%%%%%%%%%% % %%%%%%%%----------------------------------- %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%% after MNachr 2013 Q-R %%%% âîñêðåñåíüå, Ôåâðàëü 17, 2013 19:17 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% poisk voprosov po !!! %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \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{Grothendieck-Lidski\v{\i} theorem for subspaces of quotients of $L_p$-spaces } %{} \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, 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.} \email{qsrlatif87@yahoo.com} \thanks{%${ }^\maltese$ The research was supported by the Higher Education Commission of Pakistan.} \thanks{%${ }^\maltese$ AMS Subject Classification 2010: 47B06. } \thanks{${ }$ Key words: $s$-nuclear operators, eigenvalue distributions. } \begin{document} $$ {} $$ \vphantom{} \maketitle \begin{abstract} Generalizing A. Grothendieck's (1955) and V.B. Lidski\v{\i}'s (1959) trace formulas, we have shown in a recent paper that {%\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 $T.$ } \ Now, we obtain the analogues results for subspaces of quotients (equivalently: for quotients of subspaces) of $L_p$-spaces. \end{abstract} \vskip 0.75cm In the note [13], we have proved that if $p\in[1,\infty]$ and $1/s=1+|1/2-1/p|,$ then for any subspace (or quotient) of an $L_p$-space and for every $s$-nuclear operator $T$ in the space the nuclear trace of $T$ is well-defined and equals the sum of all eigenvalues of $T.$ The main fact, we are going to obtain here, is \vskip 0.1cm {\bf Theorem.}\, Let $Y$ be a subspace of a quotient (or a quotient of a subspace) of an $L_p$-space, $1\le p\le\infty.$ If $T\in N_s(Y,Y)$ $(s$-nuclear), where $1/s=1+|1/2-1/p|,$ then 1.\, the (nuclear) trace of $T$ is well defined, 2.\, $\sum_{n=1}^\infty |\lambda_n(T)|<\infty,$ where $\{\lambda_n(T)\}$ is the system of all eigenvalues of the operator $T$ (written in according to their algebraic multiplicities) and $$ \operatorname{trace}\, T= \sum_{n=1}^\infty \lambda_n(T). $$ \vskip 0.3cm Let us mention that in the proof we have to repeat some ideas of proofs from [13] (in particular, of the proof of main lemma there) as well as, simultaneously, to use the main lemma [13] itself (so, we will get a generalization of the lemma by using a part of its proof and also its statement). \centerline{\bf \S1. Preliminaries.} \vskip 0.23cm All the terminology and facts we use here can be found in [5--8]. \vskip 0.02cm Let $X,Y$ be Banach spaces. For $s\in (0,1],$ denote by $X^*\widehat\otimes_s Y$ the completion of the tensor product $X^*\otimes Y$ (considered as a linear space of all finite rank operators) with respect to the quasi-norm $$ ||z||_s:= \inf \{\left(\sum_{k=1}^N ||x'_k||^s\, ||y_k||^s\right)^{1/s}:\ z=\sum_{k=1}^N x'_k\otimes y_k\}. $$ Let $\Psi_p,$ for $p\in[1,\infty],$ be the ideal of all operators which can be factored through a subspace of a quotient of an $L_p$-space. Put $N_s(X,Y):= $ image of $X^*\widehat\otimes_s Y$ in the space $L(X,Y)$ of all bounded linear transformations under the canonical factor map $X^*\widehat\otimes_s Y\to N_s(X,Y)\subset L(X,Y).$ We consider the (Grothendieck) space $N_s(X,Y)$ of all $s$-nuclear operators from $X$ to $Y$ with the natural quasi-norm, induced from $X^*\widehat\otimes_s Y.$ Finally, let $\Psi_{p,s}$ be the quasi-normed product $N_s\circ \Psi_p$ of the corresponding ideals equipped with the natural quasi-norm $\nu_{p,s}$: if $A\in N_s\circ \Psi_p(X,Y)$ then $A=\varphi\circ T$ with $T=\beta\alpha\in \Psi_p,$ $\varphi=\delta\Delta\gamma\in N_s$ and $$ A: X \overset\alpha\to X_p\overset\beta\to Z \overset\gamma\to c_0\overset\Delta\to l_1\overset\delta\to Y, $$ where all maps are continuous and linear, $X_p$ is a subspace of a quotient of an $L_p$-space, constructed on a measure space, and $\Delta$ is a diagonal operator with the diagonal from $l_s.