\input amstex \magnification=\magstep1 \def\ds{\baselineskip 20pt plus 2pt} \def\ss{\baselineskip 15pt plus 1.5pt} \ss \def\Bbb#1{\hbox{\bf #1}} \def\N{\Bbb N} \def\Z{\Bbb Z} \def\C{\Bbb C} \def\R{\Bbb R} \def\supp{\operatorname{supp}} \def\It{\operatorname{int}} \voffset=0.4in \font\bigbold=cmbx10 scaled \magstep2 \font\sl=cmbsy10 \def\slK{\hbox{$K\textfont1=\sl$}} \def\sqr{\vcenter {\hrule height.3mm \hbox {\vrule width.3mm height 2mm \kern2mm \vrule width.3mm } \hrule height.3mm }} \def\endproof{\nobreak\hfill$\sqr$\bigskip\goodbreak} % \pageno=0 \footline={\ifnum\pageno=0\hfill\else\hss\tenrm\folio\hss\fi} \def\ans{\vrule height.1pt width80pt depth0pt} \centerline {\bigbold p-Summing Operators on} \centerline {\bigbold Injective Tensor Products of Spaces} \vskip 1truein \centerline {by} \vskip 1truein \centerline {\bf Stephen Montgomery-Smith$^{(*)}$ and Paulette Saab$^{(**)}$} \vskip1truein {\narrower\smallskip\noindent {\bf Abstract}\ \ Let $X,Y$ and $Z$ be Banach spaces, and let $\prod_p(Y,Z)\ (1\leq p<\infty)$ denote the space of $p$-summing operators from $Y$ to $Z$. We show that, if $X$ is a {\it \$}$_\infty$-space, then a bounded linear operator$T:\ X\hat \otimes_\epsilon Y\longrightarrow Z$is 1-summing if and only if a naturally associated operator$T^\#:\ X\longrightarrow \prod_1(Y,Z)$is 1-summing. This result need not be true if$X$is not a {\it \$}$_\infty$-space. For $p>1$, several examples are given with $X=C[0,1]$ to show that $T^\#$ can be $p$-summing without $T$ being $p$-summing. Indeed, there is an operator $T$ on $C[0,1]\hat \otimes_\epsilon \ell_1$ whose associated operator $T^\#$ is 2-summing, but for all $N\in \N$, there exists an $N$-dimensional subspace $U$ of $C[0,1]\hat \otimes_\epsilon \ell_1$ such that $T$ restricted to $U$ is equivalent to the identity operator on $\ell^N_\infty$. Finally, we show that there is a compact Hausdorff space $K$\ and a bounded linear operator $T:\ C(K)\hat \otimes_\epsilon \ell_1\longrightarrow \ell_2$ for which $T^\#:\ C(K)\longrightarrow \prod_1(\ell_1, \ell_2)$ is not 2-summing. \smallskip} \vfill \ans \item{$^{(*)}$} Research supported in part by an NSF Grant DMS 9001796 \item{$^{(**)}$} Research supported in part by an NSF Grant DMS 87500750 \par\noindent A.M.S.\ (1980) subject classification: 46B99 \eject \voffset=-.3in \ds \noindent {\bf Introduction}\ \ Let $X$ and $Y$ be Banach spaces, and let $X\hat \otimes_\epsilon Y$ denote their injective tensor product. In this paper, we shall study the behavior of those operators on $X\hat \otimes_\epsilon Y$ that are $p$-summing. \medskip If $X$, $Y$ and $Z$ are Banach spaces, then every $p$-summing operator $T:\ X\hat \otimes_\epsilon Y\longrightarrow Z$ induces a $p$-summing linear operator $T^\#:\ X\longrightarrow \prod_p(Y,Z)$. This raises the following question:\ \ given two Banach spaces $Y$ and $Z$, and $1\leq p<\infty$, for what Banach spaces $X$ is it true that a bounded linear operator $T:\ X\hat \otimes_\epsilon Y\longrightarrow Z$ is $p$-summing whenever $T^\#:\ X\longrightarrow \prod_p(Y,Z)$ is $p$-summing? \medskip In , it was shown that whenever $X=C(\Omega)$ is a space of all continuous functions on a compact Hausdorff space $\Omega$, then $T:\ C(\Omega)\hat \otimes_\epsilon Y\longrightarrow Z$ is 1-summing if and only if $T^\#:\ C(\Omega)\longrightarrow \prod_1(Y,Z)$ is 1-summing. We will extend this result by showing that this result still remains true if $X$ is any {\it \$}$_\infty$-space. We will also give an example to show that the result need not be true if$X$is not a {\it \$}$_\infty$-space. For this, we shall exhibit a 2-summing operator $T$ on $\ell_2\hat \otimes_\epsilon \ell_2$ that is not 1-summing, but such that the associated operator $T^\#$ is 1-summing. \medskip The case $p > 1$ turns out to be quite different. Here, the {\it \$}$_\infty$-spaces do not seem to play any important role. We show that for each$11$, we have that for every finite dimensional subspace$B$of$X$, there exists a finite dimensional subspace$E$of$X$containing$B$, and an invertible bounded linear operator$T:\ E\longrightarrow \ell^{\dim E}_\infty$such that$\parallel T\parallel\ \parallel T^{-1}\parallel\leq \lambda$. \medskip It is well known that for any Banach spaces$E$and$F$, if$T$is in$I(E,F)$, then it is also in$\prod_1(E,F)$, with$\pi_1(T)\leq \parallel T\parallel_{\text {int}}$. But$I(E,F)$is strictly included in$\prod_1(E,F)$. It was shown in [12, p. 477] that a Banach