Before I ask my question I will provide a brief introduction.

I came across the notion of Rademacher type while reading Assaf Naor's article *An introduction to the Ribe program*, which can be found here http://arxiv.org/pdf/1205.5993v1.pdf.

A Banach space $(X,||\cdot||_X)$ is said to have *Rademacher type* $p \geqslant 1$ if there exists a constant $T \in (0,\infty)$ so that for every $n \in \mathbb{N}$ and for every $x_1,\ldots,x_n \in X$ we have \begin{equation} \frac{1}{2^n}\sum_{\varepsilon_1,\ldots,\varepsilon_n \in \{-1,+1\}} \left|\left|\sum_{i=1}^n \varepsilon_i x_i \right|\right|_X \leqslant T\left(\sum_{i=1}^n ||x_i||_X^p\right)^{1/p}.\end{equation}

In the article, Naor claims that $p \leqslant 2$ is immediate by considering the case when $x_1,\ldots,x_n$ are colinear. I can prove this using probabilistic techniques as follows.

Let $(X,||\cdot||_X)$ satisfy the above inequality and consider the colinear vectors $x_1,\ldots,x_n$, where each $x_i$, $i = 1,\ldots,n$, is simply a copy of a fixed vector $x \in X$. By factoring we can cancel $||x||_X$ out from both sides, leaving \begin{equation} \frac{1}{2^n}\sum_{\varepsilon_1,\ldots,\varepsilon_n \in \{-1,+1\}} \left|\sum_{i=1}^n \varepsilon_i\right|\leqslant Tn^{1/p}.\end{equation} Now we interpret the left side as the expected distance from the origin after $n$ steps of the simple random walk in $\mathbb{Z}$. It is well known that the drift of this walk, $\mathbb{E}[|S_n|]$, satisfies $$cn^{1/2} \leqslant \mathbb{E}[S_n] \leqslant Cn^{1/2}$$ for some constants $c,C > 0$. Consequently it follows that $p \leqslant 2$.

My question is as follows: is there a more functional analytic way to conclude that the definition of Rademacher type does not make sense for $p > 2$? Since the notion of Rademacher type is a strengthening of the randomized triangle inequality, I am not too surprised that probabilistic techniques pop up, yet I am curious for an alternate approach.

Please let me know if this question is not 'research worthy' enough to be asked here, but belongs elsewhere, such as on StackExchange.

Thanks!