% typeset using TeX \input vanilla.sty \scaletype{\magstep1} \scalelinespacing{\magstep1} \def\bull{\vrule height .9ex width .8ex depth -.1ex} \title Set-functions and factorization \endtitle \author N.J. Kalton and S.J. Montgomery-Smith\footnote{Both authors were supported by grants from the National Science Foundation}\\ Department of Mathematics\\ University of Missouri-Columbia\\ Columbia, Mo. 65211 \endauthor \vskip1truecm \subheading{Abstract} If $\phi$ is a submeasure satisfying an appropriate lower estimate we give a quantitative result on the total mass of a measure $\mu$ satisfying $0\le\mu\le\phi.$ We give a dual result for supermeasures and then use these results to investigate convexity on non-locally convex quasi-Banach lattices. We then show how to use these results to extend some factorization theorems due to Pisier to the setting of quasi-Banach spaces. We conclude by showing that if $X$ is a quasi-Banach space of cotype two then any operator $T:C(\Omega)\to X$ is 2-absolutely summing and factors through a Hilbert space and discussing general factorization theorems for cotype two spaces. \vskip2truecm \subheading{1. Introduction} Let $\Cal A$ be an algebra of subsets of some set $\Omega.$ Let us say that a set-function $\phi: \Cal A\to\bold R$ is {\it monotone\/} if it satisfies $\phi(\emptyset)=0$ and $\phi(A)\le \phi(B)$ whenever $A\subset B.$ We say $\phi$ is normalized if $\phi(\Omega)=1.$ A monotone set-function $\phi$ is a {\it submeasure\/} if $$\phi(A\cup B)\le \phi(A)+\phi(B)$$ whenever $A,B\in\Cal A$ are disjoint, and $\phi$ is a {\it supermeasure\/} if $$\phi(A\cup B)\ge \phi(A)+\phi(B)$$ whenever $A,B\in\Cal A$ are disjoint. If $\phi$ is both a submeasure and supermeasure it is a (finitely additive) measure. If $\phi$ and $\psi$ are two monotone set-functions on $\Cal A$ we shall say that $\phi$ is $\psi$-continuous if $\lim_{n\to\infty}\phi(A_n)=0$ whenever $\lim_{n\to\infty}\psi(A_n)=0.$ If $\phi$ is $\psi-$continuous and $\psi$ is $\phi-$continuous then $\phi$ and $\psi$ are {\it equivalent.} A monotone set-function $\phi$ is called {\it exhaustive\/} if $\lim_{n\to\infty}\phi(A_n)=0$ whenever $(A_n)$ is a disjoint sequence in $\Cal A.$ The classical (unsolved) Maharam problem (, ,  and ) asks whether every exhaustive submeasure is equivalent to a measure. A submeasure $\phi$ is called {\it pathological} if whenever $\lambda$ is a measure satisfying $0\le\lambda\le \phi$ then $\phi=0.$ The Maharam problem has a positive answer if and only if there is no normalized exhaustive pathological submeasure. While the Maharam problem remains unanswered, it is known (see e.g.  or ) that there are non-trivial pathological submeasures. In the other direction it is shown in  that if $\phi$ is a non-trivial {\it uniformly exhaustive\/} submeasure then $\phi$ cannot be pathological. $\phi$ is uniformly exhaustive if given $\epsilon>0$ there exists $N\in\bold N$ such that whenever $\{A_1,\ldots,A_N\}$ are disjoint sets in $\Cal A$ then $$\min_{1\le i\le N}\phi(A_i)<\epsilon.