Let $\Omega$ be a probability space. Suppose $(\epsilon_i)_{1\leq i\leq n}$ is a sequence of i.i.d. Bernoulli random variables on $\Omega,$ i.e. $(\epsilon_i)_{1\leq i\leq n}$ are independent and $P(\epsilon_i=1)=P(\epsilon_i=-1)=\frac{1}{2}$ for all $1\leq i\leq n.$ Define the r.v. s on $\Omega\times\Omega$ as $\epsilon_i\otimes\epsilon_j(\omega,\omega^\prime)=\epsilon_i(\omega)\epsilon_j(\omega^\prime).$ How can I prove that $L^p$- spaces have Pisier's property $(\alpha)$ i.e. $\int_{\Omega\times\Omega}\|\sum_{i,j}\epsilon_i\otimes\epsilon_j f_{i,j}\|_{L^p}^2\equiv\|(\sum_{i,j}|f_{i,j}|^2)^{\frac{1}{2}}\|_{L^p}^2$ ?

  • $\begingroup$ By homogeneity reasons this can not be true. $\endgroup$
    – juan
    Commented Sep 5, 2017 at 15:29
  • $\begingroup$ @juan . I have edited the question. Earlier question had a typo. Thank you. $\endgroup$
    – Mathbuff
    Commented Sep 5, 2017 at 16:48
  • 3
    $\begingroup$ What is the meaning of your sign $\equiv$? What are the $f_{i,j}$'s? What are the values of $p$? $\endgroup$ Commented Nov 6, 2017 at 17:46

1 Answer 1


I suggest a paper of N. Kalton from 2006 on the Rademacher Decoupling property. Every space with Pisier's property Alpha has the decoupling property so it is more general and gets to the heart of such matters. For finite p, Lp has property alpha and therefore has decoupling. More generally, any quasi-Banach lattice with finite cotype has decoupling. Decoupling is particularly useful in obtaining non-trivial estimates for bilinear operators which map into a space with decoupling.


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