# Can one show that the dual of a quasi-Banach space separates points without explicitly identifying the dual?

I'm interested in a question regarding the identification of some duals of quasi-Banach spaces. However, I'm not familiar with the quasi-Banach literature, so I'm hoping somebody can point me in the right direction regarding a specific question.

First I'll get the definitions out of the way: a quasi-Banach space is a complete quasinormed vector space ('quasinormed' being like 'normed' except now the triangle inequality reads $||x+y|| \leq K(||x|| + ||y||)$ for some $K \geq 1$). The dual space $X^*$ of a quasi-Banach space $X$ is the set of all continuous linear maps from $X$ into the base field (let's take this to be $\mathbb{C}$), and in general $X^*$ may be trivial. We say that $X^*$ separates the points of $X$ if for every $x \in X$ there exists a $x^* \in X^*$ such that $x^*(x) \neq 0$.

If $X^*$ separates the points of $X$, we define the Banach envelope $X_c$ of $X$ to be the completion of $X$ under the norm $$||x||_{X_c} := \inf\{\epsilon > 0 : x \in \operatorname{co}(\epsilon B_X)\},$$ where $\operatorname{co}(B(0,\epsilon))$ is the convex hull of the open ball of radius $\epsilon$ centred at $0$. The fact that $||\cdot||_{X_c}$ is a norm on $X$ supposedly follows from the fact that $X^*$ separates the points of $X$ (but I'm not familiar with the proof of this fact, so don't quote me on it).

In all the literature I've seen, the Banach envelope of a specific space $X$ is identified after having already identified the dual of $X$ in order to verify that $X^*$ separates the points of $X$. I would like to use an identification of $X_c$ in order to identify the dual $X^*$ (as there is a theorem which says that $X^* = (X_c)^*$). To argue in such a way, I need to know in advance that $X^*$ separates the points of $X$.

This leads to my question(s), which explain(s) why I've taken all this space discussing Banach envelopes:

Is there a way of showing that the dual ${X^*}$ of a quasi-Banach space ${X}$ separates the points of ${X}$ without explicitly identifying ${X^*}$? Furthermore, if we know that $||\cdot||_{X_c}$ is a norm on $X$, does this guarantee that ${X^*}$ separates the points of $X$?

(sidenote: posting this question has made me realise that there's no tag explicitly dealing with quasi-Banach spaces, so I don't know how much luck I'll have here!)

• A trivial sufficient condition is that $X$ embeds continuously into some Hausdorff locally convex space $Y$. – Jochen Wengenroth Dec 8 '14 at 7:51
• @JochenWengenroth The condition is also necessary because the weak topology on $X$ is locally convex Hausdorff if $X^\ast$ separates points. – Johannes Hahn Jan 27 '15 at 10:17

I realised that this is actually not such a difficult question, so I'll answer it myself!

(My solution is somewhat inspired by Jochen Wengenroth's comment. Thanks Jochen)

From the definition of $|| \cdot ||_{X_c}$, we immediately have $||x||_{X_c} \leq ||x||_X$ for all $x \in X$. So if $||\cdot||_{X_c}$ is a norm on $X$, then the completion $X_c$ of $(X,||\cdot||_{X_c})$ is a Banach space, and $X$ embeds continuously into $X_c$. If $x \in X$, then there exists an $x^*$ in the dual $(X_c)^*$ such that $x^*(x) \neq 0$ - because $X_c$ is a Banach space. But since $X$ embeds continuously into $X_c$, we have that $x^*$ restricts to an element of $X^*$, and therefore we've found $x^* \in X^*$ such that $x^*(x) \neq 0$. So $X^*$ separates points.

This is good enough for the applications I have in mind, but I'm still interested in other characterisations when $X^*$ separates points, so feel free to contribute any more that you may have.

The following example is quite natural. Consider the Hardy class $H_p$ for some $0<p<1$. You can go to Kalton, Peck and Robert's book for general information on the quasi Banach space $H_p$, which consists of analytic functions on the disc satisfying certain conditions as one approaches the boundary.

The point I wanted to mention is that, even if the computation of the dual of $H_p$ can be very difficult, the fact that the dual is separating is nearly trivial since the evaluation'' functionals $\delta_z(f)=f(z)$ with $|z|<1$ are easily check to be bounded.

• That's a special case of the sufficient condition mentioned in my comment to the question. – Jochen Wengenroth Jan 27 '15 at 9:55
• Of course. This is just an example. By the way, your condition is both sufficient and necessary. Cheers, Félix. – Félix Cabello Sánchez Jan 27 '15 at 10:20
• Good example, thanks! It's nice to recognise specific examples where concrete functionals can be found. – user14166 Jan 27 '15 at 15:32

A natural condition that ensures that $$X^*$$ separates the points of $$X$$ is the existence of a Schauder basis. More generally, if $$X$$ has complete total biorthogonal system, then $$X^*$$ separates the points of $$X$$.