$ Thus, $A=\delta \Delta\gamma\beta\alpha$ and $A\in N_s.$ Therefore, if $X=Y,$ the spectrum of $A,$\, $sp\,(A),$ is at most countable with only possible limit point zero. Moreover, $A$ is a Riesz operator with eigenvalues of finite algebraic multiplicities and $sp\, (A)\equiv sp\, (B),$ where $B:= \alpha\delta\Delta\gamma\beta: X_p\to X_p$ is an $s$-nuclear operator, acting in a subspace of a quotient of an $L_p$-space. \vskip 0.1cm {\bf Definition.}\ Let $Y$ be a Banach space and $s\in(0,1].$ We say that $Y$ {\it possesses the property $AP_s$}\, ( the approximation property of order $s;$ written down as "$Y\in AP_s$") if for %every $X$ and any tensor element $z\in Y^*\widehat\otimes_s Y$ the operator $\tilde{z}: Y\to Y,$ associated with $z,$ is zero iff the tensor element $z$ is zero as an element of the space $Y^*\widehat\otimes Y.$ \vskip 0.1cm This is equivalent to the fact that if $z\in Y^*\widehat\otimes_s Y$ then it follows from $$ \operatorname{trace}\, z\circ R=0, \ \ \forall\, R\in Y^*\otimes Y $$ that $\operatorname{trace}\, U\circ z=0$ for every $U\in L(Y,Y^{**}).$ There is a simple characterization of the condition $Y\in AP_s$ in terms of the approximation of the identity $\operatorname{id}_Y$ on some sequences of the space $Y,$ but we omit it. We need here only some examples which are crucial for our note. For giving them, we formulate and prove the following statement, which, we hope, is interesting by itself. \vskip 0.1cm {\bf Proposition 1.}\, Let $\alpha\in [0,1/2]$ and $1/s=1+\alpha.$ For a Banach space $Y,$ suppose that $(\alpha)$\ there exist constants $C>0$ %and $d\in (0,1)$ such that for every $\varepsilon>0,$ for every natural $n$ and for every $n$-dimensional subspace $E$ of $Y$ there exists a finite rank operator $R$ in $Y$ so that $||R||\le Cn^{\alpha}$ and $||R|_E-\operatorname{id}_E||_{L(E,Y)}\le \varepsilon.$ Then $Y\in AP_s.$ \vskip 0.1cm \begin{proof} Suppose that there is an element $z\in Y^*\widehat\otimes_s Y$ such that $\operatorname{trace}\, z=b>0,$ but $\tilde{z}=0.$ Consider a representation of $z$ of the kind $$ z=\sum_{k=1}^\infty \mu_k y'_k\otimes y_k, $$ where $||y'_k||, ||y_k||=1$ and $\mu_k\ge 0,$ $\sum_{k=1}^\infty \mu_k^s<\infty.$ Without loss of generality, we can (and do) assume that the sequence $(\mu_k)$ is decreasing and that $\sum_{k=1}^\infty \mu_k\le1.$ In this situation, $\mu_k^s=o(1/k),$ so, there are $c_k>0$ with $c_k\to 0$ and $\mu_k\le c_k/k^{1/s}.$ Fix any natural $N,$ large enough, such that for all $m\ge N$ $$ \sum_{k=1}^m \mu_k \langle y'_k, y_k\rangle \ge b/2. $$ For such an $m,$ put $E:= \operatorname{span} \{y_k\}_{k=1}^m,$ and apply the condition $(\alpha)$ to find a corresponding operator $R\in Y^*\otimes Y$ for $n=m$ and $\varepsilon=b/4.$ By our assumption, $\operatorname{trace}\, R\circ z=0.$ From this, we get (for all $m\ge N):$ $$ 0 = \sum_{k=1}^m \mu_k \langle y'_k, Ry_k\rangle + \sum_{k=m+1}^\infty \mu_k \langle y'_k, Ry_k\rangle. $$ For the first sum: $$ \sum_{k=1}^m \mu_k \langle y'_k, Ry_k\rangle \ge \sum_{k=1}^m \mu_k \langle y'_k, y_k\rangle - \big| \sum_{k=1}^m \mu_k \langle y'_k, y_k- Ry_k\rangle \big| \ge b/2 - b/4=b/4. $$ For the second sum: $$ \big| \sum_{k=m+1}^\infty \mu_k \langle y'_k, Ry_k\rangle \big| \le Cm^\alpha\, \tilde{c}_m\, \int_m^\infty x^{-1/s}\, dx \le d_m\, m^\alpha m^{1-1/s}= d_m, $$ where $0\le\tilde{c}_m\to 0,$ and thus $0\le d_m\to 0.