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Let $X=(X, \|\cdot\|_X)$ be normed space with $x_1, \ldots, x_m\in X$.

Assume, $\int_{[-1,1]^m}\|\sum_{i=1}^ma_ix_i\|_Xd\mu(a)=1$, where $\mu$ is the Lebesgue measure on $[-1,1]^m$, $a \in [-1,1]^m$.

Let $r_i(t), i=1, \ldots, m$ be Rademacher sequence, i.e. $Proba(r_i=1)=Proba(r_i=-1)=1/2.$

By the 1-unconditionality of the Rademacher functions, $$ \left(\int_{[-1,1]^m}\|\sum_{i=1}^ma_ix_i\|^p_X \, d\mu(a)\right)^{1/p} = \left( \int_0^1 \int_{[-1,1]^m} \|\sum_{i=1}^m a_ir_i(t)x_i \|_X^p \, d\mu(a) \, dt\right)^{1/p} \leq \left(\int_0^1 \|\sum_{i=1}^mr_i(t) x_i\|_X^p \, dt \right)^{1/p}. $$

As I understand right, that 1-unconditionality property would not depend on the independence of the Rademacher functions. I.e. if I consider Rademacher sequence with some 'dependensy condition', say for $m=2k, k\in N$, I let $\sum_{i=1}^{2k} r_i(t) = 0$. I am wondering if the last inequality would chenged with the new condition on the Rademacher functions?

Thank you.

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Rademacher sequence is independent by definition, which is important for its 1-unconditionality. If you consider another sequence of +1,-1 variables in its place, you may be able to get K-unconditionality if the dependencies are weak. But I can't really comment on the last inequality since I don't follow the logic of your post. For example, what does the assumption $\int_{[-1,1]^m}\|\sum_{i=1}^ma_ix_i\|_Xd\mu(a)=1$ have to do with the rest of it? –  user31373 Jul 25 '12 at 19:26
Thank you. You are right. It is an extra condition. I thought it somehow used in the solution, but its not. I've got an answer to my question now. Thank you very much. –  Michael Jul 26 '12 at 6:19
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