Is there a generalised parallelogram inequality in uniformly convex spaces A version of generalized parallelogram inequality in Hilbert spaces  is $\sum_{i=1}^{n}\|\alpha_i x_i\|^2=\sum_{i=1}^{n}\alpha_i\| x_i\|^2-\sum_{i\neq j}^{n}\alpha_i\alpha_j\| x_i-x_j\|^2$
$\alpha_i\in (0,1)$. Is there a generalization of this inequality to uniformly convex Banach spaces(especially $L_p$ spaces)
 A: The parallelogram law in Hilbert spaces can be written as 
$$\mathbf E \left\| \sum_{j=1}^n \epsilon_j x_j \right\|^2
=\sum_{j=1}^n \| x_j\|^2 
\tag1 $$
where $\epsilon_j\in \{-1,1\}$ and $\mathbf E$ is the average over all $2^n$ choices of signs of $\epsilon_j$. (In probabilistic terms, $\mathbf E$ is expectation, and $\epsilon_j$ are independent Bernoulli variables.)
The natural counterparts of (1) in a Banach space $X$ are the type inequality
$$\mathbf E \left\| \sum_{j=1}^n \epsilon_j x_j \right\|^r
\le T \sum_{j=1}^n \| x_j\|^r 
\tag{T} $$
and the cotype inequality
$$\sum_{j=1}^n \| x_j\|^r\le C  \mathbf E \left\| \sum_{j=1}^n \epsilon_j x_j \right\|^r
\tag{C} $$
The constants $T$ and $C$ must be independent of the choices of vectors $x_j\in X$. 
The following are true: 


*

*$L^p$ satisfies (T) with $r=\min(p,2)$, for  $1\le p<\infty$.

*$L^p$ satisfies (C) with $r=\max(p,2)$, for  $1\le p<\infty$.

*More generally, every uniformly convex space satisfies (T) with some $1<r\le 2$ and (C) with some $2\le r<\infty$.


Reference: Asymptotic Theory of Finite Dimensional Normed Spaces by Milman and Schechtman.  
