r,$
$$
\alpha_n(\la)\le M_t n^{-n/t}
$$
(see the same book [7], p. 6).
Hence,
$$
|\alpha_n(u)|\le M_t n^{-n(1/t-1/p+1/2)}=M_t n^{-n/\omega},
$$
where $1/\omega=1/t+1/2-1/p.$
By a classical result of Hadamard (see, e.g., [7], pp. 5-6),
%!!! B. Ya. Levin, Lectures on Entire Functions, Translations of Mathematical Monographs, Volume 150
%!!! American Mathematical Society, Providence, Rhode Island 1996
the function $f(z)$ is of order $\le\omega$ and, therefore, of order $\le \nu,$
where $1/\nu=1/r+1/2-1/p$ (since $t>r$ was arbitrary).
Now, suppose that $\nu=1$ (that is, $1/r+1/2-1/p=1).$
By Hadamard (see [7], p. 26, Th. 1),
$$\det (1-zu)=e^{-az}\, \prod_i (1-z\mu_i)e^{z\mu_i}$$
(recall that $(\mu_k)$ is a sequence of all eigenvalues of $\wt{u},$ counted according to their algebraic
multiplicities).
On the other hand, as was said above,
$$
\det (1-zu)= 1-z\, \tr u+ \dots + (-1)^n z^n\alpha_n(u) + \dots,
$$
and we get (considering the expansion of the entire function $e^{-az}\, \prod_i (1-z\mu_i)e^{z\mu_i}$) %!!! expan?
that $a=\tr u.$ Therefore,
$$\det (1-zu)=e^{-z\,\tr u}\, \prod_i (1-z\mu_i)e^{z\mu_i}.$$
Now we apply Theorem 2.6 of [6] or results from [14]
%!!! W. B. Johnson, H. K"onig, B. Maurey, and J. R. Retherford, Eigenvalues of p- summing and lp-type operators in Banach spaces, J. Funct. Anal. 32:353-380 ( 1979).
% Th. 2.6 on Lp Ns kak u nas
to get that $(\mu_k)\in l_1,$ from which it follows (see, e.g., [7], p. 25-26) that
$$%\text{1. }\
\det (1-zu)=e^{-\alpha z}\, \prod_i (1-z\mu_i),\ \, \text{where }\ \alpha=\tr u-\sum_k \mu_k$$
%where $\alpha=\tr u-\sum_k \mu_k.$
and
$$
%\text{2. }\
\text{the function }\ \det (1-zu) \ \text{ is of minimal type}
$$
%Finally, since (by the same Hadamard's theorem) the function $\det (1-zu)$
%is of minimal type
(by the same Hadamard's theorem; see also [7], pp. 25-26 or the second part of the proof of Borel theorem in [7], p. 30).
Whence, $\alpha=0,$ i.e.
$\tr u=\sum_k \mu_k.$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vskip 0.33cm
\centerline{\bf \S3. Corollaries and remarks.}
\vskip 0.23cm
%\vskip 0.1cm
{\bf Corollary 1.}\it \,
Let $r, p, u$ be as in Theorem. The operator $\wt{u}: X\to X$ is equal to zero iff the tensor element $u$ is zero.
\rm
\vskip 0.1cm
The same proof as the one of Theorem, with evident changes, gives us
\vskip 0.1cm
{\bf Corollary 2.}\it \,
Let $r\in (0,1], 1\le p\le2,$ $u\in X^*\wh\ot X$ and $u$ admits a representation
$$
u=\sum_i \lambda_i x'_i\ot x_i,
$$
with $(\la_i)\in l_r,$ $(x_i)$ bounded and $(x'_i)\in l_{p'}^w(X^*).$
If $1/r+1/2-1/p=1,$ then the system $(\mu_k)$ of all eigenvalues of the operator $\wt{u}$
(written according to their algebraic multiplicities) is absolutely summable and
$$
\tr u=\sum_k \mu_k.
$$
\rm
\vskip 0.1cm
{\bf Corollary 3.}\it \,
Let $r, p, u$ be as in the previous corollary. The operator $\wt{u}: X\to X$ is equal to zero
iff the tensor element $u$ is zero.
\rm
\vskip 0.1cm
{\it Remark.}\,
For the case where $r=2/3$ and $p=\infty,$ we get 2/3-theorems of A. Grothendieck ([4]; for a simple proof
of the 2/3-theorems, see [13]). For the case $r=1$ and $p=2,$ we get the $N_{1,1,2}$-results of [10, p. 381]. %!!!
Also, Corollaries 1 and 3 are valid if we consider the operators $\wt{u}$ from $X$ to $Y,$ for any Banach $X,Y.$
\vskip 0.1cm
As was said above, in our proof we just used the ideas of A. Grothendieck from [4].
Let us mention that our Theorem could be proved by A. Grothendieck in 1955,
as well as the Lidski\v{i}'s result. Namely, in [4, Ch. II, Remark 4, p. 21], A. Grothendieck writes: %!!!
\vskip 0.1cm
"{\it Soit $0 =
\sum_m \< w_m, \Psi f_m\>= \tr \Psi A.
$$
\begin{multline}
\tr U\circ z= \tr \( \sum j^*(\mu_n)\ot Uy_n\)=
\sum \< j^*(\mu_n), Uy_n\> =\\ =
\sum \< \mu_n, jUy_n\>=
\tr jU\Phi_K^{**}\Psi=\tr (jU\Phi_K)^{**}\Psi, \label(3)
\end{multline}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vskip 0.1cm
{\bf Corollary 1.}\it \, $ (\R,\tau_p)'=(\R,\sigma)',$
where
$ \sigma=\sigma(\R, X^*\wh\ot_{p'} Y).$
Thus, the closures
of convex subsets of the space
$ \Pi_p(Y, X)$
in
$ \tau_p$
and in
$ \sigma$
are the same.
%$\quad\blacksquare$
\rm
\vskip 0.1cm
{\bf Proposition 2} [3]. \it
If the canonical mapping
$ j: X^*\wh\ot_{p'} Y\to \N_{p'}(X,Y)$
is one-to-one then
$ \Pi_p(Y,X)= \ove{Y^*\ot X}^{\,\tau_p}.$
\vskip 0.1cm
\rm
{\it Proof}
If the map
Therefore,
$ \Pi_p(Y,X)= \ove{Y^*\ot X}^{\,\tau_p}.$
\vskip 0.1cm
For a reflexive space
$ X,$
the dual space to
$ X^*\wh\ot_{p'} Y$
is equal to
$ \Pi_p(Y,X).$
Consequently, it follows from the last two statements
\vskip 0.1cm
{\bf Corollary 2.}\it \, For a reflexive space
$ X$
the canonical mapping
$ j: X^*\wh\ot_{p'} Y\to \N_{p'}(X,Y)$
is one-to-one iff the set of finite rank operators is dense in the space
$ \Pi_p(Y,X)$
in the topology
$ \tau_p$\ of $ \pi_p$-compact convergence. \rm
\vskip 0.1cm
\vskip 0.1cm
{\bf Theorem.}\it \, Let $1\le q\neq2 \le\infty.$ Then there is a $($reflexive$)$ Banach space
that fails the approximation property of type $q$
of $[1].$
\rm
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% \endcomment