$ Now, from the last three relations, we obtain: $0 \ge b/4 - d_m.$ \end{proof} %!!! style BAnCenter Let us consider some consequences of the proposition. \vskip 0.1cm {\bf Corollary 1.}\, Let $\alpha\in [0,1/2]$ and $1/s=1+\alpha.$ For a Banach space $Y,$ suppose that there exist constants $C>0$ such that for every natural $n$ and for every $n$-dimensional subspace $E$ of $Y$ there exists a finite rank operator $R$ in $Y$ so that $||R||\le Cn^{\alpha}$ and $R|_E=\operatorname{id}_E.$ Then $Y\in AP_s.$ \vskip 0.1cm {\bf Corollary 2.}\, Let $\alpha\in [0,1/2]$ and $1/s=1+\alpha.$ For a Banach space $Y,$ suppose that there exist constants $C>0$ such that for every natural $n$ and for every $n$-dimensional subspace $E$ of $Y$ there exists a finite dimensional subspace $F$ of $Y,$ containing $E$ and $Cn^\alpha$-complemented in $Y.$ Then $Y\in AP_s.$ \vskip 0.1cm {\bf Corollary 3.}\, Let $\alpha\in [0,1/2]$ and $1/s=1+\alpha.$ For a Banach space $Y,$ suppose that there exist constants $C>0$ such that for every natural $n$ and every $n$-dimensional subspace $E$ of $Y$ is $Cn^\alpha$-complemented in $Y.$ Then $Y\in AP_s.$ Moreover, every subspace of the space $Y$ has the $AP_s.$ \vskip 0.1cm It can be shown (but we do not need this in the note) that $Y\in AP_s$ iff for every Banach space $X$ the natural mapping $X^*\widehat\otimes_s Y\to L(X,Y)$ is one-to-one (for other related results see, e.g., [11], [12]). Thus, taking this in account, we get: \vskip 0.1cm {\bf Corollary 4.}\, In all above four assertions, in the case of $Y$ with mentioned properties, we have the quasi-Banach equality $X^*\widehat\otimes_s Y=N_s(X,Y),$ whichever the space $X$ was. In particular, $Y^*\widehat\otimes_s Y=N_s(Y,Y).$ \vskip 0.1cm Before giving more concrete applications of Proposition 1, let us mention the simplest example. \vskip 0.1cm {\bf Example 1.}\ Let $s\in (0,1],$ $p\in [1,\infty]$ and $1/s=1+|1/p-1/2|.$ Any subspace as well as any factor space of any $L_p$-space have the property $AP_s.$ \vskip 0.1cm We used this example in [13]. The statement of Example 1 follows from Corollary 4 and the results of D.R. Lewis (see [3], Corollary 4). As s matter of fact, one can get from the work [3] more general facts on complementability concerning $L_p$-situation. However, we prefer to consider abstract situations and to deal with spaces of nontrivial types and cotypes (partially, for using the results to be obtained in other considerations). \vskip 0.1cm 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 this references (Rademacher type p = Gauss type p and Rademacher cotype q = Gauss cotype q; so, we can apply results from G. Pisier's lecture [9], assuming that we are working with Rademacher notions). Let us collect the facts we need. \vskip 0.1cm {\bf Proposition 2.}\, Let $X$ be a Banach space and $10$ such that every finite dimensional subspace $E$ of $X$ is $D_{p.q}\, (\operatorname{dim} E)^{1/p-1/q}$-complemented in $X.$ \vskip 0.1cm Recall also the well known general fact: in any Banach space every $n$-dimensional subspace is $n^{1/2}$-complemented. We need in this note only the following immediate consequence of Proposition 2 and Corollary 3: \vskip 0.1cm {\bf Corollary 5.}\, Let $s\in (0,1],$ $p\in [1,\infty]$ and $1/s=1+|1/p-1/2|.$ If a Banach space $Y$ is isomorphic to a subspace of a quotient (or to a quotient of a subspace) of an $L_p$-space then it has the property $AP_s.$ \vskip 0.1cm In particular, we get again (cf. [10] and see [2]): \vskip 0.1cm {\bf Corollary 6 {\rm[2]}.}\, Every Banach space has the property $AS_{2/3}.$ \vskip 0.3cm \centerline{\bf \S1. Main lemma.