
 \smallskip

 {\it Remark \rm1.1}:\,
   About sharpness: it will be discussed below.
   \smallskip

   {\it Remark \rm1.2}:\,
   The statement $(***)$ has, as a matter of fact, the following  quantitative aspect:
   Given $\alpha\in [0,1/2]$ and a Banach space $X$ with the property that every
   finite dimensional subspace $F$ of $X$ is contained in a finite dimensional subspace $E\subset X,$
   which  in turn  is   $C\, (\operatorname{dim} F)^\alpha$-complemented in $X,$
   we have

   $(***)'$\,  for every sequence $(x_n)\in l_{q,\infty}^0(X),$ where $1/q=\alpha,$
   for any $\varepsilon>0$ there is a finite rank operator $R$ in $X$ so that
   $\sup_n ||Rx_n-x_n||\le\varepsilon.$

   Particular cases:

   \noindent
(i)\,
$q=2$ and $\alpha=1/2$ or $q=\infty$ and $\alpha=0;$
%(i. e.,\, "$X$ is any Banach space" or "$X$ is isomorphic to a Hilbert space");

   \noindent
(ii)\,
$(x_n)\in l_q(X), q\in [2,\infty). $ %, or $(x_n)\in c_0(X), q=\infty$ [Hilbert case].
 \smallskip

 For a while let us introduce the notions of the corresponding approximation properties
 for a Banach space $X$ (taking into account that the possibility of approximations on $c_0$-sequences
 by finite rank operators gives us the
 Grothendieck's approximation property $AP):$
 Let $0<q\le \infty$ and $1/s=1/q+1.$
 We say that $X$ has the $\widetilde {AP}_s$  [resp., the $\widetilde{AP}_{s,\infty}]$
 if for every $(x_n)\in l_q(X)$ [resp., $l_{q,\infty}^0(X)]$
 (where $l_q(X)$ means $c_0(X)$ for $q=\infty)$
 and for every $\varepsilon>0$ there exists a finite rank operator $R\in X^*\otimes X$ such that
 $\sup_n ||Rx_n-x_n||\le \varepsilon.$
 Trivially, e.g., $\widetilde{AP}_{s_2} \implies \widetilde{AP}_{s_1}$ if $s_1\le s_2.$
 Thus, $\widetilde{AP}_1 (= AP)$ implies any $\widetilde{AP}_s.$

 The statement $(*)$ (and $(**)$) says that every Banach space has the above property
 $\widetilde{AP}_{2/3}$ (and even the $\widetilde{AP}_{2/3,\infty}).$ The statement $(***)$
 gives the corresponding result for $L_p$-subspaces. Moreover, the assertion mentioned
 in Remark 1.2, shows that, for instance, any subspace of any quotient (= any quotient of any
 subspace) of a Banach space of type 2 (resp., of cotype 2) and of cotype $p,$ \, $p\in[2,\infty)$
 (resp., of type $p'),$ possesses the $\widetilde{AP}_s$
 (even the $\widetilde{AP}_{s,\infty})$  with $1/s=1+|1/2-1/p|.$




 \vskip 0.23cm

{\bf  II.}\
Let us recall that the notion of the $AP$ of Grothendieck can be reformulated in terms
of the projective tensor products "$\widehat\otimes$". Namely, a Banach space $X$ has the $AP$
iff for every Banach space $Y$ the canonical (natural) mapping $Y^*\widehat\otimes X \to L(Y,X)$
is one-to-one (or, what is the same, the natural mapping $X^*\widehat\otimes X\to L(X):=L(X,X)$
is injective).
In [7], A. Grothendieck has considered also some other tensor products (linear subspaces   %!!!R
of  "$\widehat\otimes$"'s), which we will denote
by "$\widehat\otimes_s$" for $0<s\le1$ (so that $\widehat\otimes =\widehat\otimes_1):$
For Banach spaces $X$ and $Y,$ let $Y^*\widehat\otimes_s X$ be a subspace of the projective tensor product
$Y^*\widehat\otimes X$ consisting of the tensors
 $z\in Y^*\widehat\otimes X,$ which admit representations of the form
$$
% \begin{equation}
z= \sum_{n=1}^\infty \lambda_n y'_n\otimes x_n,     %\label {(1)}           % (2.1) change!!! and below      %!!! ubral \label
%\end{equation}
$$
where $(\lambda_n)\in l_s,$\, $(y'_n)$ and $(x_n)$ are bounded sequences from $Y^*$ and $X$
respectively. With a natural "quasi-norm" (see [16])         %!!!R
the linear subspace $Y^*\widehat\otimes_s X$ of the space $Y^*\widehat\otimes X$ can be considered as
a "quasi-normed tensor product" (it is then a complete metric space [7]).

One of the nice (with a non trivial proof in [7]) theorem of Grothendieck is the fact
that the natural map from $Y^*\widehat\otimes_{2/3} X$ into $L(Y,X)$ is injective for any Banach spaces $X,Y.$
Let us compare this Grothendieck's
result with a simple assumption in Section I, where "$s=2/3$"  appeared. Clearly, it is
not a chance coincidence, and  we really have
\smallskip

{\bf Theorem 2.1.}\,
For $s\in (0,1]$ and for a Banach space $X,$ the following statements are equivalent:

$1)$\,
$X$ has the $\widetilde{AP}_{s}$ in the sense of the definition in Section I;

$2)$\,
$X$ has the $AP_s$ in the sense of the definition in [23], i.e.           %!!!R
for every Banach space $Y$ the natural mapping $Y^*\widehat\otimes_s X\to L(Y,X)$ is one-to-one.
\smallskip

Moreover, the following statement $(AP_s)$ takes place:
\smallskip

$(AP_s)$\,
A Banach space $X$ has the $AP_s,$ \, $0<s\le1,$ iff the canonical map
$X^*\widehat\otimes_s X \to L(X)$ is one-to-one (or, what is the same, there exists no tensor element
$z\in X^*\widehat\otimes_s X$ with
$\operatorname{trace}\, z=1$ and $\widetilde z=0,$ where $\widetilde z$ is the associated (with $z)$
operator from $X$ to $X$).                                 %!!! ref or maybe to prove this!!!R
\smallskip

Maybe analogous theorems and facts are  valid for the $\widetilde{AP}_{s,\infty}$ and the $AP_{s,\infty}$
from  [23] (see Section III for a discussion).                  %!!!R
\smallskip

{\it Proof}\ of  the assertion $(AP_s).$\,
Suppose that the $AP_s$-condition holds for $X,$ but there exists a Banach space $Y$ such that the natural map
$Y^*\widehat\otimes_s X\to L(Y,X)$ is not one-to-one. Take an element $z\in Y^*\widehat\otimes_s X$    which is
not zero, but generates a zero operator $\widetilde z: Y\to X.$
Then we can find an operator $U\in L(X,Y^{**})$ so that $\operatorname{trace}\, U\circ z=1.$
If $z=\sum_{k=1}^\infty \lambda_k\, y'_k\otimes x_k$ is a representation of $z$ in $Y^*\widehat\otimes_s X$\,
($(\lambda_k)\in l_s, (x_k)$ and $(y'_k)$ are bounded), then
$$
 1=\operatorname{trace}\, U\circ z= \sum_{k=1}^\infty \lambda_k \langle Ux_k, y'_k\rangle=
 \sum_{k=1}^\infty \lambda_k\, \langle x_k, U^*y'_k\rangle
$$
and $\sum_{k=1}^\infty \lambda_k\, U^*y'_k(x) x_k=0$ for every $x\in X.$
Put $x'_k:= \lambda_k U^*y'_k,$ $z_0:= \sum_{k=1}^\infty x'_k\otimes x_k\in X^*\widehat\otimes_s X.$ We have
$$
\operatorname{trace}\, z_0=1,\ \widetilde z_0\neq 0
$$
(by the assumption about  $X).$ Consider a  1-dimensional operator $R= x'\otimes x$ in $X$
with the property that $\operatorname{trace}\, R\circ z_0>0.$ Then
$$
 0<\operatorname{trace}\, R\circ z_0
 = \sum_{k=1}^\infty  \langle x'_k, x\rangle  \langle x', x_k\rangle
 = \sum_{k=1}^\infty \lambda_k\,\langle U^*y'_k, x\rangle  \langle x', x_k\rangle
 $$
 $$= \langle\sum_{k=1}^\infty \lambda_k\, \langle Ux, y'_k\rangle  x_k, x'\rangle
 = \langle x', \sum_{k=1}^\infty \lambda_k\, U^*y'_k(x) x_k\rangle =0.
$$
\smallskip

