1$, 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 [5]), 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 [3], [7] or [10]. \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 [13]. \medskip Our next result extends a result of [16] 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 [14] where similar conclusions were shown to be true for $X=A(K)$, a space of continuous affine functions on a Choquet simplex $K$ (see [2]). \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 [16] 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 [1], 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 $1

2$\ 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 if $f\in C\left([0,1]\right)$, 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 $$ where $A:\ U\rightarrow \ell^N_\infty$ is the isomorphism between $U$ and $\ell^N_\infty$. \endproof \bigskip \noindent{\bf IV Operators that factor through a Hilbert space} It is well known that {\it \$}$(X,\ell_2)=\prod_2(X, \ell_2)$\ whenever $X$\ is $C(K)$\ or $\ell_1$. One might ask whether this is true when $X = C(K,\ell_1)$. Indeed one could ask the weaker question: if $T:\ C(K, \ell_1)\longrightarrow \ell_2$\ is bounded, does it follow that the induced operator $T^\#$\ is 2-summing? We answer this question in the negative. \bigskip \noindent {\bf Theorem 12}\ \ There is a compact Hausdorff space $K$ and a bounded linear operator $T:\ C(K, \ell_1)\longrightarrow \ell_2$ for which $T^\#:\ C(K)\longrightarrow \prod_1(\ell_1, \ell_2)$ is not 2-summing. \medskip \noindent {\bf Proof:}\ \ First, we show that there is a compact Hausdorff space $K$, and an operator $R:\ C(K)\longrightarrow \ell_\infty$ that is (2,1)-summing but not 2-summing. To see this, let $K=[0,1]$, and consider the natural embedding $C[0,1] \longrightarrow L_{2,1}[0,1]$, where $L_{2,1}[0,1]$\ is the Lorentz space on $[0,1]$\ with the Lebesque measure (see [6]). By [11], it follows that this map is (2,1)-summing. To show that this map is not 2-summing, we argue in a similar fashion to [8]. For $n\in\N$, consider the functions $e_i(t) = f(t+\frac 1i \bmod 1)$\ ($1\le i\le n$), where $ f(t) = \frac 1{\sqrt t}$\ if $t\ge \frac 1n$\ and $\sqrt n$\ otherwise. Then it is an easy matter to verify that for some constant $C>0$, $$ \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 [4] (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 [10], 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 [9] or [10]) 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 {[1]} R. Bilyeu, and P. 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