space$E$is a {\it \$}$_\infty$-space if and only if for any Banach space $F$, we have that $I(E,F)=\prod_1(E,F)$. We will use this characterization of {\it \$}$_\infty$-spaces in the sequel. \medskip Finally, we note the following characterization of 1-summing operators (called right semi-integral by Grothendieck in ), which will be used later. \medskip \noindent {\bf Proposition 1}\ \ Let$E$and$F$be Banach spaces. Then the following properties about a bounded linear operator$T$from$E$to$F$are equivalent: \item {(i)}$T$is 1-summing; \item {(ii)} There exists a Banach space$F_1$, and an isometric injection$\varphi:\ F\longrightarrow F_1$, such that$\varphi\circ T:\ E\longrightarrow F_1$is an integral operator. \medskip For all other undefined notions we shall refer the reader to either ,  or . \vfill\eject \noindent {\bf II\ 1-Summing and Integral Operators} Let$X$and$Y$be Banach spaces with injective tensor product$X\hat \otimes_\epsilon Y$. For a Banach space$Z$, any bounded linear operator$T:\ X\hat \otimes_\epsilon Y\longrightarrow Z$induces a linear operator$T^\#$on$X$by $$T^\#x(y)=T(x\otimes y) \qquad (y\in Y) .$$ It is clear that the range of$T^\#$is the space {\it \$}$(Y,Z)$ of bounded linear operators from $Y$ into $Z$, and that $T^\#$ is a bounded linear operator. \medskip In this section, we are going to investigate the 1-summing operators, and the integral operators, on $X\hat \otimes_\epsilon Y$. We will use Proposition~1 to relate these two ideas together. First of all, we have the following result. \medskip \noindent {\bf Theorem 2}\ \ Let $X,Y$ and $Z$ be Banach spaces, and let $T:\ X\hat \otimes_\epsilon Y\longrightarrow Z$ be a bounded linear operator. Denote by $i:\ Z\longrightarrow Z^{**}$ the isometric embedding of $Z$ into $Z^{**}$. Then the following two properties are equivalent: \item {(i)} $T\in I (X\hat \otimes_\epsilon Y,Z)$; \item {(ii)} $\hat i \circ T\in I (X, I(Y,Z^{**}))$, where $\hat i:\ I(Y,Z)\longrightarrow I(Y,Z^{**})$ is defined by $\hat i (U)=i\circ U$ for each $U\in I(Y,Z)$. \noindent In particular, if $T^\#\in I(X, I(Y,Z))$, then $T\in I(X\hat \otimes_\epsilon Y,Z)$. \medskip {\bf Proof:}\ \ First, we show that $(X\hat \otimes_\epsilon Y)\hat \otimes_\epsilon Z^*$\ and $X\hat \otimes_\epsilon (Y\hat \otimes_\epsilon Z^*)$\ are isometrically isomorphic to one another. To see this, note that the algebraic tensor product is an associative operation, that is, $(X\otimes Y)\otimes Z^*$ and $X\otimes (Y\otimes Z^*)$ are algebraically isomorphic. Also, they are both generated by elements of the form $\sum\limits^n_{i=1}x_i\otimes y_i\otimes z_i^*$, where $x_i\in X,\ y_i\in Y$ and $z^*_i\in Z^*$. Now, if we let $B(X^*),\ B(Y^*)$ and $B(Z^{**})$ denote the dual unit balls of $X^*,\ Y^*$ and $Z^{**}$ equipped with their respective weak$^{*}$ topologies, then the spaces $(X\otimes_\epsilon Y)\otimes_\epsilon Z^*$\ and $X\otimes_\epsilon (Y\otimes_\epsilon Z^*)$\ embed isometrically into $C\left(B(X^*)\times B(Y^*)\times B(Z^{**})\right)$ in a natural way, by $$\langle \sum\limits^n_{i=1}x_i\otimes y_i\otimes z_i^*,\quad (x^*, y^*, z^{**})\rangle= \sum\limits^n_{i=1}x^*(x_i)y^*(y_i)z^{**}(z^*_i) ,$$ where $\sum\limits^n_{i=1}x_i\otimes y_i\otimes z^*_i$ is in $(X\otimes_\epsilon Y)\otimes_\epsilon Z^*$\ or $X\otimes_\epsilon (Y\otimes_\epsilon Z^*)$, and $(x^*, y^*, z^{**})$ is in the compact set $B(X^*)\times B(Y^*)\times B(Z^{**})$. Thus both spaces $(X\hat \otimes_\epsilon Y)\hat \otimes_\epsilon Z^*$ and $X\hat \otimes_\epsilon (Y\hat \otimes_\epsilon Z^*)$ can be thought of as the closure in $C\left(B(X^*)\times B(Y^*)\times B(Z^{**})\right)$ of the algebraic tensor product of $X$, $Y$ and $Z^*$. \medskip Now let us assume that $T:\ X\hat \otimes_\epsilon Y\longrightarrow Z$ is an integral operator. Then the bilinear map $\tau$ on $X\hat\otimes_\epsilon Y\times Z^*$, given by $\tau (u,z^*)=z^*(Tu)$ for $u\in X\hat \otimes_\epsilon Y$ and $z^*\in Z^*$, defines an element of $\left(X\hat \otimes_\epsilon Y\hat \otimes_\epsilon Z^*\right)^*$, that is, $$\parallel T\parallel_{\It}=\sup\left\{|\sum\limits^n_{i=1}z_i^*\left(T(x_i\otimes y_i)\right):\ \ \parallel\sum\limits^n_{i=1}x_i\otimes y_i \otimes z^*_i\parallel_\epsilon \leq 1\right\} .