{\it Proof}\ of Theorem 2.1.
We will use the assertion $(AP_s).$

$1) \implies 2).$\,
Let $z\in X^*\widehat\otimes_s X$ and $\operatorname{trace}\, z=1.$
Write $z= \sum \lambda_k\, x'_k\otimes x_k,$ where the sequences $(x'_k)$ and
$(x_k)$ are bounded and $(\lambda_k)\in l_s,$
$\lambda_k\ge0,$ $(\lambda_k)$ is non-increasing.
Then
$$
 z=\sum_{k=1}^\infty (\lambda_k^s\, x'_k)\otimes (\lambda_k^{1-s}\, x_k)
$$
(recall that $1/s=1+1/q;$ so $1-s=s/q).$
The sequence $(\lambda_k^{1-s}x_k)$ is in $l_q(X).$ By 1), for every $\varepsilon>0$ there exists
a finite rank operator $R\in X^*\otimes X$ such that
$||R(\lambda_k^{1-s}x_k)-\lambda_k^{1-s}x_k||\le\varepsilon$
for each $k\in \Bbb N.$ It follows that, for this operator $R,$
$$
  |\operatorname{trace}\, (z- R\circ z)| =
  |\sum_{k=1}^\infty \langle\lambda_k^s x'_k, \lambda_k^{1-s} x_k- R(\lambda_k^{1-s} x_k)\rangle |
  \le \sum_{k=1}^\infty \lambda_k^s ||x'_k||\cdot \varepsilon\le const\cdot \varepsilon.
$$
Hence, for small $\varepsilon>0$ we have that, for the operator $R\in X^*\otimes X,$
$$
  |\operatorname{trace}\, R\circ z| \ge 1/2
$$
and therefore $z$ generates a non-zero operator $\widetilde z.$
\smallskip

Before consider a proof of the implication $2)\implies 1)$ we will make some additional remarks.
We collect the remarks in
\smallskip

{\bf Lemma 2.1.}\
Let $s\in (0,1],$ $q\in (0,\infty],$ $1/s=1+1/q.$
For $a:=(a_k)\in l_1$ and $b:=(b_k)\in l_q$ we have
 \begin{equation}
(\sum_{k=1}^\infty |a_k b_k|^s)^{1/s}
   \le \sum_{k=1}^\infty |a_k| \cdot   (\sum_{k=1}^\infty |b_k|^q)^{1/q}.  \label {(1)}              %!!! zamenil (2) na (2.1) i nizhe raz
\end{equation}
Moreover,
$$
  ||a||_{l_1} = \sup_{||b||_{l_q}=1} (\sum_{k=1}^\infty |a_k b_k|^s)^{1/s}
$$
(if $q=\infty$,  evident changes must be made in $(1)$).
\smallskip

{\it Proof}\ of Lemma 2.1.
We may consider the case where $q\in (0,\infty).$ Putting $p:= 1/s$
(then $1/p'=1-s=s/q$ and $sp'=q)$, we obtain
$$
    \sum_{k=1}^\infty |a_k b_k|^s \le
    (\sum_{k=1}^\infty |a_k|^{sp})^{1/p} \cdot   (\sum_{k=1}^\infty |b_k|^{sp'})^{1/p'}=
    (\sum_{k=1}^\infty |a_k|)^s \cdot   (\sum_{k=1}^\infty |b_k|^q)^{s/q}.
$$

For the second part:
Let $a=(a_k)\in l_1.$ Take $b_k:= \frac{|a_k|^{1/q}}{||a||_{l_1}^{1/q}}.$ Then
$\sum_{k=1}^\infty |b_k|^{q}= \sum_{k=1}^\infty \frac{|a_k|}{||a||_{l_1}}=1$ and
$$
  (\sum_{k=1}^\infty |a_k b_k|^s)^{1/s}
  = (\sum_{k=1}^\infty \frac{|a_k|^{s/q}}{||a||_{l_1}^{s/q}}\, |a_k|^s)^{1/s}=
   (\sum_{k=1}^\infty \frac{|a_k|^{s/q+s}}{||a||_{l_1}^{s/q}})^{1/s}
$$
$$
=  (\sum_{k=1}^\infty \frac{|a_k|^{s(1+1/q)}}{||a||_{l_1}^{s/q}})^{1/s}=
 (\sum_{k=1}^\infty \frac{|a_k|}{||a||_{l_1}^{s/q}})^{1/s}=
 \frac {(\sum_{k=1}^\infty |a_k|)^{1/s}}{||a||_{l_1}^{1/q}}=
 (\sum_{k=1}^\infty |a_k|)^{1/s-1/q}= ||a||_{l_1}.
$$
\smallskip

{\it Proof}\ of Theorem 2.1 (continuation).

$2) \implies 1).$
Suppose that $X$ does not have the $\widetilde{AP_s},$ $1/s=1+1/q.$
Then there is a sequence $(x_n)\in l_q(X)$ (if $q=\infty,$ we consider a sequence from $c_0(X)=l_\infty^0(X)$)
such that there exists an $\varepsilon>0$ with the property that for any finite rank operator
$R\in X^*\otimes X$ the inequality $\sup_n ||Rx_n-x_n||>\varepsilon$ is valid.
Consider the space $C_0(K; X)$ for $K:= \{x_n\}_{n=1}^\infty \cup \{0\}.$
Every operator $U$ in $X$ can be considered as a continuous function on $K$
with values in $X$ by setting $f_U(k):= U(k)$ for $k\in K.$
In particular, for the identity map $\operatorname{id}$ in $X$ and for any $R\in X^*\otimes X$ we have
$$
  ||f_{\operatorname{id}}-f_R||_{C_0(K;X)}\ge\varepsilon.
$$
The subset $\mathcal R:=\overline{\{f_R:\ R\in X^*\otimes X\}}^{C_0(K;X)}$ of $C_0(K; X)$
is a closed linear subspace in $C_0(K; X).$ So, there exists an $X^*$-valued measure
 $\mu=(x'_k)_{k=1}^\infty\in C_0^*(K; X)= l_1(\{x_n\}_{n=1}^\infty)\cup \{0\}; X)$ such that
 $\mu|_{\mathcal R}=0$ and $\mu(f_{id})=1.$
 In other words, we can find a sequence $(x'_k)$ with $\sum_{k=1}^\infty ||x'_k||<\infty$
such that $\sum_{k=1}^\infty \langle x'_k, x_k\rangle =1$ and $\sum_{k=1}^\infty \langle x'_k, Rx_k\rangle =0$
for any $R\in X^*\otimes X.$

Define a tensor element $z\in X^*\widehat\otimes X$ by $z:= \sum_{k=1}^\infty x'_k\otimes x_k.$
Since $(x_k)\in l_q(X)$ and $(x'_k)\in l_1(X^*),$ we get from Lemma 2.1 that
$$
  (\sum_{k=1}^\infty ||x'_k||^s\, ||x_k||^s)^{1/s} \le
    \sum_{k=1}^\infty ||x'_k||\cdot (\sum_{k=1}^\infty ||x_k||^q)^{1/q}.
$$
Therefore, $z\in X^*\widehat\otimes_s X,$ $\operatorname{trace}\, z=\sum_{k=1}^\infty \langle x'_k, x_k\rangle =1$ and
$\operatorname{trace}\, R\circ z=0$ for every $R\in X^*\otimes X.$ This means that $X$ does not have the $AP_s.$
\smallskip

After Theorem 2.1 has been  proved, we can make a conclusion:
$AP_s= \widetilde{AP}_s$ for any $s\in (0,1].$ Let us mention that this equality appeared firstly
(without proofs) in [22, Lemma 2.1].



