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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}}
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                     \def\sbs{\subset}

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% Copyright 2003 by Till Tantau <tantau@cs.tu-berlin.de>.
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Z}^{d-1}_+} \newcommand{\rdd}{{\Bbb R}^{d-1}}
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    \newtheorem{rem}[theorem]{Remark}    

\title{On $\mathbb Z_d$-symmetry of spectra of linear operators in Banach spaces
}
\author{Oleg Reinov}
\institute{
 %Department of Applied Mathematics and Control Processes,
  Saint Petersburg State University}
  \date{}

\begin{document}


\frame{\titlepage
}








%\section[Introduction]{Introduction}
%\subsection[p-adic numbers]{p-adic numbers}  %
%%%%%%%%%%%%%%

%%%%%%%%%%%%     0 - BEGIN ----------------------

%______________________

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

 \frame{
 {\color{red}\bf M. I. Zelikin's Remark}
 
 All the main results below have their beginning in the following
 remark of M.I. Zelikin (Moscow State Uniersity):
 
 \begin{rem}
The spectrum of a linear operator $A: \mathbb R^k\to \mathbb R^k$ is 
central-symmetric iff the trace of any odd
power of A equals zero:
$$
\tr A^{2n-1}=0,\, n\in \mathbb N.
$$
\end{rem}
}

\frame{
 {\color{red}\bf Zelikin's Theorem}

To formulate the theorem, we need a definition:

\begin{itemize}
  \item
The spectrum of A is central-symmetric, if together with any eigenvalue $\la\neq0$ it has the
eigenvalue $-\la$ of the same multiplicity.
%\pause

It was proved in a paper by M. I. Zelikin
\smallskip

  \begin{thebibliography}{99}

  \bibitem{14} M. I. Zelikin,
A criterion for the symmetry of a spectrum",
Dokl. Akad. Nauk 418 (2008), no. 6, 737-740
             \end{thebibliography}

\item
{\bf Theorem.}\
The spectrum of a nuclear operator $A$
acting on a separable Hilbert space is central-symmetric iff
$trace\,  A^{2n - 1} = 0, \, n \in \mathbf N.$



\end{itemize}

}
%%%%%%%%%%%%%%%%%%%%%


        \frame{
 {\color{red}\bf Mityagin's $\Bbb Z_d$-symmetry}
 
 \begin{defi}
Let $T$ be an operator in $X,$
all non-zero spectral values of which 
are eigenvalues of finite
 multiplicity and have no limit point except possibly zero.
 For a fixed $d=2, 3, \dots$ and for the operator $T,$ the spectrum of $T$
 is called $\mathbb Z_d$-symmetric,
if $0\neq\lambda\in \operatorname{sp}\, (T)$ implies $t\lambda\in \operatorname{sp}\, (T)$ for every $t\in\sqrt[d]{1}$
and of the same nultiplicity.
                                \end{defi}
 
 If $d=2,$ then one has the central symmetry.
 }
         \frame{
 {\color{red}\bf Mityagin's Theorem}
 
 \begin{theor}
Let $X$ be a Banach space and $T: X\to X$ is a compact operator. Suppose that some power of $T$
is nuclear. The spectrum of $T$ is $\mathbb Z_d$-symmetric iff there is an integer $K\ge0$ such that
 for every $l> Kd$ the value $\operatorname{trace}\, T^{l}$ is well defined and
 $$\operatorname{trace}\, T^{kd+r}=0$$\, $$\text{ for all }\,  k=K, K+1, K+2, \dots \,
 \text{ and }\, r= 1, 2, \dots, d-1.$$
   \end{theor}
          \smallskip
      
      %\pause
          
In the proof, the Riesz theory of compact operators is used.
          
\begin{thebibliography}{99}

    \bibitem{7} B. S. Mityagin,
\textit{A criterion for the $\mathbb{Z}_d$-symmetry of the spectrum of a compact operator},
{J. Operator Theory}, \textbf{76}:1 (2016), 57--65.
   % doi: 10.7900/jot.2015jul24.2077
    
             \end{thebibliography}
} 

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


        \frame{
 {\color{red}\bf Our Generalization of Mityagin's Theorem}
 
  
 \begin{theor}
Let $X$ be a Banach space and $T: X\to X$ is a linear continuous  operator. Suppose that some power of $T$
is nuclear. The spectrum of $T$ is $\mathbb Z_d$-symmetric iff there is an integer $K\ge0$ such that
 for every $l> Kd$ the value $\operatorname{trace}\, T^{l}$ is well defined and
 $$\operatorname{trace}\, T^{kd+r}=0$$\, $$\text{ for all }\, k=K, K+1, K+2, \dots\, \text{ and }\, r= 1, 2, \dots, d-1.$$
   \end{theor}
          \smallskip
          
                   %\pause
                   
\begin{thebibliography}{99}

    \bibitem{7}
Oleg Reinov,
\textit{Some remarks on spectra of nuclear operators},
SPb. Math. Society Preprint 2016-09, 1-9
    
             \end{thebibliography}
         
         %\pause    
In the proof we use the Fredholm theory of A. Grothendieck:
          \smallskip
       
       \begin{thebibliography}{99}
          
    \bibitem{3} A. Grothendieck,
\textit{La th\'eorie de Fredholm},
{Bull. Soc. Math. France}, \textbf{84} (1956), 319--384.
                  \end{thebibliography}
                     
} 
                  %%%%%%%%
\frame{
 {\color{red}\bf Simplest Examples}

\begin{rem}                  
 Let $\Pi_p$ be the ideal of absolutely $p$-summing operators. 
 Then for some $n$ one has $\Pi_p^n\subset N.$ In
particular, $\Pi_2^2(C[0,1])\subset N(C[0,1]),$ but not every absolutely 2-summing operator
in $C[0,1]$ is compact.
\end{rem}
\smallskip

The p-summing operators in such spaces
 provide instances of operators to
which the last theorem  may be applied even though Mityagin's Theorem is not always
applicable.                 
 }                 
                  %%%%%%%%
                  


\frame{
 {\color{red}\bf General notation}


%\vspace{0.5cm}
 %%%%%%%%%%%%%%%%%%%%%%%

 $X, Y$  Banach spaces.
 