$$ Let us say that a monotone set-function $\phi$ satisfies an {\it upper p-estimate} where $01$ we now let $E$ be a maximal subset so that $\lambda(E)\le \delta\phi(E)$ and defining $F=\Omega\setminus E$ we obtain in this case that if $A\subset F$ then $\lambda(A)\ge \phi(A).$ In this case let $q$ be defined by $\frac1q=\frac1p-1.$ Let $\nu$ be any measure on $\Cal A$ such that $\nu(A)\ge \phi(A)$ whenever $A\subset F.$ Suppose $c_1,c_2>0$ satisfy $c_1^{-q}+c_2^{-q}=1.$ Consider the measure $$\mu(A) = c_1\lambda(A\cap E) +c_2\nu(A\cap F).$$ Then for any $A$, \align \phi(A) &\le (\phi(A\cap E)^p+\phi(A\cap F)^p)^{1/p}\\ &\le (\lambda(A\cap E)^p + \nu(A\cap F)^p)^{1/p}\\ &\le (c_1\lambda (A\cap E) +c_2\nu(A\cap F))(c_1^{-q}+c_2^{-q})\\ &= \mu(A). \endalign Hence $\mu(\Omega)\ge \gamma$ and so $$c_1\lambda(E) + c_2\nu(F) \ge \gamma.$$ Minimizing over $c_1,c_2$ yields $$\lambda(E)^p +\nu(F)^p \ge \gamma^p.\tag 3$$ In particular if we let $\nu(A) = \delta^{-1}\lambda(A\cap F)$ and set $t=\lambda(E)/\gamma$ we obtain $$t^p + (2^p-1)(1-t)^p \ge 1.$$ Since in this case $p<1$ we are led to the conclusion that $t\ge \frac12.$ Next we consider the supermeasure $\psi(A)=\phi(A\cap F)$ and deduce the existence of a measure $\nu\ge \psi$ with $\nu(\Omega)\le \gamma \psi(\Omega)=\gamma\phi(F).$ In this case (3) gives that $$\lambda(E)^p +\gamma^p\phi(F)^p \ge \gamma^p.$$ Now as $\phi(F) \le 1-\phi(E) \le 1-\delta^{-1}\lambda(E)$ we have: $$t^p + (1-\delta^{-1}\gamma t)^p \ge 1$$ and this again leads by simple calculus to the fact that $\gamma\le K_p.$ This then completes the proof if we do not require continuity of $\lambda.$ Now suppose $\phi$ is a normalized supermeasure on $\Cal A$ satisfying an upper $p$-estimate. Let $\lambda$ be a minimal measure subject to the conditions $\lambda\ge \phi$ and $\lambda(\Omega)\le K_p.$ (It follows from an argument based on Zorn's Lemma that such a minimal measure exists). Suppose $\lim_{n\to\infty}\phi(F_n)=0.$ Consider the measures $\lambda_n(A)=\lambda(A\cap E_n)$ where $E_n=\Omega\setminus F_n.$ Let $\Cal U$ be any free ultrafilter on the natural numbers and define $\lambda_{\Cal U}(A)=\lim_{\Cal U}\lambda_n(A).$ Clearly $\lambda_{\Cal U}\le \lambda$. Now for any $A$ $$\lambda_n(A) =\lambda(A\cap E_n)\ge \phi(A\cap E_n)\ge (\phi(A)^p-\phi(F_n))^{1/p}.$$ Hence $\lambda_{\Cal U}=\lambda$ by minimality. Thus $\lim_{\Cal U}\lambda(F_n)=0$ for every such ultrafilter and this means $\lim_{n\to\infty}\lambda(F_n)=0.$\bull\enddemo The following corollary is proved for more general uniformly exhaustive submeasures in . \proclaim{Corollary 2.3}Let $\Cal A$ be an algebra of subsets of $\Omega$ and let $\phi$ be a submeasure on $\Omega$ such that for some constant $c>0$ and some $q<\infty,$ we have: $$\phi(A_1\cup\cdots\cup A_n) \ge c(\phi^q(A_1)+\cdots +\phi^q(A_n))^{1/q}$$ whenever $A_1,\ldots,A_n$ are disjoint. Then there is a measure $\mu$ on $\Cal A$ such that $\mu$ and $\phi$ are equivalent.\endproclaim \demo{Proof}Define $\psi$ by $$\psi(A) =\sup(\sum_{k=1}^n\phi^q(A_k))$$ where the supremum is computed over all $n$ and all disjoint $(A_1,\ldots,A_n)$ so that $A=\cup_{k=1}^nA_k.$ It is not difficult to show that $\psi$ is a supermeasure satisfying $1/q-$upper estimate and clearly $c^q\psi\le\phi^q\le\psi.