 \vskip 0.23cm

{\bf  III.}\
Now we are going to discuss some questions around the properties $\widetilde{AP}_{s,\infty}$
and $AP_{s,\infty}.$ The $\widetilde{AP}_{s,\infty}$ was defined above. Recall the definition
of  the $AP_{s,\infty}$ from, e.g., [23]:
We say that a Banach space $X$ has the $AP_{s,\infty},$ $0<s<1,$ if for every Banach space $Y$
the natural mapping $Y^*\widehat\otimes_{s\infty} X\to L(Y,X)$ is one-to-one, where
$$
  Y^*\widehat\otimes_{s\infty} X= \{z\in Y^*\widehat\otimes X\!:\  z
  =\sum_{k=1}^\infty \lambda_k y'_k\otimes x_k,\ (x_k)\,
  \text{ and}\, (y'_k) \text{ are  bounded, }  (\lambda_k)\!\in l_{s\infty}^0\}\!.
$$
Let us consider the connections between the $AP_{s,\infty}$ and the $\widetilde{AP}_{s,\infty}.$
For a partial discussion of this we need a lemma, which follows from Lemma 2.1
by interpolation in Lorentz spaces.
\smallskip

{\bf Lemma 3.1.}\
Let $s\in (0,1), q\in (0,\infty), 1/s=1+1/q, r\in (0,\infty].$
If $a=(a_k)\in l_1,$ $b=(b_k)\in l_{qr},$ then $ab:=(a_kb_k)_{k=1}^\infty \in l_{sr}.$
In particular, for $a\in l_1$ and $b\in l_{q\infty}$ the sequence $ab$ is
in $l_{s\infty}$ (thus, evidently, in $l_{s\infty}^0).$
\smallskip

{\it Proof}\ of Lemma 3.1 consists of  the application of Lemma 2.1 and the general
interpolation theorem for the multiplication operator $\widetilde a,$ induced by a fixed
sequence $a=(a_k)\in l_1:$
$\widetilde a$ maps $(b_k)$ to $(a_kb_k).$

Namely, fix $s\in (0,1), q\in (0,\infty)$ with $1/s=1+1/q.$
Take $s_1, s_2\in (0,1)$ and $q_1, q_2\in (0,\infty)$ so that for some $\theta\in (0,1)$
we have
$$
  \frac1q = (1-\theta)\frac1{q_1} + \frac1{q_2}, \ 0<\frac1{s_2}<\frac1s < \frac1{s_1}<\infty, \
    0<\frac1{q_2}<\frac1q < \frac1{q_1}<\infty, $$
    and
  $$  \frac1{s_1}= 1+ \frac1{q_1}, \  \frac1{s_2}= 1+ \frac1{q_2}.
$$
By Lemma 2.1, $\widetilde a$ maps $l_{q_1q_1}$ into $l_{s_1s_1}$ and
 $\widetilde a$ maps $l_{q_2q_2}$ into $l_{s_2s_2}.$
 Applying, e.g., Theorem 5.3.1 from [1]          %!!!R
 or other results from the pages 113-114 in [1],       %!!!R
we get that $\widetilde a$ maps $l_{qr}$ into $l_{sr},$ $0<r\le \infty$
(note that $1/s=1+1/q= 1+(1-\theta)/q_1+\theta/q_2 = (1-\theta) +\theta +(1-\theta)/q_1 +\theta/q_2=
(1-\theta)(1+1/q_1)+\theta (1+1/q_2)=(1-\theta)/s_1+\theta/s_2$).
\smallskip

{\it Remark} 3.1:\
As a matter of fact, $l_1\cdot l_{q\infty}=l_{s1}$ in Lemma 3.1. We need now only
the above inclusion.
\smallskip

Now let $t\in (0,1],$ $p\in (0,\infty],$ $r\in (0,\infty]$ and
consider a tensor product $\widehat\otimes_{t; p,r},$ defined in the following way:
For a couple of Banach spaces $X, Y$ the tensor product
$Y^* \widehat\otimes_{t; p,r} X$ consists of those elements $z$ of the projective tensor product
$Y^*\widehat\otimes X$ which admit representations of the type
$$
  z=\sum_{k=1}^\infty a_kb_k\, y'_k\otimes x_k; \ (y'_k) \text{ and } (x_k) \text{ are bounded, }
  (a_k)\in l_t,\, (b_k)\in l_{pr}
$$
(recall that everywhere here we consider $l_{p\infty}^0$ in the case $r=\infty).$
\smallskip

{\it Remark} 3.2:\
As was noted in Remark 3.1, $l_1\cdot l_{q\infty}=l_{s1} (\subset l_{s\infty}^0\subset l_{s\infty}),$
where $0<s<1, 1/s=1+1/q.$ We have also
$$
 l_{s1}=l_1\cdot l_{q\infty}^0\ \text{and}\ l_1\cdot l_{q\infty}=l_1\cdot l_{q\infty}^0
$$
(so, for example, in the definition of $\widehat\otimes_{1; q, \infty}$
one can assume that $(a_k)\in l_1$ and $(b_k)\in l_{q \infty}^0).$
Indeed, if we use the equality $l_1\cdot l_{q \infty}=l_{s1},$ take
$d\in l_{s1}$ (assuming $d=d^*=(d^*_k)$).            %!!! nonincr... - to say?
Then $\sum_{k=1}^\infty k^{1/s}\, d^*_k/k <\infty,$ i.e. $\sum_{k=1}^\infty k^{1/q}\, d^*_k <\infty.$
Let $\varepsilon=(\varepsilon_k)$ be  a scalar sequence such that $\varepsilon_k\searrow 0$ and
$\sum_{k=1}^\infty \varepsilon_k^{-1} d^*_k k^{-1/q}<\infty.$ Put
$$
 \alpha_k:= \frac{d^*_k k^{1/q}}{\varepsilon_k}, \,  \ \beta_k:= \frac{\varepsilon_k}{k^{1/q}}.
$$
Then $\alpha:= (\alpha_k)\in l_1$ and $\beta:= (\beta_k)\in l^0_{q \infty}.$
So, $d=\alpha \beta\in l_1\cdot l_{q \infty}^0.$
Another way (not using  "$l_{s 1}$"):
Let $0<q<\infty,$ $\alpha\in l_1,$ $\beta\in l_{q \infty}$ (assuming, without loss of generality,
that $\beta=\beta^*).$
Consider a sequence $\varepsilon:= (\varepsilon_k)$ such that $\varepsilon_k\searrow 0$ and
$\frac{}{}(\alpha_k/\varepsilon_k)\in l_1.$
Put $\widetilde \alpha:= \alpha/\varepsilon=(\alpha_k/\varepsilon_k)$ and
$\widetilde \beta:=\varepsilon\beta=(\varepsilon_k\beta_k).$
Then $\widetilde \alpha\in l_1,$ $\widetilde \beta\in l^0_{q \infty}$ and
$\alpha\beta= \widetilde\alpha \widetilde\beta\in l_1\cdot l^0_{q \infty}.$
\smallskip

Let us say that $X$ has the $AP_{t; p,r},$ if  for every Banach space $Y$ and for $t, p, r$ as above
the canonical mapping $Y^*\widehat\otimes_{t; p, r} X\to L(Y,X)$ is one-to-one.

By Lemma 3.1, if $s\in (0,1)$ and $1/s=1+1/q,$ then
$\widehat\otimes_{1; q, \infty}\subset \widehat\otimes_{s, \infty}.$
Therefore, we get
\smallskip

{\bf Corollary 3.1.}\
If $s\in (0,1)$ and $1/s=1+1/q,$ then $AP_{s, \infty} \implies AP_{1; q, \infty}.$
\smallskip

Evidently, also $AP_{s,\infty}\implies AP_s$ (for $s\in (0,1)$).
\smallskip

{\bf Theorem 3.2.}\
 Let  $s\in (0,1), q\in (0,\infty)$ and $1/s=1+1/q.$ If $X$ has the $AP_{1; q,\infty},$
 then $X$ has the $\widetilde{AP}_{s,\infty}.$
In particular, $AP_{s,\infty} \implies \widetilde{AP}_{s,\infty}.$
\smallskip

{\it Proof}.\
It is enough to repeat word for word      %!!! Engl
the proof of the implication $2) \implies 1)$ of Theorem 2.1 ("continuation"),
just changing "$l_{q}(X)$" by "$l^0_{q, \infty}$" (no necessity to apply Lemma 2.1 or Lemma 3.1).
\smallskip

{\it Remark} 3.3.\
In this moment (when I am writing the text) I do not know whether the implication
"$\widetilde{AP}_{s,\infty} \implies AP_{s,\infty}$" is true, for Banach spaces. Of course, no questions
about the cases where $0<s\le 2/3$ (but the reason is only that every Banach space has
the $\widetilde{AP}_{2/3,\infty}$ and the $AP_{2/3,\infty}).$
\smallskip