 $L(X,Y)$ --- linear continuous operators.
 
 For $T: X\to Y,$
 $$||T||= \sup \{||T(x)||: \ x\in X,\, ||x||\le 1\}.$$
 $X^*= L(X, \mathbb C).$
 

 For $0<p<\infty,$
 $$l^p= \{(a_k):\ a_k\in \mathbb C,\, \sum_{k=1}^\infty |a_k|^p\le\infty\},$$
  $||(a_k)||_{l^p}= \{\sum_{k=1}^\infty |a_k|^p\}^{1/p}.$
 $$l^\infty= \{(a_k):\ ||(a_k)||_{l^\infty}=\sup_k |a_k|<\infty\};$$ 
 $$c_0= \{(a_k)\subset l^\infty;\ a_k\to 0\}.$$
           }
%%

\frame{
 {\color{red}\bf Preliminaries}


$$\mathcal F(X,Y)=\{T\in L(X,Y):\ \mathrm{ rank} T<\infty\}$$ 

$$
T\in \F(X,Y) \implies T(x)=\sum_{k=1}^n x'_k(x)y_k,
$$
where $x'_k\in X^*,\, y_k\in Y.$

If $T\in \F(X,X),$ then 
$T(x)=\sum_{k=1}^n x'_k(x)x_k\ (x'_k\in X^*,\, x_k\in X)$
and
$$\tr T:= \sum_{k=1}^n x'_k(x_k).$$
"Trace" does not depend on a representation of $T$ and
$$\tr T= \sum \mathit{eigenvalues}\, (T)$$
(written according their multuplicities).

}
         %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
         

         
         
         %%%%%%%%%%
\frame{
 {\color{red}\bf Nuclear representations}


\vspace{0.5cm}
 %%%%%%%%%%%%%%%%%%%%%%%

Also, a finite rank $T\in L(X,X).$\ Consider a {\it nuclear}\, representation
$$
 Tx=\sum_{k=1}^\infty x'_k(x) x_k, \ \sum_{k=1}^\infty ||x'_k||\, ||x_k||<\infty
$$
and 
$$
\alpha:= \sum_{k=1}^\infty x'_k(x_k).
$$

%\pause

\begin{itemize}

\item 
{\bf Question:}\
$\alpha= \tr T $?

\item
Generally, NO.

\end{itemize}

%\pause

\begin{thebibliography}{99}

  \bibitem{5} Enflo P. , A counterexample to the approximation property in Banach spaces,
  Acta Math., Volume 130,
   1973, 309--317

             \end{thebibliography}

 \smallskip



 }


 %%%
 
 \frame{
 {\color{red}\bf Nuclear operators}
 
  \begin{defi}
$T: X\to Y$ is nuclear, if 
$$\exists\ (x'_k)\subset X^*, (y_k)\subset Y:\ \sum_{k=1}^\infty ||x'_k||\, ||y_k||<\infty,
$$ 
$$
 T(x)= \sum_{k=1}^\infty x'_k(x) y_k,\ \, \forall\ x\in X.
$$ 
   \end{defi}
   \medskip
            
            %\pause
            
 {\it Remark}: 
If
$T$ is nuclear, then $T: X\to c_0\overset{\Delta}\to l_1 \to Y.$\,
$\Delta\in l^1.$
               }
  
  \frame{
 {\color{red}\bf s-Nuclear operators}
 
 Generally:
   \medskip
   
 \begin{defi}
 $T: X\to Y$ is {\it $s$-nuclear}\, $(0<s\le1),$ if 
$$\exists\ (x'_k)\subset X^*, (y_k)\subset Y:\ 
\sum_{k=1}^\infty ||x'_k||^s\, ||y_k||^s<\infty,
$$ 
$$
 T(x)= \sum_{k=1}^\infty x'_k(x) y_k,\ \, \forall\ x\in X.
$$ 
            \end{defi}

            \medskip
            
               %\pause
 {\it Remark}: 
If
$T$ is s-nuclear, then $T: X\to c_0\overset{\Delta}\to l_1 \to X,$\,
$\Delta\in l^s.$
                  }
                  
 \frame{
 {\color{red}\bf Nuclear operators: Trace and AP}
 

\begin{defi}
 Let $T\in L(X,X)$ be nuclear with
$$
 T(x)= \sum_{k=1}^\infty x'_k(x) x_k,\ \, \forall\ x\in X.
$$
If  $\sum_{k=1}^\infty x'_k(x_k)$
{\bf does not depend} on a representation, then it is
{\it the (nuclear) trace of $T$}. Notation:\, $\tr T.$
         \end{defi}
            \medskip
            
            %\pause
            
         \begin{defi}
If every nuclear $T: X\to X$ has a trace, then
{\it $X$ has the $AP.$}
 \end{defi}
 