} \vskip 0.23cm We are going to formulate to prove now the main lemma in this paper. It is interesting to note that in the proof we will use a part of the proof of Lemma from [13] as well as the statement of that Lemma itself. Let us recall the formulation of Lemma of [13]. \vskip 0.1cm {\bf Lemma 0.}\ Let $s\in (0,1],$ $p\in [1,\infty]$ and $1/s=1+|1/2-1/p|.$ Then the system of all eigenvalues (with their algebraic multiplicities) of any operator $T\in N_s(Y,Y),$ acting in any subspace $Y$ of any $L_p$-space, belongs to the space $l_1.$ The same is true for the factor spaces of $L_p$-spaces. \vskip 0.1cm The next assertion contains this lemma 0 as a particular case. \vskip 0.1cm {\bf Lemma 1.}\ Let $s\in (0,1],$ $p\in [1,\infty]$ and $1/s=1+|1/2-1/p|.$ Then the system of all eigenvalues (with their algebraic multiplicities) of any operator $T\in N_s(Y,Y),$ acting in any subspace $Y$ of any quotient of any $L_p$-space (equivalently: in any quotient $Y$ of any subspace of any $L_p$-space), belongs to the space $l_1.$ \vskip 0.1cm %%%%%% \vskip 0.2cm {\it Proof}\ of Lemma 1. \ Let $p\in [1,\infty].$ Let $Y$ be a subspace of a quotient $W$ $(= L_p/V$ for some $V\subset L_p)$ of an $L_p$-space and $T\in N_s(Y,Y)$ with an s-nuclear representation $$ T=\sum_{k=1}^\infty \mu_k y'_k\otimes y_k, $$ where $||y'_k||, ||y_k||=1$ and $\mu_k\ge 0,$ $\sum_{k=1}^\infty \mu_k^s<\infty.$ %Let-... omitted The operator $T$ can be factored in the following way: $$ T: Y\overset{A}\longrightarrow l_\infty \overset{\Delta_{1-s}}\longrightarrow l_r \overset{j}\hookrightarrow c_0 \overset{\Delta_s}\longrightarrow l_1\overset{B}\longrightarrow Y, $$ where $A$ and $B$ are linear bounded, $j$ is the natural injection, $\Delta_s\sim(\mu_k^s)_k$ and $\Delta_{1-s}\sim(\mu_k^{1-s})$ are the natural diagonal operators from $c_0$ into $l_1$ and from $l_\infty$ into $l_r,$ respectively. Here, $r$ is defined via the conditions $1/s=1+|1/p-1/2|$ and $\sum_k \mu_k^s<\infty:$ we have to have $\sum_k \mu_k^{(1-s)r}<\infty,$ for which $(1-s)r=s$ is good. Therefore, put $1/r=1/s-1,$ or $1/r=|1/p-1/2|.$ Let $\Phi: L_p\to W$ be a factor map, so that $Y\subset W.$ Denote by $Y_0,$ $Y_0\subset L_p,$ the preimage of $Y$ under the map $\Phi,$ $Y_o:=\Phi^{-1}(Y).$ Consider the operator $\Phi|_{Y_0}: Y_0\to Y$ (it is a factor map) and the following diagram: $$ Y_0\overset{\Phi|_{Y_0}}\longrightarrow Y\overset{A}\longrightarrow l_\infty \overset{\Delta_{1-s}}\longrightarrow l_r \overset{j}\hookrightarrow c_0 \overset{\Delta_s}\longrightarrow l_1\overset{B}\longrightarrow Y. $$ Since $\Phi|_{Y_0}$ is a factor map, we can find a lifting $Q:l_1\to Y$ with $B=\Phi|_{Y_0}Q: l_1\to Y_0\to Y.$ Now, we get that the operator $T$ can be factored as follows: $$ T: Y\overset{A}\longrightarrow l_\infty \overset{\Delta_{1-s}}\longrightarrow l_r \overset{j}\hookrightarrow c_0 \overset{\Delta_s}\longrightarrow l_1\overset{Q}\longrightarrow Y_0 \overset{\Phi|_{Y_0}}\longrightarrow Y. $$ Let $U_0:= Q\Delta_sj\Delta_{1-s}A: Y\to Y_0.$ Then $U_0\in N_s(Y,Y_0),$ $U:=U_0\Phi|_{Y_0}\in N_s(Y_0,Y_0)$ and $T=\Phi|_{Y_0}U\in N_s(Y,Y).$ By the principle of related operators(see [8], 6.4.3.2), $U$ and $T$ have the same eigenvalues with the same algebraic multiplicities. But $U$ acts in a subspace $Y_0$ of an $L_p$-space, so main Lemma of [13] can be applied. Therefore, by Lemma 0, Lemma 1 is proved. \vskip 0.1cm {\bf Corollary 7.} \, If $s\in (0,1],$ $p\in [1,\infty]$ with $1/s=1+|1/2-1/p|$ then the quasi-normed ideal $\Psi_{p,s}$ is of (spectral) type $l_1.$ \vskip 0.3cm \centerline{\bf \S1. Proof of Theorem} \vskip 0.23cm We prefer to give here a complete proof although we could just refer to the proof of the corresponding theorem in [13] with giving some remarks. Let $Y$ be a subspace of a quotient of an $L_p$-space and $T\in N_s(Y,Y).