$ By Theorem 2.2 we can pick a measure $\mu\ge \psi$ which is equivalent to $\psi$ and hence to $\phi.$\bull\enddemo \vskip.8truecm \subheading{3. Convexity in lattices} Let $\Omega$ be a compact Hausdorff space and suppose $\Cal B(\Omega)$ denotes the $\sigma-$algebra of Borel subsets of $\Omega$. Let $B(\Omega)$ denote the space of all real-valued Borel functions on $\Omega$. An admissible extended-value quasinorm on $B(\Omega)$ is a map $f\to \|f\|_X,$ $(B(\Omega)\to [0,\infty])$ such that:\newline (a) $\|f\|_X\le\|g\|_X$ for all $f,g\in B(\Omega)$ with $|f|\le|g|$ pointwise.\newline (b) $\|\alpha f\|_X = |\alpha|\|f\|_X$ for $f\in B(\Omega),\ \alpha\in\bold R$\newline (c) There is a constant $C$ so that if $f,g\ge 0$ have disjoint supports then $\|f+g\|_X\le C(\|f\|_X+\|g\|_X).$ \newline (d) There exists a strictly positive $u$ with $0<\|u\|_X<\infty.$ \newline (e) If $f_n\ge 0$ and $f_n\uparrow f$ pointwise, then $\|f_n\|_X\to\|f\|_X.$ \newline The space $X=\{f:\|f\|_X<\infty\}$ is then a quasi-Banach function space on $\Omega$ equipped with the quasi-norm $\|f\|_X$ (more precisely one identifies functions $f,g$ such that $\|f-g\|_X=0$). We say that $X$ is order-continuous if, in addition, we have: \newline (f) If $f_n\downarrow 0$ pointwise and $\|f_1\|_X<\infty$ then $\|f_n\|_X\downarrow 0.$ Conversely if $X$ is a quasi-Banach lattice which contains no copy of $c_0$ and has a weak order-unit then standard representation theorems can be applied to represent $X$ as an order-continuous quasi-Banach function space on some compact Hausdorff space $\Omega$ in the above sense. More precisely, if $u$ is a weak order-unit then there is a compact Hausdorff space $\Omega$ and a lattice embedding $L:C(\Omega)\to X$ so that $L[0,\chi_{\Omega}]=[0,u]$. Since $X$ contains no copy of $c_0$ we can use a result of Thomas  to represent $L$ is the form $$Lf =\int_{\Omega}fd\Phi$$ where $\Phi$ is regular $X$-valued Borel measure on $\Omega.$ This formula then extends $L$ to all bounded Borel functions. We now define the quasi-Banach function space $Y$ by $$\|f\|_Y = \sup_n\| L(\min(|f|,n\chi_{\Omega})\|_X$$ and it may be verified by standard techniques that $L$ extends to a lattice isomorphism of $Y$ onto $X$ (which is an isometry if we assume that the quasi-norm on $X$ is continuous). For an arbitrary quasi-Banach function space $X$ and $00$; it is follows from  and  that $X$ is L-convex if and only if it is geometrically convex. Let us now turn to upper and lower estimates. We say $X$ satisfies a crude upper $p$-estimate with constant $a$ if for any disjoint $f_1,\ldots,f_n$ we have $$\|f_1+\ldots f_n\|_X \le a(\sum_{i=1}^n\|f_i\|_X^p)^{1/p}$$ and we say that $X$ satisfies an upper $p$-estimate if $a=1.$ We say that $X$ satisfies a crude lower $q$-estimate with constant $b$ if for any disjoint $f_1,\ldots,f_n$ we have $$b\|f_1+\ldots f_n\|_X \ge (\sum_{i=1}^n\|f_i\|_X^p)^{1/p};$$ and $X$ satisfies a lower $q$-estimate if $b=1.$ \proclaim{Lemma 3.1}Suppose $00} t(\mu\{|f|>t\})^{1/p}<\infty.