 
 }
 
   \frame{
 {\color{red}\bf Grothendieck's $AP$}
 
 Grothendieck's Definition:
 
 \begin{defi}
$X$ has the $AP$ if $id_X$ is in the closure of $\F(X,X)$
in the topology of compact convergence:
$$
 \forall\ \e>0,\ \forall\ \mathrm{compact}\ K\subset X\
 \exists\ R\in\F(X,X):\ \, \sup_{x\in K} ||Rx-x||<\e.
$$                                                    
 \end{defi}
 
\medskip

 \begin{thebibliography}{99}
 
\bibitem{3}                                                                     %ReF
  A.~Grothendieck:                                                               %ReF
  \textit{Produits tensoriels topologiques et \'espaces nucl\'eaires},
 {Mem. Amer. Math. Soc.}, \textbf{16}(1955).
 
 \end{thebibliography}

\smallskip 

%\pause

\begin{thebibliography}{99}

  \bibitem{5} Enflo P. , A counterexample to the approximation property in Banach spaces,
  Acta Math., Volume 130,
   1973, 309--317

             \end{thebibliography}

 \smallskip
   }
            %%%
            
\frame{
 {\color{red}\bf $AP:$ Examples}
 
\begin{exam}
\begin{itemize}
\item
$AP:$\  $C(K), L_p(\mu), A, L_\infty/H^\infty$\ etc;

\item
$\forall\ p\in [1,\infty]\setminus \{2\}\ \, \exists\ X\subset l_p:\ \ X\notin AP;$
\item
$L(H)\notin AP,$\ $H^\infty$ --- not known;

\end{itemize}
\end{exam} 
 
 }            
            
            %%%
            
            \frame{
 {\color{red}\bf A characterization of $AP$}
 
 A. Grothendieck:
 
 \begin{theor}
 The following are equivalent:
 
 1)\,
 Every Banach space has the approximation property.
 
 2)\,
 If a nuclear operator $U: c_0\to c_0$ is
such that $\tr U=1,$ then
$U^2\neq0.$   
 
 \end{theor}
\medskip

%\pause

By Enflo:
          \medskip

\begin{theor}
 There exists a nuclear operator $U: c_0\to c_0$ 
such that $\tr U=1$ and $U^2=0.$   
 
 \end{theor}

}

   %%%
   
\frame{
 {\color{red}\bf Bad nuclear operators in $l^1$}

Can be obtain from Davie's

 \smallskip

 \begin{thebibliography}{99}

  \bibitem{5} Davie A.M., The approximation problem for Banach spaces, Bull. London Math. Soc., Vol 5,
  1973, 261--266

             \end{thebibliography}

\smallskip

\begin{theor}
There exists a nuclear operator $T$ in $l^1:$

(i)\,
$T$ is s-nuclear for every $s\in(2/3, 1].$ 

(ii)\,
$\tr T=1.$  

(iii)\,
$T^2=0.$  
 \end{theor}
 
 \smallskip
A proof can be found in
         \smallskip
         
 \begin{thebibliography}{99}
\bibitem{9} A. Pietsch, Operator ideals,
North-Holland, 1978. 
 \end{thebibliography} 
           \smallskip


 
} 
 
 %%%

 \frame{
 {\color{red}\bf Positive results}
 
 On the other hand:
 \medskip
 
A. Grothendieck: 
          \smallskip
          
\begin{theor}
If $T$ is 2/3-nuclear (in any $X),$ then $\tr T$ is well-defined.
Moreover, if $\tr T\neq0,$ then $T^2\neq0.$
\end{theor}
       \medskip
       
       %\pause
       
V.\,B.~Lidski\v{\i}:

          \smallskip
          
\begin{theor}
If $T: l^2\to l^2$ is 1-nuclear, then $\tr T$ is well-defined.
Moreover, if $\tr T\neq0,$ then $T^2\neq0.$
\end{theor}
       \medskip
           
Can be found in
         \smallskip
          
       \begin{thebibliography}{99}
  \bibitem{5}
 V.\,B.~Lidski\v{\i},
  \textit{Nonselfadjoint operators having a trace},
 {Dokl. Akad. Nauk SSSR}, \textbf{125}(1959), 485--487.
        \end{thebibliography} 
     \smallskip
     
or in A. Pitsch's book.

}
   %%%
  
       \frame{
 {\color{red}\bf Our aim}

 \vskip1.6cm
 
Thus, the cases of nuclear operatots in $c_0,$ $l^1$ and $l^2$ were considered
above, and these are all the cases (in the scale of $l^p$-spaces) 
that have been known to us so far.
\medskip

We are going to consider the cases where $1<p<\infty$ and {\it
to get an optimal results}\, (also in case of $c_0).$   
 }  
   %%%%
   
   \frame{
 {\color{red}\bf Main result: Our Generalization of Zelikin's Theorem}

 {\bf Our main theorems:}
\smallskip

\begin{theor}
%Theorem 1 is optimal with respect to $p$ and $r:$\,
Let $Y$ is a subspace of a quotient
(or a quotient of a subspace) of some
        $L_p(\mu)$-space, $1\le p\le\infty$ and
        $1/r=1+|1/2-1/p|.$                         
If $T: Y\to Y$ is r-nuclear, then $\tr T$ is well-defined.
For a fixed $d=2, 3, \dots,$ the spectrum of $T$
 is $\mathbb Z_d$-symmetric iff
     $$\operatorname{trace}\, T^{kd+j}=0\, \text{ for all }\, k=0, 1, 2, \dots \,
     \text{ and } \, j= 1, 2, \dots, d-1.$$
%$trace\,  T^{2n - 1} = 0, \, n \in \mathbf N.$
In particular, if $\tr T\neq0,$ then $T^2\neq0.$ 
\end{theor}
  }