\tag *$$ To show that for every $x$\ in $X$\ the operator $T^\#x$\ is in $I(Y,Z)$, with $$\parallel T^\# x\parallel_{\It}\leq \parallel x\parallel \ \parallel T\parallel_{\It} ,$$ is easy. This is because, for each $x\in X$, the operator $T^\#x$ is the composition of $T$ with the bounded linear operator from $Y$ to $X\hat \otimes_\epsilon Y$, which to each $y$ in $Y$ gives the element $x\otimes y$. \medskip If $i:\ Z\longrightarrow Z^{**}$ denotes the isometric embedding of $Z$ into $Z^{**}$, it induces a bounded linear operator $\hat i:\ I(Y,Z)\longrightarrow I(Y,Z^{**})$ given by $\hat i(U)=i\circ U$ for all $U\in I(Y,Z)$. It is immediate that $\hat i$ is an isometry. We will now show that the operator $\hat i\circ T^\#:\ X\longrightarrow I(Y,Z^{**})$\ is integral. It is well known (see [3,~p.~237]) that the space $I(Y,Z^{**})$ is isometrically isomorphic to the dual space $(Y\hat \otimes_\epsilon Z^*)^*$. Thus to show that $\hat i\circ T^\#:\ X\longrightarrow (Y\hat \otimes_\epsilon Z^*)^*$ is an integral operator, we need to show that it induces an element of $\left(X\hat \otimes_\epsilon (Y\hat \otimes_\epsilon Z^*)\right)^*$. For this, it is enough to note that, by our discussion concerning the isometry of $(X\hat \otimes_\epsilon Y)\hat \otimes_\epsilon Z^*$ and $X\hat \otimes_\epsilon (Y\hat \otimes_\epsilon Z^*)$, that $$\parallel\hat i\circ T^\#\parallel_{\It}=\sup\left\{|\sum\limits^n_{i=1}\hat i\circ T^\#x_i, y_i\otimes z^*_i|:\parallel\sum\limits^n_{i=1}x_i\otimes y_i\otimes z^*_i\parallel_\epsilon \leq 1\right\} .\tag **$$ But for each $x\in X$, $y\in Y$ and $z^*\in Z^*$, we have $$\langle \hat i\circ T^\#x, y\otimes z^*\rangle =\langle T (x\otimes y), z^*\rangle .$$ Hence, from (*) and (**), it follows that $$\parallel\hat i\circ T\parallel_{\It}=\parallel T\parallel_{\It}.$$ Thus we have shown that (i) $\Rightarrow$ (ii). The proof of (ii) $\Rightarrow$ (i) follows in a similar way. If $\hat i\circ T^\#:\ X\longrightarrow I(Y,Z^{**})$ is an integral operator, then one can show that $i\circ T:\ X\hat \otimes_\epsilon Y\longrightarrow Z^{**}$ is integral, which in turn implies that $T$ itself is integral (see [3,~p.~233]). \medskip Finally, the last assertion follows easily, since if $T^\#:\ X\longrightarrow I(Y,Z)$ is integral, then $\hat i\circ T$ is integral (see [3,~p.~232]). \endproof Since the mapping $\hat i:\ I(Y,Z)\longrightarrow I(Y,Z^{**})$ is an isometry, Proposition 1 coupled with Theorem 2 implies that, if $T:\ X\hat\otimes_\epsilon Y\longrightarrow Z$ is an integral operator, then $T^\#:\ X\longrightarrow I(Y,Z)$ is 1-summing. This result can be shown directly from the definitions. In what follows we shall present a sketch of that alternative approach. \medskip \noindent {\bf Theorem 3}\ \ Let $X$, $Y$ and $Z$ be Banach spaces, and let $T:\ X\hat \otimes_\epsilon Y\longrightarrow Z$ be a bounded linear operator. If $T$ is integral, then $T^\#:\ X\longrightarrow I(Y,Z)$ is 1-summing. If in addition $X$ is a {\it \$}$_\infty$-space, then$T:\ X\hat \otimes_\epsilon Y\longrightarrow Z$is integral if and only if$T^\#:\ X\longrightarrow I(Y,Z)$is integral. \medskip \noindent {\bf Proof:}\ \ First, we will show that, if$T:\ X\hat \otimes_\epsilon Y\longrightarrow Z$is an integral operator, then$T^\#$is in$\prod_1\left(X,I(Y,Z)\right)$with$\pi_1(T^\#)\leq \parallel T\parallel_{\text {int}}$. Let$x_1, x_2,\ldots, x_n$be in$X$, and fix$\epsilon>0$. For each$i\leq n$, there exists$n_i\in \Bbb N$,$\left(y_{ij}\right)_{j\leq n_i}$in$Y$, and$(z^*_{ij})_{j\leq n_i}$in$Z^*$, such that$\parallel \sum\limits^{ni}_{j=1}y_{ij}\otimes z^*_{ij}\parallel_\epsilon \leq 1$, and $$\parallel T^\#x_i\parallel_{\text {int}}\leq \sum\limits^{n_i}_{j=1}z^*_{ij}\left(T(x_i\otimes y_{ij})\right)+\dfrac \epsilon {2^i}.$$ Since$T$is an integral operator, and $$\parallel \sum\limits^n_{i=1}\sum\limits^{n_i}_{j=1}x_i\otimes y_{ij}\otimes z^*_{ij}\parallel_\epsilon \leq \sup \left\{\sum\limits^n_{i=1}|x^*(x_i)|:\ \parallel x^*\parallel \leq 1, x^*\in X^*\right\},$$ it follows that $$\sum\limits^n_{i=1}\sum\limits^{n_i}_{j=1} z^*_{ij} \left(T(x_i\otimes y_{ij})\right)\leq \parallel T\parallel_{\text {int }}\sup\left\{\sum\limits^n_{i=1}|x^*(x_i)|:\ \parallel x^*\parallel \leq 1,\ x^*\in X^*\right\}.$$ Therefore $$\sum\limits^n_{i=1}\parallel T^\#x_i\parallel_{\text {int }}\leq \parallel T\parallel_{\text {int }}\sup\left\{\sum\limits^n_{i=1}|x^*(x_i)|:\ x^*\in X^*, \parallel x^*\parallel \leq 1\right\}+ \epsilon.$$ \medskip Now, if in addition$X$is a {\it \$}$_\infty$-space, then by [12,~p.