$$\proclaim{Theorem 3.2}Let X be a p-normable quasi-Banach function space which satisfies a crude lower q-estimate. Then: \newline (i) X is r-convex for 0t)^{q/p}dt\right)^{1/q}$$. It is then clear that if$p\le q$then$L_{p,q}$satisfies an upper$p$and a lower$q$-estimate. If$p>q$then$L_{p,q}$has an upper$q$and a lower$p-$estimate. Suppose$pq$we define$\Lambda_{p,q}$by $$\|f\|_{\Lambda_{p,q}} = \inf(\sum_{i=1}^n (\sup_{\omega\in A_i}|f(\omega)|^q)\mu(A_i)^{q/p})^{1/q}$$ where the infimum is again computed over all Borel partitions of$\Omega.$Proposition 3.4 now has an analogue whose proof is very similar and we omit most of the details. \proclaim{Proposition 3.4}Suppose$00\mu-$a.e. and if$g\in X$, $$\|gf^{-1}\|_{\Lambda_{p,q}(\mu)} \le abK_{q/p}^{-1/p}\|g\|_X.$$ \newline (ii) There is a probability measure$\lambda$on$\Omega$such that$f>0\lambda-$a.e. and if$g\in X$$$\|g\|_X \le abK_{p/q}^{1/q}\|gf^{-1}\|_{\Lambda_{q,p}}(\lambda).$$ \endproclaim \demo{proof}We first introduce an equivalent quasinorm$\|\,\|_Y$with an exact upper$p$and lower$q$-estimate as in Lemma 3.1 so that$\|g\|_X\le\|g\|_Y\le ab\|g\|_X$for all$g$. (i)As in Theorem 3.2 we consider the submeasure$\phi(A)=\|f\chi_A\|_Y^p.$There is a probability measure$\mu$such that $$0\le\mu\le K_{q/p}^{-1} \frac{\phi(A)}{\phi(\Omega)}$$ for all Borel sets$A.$Then for$g\in X,$and any Borel partition$\{A_1,\ldots,A_n\}$of$\Omega,\align (\sum_{i=1}^n(\inf_{A_i}|gf^{-1}|^q)\mu(A_i)^{q/p})^{1/q} &\le K_{q/p}^{-1/p}\phi(\Omega)^{-1/p} (\sum_{i=1}^n (\inf_{A_i}|gf^{-1}|^q)\|f\chi_A\|_Y^{q})^{1/q} \\ &\le K_{q/p}^{-1/p}\|f\|_Y^{-1}\|g\|_Y\\ &\le K_{q/p}^{-1/p}ab\|g\|_X. \endalign Thus (i) follows. The proof of (ii) is very similar. In this case we consider the supermeasure\phi(A)=\|f\chi_A\|_Y^q$. There is a probability measure$\lambda$on$\Omega$such that $$0\le K_{p/q}^{-1}\frac{\phi(A)}{\phi(\Omega)} \le \lambda$$ for all Borel sets$A.$Thus for$g\in X,$and any Borel partition$\{A_1,\ldots,A_n\}$of$\Omega,\align \|g\|_X &\le \|g\|_Y\\ &\le (\sum_{i=1}^n(\sup_{A_i} |gf^{-1}|^p)\|f\chi_{A_i}\|_Y^p)^{1/p}\\ &\le (\sum_{i=1}^n(\sup_{A_i} |gf^{-1}|^p) \phi(A_i)^{p/q})^{1/p}\\ &\le K_{p/q}^{1/q}\|f\|_Y (\sum_{i=1}^n (\sup_{A_i}|gf^{-1}|^p) \lambda(A_i)^{p/q})^{1/p} \endalign and the result follows.\bull\enddemo \proclaim{Theorem 3.6}Suppose00$a.e. and such that if$g\in X$$$\|gf^{-1}\|_{\Lambda_{p,q}} \le K_{q/p}^{-1/p}\|g\|_X.$$ Notice that$K_{q/p}^{-1}\le e^{c_1\theta}$for some$c_1$. We also have$q/p\le e^{\theta}$and$(q-r)/(p-r)\le c_2\theta$where$c_2$depends only on$p,r.$Let$w_i=f_if^{-1}$. We note first that for any$w\ge 0$in$B(\Omega)we have \align \int w^rd\mu &= \int_0^1 w^*(t)^rdt\\ &\le (\frac{q}{p})^{-r/q}\|w\|_{L_{p,q}}^r(\int_0^1 t^{-\frac{r(q-p)}{p(q-r)}}dt)^{1-r/q}\\ &\le \frac{p}{q}(\frac{q-r}{p-r})^{1-r/q}\|w\|_{L_{p,q}}^r\\ &\le e^{c_3\theta}\|w\|_{L_{p,q}}^r \endalign wherec_3=c_3(r,p).$Thus $$\int w^r d\mu \le \theta^{-r\theta/q}e^{c_4\theta}\|w\|_{ \Lambda_{p,q}}^r$$ where$c_4=c_4(r,p).$Applying this to the$w_i$and summing we have $$1 \le \theta^{-r\theta/p}e^{c_5\theta}\sum_{i=1}^n\|f_i\|_X^r,$$ with$c_5$depending only on$r,p.$The result now follows.\bull\enddemo \demo{Example} We show that the estimate in the previous theorem is essentially best possible. For convenience we consider the case$p=1$and$q=1+\theta.