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I have a second question about Chapter 9 in Milman and Schechtman's book "Asymptotic theory of finite dimensional normed spaces" (first question here). It's about the proof of Theorem 9.7 (pg. 55). Here's a scan of the theorem and first paragraph of the proof.

Theorem 9.7

I don't understand the first sentence of the proof. By definition, a normed space $X$ is said to have Gaussian type $p$ constant $\alpha$ if for all $x_1, \dots, x_n \in X$, $\alpha$ is the smallest value such that: $\left(\mathbb{E}\|\sum_{i=1}^n g_i x_i\|^2\right)^{1/2} \leq \alpha \left(\sum_{i=1}^n \|x_i\|^p\right)^{1/p} $ where the $g_i$'s are independent gaussian variables. Hence, if the type 2 constant is $\alpha$, it would imply that: $\mathbb{E}\|\sum_{i=1}^n g_i x_i\| \leq \alpha \left(\sum_{i=1}^n \|x_i\|^2\right)^{1/2} $ But this inequality is in the opposite direction as that shown in the proof...

I initially thought that it's a typo, but actually, the inequality in the given direction is crucially needed later on in the proof. I'm probably missing an easy point, but somehow, I'm having trouble getting it. Thanks!

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Notice the role of $\alpha$ in the claimed inequality for $k$ - it provides a lower bound there. Thus, the essence of the assumption is that the type $p$ inequality does not hold for any smaller value of $\alpha$. Which is the same as saying that the reverse inequality holds for some collection of $n$ vectors.