~477], the operator $T^\#$ is indeed integral. \endproof \medskip \noindent {\bf Remark 4}\ \ If $X=C(\Omega)$ is a space of continuous functions defined on a compact Hausdorff space $\Omega$, one can deduce a similar result to Theorem 3 from the main result of . \medskip Our next result extends a result of  to {\it \$}$_\infty$-spaces, where it was shown that whenever$X=C(\Omega)$, a space of all continuous functions on a compact Hausdorff space$\Omega$, then a bounded linear operator$T:\ C(\Omega)\hat \otimes_\epsilon Y\longrightarrow Z$is 1-summing if and only if$T^\#:\ C(\Omega)\longrightarrow \prod_1(Y,Z)$is 1-summing. This also extends a result of  where similar conclusions were shown to be true for$X=A(K)$, a space of continuous affine functions on a Choquet simplex$K$(see ). \medskip We note that one implication follows with no restriction on$X$. If$X$,$Y$and$Z$are Banach spaces, and$T:\ X\hat \otimes_\epsilon Y\longrightarrow Z$is a 1-summing operator, then$T^\#$takes its values in$\prod_1(Y,Z)$. This follows from the fact that for each$x\in X$, the operator$T^\#x$is the composition of$T$with the bounded linear operator from$Y$into$X\hat \otimes_\epsilon Y$which to each$y$in$Y$gives the element$x\otimes y$in$X\hat \otimes_\epsilon Y$, and hence $$\pi_1(T^\#x)\leq \parallel x\parallel \pi_1(T).$$ Moreover, one can proceed as in  to show that$T^\#:\ X\longrightarrow \prod_1(Y,Z)$is 1-summing. \medskip \noindent {\bf Theorem 5}\ \ If$X$is a {\it \$}$_\infty$ space, then for any Banach spaces $Y$ and $Z$, a bounded linear operator $T:\ X\hat \otimes_\epsilon Y\longrightarrow Z$ is 1-summing if and only if $T^\#:\ X\longrightarrow \prod_1(Y,Z)$ is 1-summing. \medskip \noindent {\bf Proof:}\ \ Let $T:\ X\hat \otimes_\epsilon Y\longrightarrow Z$ be such that $T^\#:\ X\longrightarrow \prod_1(Y,Z)$ is 1-summing. Since $X$ is a {\it \$}$_\infty$-space, it follows from [14,~p.~477] that$T^\#:\ X\longrightarrow \prod_1(Y,Z)$is an integral operator. Let$\varphi$denote the isometric embedding of$Z$into$C\left(B(Z^*)\right)$, the space of all continuous scaler functions on the unit ball$B(Z^*)$of$Z^*$with its weak$^*$-topology. This induces an isometry $$\hat \varphi:\ {\tsize\prod\nolimits}_1(Y,Z)\longrightarrow {\tsize\prod\nolimits}_1\left((Y,C(B(Z^*))\right),$$ $$\hat \varphi (U)=\varphi\circ U\qquad \text { for all }U\in {\tsize \prod\nolimits}_1(Y,Z).$$ Now, it follows from [15,~p.~301], that$\prod_1\left(Y,C(B(Z^*))\right)$is isometric to$I\left(Y, C(B(Z^*))\right)$. Hence we may assume that$\hat \varphi \circ T^\#:\ X\longrightarrow I\left(Y,C(B(Z^*))\right)$is an integral operator. Moreover, it is easy to check that$(\varphi\circ T)^\#=\hat \varphi \circ T^\#$. By Theorem 2 the operator$\varphi \circ T:\ X\hat \otimes_\epsilon Y\longrightarrow C(B(Z^*))$is an integral operator, and hence$T$is in$\prod_1\left(X\hat \otimes_\epsilon Y, Z\right)$by Proposition 1. \endproof In the following section we shall, among other things, exhibit an example that illustrates that it is crucial for the space$X$to be a {\it \$}$_\infty$-space if the conclusion of Theorem 5 is to be valid. \bigskip \noindent {\bf III\ \ 2-summing Operators and some Counter-examples.} In this section we shall study the behavior of 2-summing operators on injective tensor product spaces. As we shall soon see, the behavior of such operators when $p=2$ is quite different from when $p=1$. For instance, unlike the case $p=1$, the {\it \$}$_\infty$-spaces don't seem to play any particular role. In fact, we shall exhibit operators$T$on$C[0,1]\hat \otimes_\epsilon \ell_2$which are not 2-summing, yet their corresponding operators$T^\#$are. We will also give other interesting examples that answer some other natural questions. \medskip We will present the next theorem for$p=2$, but the same result is true for any$1\leq p<\infty$, with only minor changes. \medskip \noindent {\bf Theorem 6}\ \ Let$X,Y$and$Z$be Banach spaces. If$T:\ X\hat \otimes_\epsilon Y\longrightarrow Z$is a 2-summing operator, then$T^\#:\ X\longrightarrow \prod_2 (Y,Z)$is a 2-summing operator. \medskip \noindent {\bf Proof:}\ \ If$T:\ X\hat \otimes_\epsilon Y\longrightarrow Z$is 2-summing, then using the same kind of arguments that we have given above, it can easily be shown that for each$x\in X$, that$T^\#x\in \prod_2(Y,Z)$, with$\pi_2(T^\#x)\leq \pi_2(T)\parallel x\parallel$. Now we will show that$T^\#:\ X\longrightarrow \prod_2 (Y,Z)$\ is 2-summing. Let$(x_n)$be in$X$such that$\sum\limits_n|x^*(x_n)|^2<\infty$\ for each$x^*$in$X^*$. Fix$\epsilon>0$. For each$n\geq 1$, let$(y_{nm})$be a sequence in$Y$such that $$\sup\left\{\Bigg(\sum\limits^\infty_{m=1}|y^*(y_{nm})|^2\Bigg)^{1/2}:\ \parallel y^*\parallel \leq 1, y^*\in Y^*\right\}\leq 1 ,$$ and $$\pi_2\left(T^\#x_n\right)\leq \left(\sum\limits^\infty_{m=1}\parallel T(x_n\otimes y_{nm})\parallel^2\right)^{1/2}+\dfrac \epsilon {2^n}.