   \frame{
 {\color{red}\bf Main result: Sharpness}

 Theorem is optimal with respect to $p$ and $r:$
 \smallskip
 
\begin{theor}
  Let  $p\in [1,\infty],  p \neq2, $ $1/r= 1+|1/2-1/p|.$
There exists a nuclear operator $V$ in $l^p$  $($in $c_0$ for $p=\infty)$ such that

$1)$\,
$V$ is $s$-nuclear for each $s\in (r, 1];$

$2)$\,
$V$ is not $r$-nuclear;

$3)$\,
$\operatorname{trace}\, V=1$ and $V^2=0.$
\end{theor}


         % \smallskip
          
Note that for $p=\infty$ we have $r=2/3$  and
for $p=2$ we have $r=1.$  

 
 }
    %%%
   
\frame{
 {\color{red}\bf Our auxiliary theorem}
 
  $L_c(X,Y)$ --- \, $L(X,Y)$ with topology of compact convergence.
 
 %The main thing in constructing an operator in $l^p$ $(1<p\neq2<\infty)$
 %is by using Enflo's construction
% to get a subspace of $l^p$ with:
 Main ingredient for getting $V$ above: % from the title of the talk:
 
 \begin{theor}
% ${(\star)}$\,               
    {\it
 Let $r\in [2/3,1), p\in(2,\infty], 1/r=3/2-1/p.$
There exist a subspace $Y_p$ of the space $l_p$ 
$(c_0$ if $p=\infty),$
 a linear continuous functional $\Psi$ on $L_c(Y_p,Y_p)$ and
 systems $(y_k)\subset Y^*,$ $(y_k)\subset Y$
%$w_p\in Y_p^*\widehat\otimes_1 Y_p$ 
such that
$$
 \sum_{k=1}^\infty ||y'_k||^s\,||y_k||^s <\infty \ \forall\ s>r,
$$ 
$$
  \Psi(U)=\sum_{k=1}^\infty y'_k(Uy_k)\ \forall\ U\in L(Y_p,Y_p),
$$
$$
 \Psi(R)=0\ \forall\ R\in \F(Y_p,Y_p).
$$
Moreover, such situation is impossible for $s=r.$
}
\end{theor}
%\smallskip

} 
            
  %%%
%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\frame{
\LARGE
{\color{red}Thank you for your attention!}
}
\end{document}


%%%%%%%%%%%%%%%%%%%%%
       % REFS
         \frame{
 {\color{red}\bf General notation}
 
 \begin{thebibliography}{99}
 
       \bibitem{1} M. I. Zelikin:
 \textit{A criterion for the symmetry of a spectrum},
Dokl. Akad. Nauk, \textbf{418} (2008), no. 6, 737-740.

      \bibitem{2}
 V.\,B.~Lidski\v{\i}:
  \textit{Nonselfadjoint operators having a trace},
 {Dokl. Akad. Nauk SSSR}, \textbf{125} (1959), 485--487.

\bibitem{3}
 A.~Grothendieck:
 \textit{Produits tensoriels topologiques et \'espaces nucl\'eaires},
{Mem. Amer. Math. Soc.}, \textbf{16} (1955).

\bibitem{1}
Oleg Reinov,
\textit{Some remarks on spectra of nuclear operators},
SPb. Math. Society Preprint 2016-09, 1-9
              \end{thebibliography}
 
}
%%%%%%%%%%       
  
  %%%
General Reference for Definitions, main Questions under consideration etc is
\medskip

 \begin{thebibliography}{99}
 
\bibitem{3}                                                                     %ReF
  A.~Grothendieck:                                                               %ReF
  \textit{Produits tensoriels topologiques et \'espaces nucl\'eaires},
 {Mem. Amer. Math. Soc.}, \textbf{16}(1955).
 
 \end{thebibliography}

\smallskip

%\pause

Our considerations will revolve around the following  question of  A.~Grothendieck:

\begin{quest}
{\it Suppose $T$ is a (bounded linear) operator acting in a Banach space $X.$ 
 Is it true that if $T^*$ is nuclear then $T$ is nuclear too}?
%}
\end{quest}

\begin{itemize}
\item
A. Grothendieck has shown that 
\end{itemize}

\begin{theor}
The answer is positive if  $X^*$ possesses
the approximation property (AP).
\end{theor}
}

\frame{
 {\color{red}\bf Introduction --- $AP_s$}


\vspace{0.5cm}
 %%%%%%%%%%%%%%%%%%%%%%%
 $T\in \mathcal F(X)\subset L(X),$
$
 Tx=\sum_{k=1}^m \langle f'_k,x\rangle g_k;
$
$
\tr T=\sum_{k=1}^m \langle f'_k,g_k\rangle
$
--- does not depend on a representation.

%\pause

Also, $T\in L(X).$\ Consider a {\it nuclear}\, representation
$$
 Tx=\sum_{k=1}^\infty \langle x'_k,x\rangle x_k, \ \sum_{k=1}^\infty ||x'_k||\, ||x_k||<\infty
$$
and 
$$
\alpha:= \sum_{k=1}^\infty \langle x'_k,x_k\rangle.
$$

%\pause

\begin{itemize}

\item 
{\bf Question:}\
$\alpha= \tr T $?

