I have a quick question which is probably supposed to be obvious, but for some reason I just don't see it: How does one re-norm a quasi-Banach space to produce a $p$-Banach space ($0<p\leq 1$) with the same topology?

A quasi-Banach space is, of course, just like a Banach space except the triangle inequality requirement for the norm is relaxed to $\|x+y\|\leq\kappa(\|x\|+\|y\|)$ for some $1\leq\kappa<\infty$, producing a "quasi-norm" (the topology remains the $\varepsilon$-ball topology, which is metrizable even if not always metric, so that we can talk about completions). A special case of quasi-Banach space are "$p$-Banach spaces" ($0<p\leq 1$) which satisfy $\|x+y\|^p\leq\|x\|^p+\|y\|^p$ (and then we can take $\kappa=2^{1/p-1}$).

This paper claims that every quasi-Banach space admits an equivalent $p$-norm for some $0<p\leq 1$. Here is a quotation for context:

The Aoki-Rolewicz’s Theorem states that any quasi-Banach space $\mathbb{X}$ is $p$-convex for some $0<p\leq 1$, i.e., there is a constant $C$ such that

$$\left\|\sum_{j=1}^nf_j\right\|\leq C\left(\sum_{j=1}^n\|f_j\|^p\right)^{1/p},\;\;n\in\mathbb{N},\;\;f_j\in\mathbb{X}.$$

This way, $\mathbb{X}$ becomes a quasi-Banach space under a suitable renorming.

I'm sorry if I'm missing something obvious, but what is that "suitable renorming", exactly?



1 Answer 1



$$\|x\|^\prime = \inf\Big\{ \big(\sum_{i=1}^n \|x_i\|^p\big)^{1/p}\colon \sum_{i=1}^n x_i = x, x_i\in X, n\in \mathbb N \Big\}\qquad (x\in X)$$

is the standard $p$-convex renorming. The hardish part is to find a suitable $p$. You will find more details in Kalton & Peck's An F-space sampler.


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