$$ Then $$\left[\pi_2\left(T^\#x_n\right)\right]^2\leq \sum\limits^\infty_{m=1}\parallel T\left(x_n\otimes y_{nm}\right)\parallel^2+\dfrac \epsilon {2^{n-1}}\left(\sum\limits^\infty_{m=1}\parallel T\left(x_n\otimes y_{nm}\right)\parallel^2\right)^{1/2}+\dfrac {\epsilon^2}{2^{2n}} .\qquad$$ Now, consider the sequence$(x_n\otimes y_{nm})$in$X\hat \otimes_\epsilon Y$. For each$\xi \in \left(X\hat \otimes_\epsilon Y\right)^*\simeq I(X,Y^*)we have that \aligned \sum\limits_{m,n}\left|\xi(x_n)(y_{nm})\right|^2&=\sum\limits^\infty_{n =1} \sum\limits^\infty_{m=1}|\xi (x_n)(y_{nm})|^2\\ &\leq \sum\limits^\infty_{n=1}\parallel \xi (x_n)\parallel ^2. \endaligned Since\xi \in I(X,Y^*)$, it follows that$\xi\in \prod_2(X,Y^*)$, and so $$\sum\limits^\infty_{n=1}\parallel \xi (x_n)\parallel^2<\infty.$$ Hence we have shown that for all$\xi \in (X\hat \otimes_\epsilon Y)^*$, $$\sum\limits_{m,n} \left|\xi (x_n)(y_{nm})\right|^2<\infty.$$ Since$T\in \prod_2\left(X\hat \otimes_\epsilon Y, Z\right)$, we have that $$\sum\limits_{m,n}\parallel T(x_n\otimes y_{nm})\parallel^2<\infty,$$ and therefore $$\sum\limits_n\left[\pi_2\left(T{^\#}\!\!x_n\right)\right]^2<\infty.$$ \endproof \noindent {\bf Remark 7}\ \ The above result extends a result of , where it was shown that if$T:\ X\hat \otimes_\epsilon Y\longrightarrow Z$is$p$-summing for$1\leq p<\infty$, then$T^\#:\ X\longrightarrow${\it \$}$(Y,Z)$ is $p$-summing. \medskip Now we shall give the example that we promised at the end of section~II. \medskip \noindent {\bf Theorem 8}\ \ There exists a bounded linear operator $T:\ \ell_2\hat\otimes_\epsilon \ell_2\longrightarrow \ell_2$ such that $T$ is not 1-summing, yet $T^\#:\ \ell_2\longrightarrow \pi_1(\ell_2, \ell_2)$ is 1-summing. \medskip \noindent{\bf Proof:}\ \ First, we note the well known fact that $\ell_2\hat \otimes_\epsilon \ell_2 = \slK(\ell_2, \ell_2)$, the space of all compact operators from $\ell_2$ to $\ell_2$. Now we define $T$\ as the composition of two operators. \medskip Let $P:\ \slK(\ell_2, \ell_2)\longrightarrow c_0$ be the operator defined so that for each $K\in \slK(\ell_2, \ell_2)$, $$P(K)=\left(K(e_n)(e_n)\right),$$ where $(e_n)$ is the standard basis of $\ell_2$. It is well known [10, p.145] that the sequence $(e_n\otimes e_n)$ in $\ell_2\hat \otimes_\epsilon \ell_2$ is equivalent to the $c_0$-basis, and that the operator $P$ defines a bounded linear projection of $\slK(\ell_2, \ell_2)$ onto $c_0$. \medskip Let $S:\ c_0\longrightarrow \ell_2$ be the bounded linear operator such that for each $(\alpha_n)\in c_0$ $$S(\alpha_n)=\left(\dfrac {\alpha_n}n\right).$$ It is easily checked [7,~p.~39] that $S$ is a 2-summing operator that is not 1-summing. \medskip Now we define $T:\ \slK(\ell_2, \ell_2)\longrightarrow \ell_2$ to be $T=S\circ P$. Thus $T$ is 2-summing but not 1-summing. It follows from Theorem 6 that the induced operator $T^\#:\ell_2\longrightarrow \prod_2(\ell_2, \ell_2)$ is 2-summing. Since $\ell_2$ is of cotype 2, it follows from [10,~p.~62], that for any Banach space $E$, we have $\prod_2(\ell_2, E)=\prod_1(\ell_2, E)$, and that there exists a constant $C>0$ such that for all $U\in \prod_2(\ell_2, E)$\ we have $$\pi_1(U)\leq C\pi_2(U).$$ This implies that $T^\#$ is 1-summing as an operator taking its values in $\prod_1(\ell_2, \ell_2)$. \endproof \medskip \noindent {\bf Remark 9}\ \ We do not need to use Theorem~6 to show that $T^\#$\ is 1-summing in the example above. Instead, we can use the following argument. First note that $T^\#$ factors as follows: $$\matrix \ell_2&@>{T^\#}>>&\pi_1(\ell_2, \ell_2)\\ @VAVV\\ \ell_2&\nearrow B\endmatrix$$ Here $A:\ \ell_2\rightarrow \ell_2$ is the 1-summing operator defined by $$A(\alpha_n)=\left(\dfrac {\alpha_n}n\right),$$ for each $(\alpha_n)\in \ell_2$, and $B:\ \ell_2\longrightarrow \pi_1(\ell_2, \ell_2)$ is the natural embedding of $\ell_2$ into the space $\pi_1(\ell_2, \ell_2)$\ defined by $$B(\beta_n)(\gamma_n)=(\beta_n\gamma_n)$$ for each $(\beta_n)$, $(\gamma_n)\in \ell_2$. \bigskip Now we will give two examples concerning the case when $p>1$. We will show that we do not have a converse to Theorem~8, even when the underlying space $X$ is a {\it \$}$_\infty$-space. \medskip First, let us fix some notation. In what follows we shall denote the space$\ell_p(\Z)$\ by$\ell_p$, and call its standard basis$\{e_n:n\in \Z\}$. Thus if$x=\left(x(n)\right)\in \ell_p$, then$x(n)=\langle x, e_n\rangle$, and $$\parallel x\parallel_{\ell_p}=\left(\sum\limits^\infty_{n=1}|\langle x, e_n\rangle|^p\rangle\right)^{\frac 1p}.