%%%%%%%%%% % \input trtrtrtr.tex %%%%%%%%----------------------------------- %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%% after Kalinin 1985 Primenen. FA v t. pribl. %%%% 28.03.2012 9:16:24 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% poisk voprosov po !!! \def\nor#1{||{#1}||} % \def\md#1{|{#1}|} % \def\sp#1#2{\(#1,#2\)} % \def\ove#1{\overline{#1}} % \def\ovs#1#2{\overset{#1}\to{#2}} \def\({\left(} \def\al{\alpha} \def\lee{\leqslant} \def\){\right)} \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\det{\operatorname{det}\,} %%%%%%%%%%%%% after 30.01.00 02:44:40 Sat: \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\reg{\operatorname{reg}} \def\dual{\operatorname{dual}} \def\sbs{\subset} %%%%%%%%%%%%%%%%%%%%%%%%% % \def\{\quad\blacksquare} \def\med{\medpagebreak} \def\QQ{$\quad\square$} \def\small{\smallpagebreak} \def\Q{\quad\blacksquare} \def\bigp{\bigpagebreak} \def\f{\vec} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \documentclass[12pt,oneside]{amsproc} \usepackage[T2A]{fontenc} %\usepackage[cp1251]{inputenc} \usepackage[english]{babel} %\usepackage[russian]{babel} \usepackage{amssymb} \usepackage{amsthm} \textheight=22cm \textwidth=15cm \oddsidemargin=5mm \topmargin=-5mm \title{Some more remarks on Grothendieck-Lidski\v{\i} trace formulas} %{} \author{Oleg Reinov} \address{Department of Mathematics and Mechanics, St. Petersburg State University, Saint Petersburg, RUSSIA.\newline \phantom{Ao} Abdus Salam School of Mathematical Sciences, Government College University, Lahore, PAKISTAN. } %\address{Abdus Salam School of Mathematical Sciences, 68-B, New Muslim Town, Lahore 54600, PAKISTAN.} \email{orein51@mail.ru} %\author{Qaisar Latif} %\address{Abdus Salam School of Mathematical Sciences, 68-B, New Muslim Town, Lahore 54600, PAKISTAN.} \thanks{%${ }^\maltese$ The research is supported by the Higher Education Commission of Pakistan and by grant 12-01-00216 of RFBR.} \thanks{%${ }^\maltese$ AMS Subject Classification 2010: 47B06. } \thanks{${ }$ Key words: $(s,p)$-nuclear operators, eigenvalue distributions. } \begin{document} $$ {} $$ \vphantom{} \maketitle \begin{abstract} {\bf Theorem:}\, {\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 corresponding operator $\wt{u}$ $($written according to their algebraic multiplicities$)$ is absolutely summable and $$ \tr u=\sum_k \mu_k. $$ } One of the main aim of these notes is not only to give a proof of the theorem but also to show that it could be obtained by A. Grothendieck in 1955. \end{abstract} \vskip 0.75cm In 1955, A. Grothendieck [4] has shown that if the linear operator $T$ in a Banach space is $2/3$-nuclear then the trace of $T$ is well defined and is equal to the sum of all eigenvalues $\{\mu_k(T)\}$ of $T.$ \, V.B. Lidski\v{\i} [8], in 1959, proved his famous theorem on the coincidence of the trace of the $S_1$-operator in an (infinite dimensional) Hilbert space with its spectral trace $\sum_{k=1}^\infty \mu_k(T).$ % Any Banach space is a subspace of an $L_\infty(\nu)$-space, as well as any Hilbert space % is a (subspace of) $L_2(\nu)$-space. Also, any Banach space is a factor space of an $L_1(\nu)$-space, % as well as any Hilbert space is a (factor space of) $L_2(\nu)$-space. %We obtain the following generalization of these theorems: %{\it %for $p\in[1,\infty]$ and $s\in (0,1]$ with $1/s=1+|1/2-1/p|,$ %and for every $s$-nuclear operator $T$ in every subspace of any $L_p(\nu)$-space %the trace of $T$ is well defined and equals the sum of all eigenvalues of $T.$} %Note that for $p=2$ one has $s=1,$ and for $p=\infty$ one has $s=2/3.$ \vskip 0.08cm %in 1955, A. Grothendieck [1] has shown that if the linear operator $T$ %in a Banach subspace of an $L_\infty$-space is $2/3$-nuclear then %the trace of $T$ is well defined and is equal the sum of all eigenvalues $\{\mu_k(T)\}$ %of $T.$ %V.B. Lidski\v{\i} [2], in 1959, proved his famous theorem on the coincidence of the trace of an $S_1$-operator %in $L_2(\nu)$ with its spectral trace $\sum_{k=1}^\infty \mu_k(T).