Let $0<r<1$ and  $0< w\le \infty,$ or  $r=1$  and  $0< w\le 1.$
For Banach spaces $X, Y$ denote by $Y^*\widehat\otimes_{(r,w)} X$ the subset of
$Y^*\widehat\otimes X$ consisting of tensors $z$ such that
$$
  z=\sum_{k=1}^\infty \lambda_k\, y'_k\otimes x_k,\ \text{where} \ (y'_k) \text{ and } (x_k)
  \text{ are bounded and }
  (\lambda_k)\in l_{rw}.
$$

As was noted in Remark 3.1, if  $s\in (0,1), q\in (0,\infty), 1/s=1+1/q,$ then
$l_1\cdot l_{q \infty}= l_{s1}$ (in the sense of the product in Lemma 3.1).
In general case, where $0< q_1, q_2, t_1, t_2 \le \infty,$
one has
%$$
\begin{equation}
 l_{q_1 t_1}\cdot l_{q_2 t_2}= l_{s,t}\ \text{provided that:}\ \frac1{q_1}+\frac1{q_2}=
 \frac1s\ \text{and}\   % !!! ispravil i + PiOpI
   \frac1{t_1}+\frac1{t_2}=\frac1t  \label{(2)}
 \end{equation}
 (cf. [17], 2.1.13 Proposition).                  %!!!Ref
%$$
We can introduce a new definition of approximation properties, which are connected
with Lorentz sequence spaces, namely:
Let  $0<r<1$ and  $0< w\le \infty.$ or  $r=1$  and  $0< w\le 1.$
A Banach space $X$ has the $AP_{(r,w)},$  if for every Banach space $Y$
the natural map $Y^*\widehat\otimes_{(r,w)} X\to L(Y,X)$ is one-to-one.

It follows (from Remark 3.1 or from (2)) that $AP_{1; q, \infty}= AP_{(s, 1)}$ (for $s\in(0,1)$ and  $1/s=1+1/q)$
and, more generally,  $AP_{t;p,r} = AP_{(s,u)}$ for $1/t+1/p=1/s$ and $1/t+1/r=1/u$\, $(t\in(0,1]).$

Therefore, we have (for $s\in(0,1)$)
$$
 AP_{s,\infty} \implies AP_{(s,1)} \implies \widetilde{AP}_{s,\infty}.
$$
Moreover, taking into account the equality $\widehat\otimes_{1;q,\infty}= \widehat\otimes_{(s,1)}$
and applying the arguments from the proof of the implication
"$\widetilde{AP}_{s}\implies AP_s$" of Theorem 2.1, we easily get
\smallskip

{\bf Theorem 3.3.}\
$AP_{(s,1)}=\widetilde{AP}_{s,\infty}.$
\smallskip

{\it Proof}.\
As was mentioned above, $AP_{(s,1)}\implies \widetilde{AP}_{s,\infty}.$
Let $X$ has the $\widetilde{AP}_{s,\infty},$ i.e.
for every sequence $(x_n)\in l_{q,\infty}^0$ (where $1/s=1+1/q)$ and every $\varepsilon>0$
there exists a finite rank operator $R\in X^*\otimes X$ such that
$\sup_n ||Rx_n-x_n||<\varepsilon.$ Since $AP_{(s,1)}= AP_{1; q,\infty},$ it is enough
to show that if $Y$ is a Banach space, $z\in Y^*\widehat\otimes_{1; q,\infty} X$ and $z\neq0,$
then the corresponding operator $\widetilde z: Y\to X$ is not zero too.

Let $z=\sum_{k=1}^\infty a_kb_k\, y'_k\otimes x_k$ be a representation of $z$ with
$(x_k), (y'_k)$ bounded, $(a_k)\in l_1,$ $(b_k)\in l^0_{q\infty}$ and $b_k\searrow0.$
Then $(\widetilde x_k:=b_kx_k)\in l^0_{q \infty}$ and, for an $\varepsilon>0$ small enough (to be chosen),
we can find an operator $R\in X^*\otimes X$ with the property that
$\sup_n ||R\widetilde x_n-\widetilde x_n||\le \varepsilon.$
Since $z\neq0,$ we can find an operator  $V\in L(Y^*, X^*)$ such that
$\sum_{k=1}^\infty a_k\, \langle Vy'_k, \widetilde x_k\rangle =1.$
Now, when $V$ is chosen, we have
$$
  1= \sum_{k=1}^\infty a_k\, \langle Vy'_k, \widetilde x_k-R\widetilde x_k\rangle
  +\sum_{k=1}^\infty a_k\, \langle Vy'_k, R\widetilde x_k\rangle
  $$
  $$\le \varepsilon\, ||(a_k)||_{l_1}\,  ||V||\cdot const +
   |\sum_{k=1}^\infty a_kb_k\, \langle R^*Vy'_k,  x_k\rangle |,
$$
and, if $\varepsilon$ is small enough, we get for the finite rank operator
$R^*V: Y^*\to X^*$ that
$$
 |\operatorname{trace}\, z^t\circ (R^*V)|= |\operatorname{trace}\, (R^*V)\circ z^t|=
 |\sum_{k=1}^\infty a_kb_k\, \langle R^*Vy'_k, x_k\rangle |>0.
$$
The last sum is the nuclear trace of the tensor element $\sum_{k=1}^\infty a_kb_k\, R^*Vy'_k\otimes x_k,$
which is a composition $R\circ z_0$ of the finite rank operator $R$ and the tensor element
$\sum_{k=1}^\infty a_kb_k\, Vy'_k\otimes x_k,$ that
belongs to the tensor product $X^*\widehat\otimes_{1;q,\infty} X.$
It follows that both $z_0$ and $z$ generate the non-zero operators $\widetilde z_0$ and $\widetilde z.$
\smallskip

{\it Remark} 3.4.\
Because of the equality  $\widehat\otimes_{1;q,\infty}= \widehat\otimes_{(s,1)},$
it follows from the proof of Theorem 3.3
that $X$ has the $AP_{(s,1)}$ iff the canonical mapping $X^*\widehat\otimes_{(s,1)}X \to L(X)$ is one-to-one
(just like in the case of the classical Grothendieck approximation property).
\smallskip

{\it Remark} 3.5.\
Of course, it follows from Theorem 3.3 that every Banach space has the $AP_{(2/3,1)},$ but
it is trivial because of the implication
$$
 AP^0_{(2/3,\infty)} \equiv AP_{2/3,\infty} \implies AP_{(2/3,w)} \ \text{for any } w<\infty
$$
(and, again, since every $X$ has the $AP_{2/3,\infty}!).$
\smallskip

Our question in Remark 3.3 can be reformulated now as:

$({}^*)$\ Is it true that the $AP_{(s,1)}$ implies the $AP_{s,\infty}?$



 \vskip 0.23cm

{\bf  IV.}\
Let us consider an application of the previous considerations.
Now we know, in particular, that every Banach space has the $AP_{(2/3,1)}.$
On the other hand, the corresponding operator ideal $N_{(2/3,1)}$
(related to the Lorentz space $l_{2/3\, 1 })$ has the eigenvalue type $l_{1}$
(see, e.g., [8, p. 243]). Since the continuous trace is unique on
$\widehat\otimes_{(2/3,1)}$ and $\widehat\otimes_{(2/3, 1)}= N_{(2/3, 1)},$
it follows from White's results [32] that for each Banach space $X$ and for every operator
$T\in N_{(2/3,1)}(X,X)$ the (nuclear) trace of $T$ is well defined and equals the sum of all
eigenvalues of $T:$
$$
  \operatorname{trace}\, T=
   \sum_{k=1}^\infty \mu_k(T)\ \text{(eigenvalues)}\, \forall\, X,\ \forall\, T\in N_{(2/3,1)}(X)
$$
(on the right is the so-called "spectral sum" of $T).$ More precisely, the last statement follows
from Theorem 4.1 below.