\item
Generally, NO.

\end{itemize}

%\pause

\begin{thebibliography}{99}

  \bibitem{5} Enflo P. , A counterexample to the approximation property in Banach spaces,
  Acta Math., Volume 130,
   1973, 309--317

             \end{thebibliography}

 \smallskip

{\it Remark}: 
If
$T$ is nuclear, then $T: X\to c_0\to l_1 \to X.$

 }

 \frame{

{\it Remark}:
$\alpha= \tr T $ for all $T$ and all nuclear representations of $T$ means $X\in AP.$

%\pause

 \smallskip

Now, let in a nuclear representation of $T$

$$
 Tx=\sum_{k=1}^\infty \langle x'_k,x\rangle x_k,$$
 we have, for $s\le1,$
$$ \sum_{k=1}^\infty ||x'_k||^s\, ||x_k||^s<\infty
$$
($T$ is an {\it $s$-nuclear operator}).
%\pause

Again:

\begin{itemize}

\item 
{\bf Question:}\
$\alpha= \tr T $?
\smallskip

\item
Generally, NO --- for every $s\in (2/3, 1]$.

\end{itemize}

%\pause

 \smallskip

 \begin{thebibliography}{99}

  \bibitem{5} Davie A.M., The approximation problem for Banach spaces, Bull. London Math. Soc., Vol 5,
  1973, 261--266

             \end{thebibliography}

}

\frame{
 {\color{red}\bf $s=2/3$ and $AP_s$}
 
BUT:\ 

\begin{itemize}
  \item
 if $s=2/3,$ then for any above $s$-nuclear representation of $T$
$$
\tr T= \sum_{k=1}^\infty \langle x'_k,x_k\rangle.
$$

\end{itemize}
  
  %\pause
  
 \begin{thebibliography}{99}
 
\bibitem{3}                                                                     %ReF
  A.~Grothendieck:                                                               %ReF
  \textit{Produits tensoriels topologiques et \'espaces nucl\'eaires},
 {Mem. Amer. Math. Soc.}, \textbf{16}(1955).
 
 \end{thebibliography}

%\pause

\begin{itemize}
  \item
{\it Definition}: If
$\alpha= \tr T $ for all $T$ and all s-nuclear representations of $T,$ then $X\in AP_s.$
 
 \end{itemize}
  
  Note that $AP=AP_1.$ Every Banach space has the $AP_{2/3}.$
}

\frame{
 {\color{red}\bf Retreat for Examples}

\begin{exam}
\begin{itemize}
\item
$AP:$\  $C(K), L_p(\mu), A, L_\infty/H^\infty$\ etc;

\item
$\forall\ p\in [1,\infty]\setminus \{2\}\ \, \exists\ X\subset l_p:\ \ X\notin AP;$
\item
$L(H)\notin AP,$\ $H^\infty$ --- not known;

\item
$AP_s:$\  if $s\in [2/3,1], 1/p+1/2=1/s,$ \, then $\forall\ X\subset L_p/E$\, $(E\subset L_p);$

\item
$AP_{2/3}:$\ all.
\end{itemize}
\end{exam}
}

\frame{
 {\color{red}\bf Comparing nuclearity of $T$ and $T^*$}

Now, let $T: X\to X,$\ $\pi: X\hookrightarrow X^{**},$\ so 
$$
\pi T: X\to X \hookrightarrow X^{**}
$$
($T$ is f. r. or any).

\begin{itemize}
  \item 
  {\bf Question:}\
  Suppose 
  $$\pi Tx= \sum_{k=1}^\infty \langle x'_k,x\rangle x''_k,
\  \sum_{k=1}^\infty ||x'_k||\, ||x''_k||<\infty.
$$
Is it true:
  $$ Tx= \sum_{k=1}^\infty \langle y'_k,x\rangle y_k,$$
with $ \sum_{k=1}^\infty ||y'_k||\, ||y_k||<\infty$?

\smallskip

\item
Generally, NOT.\, If $T$ is f. rank, then YES (evidently).
\smallskip

\begin{thebibliography}{99}

  \bibitem{6} Figiel T., Johnson W.B., The approximation property does not imply  the bounded
   approximation property, Proc. Amer. Math. Soc. 41, 197--200 (1973)


             \end{thebibliography}

\end{itemize}

}

\frame{
 {\color{red}\bf Comparing nuclearity of $T$ and $T^*$}


BUT ("quantitative" question):

\begin{itemize}
\item
Let, {\it for a finite rank}\, $T,$
  $$\pi Tx= \sum_{k=1}^\infty \langle x'_k,x\rangle x''_k,$$
$$ \sum_{k=1}^\infty ||x'_k||\, ||x''_k||<1.
$$
Is it true:
  $$ Tx= \sum_{k=1}^\infty \langle y'_k,x\rangle y_k,$$
$$ \sum_{k=1}^\infty ||y'_k||\, ||y_k||<1?
$$
(i.e., if the nuclear norm $\nu(\pi T)<1,$ then $\nu(T)<1$?)

