Banach space of Type $p<1$? Let $X$ be a Banach space. We say that it has type $p$ if there exist $T>0$ such that for any $n\in\Bbb N$ and $x_1,\dots,x_n\in X$,
$$
\Bbb E_{\varepsilon}\left|\left|\sum_{i=1}^n\varepsilon_i x_i\right|\right|^p \le T \sum_{i=1}^n ||x_i||^p,
$$
where the expectation is taken over all $\varepsilon\in\{-1,1\}^n$.
In all cases I've seen, the literatures only discuss the case where $p\ge 1$. What could be said about that case $0<p<1$?
 A: The reason it is not discussed is that with $0<p\le 1$ this inequality holds in every normed space, and therefore does not carry any information about the space. Indeed, for $p=1$ we have
$$\left\|\sum_{i=1}^n\varepsilon_i x_i\right\| \le 
\sum_{i=1}^n\|\varepsilon_i x_i\| =
\sum_{i=1}^n\|x_i\|
$$
so $T=1$ works. 
Concerning $0<p<1$, note that by the Khinchine-Kahane inequality (ref)  for $p,q>0$ there exists a constant $C_{p,q}$ such that
$$\left(\Bbb E_{\varepsilon}\left\|\sum_{i=1}^n\varepsilon_i x_i\right\|^p\right)^{1/p}
\le C_{p,q} \left(\Bbb E_{\varepsilon}\left\|\sum_{i=1}^n\varepsilon_i x_i\right\|^q \right)^{1/q}
$$
which means that the type-p inequality can be equivalently written as 
$$\Bbb E_{\varepsilon}\left\|\sum_{i=1}^n\varepsilon_i x_i\right\| \le T \left(\sum_{i=1}^n \|x_i\|^p\right)^{1/p}$$
The right hand side is a decreasing function of $p$ (exercise), so the validity of the type-$p$ inequality for one $p$ implies it for all $p' \in (0,p)$. This is why one is concerned with the supremum of exponents $p$ for which the type inequality holds.