Let us explain in more details how we  apply a result of  M.C. White. To do this, we formulate and
 prove a theorem which is almost an immediate consequence of  White's theorem.
 \smallskip

 {\bf Theorem 4.1.}\
Let $A$ be a quasi-Banach operator ideal, $X$ be a Banach space, for which the set of all finite rank
operators is dense in the space $A(X).$ Suppose that the natural functional "$\operatorname{trace}\,$" is bounded
on the subspace of all finite rank operators of $A(X)$ (and, therefore, can be extended to a continuous
functional "$\operatorname{trace}_A$" on the whole space $A(X)$).
  If the quasi-Banach operator ideal $A$ is of eigenvalue type $l_1,$ then
 the spectral trace (= "spectral sum") is continuous  on the space $A(X)$ and for any
 operator $T\in A(X)$ we have
 $$
   \operatorname{trace}_A(T) = \sum_{n=1}^{\infty} \mu_n(T).
 $$
 where $(\mu_n(T))_{n=1}^\infty$ is  the sequence
 of all eigenvalues of $T$ counted according to their
multiplicities.
 \smallskip

  {\it Proof}\ of Theorem 4.1.    \
  Let $T\in A(X).$
 By the assumption, the sequence $\{\mu_n(T)\}_{n=1}^\infty$ of all eigenvalues of $T$ (counted according to their
multiplicities) is in $l_1.$
Since the quasi-normed ideal $A$ is  of spectral (= eigenvalue)
type $l_1,$  we can apply the main result from the paper [32] of M.C. White, which asserts:  %that:

$({}^*{}_*{}^*)$\,  {\it If $J$ is a quasi-Banach operator ideal with eigenvalue type $l_1,$ then
the spectral sum is a trace on the ideal $J$}.

Recall   (see  [18], 6.5.1.1, or Definition 2.1 in [32]) that a {\it trace}\ on an operator ideal $J$
is a class of complex-valued functions, all of which they write as $\tau,$ one for each component
$J(E,E)$ (where $E$ is a Banach space) so that

(i)\  $\tau(e'\otimes e)= \langle  e',e\rangle$ for all $e'\in E^*, e\in E;$

(ii)\ $\tau(AU)=\tau(UA)$ for all Banach spaces $F$ and operators $U\in J(E,F)$ and    $A\in L(F,E); $

(iii)\ $\tau(S+U)=\tau(S) +\tau(U)$ for all $S,U\in J(E,E);$

(iv)\ $\tau(\lambda U)= \lambda \tau(U)$ for all $\lambda\in \Bbb C$ and $U\in J(E,E).$

Our operator $T$  belongs to the space $A(X,X)=A(X)$ and
$A$ is of eigenvalue type $l_1.$ Thus, the assertion $({}^*{}_*{}^*)$ implies that
the spectral sum $\mu$ defined by
$\mu(U):= \sum_{n=1}^\infty \mu_n(U)$   for $U\in A(E,E)$
is a trace on $A.$

By the principle of uniform boundedness (see [17], 3.4.6 (page 152), or [15]),
there exists a constant $C>0$ such that
$$
 |\mu(U)|\le ||\{\mu_n(U)\}||_{l_1} \le C\, a(U)
$$
for all Banach spaces $E$ and operators $U\in A(E,E). $

Now, remembering
that all operators in $A(X)$ can be approximated by finite rank operators and
taking in account the conditions (iii)--(iv) for $\tau=\mu$,
we obtain that the $A$-trace, i.e. $\operatorname{trace}_A T,$
of our operator $T$ coincides
with $\mu (T)$ (recall that the continuous trace is uniquely defined in
such a situation, that is on the space $A(X);$ cf. [18], 6.5.1.2).
\smallskip

 Since $\widehat\otimes_{1;2,\infty}= \widehat\otimes_{(2/3,1)}$ (see Theorem 3.3), we can reformulate the result,
 which we considered in the very beginning of this section, as
 \smallskip

 {\bf Corollary 4.1.}\
 For each Banach space $X$ and for every operator
 $T\in N_{1;2,\infty}(X,X)$ the (nuclear) trace of $T$ is well defined and equals the sum of all
 eigenvalues of $T:$
 $$
   \operatorname{trace}\, T= \sum_{k=1}^\infty \mu_k(T)\ \text{(eigenvalues)}\, \forall\, X,\
   \forall\, T\in N_{(1;2,\infty)}(X).
 $$
 \smallskip

 {\it Remark \rm4.1}:\,
 Recall that A. Grothendieck [7] has obtained the last assertion  for the
 case of $2/3$-nuclear operators, i.e. for the ideal
 $N_{2/3}= N_{(2/3,\, 2/3)}$ (note that $l_{2/3}\subset l_{2/3\, 1}).$
% \smallskip



  \vskip 0.23cm

 {\bf  V.}\
 The discussion in Section I shows that, for $p\in[1,\infty],$
 any subspace of any quotient (= any quotient of any
subspace) of  an $L_p$-space possesses the $\widetilde{AP}_s$
 (even the $\widetilde{AP}_{s,\infty})$  with $1/s=1+|1/2-1/p|.$
 We apply now these facts together with  White's theorem for proving
 some more theorems concerning the distributions of eigenvalues of the nuclear operators.
 Below we will use Theorem 2.1 and, therefore, the fact that  any subspace of any quotient
 of  an $L_p$-space possesses the ${AP}_s$ (where $p,s$ as above).
 Thus, for such Banach spaces $X,$ we can identify the tensor product $X^*\widehat\otimes_s X$
 with its canonical image in the space $L(X)=L(X,X),$ i. e., with the space
 $N_s(X)$ of all $s$-nuclear operators in $X,$ equipped with the quasi-norm induced from
 $X^*\widehat\otimes_s X.$

 We  give below  relatively simple proofs of some recent results from the papers
 [25] and [28].
 \smallskip

 {\bf Theorem 5.1.}\
 Let $X$ be a subspace of an $L_p$-space,
 $1\le p\le \infty.$ If $T\in N_s(X,X),$\, where
 $1/s=1+|1/2-1/p|,$   \,
 then

 1.\, the (nuclear) trace  of $T$ is well defined,

 2.\, $\sum_{n=1}^\infty |\mu_n(T)|<\infty,$ where
 $\{\mu_n(T)\}$ is the system of all eigenvalues of the operator $T$
 (written in according to their algebraic multiplicities)

 and
 $$
  \operatorname{trace}\, T= \sum_{n=1}^\infty \mu_n(T).
 $$

  \vskip 0.2cm

 {\it Proof}.\
 Let $X$ be a subspace  of an $L_p$-space $L_p(\mu)$
 and $T\in N_s(X,X)$ with an s-nuclear representation
 $$
  T=\sum_{k=1}^\infty \lambda_k x'_k\otimes x_k,
 $$
 where $||x'_k||, ||x_k||=1$ and $\lambda_k\ge 0,$  $\sum_{k=1}^\infty \lambda_k^s<\infty.$
 By Hahn-Banach, we can find  the functionals $\widetilde x'_k\in L^*_p(\mu)$ \, $(k=1,2,\dots)$ with
 the same norms as for the corresponding functionals $x'_k$ and so that
 $\widetilde x'_k|_X=x'_k$ for every $k.$
 Denote by $\widetilde T$  the operator
 $$
  \widetilde T: L_p(\mu)\to X,\ \widetilde T:= \sum_{k=1}^\infty \lambda_k \widetilde x'_k\otimes x_k,
 $$
and by $j: X\to L_p(\mu)$ the natural injection. Since the space $X$ has the property $AP_s,$
we have $N_s(L_p(\mu), X)= L^*_p(\mu)\widehat\otimes_s X$ and, therefore,
the nuclear traces of the operators $j\widetilde T$ and $\widetilde Tj$ are well defined.
 We have a diagram
 $$
  X\overset{j}\to L_p(\mu) \overset{\widetilde T}\to X \overset{j}\to L_p(\mu),                %ooo!!! ovs check.......... i nizhe
 $$
 in which $\widetilde Tj = T\in N_s(X).$
Hence, the complete systems of eigenvalues of the operators  $T=\widetilde Tj $
and $j\widetilde T\in N_s(L_p(\mu))$ coincide.
Applying Theorem 2.b.13 from [10] (see also [25]), we obtain that
the sequence $(\mu_k(j\widetilde T))$ is summable. Therefore, we  have
$\mu_k(T)\in l_1$
and we can apply Theorem 4.1. But we apply the theorem firstly for the simplest case
(later on we will continue the proof of our theorem 5.1).
 \smallskip

 The first assertion of the next theorem is due to A. Grothendieck [7],          %!!!R
 the second one was proved by H. K\"onig in [11]. Surprisingly,                %!!!R
 but we could not find anywhere the main statement of the theorem about coincidence
 of the nuclear and spectral traces, neither in the monographs, nor in the mathematical journals.
 So we have no reference for this statement and have to formulate and to prove the next theorem
 here. Let us remark that, in any case, this theorem was proved (as a partial case of the proved there
 our Theorem 5.1) in [25].                              %!!!R
 \smallskip