\end{itemize}

\smallskip

Let us give an estimation of type $\nu(T)\le C\, \nu(\pi T).$

}



\frame{
 {\color{red}\bf Comparing nuclearity of $T$ and $T^*$ --- an estimation}


     \begin{itemize}
  \item
  {\bf An estimation.}\
Let $T\in \mathcal F(X), \dim T(X)=N.$  
$$
\pi T: X\to X\hookrightarrow X^{**},\ T(X)\subset X^{**}.
$$
$\exists \ P: X^{**}\overset{\text{onto}}\to T(X)$ with $||P||\le \sqrt N.$
\smallskip

\begin{thebibliography}{99}

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

\end{thebibliography}

\pause

\smallskip

We have:
$$
T=iP\pi T: X \overset{{T}}\to X \overset{{\pi}}\hookrightarrow X^{**} \overset{P}\to T(X)\overset{i}\hookrightarrow X.
$$
Therefore, 
$$
\nu(T)=\nu(iP\pi T)\le ||iP||\, \nu(\pi T)\le \sqrt N\, \nu(\pi T).
$$
OR:

\item
{\it If $\nu(\pi T)<1,$ then $\nu(T)< \sqrt N.$}

 \end{itemize}
 
\smallskip

It is sharp.

}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\frame{
 {\color{red}\bf Comparing nuclearity of $T$ and $T^*$ --- a theorem}

\begin{theor}
%{\bf Theorem}\, 
(O. Reinov)

      \begin{itemize}
  \item
 If $T$ is an n-dimensional operator in a Banach space with nuclear norm
$\nu(T)=1,$ then $\nu(T^*)\ge n^{-\frac12}.$

\item
There exist a separable Banach space $X$ with the AP,
a sequence of operators $(z_n), z_n: X\to X,$ and a constant $C>0$ such that
$\dim z_n=n,  \nu(z_n)\ge \tr z_n=1$ and $\nu(z_n^*)\le C\, n^{-\frac12}.$

\end{itemize}

\end{theor}
%\smallskip


\pause
%\vspace{1cm}

{\it Remark}:\,
The last space $X$ has the AP, but does not have the BAP
(roughly speaking, BAP is a property of a space X meaning that the second statement is not true). 
The first example  was given in

\begin{thebibliography}{99}

  \bibitem{6} Figiel T., Johnson W.B,
The approximation property does not imply  the bounded
   approximation property, Proc. Amer. Math. Soc., 41
(1973), 197--200

             \end{thebibliography}

If we will have time, we will give a sketch of the proof. 

 
}

\frame{
 {\color{red}\bf $s$-nuclear operators -- Applications  de puissance p.\'eme sommable}

\begin{itemize}
  \item
 Recall that an operator $T:X\to Y$ is $s$-nuclear $(0<s\le1)$ if it is of the form
$$
 Tx=\sum_{k=1}^\infty \langle x'_k,x\rangle y_k
$$
for all $x\in X,$ where $(x'_k)\subset X^*, (y_k)\subset Y,\, \sum_k ||x'_k||^s\,||y_k||^s<\infty.$ We use
the notations $N_s(X,Y)$ and $\nu_s(T)$ for
$\inf   (\sum_k ||x'_k||^s\,||y_k||^s)^{1/s}.$
 


 
\pause

In 2014, answering  a question of A.~Hinrichs and A.~Pietsch
 (2010,  [Problem 10.1]), we have found some sharp conditions for a operator 
in Banach spaces to be nuclear, if its adjoint is s-nuclear 
($0< s < 1$). 


  \item
\begin{thebibliography}{99} 

  \bibitem{6} A. Hinrichs, A. Pietsch, $p$-nuclear operators in the sense of Grothendieck,
 Math. Nachr., Volume 283, No. 2 (2010), 232--261.
 
 
             \end{thebibliography}
             
 Recall a part of the conditions.

\end{itemize}
}


\frame{
 {\color{red}\bf When s-nuclearity of $T^*$ implies nuclearity of $T$?}


\begin{theor}
Let $\, s\in (0,1],$ $ T\in L(X)$
and assume that 
$\, X^*\in \,AP_s. $ 
If $ \pi T\in N_s(X, X^{**}),$
then $T\in N_1(X,X).$ 
\end{theor}
\smallskip

In other words, under these conditions,
from the $ s$-nuclearity of the conjugate operator $ T^*,$ it follows
that the operator $ T$ is nuclear.

\begin{thebibliography}{09}

%\bigskip
\medskip

  \bibitem{21} O.~I. Reinov, On linear operators with $s$-nuclear adjoints, $0<s\le1,$
  J. Math. Anal. Appl., Volume 415 (2014) 816-824.
 \end{thebibliography}
 
It was shown also that the above condition is sharp. 
Now, we present  some finite dimensional analogues 
of these results.

But before, we give a new "quantitative" version of the above theorem.
}


\frame{
 {\color{red}\bf When s-nuclearity of $T^*$ implies nuclearity of $T$?}


\begin{theor}
Let $\, s\in (0,1],$ $ T\in L(X)$
and assume that 
$\, X^*\in \,AP_s. $ 
If $ \pi T\in N_s(X, X^{**})$ and $\nu_s(\pi T)<1,$
then $T\in N_1(X,X)$ and $\nu(T)<1.$
\end{theor}
\smallskip

\pause

In other words, under these conditions,

\begin{itemize}
  \item
From the fact that 
 the conjugate operator $ T^*$ 
 lies in the "unit" ball of the space of all s-nuclear operators,
 it follows
that the operator $ T$ is nuclear and belongs to the unit ball
of the space of all nuclear operators, too..
\end{itemize}
  


}



%----------------------


\frame{
 {\color{red}\bf Finite rank operators}

The last theorem can be applied, in particular, to the case of finite rank operators.
So, in the case where $X^*$ has the $AP_s,$ the situation, like 
a $n^{-1/2}$-situation in one of the above theorem, is not possible.
However, in general case things are not so good.

