Type and cotype

Remarks

Example: a Hilbert space $H$ has type $2$ and cotype $2$ with $C = 1$. Expanding the squares, the cross terms $\pm 2 \operatorname{Re} \ip{x_i}{x_j}$ cancel in the average over signs. So $R(x_1, \dots, x_n)^2 = \sum_i \norm{x_i}^2$ exactly.

Non-example: $\ell^1$ has no type $p > 1$. For the unit vectors $e_1, \dots, e_n$, every sign sum has norm $n$, so $R = n$. The right side of the type inequality is $C n^{1/p}$, and $n \le C n^{1/p}$ fails for large $n$.

Every Banach space has type $1$, by the triangle inequality. Type $p$ implies type $p'$ for $p' < p$, and cotype $q$ implies cotype $q'$ for $q' > q$. By the Kahane inequality, the quadratic mean in $R$ can be replaced by the plain mean of the norm, with other constants. For $1 \le p < \infty$, the space $L^p$ has type $\min(p, 2)$ and cotype $\max(p, 2)$.