TITLE: A basic question on Type and Cotype theory
QUESTION [5 upvotes]: I'm studying basic theory of type and cotype of banach spaces, and I have a simple question. I'm using the definition using averages. All Banach spaces have type 1, that was easy to prove, using the triangle inequality. But I'm having a hard time trying to show that all Banach spaces have cotype $\infty$.
What I'm trying to show is that there existsc $C>0$ such that, for every $x_1, \dotsc, x_n$  in a Banach space $X$,
$$\left( \frac {{\displaystyle \sum\limits_{\varepsilon_i = \pm 1}} \lVert \sum^n_{i=1} \varepsilon_i x_i\rVert} {2^n} \right)  \ge 
C \max_{1\le i \le n} \lVert x_i \rVert $$
How is it done ? This is supposed to be trivial, as the literature keeps telling me "it's easy to see".
Thanks !

REPLY [4 votes]: The argument is by induction:
It is trivial for $n=1$. For the case $n=2$ note that we have, by the triangle inequality and the fact that $\|z\|=\|-z\|$,
$$
\| x-y \| + \|x+y\| \geq 2\max\{ \| x\|, \| y\|\},
$$
so that the inequality in this case follows with $C=1$. For the general case consider a vector $\bar{\varepsilon}=(\varepsilon_2,\ldots,\varepsilon_n) \in \{0,1\}^{n-1}$ and $\bar{x}=(x_2,\ldots,x_n)$ and the natural dot product 
$$
\bar{\varepsilon}\cdot \bar{x}= \sum_{j=2}^n \varepsilon_ix_i\in X.
$$
Then the left hand side of the desired inquality (which we call $A$) can be rewritten as
$$
A=\frac{\sum_{\bar{\varepsilon}} \sum_{\varepsilon_1=\pm1} \| \varepsilon_1x_1 + \bar{\varepsilon}\cdot \bar{x}\|}{2^n}.
$$
Notice that, if $y=\bar{\varepsilon}\cdot \bar{x}$ then, by the argument for $n=2$,
$$
\sum_{\varepsilon_1=\pm1} \| \varepsilon_1x_1+y\| \geq 2\max\{ \| x_1\|,\|y\|\},
$$
so that, plugging into the previous inequality, and recalling the obvious inequality $\max\{ \sum_j a_j, \sum_jb_j\} \leq \sum_j \max\{ a_j,b_j\}$, we obtain
$$
A\geq \max\left\{ \| x_1\|, \frac{\sum_{\bar{\varepsilon}} \| \bar{\varepsilon}\cdot \bar{x}\|}{2^{n-1}} \right\} \geq \max_{1\leq i\leq n} \| x_i\|.
$$
This is what you want, with $C=1$.
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