$$ For$f\in L_p [0,1]$, we let $$\parallel f\parallel_{L_p}=\left(\int^1_0|f(t)|^pdt\right)^{\frac 1p}.$$ If$\Omega$is a compact Hausdorff space, and$Y$is a Banach space, then$C(\Omega,Y)=C(\Omega)\hat \otimes_\epsilon Y$will denote the Banach space of continuous$Y$-valued functions on$\Omega$under the supremum norm. \medskip We recall that since$\ell_2$is of cotype 2, we have that$\prod_2(\ell_2,\ell_2)=\prod_1(\ell_2, \ell_2)$. We also recall that, if$u=\sum\limits^\infty_{n=1}\alpha_ne_n\otimes e_n$is a diagonal operator in$\prod_2(\ell_2, \ell_2)$, then $$\pi_2(u)=\left(\sum\limits^\infty_{n=1}|\alpha_n|^2\right)^{\frac 12}=\text { the Hilbert-Schmidt norm of u.}$$ \bigskip \noindent {\bf Theorem 10}\ \ For each$12$\ follows by the same argument. For each$n\in \Z$, let$\epsilon_n(t):\ [0,1]\rightarrow \C,\ \epsilon_n(t)=e^{2\pi \It}$denote the standard trigonometric basis of$L_2[0,1]$. If$f\in L_1[0,1]$, let$\hat f(n)=\int^1_0f(t)\epsilon_n(t)dt$denote the usual Fourier coefficient of$f$. For each$\lambda=(\lambda_n)$, where$|\lambda_n|\leq 1$for all$n\in \Z$, define the operator $$T_\lambda:\ C\left([0,1], \ell_2\right)\longrightarrow \ell_2$$ such that for$\varphi\in C\left([0,1], \ell_2\right)$\ we have $$T_\lambda\varphi=\left(\lambda_n\ \langle \hat \varphi (n), e_n\rangle\ \right).$$ Here$\hat \varphi (n)=\text { Bochner --}\int^1_0\varphi(t)\epsilon_n(t)dt$. \medskip The operator$T_\lambda$is a bounded linear operator, with$\parallel T_\lambda \varphi \parallel_{\ell_2}\leq \parallel \varphi \parallel$. To see this, note that for$\varphi\in C\left([0,1], \ell_2\right)\ we have \aligned \parallel T_\lambda \varphi \parallel^2_{\ell_2}&=\sum\limits_n |\lambda_n|^2|\,\langle \hat \varphi (n), e_n\rangle\,|^2\\ &\leq \sum\limits_n|\,\langle \hat \varphi (n), e_n\rangle\, |^2\\ &\leq \sum\limits_n\int^1_0|\,\langle \varphi (t), e_n\rangle\, |^2dt\\ &=\int^1_0\parallel \varphi (t)\parallel^2_{\ell_2}dt\\ &\leq \sup\limits_t\parallel \varphi (t)\parallel^2_{\ell_2}.\endaligned Now, note that iff\in C\left([0,1]\right)$, and$x\in \ell_2$, then $$T_\lambda (f\otimes x)=\left(\lambda_n \hat f(n)\langle x, e_n\rangle\right),$$ and hence the operator$T_\lambda^\#:\ C[0,1]\rightarrow ${\it \$}$(\ell_2, \ell_2)$ is such that $$T^{\#}_\lambda f (x)=\left(\lambda_n\hat f(n) \langle x, e_n \rangle \right).$$ Thus $$\pi_2(T^\#_\lambda f)=\left(\sum\limits_n |\lambda_n|^2|\hat f(n)|^2\right)^{\frac 12} .$$ Hence, by H\"older's inequality, $$\pi_2(T^\#_\lambda f) \leq \parallel (\lambda_n)\parallel_{\ell_r}\parallel (\hat f (n))\parallel_{\ell_q},$$ where $\dfrac 1r+\dfrac 1q=\dfrac 12$. By the Hausdorff-Young inequality, we have that $$\parallel (\hat f (n))\parallel_{\ell_q}\leq \parallel f\parallel_{L_p},$$ where $1\leq p\leq 2$ and $\dfrac 1p+\dfrac 1q=1$. Thus $$\pi_2(T^\#_\lambda f)\leq \parallel (\lambda_n)\parallel_{\ell_r}\ \parallel f\parallel_{L_p},$$ for $1\leq p\leq 2,\ 2\leq r\leq \infty$ and $\dfrac 1p=\dfrac 1r+\dfrac 12$. This shows that if $\parallel (\lambda_n)\parallel_{\ell_r}<\infty$, then \item {(1)} $T^\#_\lambda\left(C[0,1]\right)\subseteq \pi_2 (\ell_2, \ell_2)=\pi_p(\ell_2, \ell_2)$; \item {(2)} $T^\#_\lambda:\ C[0,1]\longrightarrow \pi_p(\ell_2, \ell_2)$ is $p$-summing. \medskip Now, let $U\subset C\left([0,1], \ell_2\right)$ be the closed linear span of $\{\epsilon_i\otimes e_i,\ a_i\in \Z\}$. Then $U$ is isometrically isomorphic to $\ell_2$. This is because \aligned \parallel \sum\limits_i\mu_i\epsilon_i\otimes e_i\parallel &=\sup\limits_{t\in [0,1]}\parallel \left(\mu_n\epsilon_n(t)\right)\parallel_{\ell_2}\\ &=\parallel \left(\mu_i\epsilon_i (t_0)\right)\parallel_{\ell_2},\endaligned for some $t_0 \in [0,1]$, and hence $$\parallel \sum\limits_i\mu_i\epsilon_i\otimes e_i\parallel=\left(\sum\limits_i|\mu_i|^2\right)^{\frac 12}.