$The conclusion of the theorem is that, for$00$a similar conclusion is attained in the case$r=0.$We establish a converse by considering only the case$r=0.$For$\theta>0$we let$X=X_{\theta}=\Lambda_{1,q}[0,1]$and we let$\kappa(\theta)=M^{(0)}(X).$We will set$\phi(\theta)=\exp(-|\log\theta|^{1/2})$so that$\lim_{\theta\to0}\phi(\theta)=0.$We further set$\psi(\theta)=(2^{1/q}-1)^{-2}$; then$\psi(\theta)=1+4\theta\log2 +O(\theta^2).$We will consider the function$f=f_{\theta}\in X$defined by$f(t)=t^{-1}\chi_{[1-\phi,1]}.$It follows easily from the definition of$M^{(0)}(X)$that $$\exp(-\frac1\phi\int_{1-\phi}^1\log t\, dt)\|\chi_{[1-\phi,1]}\|_X \le \kappa \|f\|_X.$$ To obtain this one derives the integral version of geometric convexity and applies it to suitable rotations of$f$. Thus if$\beta(\theta)=1+(1-\phi)\phi^{-1}\log(1-\phi)$then $$e^{\beta}\phi \le \kappa\|f\|_X.$$ Turning to the estimation of$\|f\|_X$we note that$\|f\|_X^q$is the supremum of expressions of the form $$\sum_{j=1}^n (\tau_j-\tau_{j-1})^q\tau_j^{-q}$$ where$1-\phi=\tau_0<\tau_1<\cdots<\tau_n=1.$Now if$\tau_j>\psi\tau_{j-1}$it can be checked that this expression is increased by interpolating$(\tau_j\tau_{j-1})^{1/2}$into the partition. We there fore may suppose that we consider only partitions where$\tau_j\le \psi\tau_{j-1}.In this case we estimate: \align (\tau_j-\tau_{j-1})^q\tau_j^{-q} &\le (\tau_j-\tau_{j-1})(\tau_j-\tau_{j-1})^{\theta}\tau_{j-1}^{-1+\theta}\\ &\le (\psi-1)^{\theta} (\tau_j-\tau_{j-1})\tau_{j-1}^{-1} \endalign and after summing we get the estimate $$\|f\|_X^q \le (\psi-1)^{\theta}|\log(1-\phi)|.$$ Thus $$\kappa \ge e^{\beta}\phi(\psi-1)^{-\theta/q}|\log(1-\phi)|^{-1/q}.$$ Now for small\theta$we can estimate$|\log(1-\phi)|\le (1+\phi)\phi.$Thus $$\kappa^q \ge e^{q\beta}\phi^{\theta}(\psi-1)^{-\theta}(1+\phi)^{-1}$$ so that $$\liminf_{\theta\to 0}\frac{\log\kappa}{\theta|\log\theta|} \ge \liminf_{\theta\to 0}\frac{(\log\phi-\log(\psi-1))}{|\log\theta|}\ge 1.$$ Thus we conclude from this calculation and the theorem that $$\log\kappa(\theta) = -\theta\log\theta + o(\theta|\log\theta|)$$ as$\theta\to 0.$\bull\enddemo \vskip.8truecm \subheading{4. The factorization theorems of Pisier} We next show how the results of Proposition 3 quickly give extensions of some factorization theorems due to Pisier . Our approach is valid for quasi-Banach spaces since it does not depend on any duality. \proclaim{Theorem 4.1}Suppose$0 1 ,$$and let g=g_1+\ldots g_m. Then there are f_{ik}\ such that$$ \sum_{k=1}^m \left(\sum_{i=1}^n \|T(f_{ik}g_k)\|^q \right)^{t/q} = 1 $$and$$ \| \sum_{i=1}^n f_{ik} \|_s \le 1 .$$We let$$ \alpha_k = \left(\sum_{i=1}^n \|T(f_{ik}g_k)\|^q \right)^{t/qs} $$and f'_{ik} = \alpha_k f_{ik}. Then we have that$$ \sum_{k=1}^m \sum_{i=1}^n \|T(f'_{ik} g)\|^q \ge 1, $$and$$ \| \sum_{k=1}^m \sum_{i=1}^n f'_{ik} \|_s^s \le \sum_{k=1}^m \alpha_k^s \le 1 ,$$that is, \| g\|_Z \ge 1 . We also notice that \chi_{\Omega}\in Z. Thus by Propositions 3.4 and 3.5, there is a probability measure \lambda so that for some C_3 and all bounded g$$ \|g\|_Z \le C_3 \|g\|_{L_{t,r}(\lambda)} .