Indeed, we can prove (getting one way estimation):

\begin{theor}
Let $s\in [2/3, 1].$
      \begin{itemize}
  \item
 If $T$ is an n-dimensional operator in a Banach space with nuclear norm
$\nu(T)=1,$ then $\nu_s(T^*)\ge n^{1/s-3/2}.$
\end{itemize}
\end{theor}
\smallskip

The proof is not so simple as in the case $s=1$ (above). We do not give it,
but consider the limit cases $s=1$ and $s=2/3.$ But before this, let us mention what we
use in the general proof.



}


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



\frame{
 {\color{red}\bf Finite rank operators (continued)}

In the proof we use, in particular,

\begin{itemize}
  \item
factorization Grothendieck technique;

\item
duality results for operator ideals;

\item
finite dimensional estimations of p-summing operators;

\item
duality due to A. Grothendieck.

\end{itemize}

Now, let $s=1.$ Then we have $1/s-3/2=-1/2,$ and this is the case, which was
considered before.

Finally, let $s=2/3,$ so that $1/s-3/2=0$ (in Estimation, $n^0=1).$
Write $\pi T$ as
$$
 \pi T=\sum_{k=1}^\infty  \mu_k\, x'_k\otimes x''_k,
$$
 where $\mu_k\ge0, \, (x'_k)\subset X^*, (x''_k)\subset X^{**},\, \sum_k \mu_k^{2/3}<1,\  ||x'_k||=1,\,||y_k||=1.$ 

Putting $\alpha_k:=\mu_k^{2/3}$ and $\beta_k:= \mu_k^{1/3},$ we factorize $\pi T$ as
$$
\pi T:  X\overset{A}\to l_\infty \overset{\Delta_\alpha}\to l_1 \overset{j} \to l_2 \overset{\Delta_\beta}\to l_1
\overset{B}\to X^{**}.
$$

}

\frame{
 {\color{red}\bf Finite rank operators - $\nu_{2/3}$ (continued)}


Putting $\alpha_k:=\mu_k^{2/3}$ and $\beta_k:= \mu_k^{1/3},$ we factorize $\pi T$ as
$$
\pi T:  X\overset{A}\to l_\infty \overset{\Delta_\alpha}\to l_1 \overset{j} \to l_2 \overset{\Delta_\beta}\to l_1
\overset{B}\to X^{**}.
$$
Here,

\begin{itemize}
  \item
$Ax:= (\<x'_k, x\>)\in l_\infty;$


\item
$\Delta_\alpha (a_k):= (\alpha_k\, a_k)\in l_1;$ 

\item
$j$ is the natural embedding;

\item
$\Delta_\beta (b_k):= \sum \beta_k b_k\, x''_k \in X^{**}.$

\end{itemize}

$\nu(j \Delta_\alpha A)<1;$\
$\pi T(X)\subset X\subset X^{**},$

Take a projector $P: l_2 \underset {\text{onto}}\to E:=(B \Delta_\beta)^{-1}(T(X))$ with norm 1.

Then 
$$T: 
X\overset{A}\to l_\infty \overset{\Delta_\alpha}\to l_1 \overset{j} \to l_2 \overset{P}\to E 
\overset{B\Delta_\beta|_{E}}\to X.
$$


}

\frame{
 {\color{red}\bf Finite rank operators - $\nu_{s}$}

{\it Remark}:\
One can consider the last theorem as an "interpolation theorem" between $s=2/3$ and $s=1.$

\pause

\medskip

The following can be seen as preparation for getting the sharpness of the last theorem.

\begin{theor}
%Let $s\in (2/3, 1]$      %, q\in [2,\infty),$ $1/q= 3/2 -1/s.$
  There exist a subspace $Y$ of the space $c_0$ and a finite rank operators $z_n, n=1,2,\dots,$ in $Y$
such that 

    \begin{itemize}
  \item
  $Y$  does not have the $AP_r$ for every $r\in (2/3, 1];$
  
  \item
  $\dim z_n=n$\, and $\tr z_n = 1,$ $n=1,2, \dots;$
  
  \item
 for all $s\in (2/3,1],\, \delta>0$\ \,  $\exists\ C_\delta>0:$\ $\nu_s(z_n)\le C_\delta\, n^{1/s-3/2+\delta}.$
  
\end{itemize}


\end{theor}
\smallskip

[
RAPPEL:\   If $T$ is an n-dimensional operator in a Banach space with nuclear norm
$\nu(T)=1,$ then $\nu_s(T^*)\ge n^{1/s-3/2}.$
]
}


%\end{document}

\frame{
 {\color{red}\bf Finite rank operators - $\nu_{s}$ (continued)}


\begin{theor}
Let $s\in (2/3, 1], q\in [2,\infty),$ $1/q= 3/2 -1/s.$
  There exist a separable reflexive Banach space $Y$ and a finite rank operators $z_n, n=1,2,\dots,$ in $Y$
such that 

    \begin{itemize}
  \item
  $Y$  (as well as $Y^*)$ has the $AP_r$ for every $r<s;$
  
  \item
  $Y$ does not have the $AP_s;$
  
  \item
  $\dim z_n=n$\, and   $\tr z_n = 1,$ $n=1,2, \dots;$
  
  \item
  $\nu_s(z_n)\le \frac{C}{log (n+1)}.$
  
\end{itemize}

Moreover, $Y\subset \(\sum_N l^{3\cdot 2^N}_{q_N}\)_{l_q},$ where $q_N\searrow q.$
\end{theor}
\smallskip

\pause

{\it Remark}:\
We have a nice "by-product consequence" of Theorem.