$ In 1970's and in early 1980's, the interest to the trace formulas (and, generally, to the distribution of eigenvalues of some classes of operators) has been increased (A. Pietsch, H. K{\"o}nig and others). The trace formula was established for such ideals of operators as $\mathfrak L_1^{app},$\, $\mathfrak{P}_2\circ \mathfrak{P_2},$\, $(\mathfrak{P}_2)^{app}_{2,1}$\, $\mathfrak L_1^{gel},$\, $\mathfrak L_1^{kol},$\, $\mathfrak L_1^{weil},$\, $\mathfrak L_1^{ent}$\, (see [12], p. 404). In the book [10] by A. Pietsch, one can find a generalization of Grothendieck-Lidski\v{i} theorem to the case of the quasinormed operator ideal $N_{1,1,2}$ of the so called (1,1,2)-nuclear operators (see [10], Th. 27.4.11). %!!! In 1996, M. White [15] has obtained a very general theorem on the spectral trace for a wide classes of quasi-normed operator ideals. What about concrete Banach spaces, it was shown recently by Oleg Reinov and Qaiser Latif [14] that the Grothendieck-Lidski\v{i} formula can be "interpolated" between $L_\infty$--$L_2$ (or, between $L_1$--$L_2)$ cases. More precisely, they have shown that for $p\in[1,\infty]$ and $s\in (0,1]$ with $1/s=1+|1/2-1/p|,$ and for every $s$-nuclear operator $T$ in every subspace of any $L_p(\nu)$-space the trace of $T$ is well defined and equals the sum of all eigenvalues of $T.$ The same is true for quotients of $L_p(\nu)$-spaces. Note that for $p=2$ one has $s=1,$ and for $p=\infty$ one has $s=2/3.$ In this note, we are going to give some more examples of such a kind %!!! (see Theorem below). Let us mention that the proof of the theorem consists of a "reconsideration" of some Grothendieck's arguments from [4], Chap. II, of proving by him his famous trace formula in the case of 2/3-nuclear operators. In the case where $X=H$ is a Hilbert space and $T\in S_1(H)$ (nuclear case; so, $p=2$ in the theorem below), our theorem gives the Lidski\v{i} formula; in the case where $X$ is any Banach space and $T$ is 2/3-nuclear (so, $p=+\infty$ in the theorem below), we obtain the Grothendieck 2/3-theorem (with an analogues proof!). If $X$ is any and $p=2$ in our theorem, we obtain the above mentioned $N_{1.1.2}$-result. We give also (after the proof of the theorem) some new consequences and make some remarks on Grothendieck's considerations in Chapter II of his famous work [4]. Let us note now only that, in particular, A. Pietsch writes (concerning Lidski\v{i}'s 1959 formula) in the book [12], p. 404: %!!! "a remark in [GRO1, Chap. II, p. 13] indicates that by 1955, Grothendieck was aware of this fact". We will give a citation from [4], which shows that A. Grothendieck (in 1955) was aware of a more stronger result than the Lidski\v{i} theorem (but the result was given there without any proof). Surely, that work by A. Grothendieck was unknown to V. Lidski\v{i}, so, the famous Lidski\v{i}'s formula is Lidski\v{i}'s formula forever. % We will discuss some of the corresponding results and will give some generalizations % of Grothen\-dieck-Lidski\v{\i} trace formulas, "interpolating"\, between subspaces of $L_\infty$- % and $L_2$-spaces. \vskip 0.08cm % We were just analyzing some proofs given by A. Grothendieck [Gr1, Ch.2] for his trace formula %for $2/3$-nuclear operators and have obtained a generalization below. \vskip 0.33cm \centerline{\bf \S1. Preliminaries and a theorem} \vskip 0.23cm %{\bf I.}\. All the terminology and facts (now classical), given here without any explanations, can be found in [1, 2, 4, 7, 10, 11]. %!!! \vskip 0.02cm Let $X,Y$ be Banach spaces. For the Banach dual of $X,$ we use the notation $X^*.$ If $x\in X$ and $x'\in X^*,$ then we use the notation $\$ for $x'(x).$ Denote by $X^*\wh\ot Y$ the completion of the tensor product $X^*\ot Y$ (considered as a linear space of all finite rank operators from $X$ to $Y$) with respect to the projective norm $$ ||w||:= \inf \{\(\sum_{k=1}^N ||x'_k||\, ||y_k||\):\ w=\sum_{k=1}^N x'_k\ot y_k\} $$ (see, e.g., [4], [1]). For $X=Y,$ the natural linear continuous functional "trace" on $X^*\ot X$ has a unique continuous extension to the space $X^*\wh\ot X,$ which we still will denote by "trace". Put $N(X,Y):= $ image of $X^*\wh\ot Y$ in the space $L(X,Y)$ of all bounded linear transformations under the canonical factor map $X^*\wh\ot Y\to N(X,Y)\sbs L(X,Y).