$$Now we show that  (ii) \Rightarrow (iv) . We notice that \lambda is \mu-continuous, and so by the Radon-Nikodym theorem, we can find w=d\lambda/d\mu. Then for any measurable set A we have:$$ \|T(\chi_A w^{1/s}\| \le C_2 \|\chi_A w^{1/s}\|_{L_{s}(\mu)} \lambda(A)^{1/q-1/s} = C_2 \lambda(A)^{1/q}.$$Now define a function space W by$$ \|f\|_W =\sup_{|g|\le |f|, g\in B_{\infty}}\|T(gw^{1/s})\|.$$Then W has an upper r-estimate and$$ \|\chi_A\|_W \le C_2 \lambda(A)^{1/q} .$$Hence by Proposition 3.4$$ \|f\|_W \le C_4 \|f\|_{L_{q,r}(\lambda)}$$for some C_4. \bull\enddemo To conclude the paper let us observe that Theorem 4.1 can be used to extend other factorization results to non-locally convex spaces. Let us first recall that an operator T:X\to Y where X is a Banach space and Y is a quasi-Banach space is called 2-absolutely summing if there is a constant C so that for x_1,\ldots,x_n\in X we have$$ (\sum_{i=1}^n \|Tx_i\|^2)^{1/2}\le C\max_{\|x^*\|\le 1}(\sum_{i=1}^n|x^*(x_i)|^2)^{1/2}.$$A quasi-Banach space X is of cotype p if there is a constant C so that if x_1,\ldots,x_n\in X then$$ \text{Ave }_{\epsilon_i=\pm1}\|\sum_{i=1}^n\epsilon_ix_i\| \le C(\sum_{i=1}^n\|x_i\|^p)^{1/p}.$$For the following theorem for Banach spaces see , p. 62. \proclaim{Theorem 4.3}Suppose \Omega is a compact Hausdorff space and Y is a quasi-Banach space with cotype two. Suppose T:C(\Omega)\to Y is a bounded operator. Then T is 2-absolutely summing and hence there is a probability measure \mu on \Omega and a constant C so that \|Tf\|\le C\|f\|_{L_2(\mu)} for f\in C(\Omega).\endproclaim \demo{Proof}We may assume that X is an r-Banach space where 00 there exists C_0(\epsilon) so that, for any probability measure \nu, if T:X\to L_p(\nu) is strongly approximable then there exists a Borel subset E of \Omega so that \nu(E)>1-\epsilon and \int_E |Tx|\,d\mu \le C_0\|T\|\|x\| for x\in X. Let us fix \epsilon=1/2 and let C_0=C_0(1/2). Suppose x_1,\ldots,x_n\in X and let f=\sum_{i=1}^n|Tx_i|. If \|f\|_p>0, define \nu=\|f\|_p^{-p}|f|^p\mu. Consider the operator S:X\to L_p(\nu) defined by Sx= |f|^{-1}Tx (set Sx(\omega)=0 when f(\omega)=0.) Then \|S\| \le \|f\|_p^{-1}\|T\|. Thus there is a Borel subset E of \Omega with \nu(E)>1/2 and so that \int_E|Sx|\,d\nu\le C_0\|f\|_p^{-1}\|T\|\|x\| for x\in X. Now$$ \align \frac12 \le \nu(E) &= \int_E \sum_{i=1}^n|Sx_i|\, d\nu\\ &\le C_0 \|f\|_p^{-1}\|T\| \sum_{i=1}^n\|x_i\|. \endalign $$and so we obtain an inequality$$ (\int (\sum_{i=1}^n |Tx_i|)^pd\mu)^{1/p}\le 2C_0\|T\|\sum_{i=1}^n\|x_i\|, where $C_1$ depends only on $p,X.$ Now by the factorization results of Maurey  (see  p. 264) we obtain the desired conclusion.\bull\enddemo \proclaim{Theorem 4.5}Let $X$ be a quasi-Banach space such that $X^*$ has cotype two. Suppose $\Omega$ is a compact Hausdorff space and $\mu$ is a $\sigma$-finite measure on $\Omega.$ Then for $00$ there exists a finite-rank operator $F:X\to F$ with $\|Tx-Fx\|<\epsilon$ for $x\in K.$ Pisier has shown that if $X,Y$ are Banach spaces such that $X^*$ and $Y$ have cotype two and $T:X\to Y$ is an approximable linear operator then $T$ factorizes through a Hilbert space (see , ). Does the same result hold if we assume $X,Y$ are quasi-Banach spaces? 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