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\def\AP{\operatorname{AP}}
\def\BAP{\operatorname{BAP}}
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\def\I{\operatorname{I}}
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\def\QN{\operatorname{QN}}
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% Copyright 2003 by Till Tantau <tantau@cs.tu-berlin.de>.
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% The purpose of this example is to show how \part can be used to
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\newtheorem{quest}[theorem]{Question}
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\newtheorem{exam}[theorem]{Examples}


\title{Finite dimensional aspect of existence of non-nuclear operators with s-nuclear adjoints
}
\author{Oleg Reinov}
\institute{
 %Department of Applied Mathematics and Control Processes,
  Saint Petersburg State University}
  \date{}

\begin{document}


\frame{\titlepage
%%\pause
%\vspace{-1.3cm}%\hspace{3cm}
%\includegraphics[width=4cm]{3-adic_disk.png}
}
%\section{p-Adic Multiresolution analysis}








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

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


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

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}

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\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}

}




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 {\color{red}\bf Reference}


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\begin{thebibliography}{09}

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



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\end{thebibliography}

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