0$ such that $\normo{X+Y}_\cL \le K(\normo X_\cL + \normo Y_\cL)$ for all $X,Y \in \cL$. Notice that if $X^*(2t) \le Y^*(t)$, and $Y \in \cL$, then $X$ may be written as the sum of two disjoint random variables $Y_1$ and $Y_2$ with $Y_1^*(t),Y_2^*(t) \le Y^*(t)$, and hence $\normo X_\cL \le 2K \normo Y_\cL$. Given two rearrangement invariant spaces $\cL$ and $\cM$, we will say that $\cL$ embeds into $\cM$ if there is a positive constant $c$ such that if $X \in \cL$, then $X \in \cM$ and $\normo X_\cM \le c \normo X_\cL$. We will call the least such $c$ the {\em embedding constant\/} of $\cL$ into $\cM$. \begin{thm} Let $p_0>0$, and let $\cL$ be a rearrangement invariant space such that $\cL$ embeds into $L_p$, and $L_q$ embeds into $\cL$, where $q \ge p \ge p_0$. Then there is a positive constant $c$, depending only upon the quasi-constant of $\cL$, the embedding constants, $p_0$ and $q/p$, such that for any sequence of Banach valued independent random variables $(X_n)$ $$ c^{-1}(\normo U_p + \normo\ell_\cL) \le \normo U_\cL \le c(\normo U_p + \normo\ell_\cL) .$$ \end{thm} \Proof Let us first obtain the left hand side inequality. It follows by hypothesis that $\normo U_\cL \ge c_1^{-1} \normo U_p$, where $c_1$ is the embedding constant of $\cL$ into $L_p$. Furthermore, $U \ge {\frac{1}{ 2}} M$, and by Proposition~2.1, $\ell(t) \le M^*(2t)$. Hence $ \normo U_\cL \ge (4K)^{-1} \normo\ell_\cL$, where $K$ is the quasi-constant of $\cL$. Now let us obtain the right hand inequality. By Corollary~\ref{klass-nowicki-decr}, we have that there is a universal positive $c_2$ for $0 \le t \le 1$ $$ U^*(t) I_{0\le t \le 2^{-2q/p}} \le c_2 {\frac{2q }{ p}} (U^*(t^{p/2q}) + M^*(t/2)). $$ Now $U^*(t) I_{0\le t \le 2^{-2q/p}} \ge U^*(2^{2q/p} t)$, and hence $$ \normo U_{\cL} \le (2K)^{\lceil 2q/p \rceil} c_2 K {\frac{2q}{ p}} (\snormo{t\mapsto U^*(t^{p/2q})}_\cL + \normo M_\cL) .$$ To finish the proof, suppose that $\normo U_p = \lambda$. Then it is easily seen that $U^*(t) \le \lambda t^{-1/p}$. Thus, if $c_3$ is the embedding constant of $L_q$ into $\cL$, then \begin{eqnarray*} \snormo{t\mapsto U^*(t^{p/2q})}_\cL &\le& c_3 \snormo{t\mapsto U^*(t^{p/2q})}_q \\ &=& c_3 \left(\int_0^1 (U^*(t^{p/2q}))^q \, dt \right)^{1/q} \\ &\le& c_3 \lambda \left(\int_0^1 t^{-1/2} \, dt\right)^{1/q} \\ &=& 2^{1/q} c_3 \lambda . \end{eqnarray*} \qed \begin{thebibliography}{99} \bibitem{Bi-95} Billingsley, P. 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