$ By Corollary 5, we may (and do) identify the space $N_s(Y,Y)$ with the corresponding tensor product $Y^*\widehat\otimes_s Y,$ which, in turn, is a subspace of the projective tensor product $Y^*\widehat\otimes Y.$ Thus, the nuclear trace of $T $ is well defined, and we have to show that this trace of $T$ is just the spectral trace (= spectral sum) $\sum_{n=1}^{\infty} \lambda_n(T).$ By Lemma, the sequence $\{\lambda_n(T)\}_{n=1}^\infty$ of all eigenvalues of $T,$ counting by multiplicities, is in $l_1.$ Since the quasi-normed ideal $\Psi_{p,s}$ is of spectral (= eigenvalue) type $l_1$ (see Corollary 7), we can apply the main result from the paper [14] 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 that ideal $J$}. For the sake of completeness and to simplify the understanding, we (as in the paper [13]) give here some information about "trace" on an operator ideal. Namely, recall (see [8], 6.5.1.1, or Definition 2.1 in [14]) that a {\it trace}\ on an operator ideal $J$ is a class of complex-valued functions, all of which one writes 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 $\lambda\in \mathbb C$ and $U\in J(E,E).$ Our operator $T,$ evidently, belongs to the space $\Psi_{p,s}(Y,Y)$ and, as was said, $\Psi_{p,s}$ is of eigenvalue type $l_1.$ Thus, the assertion $(**)$ implies that the spectral sum $\lambda,$ defined by $\lambda(U):= \sum_{n=1}^\infty \lambda_n(U)$ for $U\in \Psi_{p,s}(E,E),$ is a trace on $\Psi_{p,s}.$ By principle of uniform boundedness (see [7], 3.4.6 (page 152), or [5]), there exists a constant $C>0$ with the property that $$ |\lambda(U)|\le ||\{\lambda_n(U)\}||_{l_1} \le C\, \nu_{p,s}(U) $$ for all Banach spaces $E$ and operators $U\in \Psi_{p,s}(E,E). $ Now, remembering that all operators in $\Psi_{p,s}$ can be approximated by finite rank operators and taking in account the conditions (iii)--(iv) for $\tau=\lambda$, we obtain that the nuclear trace of our operator $T$ coincides with $\lambda (T)$ (recall that the continuous trace is uniquely defined in such a situation; see [8], 6.5.1.2). % \newpage %%%%%%%%%% ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ complementability %\bigskip %\bigskip %\medskip \begin{thebibliography}{0} \bigskip \medskip \bibitem{1} J. 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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{10} O.I. Reinov: \textit{A simple proof of two theorems of A. Grothendieck}, Vestn. Leningr. Univ. 7 (1983), 115-116. %O. I. Reinov, \A simple proof of two theorems of A. Grothendieck, Vestn. Leningr. Univ., 7, 115-116 (1983). \bibitem{11} O.I. Reinov: \textit{Disappearance of tensor elements in the scale of p-nuclear operators}, Theory of operators and theory of functions (LGU) 1(1983), 145-165. \bibitem{12} O.I. Reinov: \textit{Approximation properties $AP_s$ and p-nuclear operators (the case $0 {\rangle} %\def\wt{\widetilde} % \def\sbs{\subset} %\def\tr{\operatorname{trace}\,} %%%%%%%%%%%%% after 30.01.00 02:44:40 Sat: %\def\Gr{\operatorname{Gr}} %\def\AP{\operatorname{AP}} %\def\BAP{\operatorname{BAP}} %\def\N{\operatorname{N}} %\def\I{\operatorname{I}} %\def\id{\operatorname{id}} %\def\L{\operatorname{L}} %\def\QN{\operatorname{QN}} %\def\J{\operatorname{J}} %\def\R{\operatorname{R}} %\def\reg{\operatorname{reg}} %\def\dual{\operatorname{dual}} % \def\sbs{\subset} %%%%%%%%%%%%%%%%%%%%%%%%% % \def\{\quad\blacksquare} %\def\med{\medpagebreak} % \def\QQ{$\quad\square$} \def\small{\smallpagebreak} % \def\Q{\quad\blacksquare} \def\bigp{\bigpagebreak} %\def\f{\vec} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%