$ We consider the (Grothendieck) space $N(X,Y)$ of all nuclear operators from $X$ to $Y$ with the natural norm, induced from $X^*\wh\ot Y.$ For a tensor element $u\in X^*\wh\ot Y,$ we denote by $\wt{u}$ the corresponding nuclear operator from $X$ to $Y.$ For $q\in (0,+\infty],$ we denote by $l_q^w(X)$ the space of all weakly $q$-summable sequences $(x_i)\sbs X$ (see, e.g., [9], [10]) with a quasi-norm $$ \e_q((x_i)) := \sup \{\(\sum_i |\|^q\)^{1/q}:\ x'\in X^*,\, ||x'||\le1\} $$ (in the case where $q=\infty,$ we suppose $(x_i)$ to be just bounded and tending to zero, i.e., $\e_\infty((x_i))=\sup_i ||x_i||$). We are going to prove {\bf Theorem.}\, {\it Let $r\in (0,1], 1\le p\le2,$ $u\in X^*\wh\ot X$ and $u$ admits a representation $$ 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. $$ } \vskip 0.1cm We obtained this result rather casually, just analyzing the arguments, given by A. Grothendieck [4, Ch. II] for getting his trace formula for $2/3$-nuclear operators, and noting that Hadamard's inequality for determinants may be improved in some $L_p$ situations (this idea appeared after considerations again of arguments from [14] and the facts that the Hilbert spaces are the best Banach spaces, but the Banach spaces of type $L_p$ for $p\in(1,\infty)$ are, maybe, worse than an H but better that any $X$ (or, the same, in a sence, than $L_\infty$)). % and have obtained a generalization below. In the proof of Theorem, we shall use, in particular, the "related operators theorem" [10, p. 375], %!!! namely, in the following situation. If $u$ is as in Theorem then it is easy to see that it admits a factorization $$ u=AB:\ X\to l_p\to X, $$ where $B$ is $s$-nuclear (is generated by "un noyau de Fredholm de puissance s.\`{e}me sommable dans $X^*\wh\ot l_p$" in terms of [4]), $A$ maps the unit vector basis of $l_p$ to the sequence $(x_i)$ (which is weakly $p'$-summable). Therefore, the set of all eigenvalues of $u$ is the same as the set of all (with their algebraic multiplicities) eigenvalues of the operator $BA,$ which maps $l_p$ into $l_p$ (so, results of [6] and [14] may be applied). %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \vskip 0.33cm \centerline{\bf \S2. Proofs} \vskip 0.23cm %{\bf II.}\, Let $u$ be an element of the projective tensor product $X^*\wh\ot X.$ It can be represented in the form $$ u=\sum_i \lambda_i x'_i\ot x_i, $$ where $(\la_i)\in l_1$ and $||x'_i||\le1,$ $||x_i||\le1$ (see [4], [1]). Recall that the Fredholm determinant $\det (1-zu)$ of $u$ (see [4], [5], [10], [11]) is an entire function $$ \det (1-zu)= 1-z\, \tr u+ \dots + (-1)^n z^n\alpha_n(u) + \dots, $$ all zeros of which are exactly (according to their multiplicities) the inverses of nonzero eigenvalues $(\mu_k)$ of the operator $\wt{u},$ associated with the tensor element $u.$ If $u$ has a form $u=\sum_i \lambda_i x'_i\ot x_i$ as above, the coefficients $\alpha_n(u)$ in the previous formula are defined explicitly by $$ \alpha_n= \sum_{i_1<\dots)_{1\le \alpha,\beta\le n} $$ (see [4, Chap. II, p.13, (5bis)], [5]). Suppose now, that $u$ has a representation $$ u=\sum_i \lambda_i x'_i\ot x_i, $$ with $(\la_i)\in l_r,$ $\la_i\ge0,$ $r\in(0,1],$ $||x'_i||\le1,$ $(x_i)\in l_{p'}^w(X),$ $ \e_{p'}((x_i))\le1$ (here $1\le p\le2).$ We have: $$ f(z):= \det (1+zu)= \sum_{n=0}^\infty \alpha_n(u)\, z^n, $$ where $\alpha_n(u)$ are as above; therefore, taking in account that for every $\alpha=1,\dots, n$ $$ \(\sum _{\beta=1}^n |\|^{p'}\)^{1/p'}\le1 $$ and thus $$ \(\sum _{\beta=1}^n |\|^{2}\)^{1/2}\le n^{1/p-1/2}, $$ by Hadamard's inequality for determinants (see, e.g., [16], 8.7.4 Problems and Exercises, Ex. 9c), or [2, p. 1018]), %!!! %!!! Vladimir A. Zorich, Mathematical Analysis I, Springer-Verlag Berlin Heidelberg 2004, 597 pages %!!! Dunford, N., Schwartz, J. T. [DUN] Linear operators, vol. I, New York—London 1958 we get $$ |\alpha_n(u)|\le n^{n(1/p-1/2)}\, \alpha_n(\lambda), $$ where $$ \alpha_n(\la)=\sum_{i_1<\dotsr,$ $$ \alpha_n(\la)\le M_t n^{-n/t} $$ (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 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % \endcomment