}

\frame{
 {\color{red}\bf An unexpected application}


For $q=2$ (that is, $s=1)$, the space $Y$ is a subspace of the space of type
$\left(\sum_j l_{p_j}^{k_j}\right)_{l_2}$ with $p_j\searrow 2$ and $k_j\nearrow \infty.$ 
Every such space is an asymptotically Hilbertian space (for definitions and some discussion,
see 
%\end{thebibliography}
\begin{thebibliography}{99}
\bibitem{1}  P.~G. Casazza,  C.~L. Garc\'{\i}a,  W.~B. Johnson,
An example of an asymptotically Hilbertian space  which fails the approximation property,
Proc. Amer. Math. Soc., Volume 129, No. 10 (2001), 3017-3024.
\end{thebibliography}
). So, we got:
\smallskip

\begin{coro}%{\bf Corollary.}\
There exists an asymptotically Hilbertian space without the Grothendieck approximation property.
 \end{coro}

}

\frame{
 {\color{red}\bf Reminding.}


RAPPEL:\  


A Banach space $X$ is said to be {\it asymptotically
Hilbertian} provided there is a constant K so that for every m there exists n so
that X satisfies:
 there
is an n-codimensional subspace $X_m$ of X so that every m-dimensional subspace
of X
m is K-isomorphic to 
$l_2^m.$


}


%----------------------------------------------------------------------------------------------
\frame{
 {\color{red}\bf An unexpected application (continued)}

 First example of an asymptotically Hilbertian space without the Grothendieck approximation property
  was constructed (by O. Reinov) in 1982 in
 
 % \end{thebibliography}
\begin{thebibliography}{99}
  \bibitem{10} O. I. Reinov, Banach spaces without approximation property, 
Functional Analysis and Its Applications, Volume 16,
No. 4  (1982), 315-317. 
\end{thebibliography}
 where A. Szankowski's results were used
 
 Later,  in 2000, by applying Per Enflo's example in a version of  A.M. Davie,        %R!!!
 P.~G. ~Casazza,   C.~L. ~Garc\'{\i}a and   W.~B. ~Johnson 
 gave another example of an asymptotically Hilbertian space  which fails the approximation property.

 We here, not being searching for an example of such a space, have got it (accidentally)
 by using the construction from  
 \begin{thebibliography}{99}
\bibitem{9} A. Pietsch, Operator ideals,
North-Holland, 1978. 
 \end{thebibliography}
}

\frame{
 {\color{red}\bf $\nu_s(\text{"adjoint operator"})$}

Recalling:
 \begin{itemize}
 \item
 Let $\, s\in (0,1],$ $ T\in L(X)$
and assume that 
$\, X^*\in \,AP_s. $ 
If $ \pi T\in N_s(X, X^{**})$ and $\nu_s(\pi T)<1,$
then $T\in N_1(X,X)$ and $\nu(T)<1.$
  \item
 Let $\, s\in (2/3,1].$ If $T$ is an n-dimensional operator in a Banach space with nuclear norm
$\nu(T)=1,$ then $\nu_s(T^*)\ge n^{1/s-3/2}.$
%\item

\end{itemize}

We have, finishing a talk and in particular:

\begin{theor}
 There exist a Banach space $W$  and a finite rank operators $z_n, n=1,2,\dots,$ in $W$
such that 

    \begin{itemize}
    \item
    $W\in AP;$
  \item
  $W^*$  does not have the $AP_r$ for every $r\in (2/3, 1];$
  
  \item
  $\dim z_n=n$\, and $\tr z_n = 1,$ $n=1,2, \dots;$
  
  \item
 for all $s\in (2/3,1],\, \delta>0$\ \,  $\exists\ C_\delta>0:$\ $\nu_s(z^*_n)\le C_\delta\, n^{1/s-3/2+\delta}.$
  
\end{itemize}
\end{theor}

}



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

\frame{
\LARGE
{\color{red}
Thank you for your attention!
}
}
\end{document}


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




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

\frame{
 {\color{red}\bf Example we use}

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

\pause

        \item
%\vskip 0.1cm
For $q=2$ (that is, $r=1)$, the space $Y_0$ is a subspace of a space of  the type
$\left(\sum_j l_{p_j}^{k_j}\right)_{l_2}$ with $p_j\searrow 2$ and $k_j\nearrow \infty.$
Every such space is an asymptotically Hilbertian space (for definitions and some discussion,
see

   \begin{thebibliography}{99}

  \bibitem{1}  P.~G. Casazza,  C.~L. Garc\'{\i}a,  W.~B. Johnson,
An example of an asymptotically Hilbertian space  which fails the approximation property,
Proc. Amer. Math. Soc,, Volume 129, No. 10 (2001), 3017-3024.


             \end{thebibliography}
).

\end{itemize}

}




%%%%%%%%%%       ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
\frame{
 {\color{red}\bf Reference}


%\bigskip
%\bigskip
%\medskip

\begin{thebibliography}{09}

%\bigskip
\medskip

  \bibitem{21} O.~I. Reinov, On linear operators with $s$-nuclear adjoints, $0<s\le1,$
  J. Math. Anal. Appl., Volume 415 (2014) 816-824.



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

\end{thebibliography}

}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\frame{
\LARGE
Thank you for your attention!
}
\end{document}


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