 {\bf Theorem $\mathbf5.1'.$}\
Let $L$ be  an $L_p$-space,
$1\le p\le \infty.$ If $T\in N_s(L, L),$\, where
$1/s=1+|1/2-1/p|,$   \,
then

1.\, the (nuclear) trace  of $T$ is well defined,

2.\, $\sum_{n=1}^\infty |\mu_n(T)|<\infty,$ where
$\{\mu_n(T)\}$ is the system of all eigenvalues of the operator $T$
(written in according to their algebraic multiplicities)

and
$$
 \operatorname{trace}\, T= \sum_{n=1}^\infty \mu_n(T).
$$

 \vskip 0.2cm

{\it Proof}.\
As we have said above, the assertions 1 and 2 are well known.
To prove the last equality, consider the Banach operator ideal $\mathcal L_p$
of all operators which can be factored through $L_p$-spaces. Then the product
$\mathcal L_p\circ N_s$ is a quasi-Banach operator ideal of spectral (=eigenvalue)
type $l_1$ (e.g., by the assertion 2, proved earlier by H. K\"onig [11]).       %!!!R
Now it is enough to apply Theorem 4.1 to finish the proof.
\smallskip

{\it Proof \, of Theorem }5.1 \, (continuation).
As we have  said, the complete systems of eigenvalues of the operators  $T=\widetilde Tj $
and $j\widetilde T\in N_s(L_p(\mu))$ coincide. By Theorem 5.1',
$$
\operatorname{trace}\, j\widetilde T =
\sum_{k=1}^\infty \lambda_k \, \langle \widetilde x'_k, jx_k\rangle  = \sum_{n=1}^\infty \mu_n(j\widetilde T),
$$
the last sum is equal to
$$
 \sum_{n=1}^\infty \mu_n(T)
$$
 and the sum in the middle is
 $$
   \sum_{k=1}^\infty \lambda_k \, \langle \widetilde x'_k, jx_k\rangle  =
    \sum_{k=1}^\infty \lambda_k\, \langle x'_k, x_k\rangle =\operatorname{trace}\, T.
 $$
 The (nuclear) trace of the operator $T$ is well defined, because the space $X$ has the $AP_s.$
 Therefore,
 $$
   \operatorname{trace}\, T=  \sum_{n=1}^\infty \mu_n(T),
 $$
 and we are done.
  \smallskip

 If $Y$ is a quotient of an $L_p$-space, then, for a compact operator                                                         %!!!
 $U\in L(E,E),$ the adjoint    $U^*$ is also a compact operator and these two  operators
 have the same eigenvalues $\mu\neq0$ with the same multiplicities
 (see, e.g., [17], Theorem 3.2.26, or [5], Exercise VII.5.35).                 %%!!!R from MNwe
 Also, any quotient of an $L_p$-space has the $AP_s$ (where $p,s$ are as above).
 So, it follows immediately from the just proved Theorem 5.1
  \smallskip

  {\bf Corollary 5.1.}\
   Let $Y$ be a quotient  of an $L_p$-space,
 $1\le p\le \infty.$ If $T\in N_s(Y,Y),$\, where
 $1/s=1+|1/2-1/p|,$   \,
 then

 1.\, the (nuclear) trace  of $T$ is well defined,

 2.\, $\sum_{n=1}^\infty |\mu_n(T)|<\infty,$ where
 $\{\mu_n(T)\}$ is the system of all eigenvalues of the operator $T$
 (written in according to their algebraic multiplicities)

 and
 $$
  \operatorname{trace}\, T= \sum_{n=1}^\infty \mu_n(T).
 $$

  \vskip 0.2cm

  We used above some facts from Section I. After Theorem 5.1 and its consequence have been  proved,
  we are ready to present a simple prove of the corresponding result on the subspaces of quotients
  of the $L_p$-spaces (recall that, again, all such spaces have the $AP_s$ with $s$ and $p$  satisfying
  the same conditions).
  \smallskip

   {\bf Theorem 5.2.}\
  Let $W$ be  a quotient of a subspace (= a subspace of  a quotient) of
  an $L_p$-space,
  $1\le p\le \infty.$ If $T\in N_s(W,W),$\, where
  $1/s=1+|1/2-1/p|,$   \,
  then

  1.\, the (nuclear) trace  of $T$ is well defined,

  2.\, $\sum_{n=1}^\infty |\mu_n(T)|<\infty,$ where
  $\{\mu_n(T)\}$ is the system of all eigenvalues of the operator $T$
  (written in according to their algebraic multiplicities)

  and
  $$
   \operatorname{trace}\, T= \sum_{n=1}^\infty \mu_n(T).
  $$

   \vskip 0.2cm

  {\it Proof}.\
  Let $L_p(\mu)$ be an $L_p$-space. Take Banach subspaces
  $X_0\subset X\subset L_p(\mu)$ and consider the quotient $X/X_0.$
  If $T\in N_s(X/X_0, X/X_0)$ (=$(X/X_0)^*\widehat\otimes_s X/X_0),$ then
  $T$ admits a factorization of the type
  $$
    X/X_0 \overset{A}\to c_0 \overset{D}\to l_1 \overset{B}\to  X/X_0,       %ooo
  $$
  where $A,B$ are continuous and $D$ is a diagonal operator with a diagonal from $l_s.$

  Denoting by $\varphi: X\to X/X_0$ the factor map from  $X$ onto $X/X_0$ and
  taking a lifting $\Phi: l_1\to X$ for $B$ with $B=\varphi\Phi,$
  we obtain that the maps $\varphi \Phi DA: X/X_0\to X/X_0$ and $\Phi DA\varphi: X\to X$
  have the same eigenvalues $\mu\neq0$ with the same multiplicities:
  $$
    X\overset{\varphi}\to X/X_0 \overset{A}\to c_0 \overset{D}\to l_1 \overset{\Phi}\to X \overset{\varphi}\to  X/X_0,         %ooo
  $$
  The spaces $X$ and $X/X_0$ have the $AP_s.$ Therefore, we have (cf. the proof of Theorem 5.1)
  $$
    \operatorname{trace}\, \varphi \Phi DA = \operatorname{trace}\, \Phi DA\varphi.
  $$
  Since $X$ is a subspace of  $L_p(\mu),$ we have, by Theorem 5.1,
  $$
   \operatorname{trace}\, \Phi DA\varphi= \sum_{n=1}^\infty \mu_n(\Phi DA\varphi).
  $$
  Therefore,
  $$
    \operatorname{trace}\, T = \operatorname{trace}\, BDA = \operatorname{trace}\, \varphi \Phi DA
    = \sum_{n=1}^\infty \mu_n(\Phi DA\varphi) $$
    $$ =
     \sum_{n=1}^\infty \mu_n(\varphi\Phi DA) = \sum_{n=1}^\infty \mu_n(BDA) =\operatorname{trace}\, T.
  $$
 % \smallskip


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


   \vskip 0.23cm

  {\bf  VI.}\
    As is well known, in the classical case of  the   Grothendieck approximation property $AP,$
  if $X^*$ has the $AP$ then the space $X$ also has this property. We will show now that the same
  is true for all approximation properties which are under consideration in this paper.