$$ Moreover $$T_\lambda(\epsilon_i\otimes e_i)=\lambda_ie_i\qquad \text { for all }i\in \Z,$$ Therefore, we have the following commuting diagram $$\matrix U&@>{T_{\lambda |U}}>>&\ell_2\\ @VQVV&\nearrow {S_\lambda}&\\ \ell_2\endmatrix$$ where $Q:\ U\rightarrow \ell_2$ is the isomorphism from $U$ onto $\ell_2$ such that $Q(\epsilon_n\otimes e_n)=e_n$ for all $n\in \Z$, and $S_\lambda:\ \ell_2\longrightarrow \ell_2$ is the operator given by $S_\lambda(e_n)=\lambda_n e_n$. So to show that $T_\lambda$ is not $p$-summing, it is sufficient to show that one can pick $\lambda=(\lambda_n)$ such that $S_\lambda$ is not $p$-summing. To do this, we consider two cases. If $p=2$, we take $\lambda_n=1$ for all $n\in \Z$. Then the map $S_\lambda$ induced on $\ell_2$ is the identity map which is not $s$-summing for any $s<\infty$. If 1{T_{|U}}>> T(U)\\ @VAVV @AA{A^{-1}}A\\ \ell^N_\infty @>{id_{\ell^N_\infty}}>> \ell^N_\infty\endCD $$where A:\ U\rightarrow \ell^N_\infty is the isomorphism between U and \ell^N_\infty. \endproof \bigskip \noindent{\bf IV Operators that factor through a Hilbert space} It is well known that {\it \}(X,\ell_2)=\prod_2(X, \ell_2)\ whenever X\ is C(K)\ or \ell_1. One might ask whether this is true when X = C(K,\ell_1). Indeed one could ask the weaker question: if T:\ C(K, \ell_1)\longrightarrow \ell_2\ is bounded, does it follow that the induced operator T^\#\ is 2-summing? We answer this question in the negative. \bigskip \noindent {\bf Theorem 12}\ \ There is a compact Hausdorff space K and a bounded linear operator T:\ C(K, \ell_1)\longrightarrow \ell_2 for which T^\#:\ C(K)\longrightarrow \prod_1(\ell_1, \ell_2) is not 2-summing. \medskip \noindent {\bf Proof:}\ \ First, we show that there is a compact Hausdorff space K, and an operator R:\ C(K)\longrightarrow \ell_\infty that is (2,1)-summing but not 2-summing. To see this, let K=[0,1], and consider the natural embedding C[0,1] \longrightarrow L_{2,1}[0,1], where L_{2,1}[0,1]\ is the Lorentz space on [0,1]\ with the Lebesque measure (see ). By , it follows that this map is (2,1)-summing. To show that this map is not 2-summing, we argue in a similar fashion to . For n\in\N, consider the functions e_i(t) = f(t+\frac 1i \bmod 1)\ (1\le i\le n), where f(t) = \frac 1{\sqrt t}\ if t\ge \frac 1n\ and \sqrt n\ otherwise. Then it is an easy matter to verify that for some constant C>0,$$ \left( \sum_{i=1}^n |e^*(e_i)|^2 \right)^{\frac12} \le C \sqrt{\log n} $$for every e^*\ in the unit ball of C[0,1]^*, whereas$$ \left( \sum_{i=1}^n \| e_i \|_{L_{2,1}[0,1]}^2 \right)^{\frac12} \ge C^{-1} \log n . $$Finally, since L_{2,1}[0,1]\ is separable, it embeds isometrically into \ell_\infty. \medskip Define T:\ C(K, \ell_1)\rightarrow \ell_2 as follows:\ \ for \varphi=(f_n)\in C(K,\ell_1), let$$ T(f_n)=\sum\limits_n Rf_n (n)e_n. $$Then T is bounded, for$$ \aligned \parallel T(f_n)\parallel_2&=\left(\sum\limits_n|Rf_n(n)|^2\right)^{\frac 12}\\ &\leq \left(\sum\limits_n \parallel Rf_n\parallel^2_{\ell_\infty}\right)^{\frac 12}\\ &\leq \pi_{2,1}(R)\sup\limits_{t\in K}\sum\limits_n|f_n(t)|.\endaligned $$Thus$$ \parallel T\parallel \leq \pi_{2,1}(R). $$But T^\#:\ C(K)\longrightarrow {\it \}(\ell_1, \ell_2) is not 2-summing, because for each f\in C(K), the operator T^\#f:\ \ell_1\longrightarrow \ell_2 is the diagonal operator \sum\limits_nRf(n)e_n\otimes e_n. Hence the strong operator norm of T^\#f is$$ \parallel T^\#f\parallel =\sup\limits_n|Rf(n)|=\parallel Rf\parallel_{\ell_\infty}.$Thus$T^\#:\ C(K)\longrightarrow${\it \$}$(\ell_1, \ell_2)$ is not 2-summing, because $R:\ C(K)\longrightarrow \ell_\infty$ is not 2-summing. \endproof \bigskip \noindent {\bf Discussions and concluding remarks} \noindent{\bf Remark 13}\ \ Theorem 12 shows that if $X$ and $Y$ are Banach spaces such that {\it \$}$(X, \ell_2)=\prod_2(X, \ell_2)$and {\it \$}$(Y, \ell_2)=\prod_2(X, \ell_2)$, then $X\hat \otimes_\epsilon Y$ need not share this property. This observation could also be deduced from arguments presented in  (use Example~3.5 and the proof of Proposition~3.6 to show that there is a bounded operator $T:(\ell_1\oplus\ell_1\oplus\ldots\oplus\ell_1)_{\ell_\infty} \longrightarrow \ell_2$\ that is not $p$-summing for any $p<\infty$). \medskip \noindent {\bf Remark 14}\ \ In the proof of Theorem 2 we showed that the injective tensor product is an associative operation, that is, if $X, Y$ and $Z$ are Banach spaces, then $(X\hat\otimes_\epsilon Y)\hat \otimes_\epsilon Z$ is isometrically isomorphic to $X\hat \otimes_\epsilon (Y\hat \otimes_\epsilon Z)$. It is not hard to see that the same is true for the projective tensor product. However, we can conclude from Theorem~12 that what is known as the $\gamma_2^*$-tensor product is not an