  Denote by $\widehat\otimes_\alpha$ any of the tensor product
     $\widehat\otimes_s,$    $\widehat\otimes_{s,\infty},$    $\widehat\otimes_{t; p,r}$,      $\widehat\otimes_{(r,w)}$
      with the parameters (see above), for which all these tensor products are the  linear subspaces
    of the projective tensor product $\widehat\otimes.$ Also, let us say that a Banach space $X$
   has the $AP_\alpha,$ if it is possesses the corresponding approximation property (i.e., $AP_s,$
   $AP_{s,\infty}$ etc.).
\smallskip

We need the following  auxiliary result which may be of its own interest (compare with Remark 3.4).
\smallskip

{\bf Proposition 6.1}
A Banach space $X$ has the $AP_\alpha$ iff  the canonical map $X^*\widehat\otimes_\alpha X\to L(X)$ is
one-to-one.
\smallskip

{\it Proof}.\
Suppose that the canonical map $X^*\widehat\otimes_\alpha X\to L(X)$ is one-to-one,
but there exists a Banach space $Y$ such that the natural map
$Y^*\widehat\otimes_\alpha X\to L(Y, X)$ is not injective. Let $z\in Y^*\widehat\otimes_\alpha X$
be such that $z\neq0$ and the associated operator $\widetilde z$ is a 0-operator.
Then we can find an operator $V$ from $L(Y^*,X^*)$ (the dual space to the
projective tensor product $Y^*\widehat\otimes X)$ so that $\operatorname{trace}\, V\circ z^t=1,$
where, as usual, $z^t$ is the transposed tensor element, $z*t\in X\widehat\otimes Y^*.$
Since $V\circ z^t\in X\widehat\otimes X^*$ and   $\operatorname{trace}\, V\circ z^t=1,$ the tensor element
$(V\circ z^t)^t$ (which, evidently, belongs to $X^*\widehat\otimes_\alpha X)$
is not zero. On the other hand, the operator induced by this element must be
a 0-operator. Contradiction.
\smallskip

  %\noindent
  {\bf Proposition 6.2.}\
With the above understanding, if the dual space $Y^*$ has the $AP_\alpha,$ then
$Y$ has the $AP_\alpha$  too.
   \medskip

  %\noindent
  {\it Proof}.\,
  We use Proposition 6.1.
  As it is known [7], 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_\alpha Y $ is  a linear subspace of  $Y^*\widehat\otimes Y,$
    as well as  $Y^*\widehat\otimes_\alpha Y^{**}$ is a linear subspace of  $Y^*\widehat\otimes Y^{**}.$
    Therefore, the natural map  $Y^*\widehat\otimes_\alpha Y \to Y^*\widehat\otimes_\alpha Y^{**} $
    is one-to-one. Now if $Y^*$ has the $AP_\alpha,$ then the canonical map
     $Y^{**}\widehat\otimes_\alpha Y^* \to L(Y^*,Y^*)$ is one-to-one.
     Since we can identify the tensor product $Y^{**}\widehat\otimes_\alpha Y^*$ with
     the tensor product $Y^{*}\widehat\otimes_\alpha Y^{**}$ (because of the "symmetries"
      in the definitions of the corresponding tensor products), it follows that
     the natural map $Y^*\widehat\otimes_\alpha Y \to L(Y,Y)$ is one-to-one.
     Thus, if $Y^*$ has the $AP_\alpha,$ then $Y$ has the $AP_\alpha$ too.
     \smallskip

   %  We will use Lemma 4 in the proof of Theorem 1 in the next section.

      % \smallskip

     %%%%%%%%   Zamechanie o tochnosti - perepravit' !!!
  %   \noindent
     {\it Remark \rm6.1}:\   The inverse statement is not true. For example,                  %!!! rem No...
    if $s\in(2/3,1],$ then 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 [21] or [23]).   %!!! Ref me or me!!!
      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 [13, 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

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




    \vskip 0.23cm

   {\bf  VII.}\
   Let us  consider some more notions of the approximation properties associated with some other
   tensor products.
   For Banach spaces $X$ and $Y$ and $r\in (0,1], p\in [1,2],$ define  a quasi-norm  $|| \cdot||_{N_{[r,p]}}$
   on the tensor product  $X^*\otimes Y$ by

    \begin{equation}\nonumber                                               %!!! 888 nuzhna numeracija
   \|u\|_{N_{[r,p]}}:=\inf\left\{\|(x_{i}')_{i=1}^{n}\|_{\ell_{r}(X^*)}\cdot\|(y_{i})_{i=1}^{n}
   \|_{\ell_{p'}^{w}(Y)}:\ u=\sum_{i=1}^{n}x_{i}'\otimes y_{i}\right\}              \label {(3)}
   \end{equation}
 Here we denote, as usual, by $l_r(X^*)$ and $l_q^w(Y)$ the spaces of $r$-absolutely
 summable and weakly $q$-summable sequences, respectively.

 Denote by   $X^{*}\widehat\otimes_{[r,p]}Y$  the completion of the space
 $\left(X^{*}\otimes Y,\ \|\cdot\|_{N_{[r,p]}}\right).$
 We have a natural continuous injection
  $$j_{[r,p]}: X^{*}\widehat\otimes_{[r,p]}Y\rightarrow X^{*}\widehat\otimes Y$$ with $||j_{[r,p]}||\le1.$

  Every element   $u\in X^{*}\widehat\otimes_{[r,p]}Y$ has a representation of the type
  $u=\sum_{i=1}^{\infty}x_{i}'\otimes y_{i},$
where $(x_{i}')_{i=1}^{\infty}\in\ell_{r}(X^{*})$ and $(y_{i})_{i=1}^{\infty}\in \ell_{p'}^{w}(Y)$.
     Consider the natural mappings
     $$
             X^{*}\widehat\otimes_{[r,p]} Y   \overset{{j}_{[r,p]}}\to  X^{*}\widehat\otimes Y  \overset{{j }}\to  L(X,Y).
      $$
      The image of the tensor product  $X^{*}\widehat\otimes_{[r,p]} Y $ under the composition
      $\widetilde{j}_{[r,p]}:={j}\circ  j_{[r,p]}$
      is denoted by  $N_{[r,p]}(X,Y)$. This is a quasi-Banach space of the $(r,p)$-nuclear operators
      (the quasi-norm is induced from the
      tensor product $X^{*}\widehat\otimes_{[r,p]} Y).$
      It is not difficult to see that every operator $T\in N_{[r,p]}(X,Y)$ admits a factorization of the kind
%      $$
        \begin{equation}
        X \overset{A}\to c_0 \overset{D_r}\to l_1 \overset{i}\to l_p \overset{B}\to Y,     \label {(3)}      %ooo
       \end{equation}
    %  $$
      where $A,B$ are compact, $i$ is the injection, $D_r$ is a diagonal operator with a diagonal from $l_r.$
        Also, every operator, which can be factored in such a way, is in $N_{[r,p]}(X,Y).$
      \smallskip

      By the analogous way, we define the tensor product $X^{*}\widehat\otimes^{[r,p]}Y$
      and the quasi-normed operator ideals
      $N^{[r,p]}(X,Y).$ Namely,
      $X^{*}\widehat\otimes^{[r,p]}Y$ is a linear subspace of the projective tensor product
      $X^*\widehat\otimes Y,$
      consisting of tensor elements $z$ which admit a representation
      $$
        u=\sum_{i=1}^{\infty}x_{i}'\otimes y_{i},
        $$
        where $(x_{i}')_{i=1}^{\infty}\in\ell_{p'}^w(X^{*})$ and $(x_{i})_{i=1}^{\infty}\in \ell_{r}(Y).$
        Its canonical image in $L(X,Y)$ is the quasi-normed space $N^{[r,p]}(X,Y).$
        It is not difficult to see that every operator $T\in N^{[r,p]}(X,Y)$ admits a factorization of the kind
        $$
          X \overset{A}\to l_{p'} \overset{D_r}\to c_0 \overset{i}\to l_1 \overset{B}\to Y,       %ooo
        $$
        where $A,B$ are compact, $i$ is the injection, $D_r$ is a diagonal operator with a diagonal from $l_r.$
        Also, every operator, which can be factored in such a way, is in $N^{[r,p]}(X,Y).$

        It is clear that $T\in  N_{[r,p]}(X, Y)$ implies $T^*\in N^{[r,p]}(Y^*, X^*)$   and
$T\in  N^{[r,p]}(X,Y)$ implies $T^*\in N_{[r,p]}(Y^*, X^*).$
Inverse is not true (see, e.g., Example 8.3 below). %!!! Check!!!