associative operation. \medskip If $E$ and $F$ are Banach spaces, and $T:\ E\longrightarrow F$ is a bounded linear operator, following , we say that $T$ {\bf factors through a Hilbert space} if there is a Hilbert space $H$, and operators $B:\ E\longrightarrow H$ and $A:\ H\longrightarrow F$ such that $T=A\circ B$. We let $\gamma_2(T)=\inf\{\parallel A\parallel\ \parallel B\parallel \}$, where the infimum runs over all possible factorization of $T$, and denote the space of all operators $T:\ E\longrightarrow F$ that factor through a Hilbert space by $\Gamma_2(E,F)$. It is not hard to check that $\gamma_2$ defines a norm on $\Gamma_2(E,F)$, making $\Gamma_2(E,F)$\ a Banach space. We define the $\gamma_2^*$-norm $\parallel\quad \parallel_*$ on $E\otimes F$ (see  or ) in which the dual of $E\otimes F$ is identified with $\Gamma_2(E,F^*)$, and let $E\hat \otimes_{\gamma^*_2}F$ denote the completion of $(E\otimes F, \parallel\quad \parallel_*)$. \medskip The operator $T:\ C(K)\hat \otimes_{\gamma^*_2}\ell_1\longrightarrow \ell_2$\ exhibited in Theorem 12, induces a bounded linear functional on $\left[(C(K)\hat \otimes_{\gamma^*_2}\ell_1)\hat \otimes_{\gamma^*_2}\ell_2\right]^*$. Now we see that if $C(K)\hat \otimes_{\gamma^*_2}(\ell_1\hat \otimes_{\gamma^*_2}\ell_2)$ were isometrically isomorphic to $(C(K)\hat \otimes_{\gamma^*_2}\ell_1)\hat \otimes_{\gamma^*_2}\ell_2$, then the operator $T^\#:\ C(K)\rightarrow${\it \$}$(\ell_1, \ell_2)$would induce a bounded linear functional on$\left[C(K)\hat \otimes_{\gamma^*_2}(\ell_1\hat \otimes_{\gamma^*_2}\ell_2)\right]^*$, showing that$T^\#\in \Gamma_2\left(C(K),\text {\it \$}(\ell_1, \ell_2)\right)$, implying that $T^\#$ would be 2-summing [10,~p.~62]. This contradiction shows that $C(K) \hat \otimes_{\gamma^*_2}(\ell_1\hat \otimes_{\gamma^*_2}\ell_2)$ and $\left(C(K)\hat \otimes_{\gamma^*_2}\ell_1\right)\hat \otimes_{\gamma^*_2}\ell_2$ cannot be isometrically isomorphic. \medskip Another example showing that the $\gamma_2^*$-tensor product is not associative was given by Pisier (private communication). \vfill\eject \centerline {\bf Bibliography} \item {} R. Bilyeu, and P. Lewis, {\it Some Mapping Properties of Representing Measures}, Ann. Math Pure Appl. CIX (1976) p. 273--287. \item {} G. Choquet, {\bf Lectures on Analysis}, Vol. {\bf II}, Benjamin, New York, (1969). \item {} J. Diestel, and J.J. Uhl Jr., {\bf Vector Measures}, Math Surveys, {\bf 15}, AMS, Providence, RI (1977). \item{} T. Figiel, J. Lindenstrauss, and V. Milman, {\it The dimension of almost spherical sections of convex bodies}, Acta Mathematica, {\bf 139}, (1977), p.~53--94. \item {} A. Grothendick, {\bf Produits tensoriels topologiques et espaces nucl\'eaires}, Mem. A.M.S. {\bf 16}, (1955). \item {} R.A. Hunt, {\it On $L(p,q)$\ spaces}, L'Enseignement Math.\ (2), {\bf 12}, (1966), p.~249--275. \item {} G.J.O. Jameson, {\bf Summing and Nuclear Norms in Banach Space Theory}, LMSST {\bf 8}, Cambridge University Press (1987). \item {} G.J.O. Jameson, {\it Relations between summing norms of mappings on $\ell_\infty$}, Math.~Z, {\bf 194}, (1987), p.~89--94. \item {} S. Kwapien, {\it On operators factorizable through $L_p$-spaces}, Bull.\ Soc.\ Math.\ France, M\'em 31--32, (1972), p.~215--225. \item {} G. Pisier, {\bf Factorization of Linear Operators and Geometry of Banach Spaces}, AMS CBMS {\bf 60}, Providence RI (1986). \item {} G. Pisier, {\it Factorization of operators through $L_{p\infty}$\ or $L_{p1}$\ and non commutative generalizations}, Math.\ Ann., {\bf 276}, (1986), p.~105--136. \item {} J.R. Retherford, and C. Stegall, {\it Fully Nuclear and Completely Nuclear Operators with applications to {\it \$}$_1$and {\it \$}$_\infty$ spaces}, T.A.M.S., {\bf 163}, (1972) p. 457--492. \item {} P. Saab, {\it Integral Operators on Spaces of Continuous Vector Valued Functions}, Proc.\ Amer.\ Math.\ Soc.\ (to appear). \item {} B. Smith, {\it Some Bounded Linear Operators On the Spaces $C(\Omega,E)$ and $A(K,E)$}, Ph.D. Dissertation, The University of Missouri-Columbia, 1989. \item {} C. Stegall, {\it Characterization of Banach spaces whose duals are $L_1$ spaces}, Is. J. of Math, {\bf 11}, (1972) p. 299--308. \item {} C. Swartz, {\it Absolutely summing and dominated operators on spaces of vector-valued continuous functions}, T.A.M.S., {\bf 179}, (1973) p. 123--132. \vfill \centerline{\vbox{\halign{#\hfill\cr University of Missouri\cr Dept.\ of Math.\cr Columbia, MO 65211\cr}}} \eject \bye