  Now we can define the notions of the corresponding approximation properties by the usual way.
  We say that he space $X$ has the $AP_{[r,p]}$ (respectively, the $AP^{[r,p]})$ if
  for every Banach space $Y$ the natural mapping $ Y^{*}\widehat\otimes_{[r,p]} X\to L(Y,X)$
  (respectively, $ Y^{*}\widehat\otimes^{[r,p]} X\to L(Y,X)$)     is one-to-one.
  It can be seen that
  a Banach space $X$ has the $AP_{[r,p]}$ (or $AP^{[r,p]})$ iff  the canonical map $X^*\widehat\otimes_{[r,p]} X\to L(X)$
  (or $X^*\widehat\otimes^{[r,p]} X\to L(X)$) is
  one-to-one (the proof is essentially the same as the proof of Theorem 6.1).
  Also, if $X^*$ has the $AP_{[r,p]}$ (or $AP^{[r,p]})$   then
$X$ has the $AP^{[r,p]}$ (or $AP_{[r,p]})$ (the proof is the same as in Theorem 6.2).
 \smallskip


{\bf  Theorem 7.1.}\
Let   $1/r-1/p=1/2.$ Every Banach space has the properties $AP_{[r,p]}$
and $AP^{[r,p]}.$
\smallskip


{\it Proof}.\
Suppose that $X\notin AP_{[r,p]}$  where $1/r-1/p=1/2.$ Let $z\in X^*\widehat\otimes_{[r,p]} X$
be an element such that
    $\operatorname{trace}\, z=1, \tilde z=0.$
   Since $z=\sum x'_k\otimes x_k,$ where  $(x'_k)\in l_r(X^*) $ and $ (x_k) $
   is weakly $ p'$-summable, the operator $ \tilde z $ can be factored as
     $$
    \tilde z:\ X\overset{A}\to l_\infty \overset{\Delta}\to l_1 \overset{j}\to l_p \overset{V}\to X,
    $$
   where all the operators are continuous, $ j $ is an injection, $ \Delta $ is a diagonal operator
   with a diagonal from $ l_r. $
Since $\tilde z=0,$ we have $V|_{j\Delta A(X)}=0.$ Consider $S:= j\Delta AV: l_p\to l_p.$
Evidently, $S^2=0$ and  $\operatorname{trace}\, S=\operatorname{trace}\, z=1.$
    Since $S\in N_r(l_p,l_p),$  its nuclear trace equals the sum  of all its eigenvalues
    (see  Theorem $\mathrm5.1'$ above).
    This contradicts the fact that $S^2=0.$
    Now, let $Y$ be another Banach space and put $X:= Y^*.$ We have shown that $X$ has the $AP_{[r,p]}.$
    Therefore (see remarks before the formulation of Theorem 7.1), $Y$ has the $AP^{[r,p]}.$
   \vskip0.1cm

   %%%

    We are ready to apply the above results to the investigation of eigenvalues problems
    for $N_{[r,p]}$-  and $N^{[r,p]}$-operators. The first theorem below was proved in [26]
    by using Fredholm Theory. The same proof can be applied for the second theorem. Below
    we present very different simple proofs of them.
\smallskip

{\bf  Theorem 7.2.}\
Let   $1/r-1/p=1/2.$
For every Banach space $X$ and every operator $T\in N_{[r,p]}(X)$, $\operatorname{trace}\,T$
 is well defined and
if $(\mu_{i})_{i=1}^{\infty}$ is a system of all eigenvalues of $T,$ then $(\mu_{i})_{i=1}^{\infty}\in l_1$ and
     \begin{equation}\nonumber
     \operatorname{trace}\,T=\sum_{i=1}^{\infty}\mu_{i}.
     \end{equation}
  \smallskip

{\bf  Theorem 7.3.}\
Let   $1/r-1/p=1/2.$
For every Banach space $X$ and every operator $T\in N^{[r,p]}(X)$, $\operatorname{trace}\,T$  is well defined and
if $(\mu_{i})_{i=1}^{\infty}$ is a system of all eigenvalues of $T,$ then $(\mu_{i})_{i=1}^{\infty}\in l_1$ and
     \begin{equation}\nonumber
     \operatorname{trace}\,T=\sum_{i=1}^{\infty}\mu_{i}.
     \end{equation}
  \smallskip

Both theorems can be proved by the analogues methods and
the proofs are almost the same as the proof of Theorem 5.2 (by using Theorem 7.1).
%So we omit it here.
On the other hand, Theorem 7.3 is a consequence of Theorem 7.2 and vice versa.
Let us give, firstly, a simple proof of Theorem 7.2 and then deduce Theorem 7.3.

{\it Proof}.\,
Suppose that $T\in N_{[r,p]}(X)$ and consider a  factorization (3) (in which $Y=X)$
of the operator $T.$ The sequence $(\mu_k)$ of all eigenvalues of $T$ is the same as
the sequence  of all eigenvalues of the operator $iD_rAB,$ which maps $l_p$ into $l_p.$
The last operator is $r$-nuclear, where $p\in [1,2]$ and $1/r=1/p+1/2.$ By Theorem $\mathrm5.1',$
$(\mu_k)\in l_1$ and $\operatorname{trace}\, iD_rAB= \sum \mu_k.$ Since the trace of $T$ is well defined
(Theorem 7.1), it is clear that $\operatorname{trace}\, T=\operatorname{trace}\, iD_rAB=\sum \mu_k.$

Now, let $T\in N^{[r,p]}(X).$ Then $T^*\in N_{[r,p]}(X^*)$ and $\operatorname{trace}\, T^*=\operatorname{trace}\, T$
(apply Theorem 7.1). Since the operators $T$ and $T^*$ have the same systems of
eigenvalues, Theorem 7.3 follows from  just proved statement of Theorem 7.2.
    \smallskip

% {\it Proof}.\




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




   \vskip 0.23cm

  {\bf  VIII.}\
The next examples are taken from [26], where one can find the corresponding
proofs. They show that all the above affirmative results
concerning approximation properties and trace-formulas are sharp.

  \smallskip

  {\bf Example 8.1.}\
  Let $r\in(2/3,1], p\in(1,2], 1/r-1/p=1/2.$
  %There exist Banach spaces such that for each r
  There exist Banach spaces  $E$ and $V,$ $z_0\in E^*\widehat\otimes V, S\in L(V,E)$
  so that for every $p_0\in [1,p)$

  1)\, $z_0\in E^*\widehat\otimes_{[r,1]} V;$

  2)\, $V$ has a basis;

  3)\, $V$ is the space of type $p_0$ and of cotype $2;$

  4)\, $S\circ z_0\in E^*\widehat\otimes_{[r,p_0]} E;$

  5)\, $\operatorname{trace}\, S\circ z_0=1;$

  6)\, the corresponding operator  $\widetilde{S\circ z_0}$  is a 0-operator and,
  therefore, has no nonzero eigenvalues.
    \vskip0.1cm


  %%%%%%%

  {\bf Example 8.2.}\
  Let $r\in[2/3,1), p\in[1,2), 1/r-1/p=1/2.$
  %There exist Banach spaces such that for each r
 There exist Banach spaces $E$ and $V,$ $z_0\in E^*\widehat\otimes V, S\in L(V,E)$
  so that for every $\epsilon>0$

  1)\, $z_0\in E^*\widehat\otimes_{[r+\epsilon,1]} V;$

  2)\, $V$ has a basis;

  3)\, $S\circ z_0\in E^*\widehat\otimes_{[r+\epsilon,p]} E;$

  4)\, $\operatorname{trace}\, S\circ z_0=1;$

  5)\, the corresponding operator  $\widetilde{S\circ z_0}$  is a 0-operator and
 therefore, has no nonzero eigenvalues.
  \vskip0.1cm


  {\bf Example 8.3.}\
  Let $r\in(2/3,1]$, $p\in(1,2], 1/r-1/p=1/2.$
 There exist two separable Banach spaces $X$ and $Z$ so that

  (i)\, $Z^{**}$ has a basis;

  (ii)\, $\exists \, V\in X^*\widehat\otimes Z^{**}: \ V=\sum_{k=1}^\infty x'_k\otimes z''_k; $\ $(x'_k)$  weakly
  $p'_0$-summable for each $p_0\in [1,p);$ $(z''_k)\in l_r(Z^{**});$

  (iii)\, $V(X)\subset Z; $  the operator $V$ is not nuclear as a map from $X$ into $Z.$

 Moreover, there exists an operator $U:Z^{**}\to Z$ such that

  $(\alpha)$\, $\pi_ZU\in N^{[r,p_0]}(Z^{**},Z^{**})=Z^{***}\widehat\otimes^{[r,p_0]} Z^{**},\ \forall \, p_0\in[1,p);$

  $(\beta)$\, $U$ is not nuclear as a map from $Z^{**}$ into $Z;$

  $(\gamma)$\, $\operatorname{trace}\, \pi_ZU=1;$

  $(\delta)$\, $\pi_ZU: Z^{**}\to Z^{**}$ has no nonzero eigenvalues.

\medskip
{\bf Acknowledgement.}\,
The author would like to thank the referee for helpful remarks and Alexander Alenitsyn
for his technical help.

  \vskip1.1cm


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

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