# Existence of a Gelfand triple involving the Arens–Eells space (aka Lipschitz free space)

$$\DeclareMathOperator\Lip{Lip}\DeclareMathOperator\AE{AE}$$Background

Gelfand triples. Let $$\mathcal B$$ be a Banach space, $$\mathcal B^*$$ its dual space, and $$\mathcal H$$ a Hilbert space. The triple $$(\mathcal B,\mathcal H, \mathcal B^*)$$ is called a Gelfand triple if the following embeddings are continuous $$\mathcal B \hookrightarrow \mathcal H \hookrightarrow \mathcal B^*.$$

An example that I am familiar with is the triple $$(BV(\Omega), L^2(\Omega), BV^*(\Omega))$$, where $$\Omega \subset \mathbb R^2$$ is bounded and $$BV(\Omega)$$ is the space of functions of bounded variation.

Arens-Eells space. Let $$X$$ be a compact pointed metric space with base point $$e$$. An elementary molecule is defined as follows (Nik Weaver, Lipschitz Algebras, 2nd ed.) $$m_{pq} := \delta_p - \delta_q,$$ where $$\delta_p, \delta_q$$ are delta-functions placed at $$p,q$$.

The Arens-Eells space $$\AE(X)$$ (also known as the Lipschitz-free space) is the completion of the linear span of elementary molecules with respect to the Arens-Eells norm $$\|{m}\|_{\AE} := \inf \left\{\sum_{i=1}^n |{a_i}| d(p_i,q_i) \colon m = \sum_{i=1}^n a_i m_{p_iq_i} \right\},$$ where $$d(p,q)$$ is the distance between $$p,q \in X$$.

The dual of the Arens-Eells space is the $$\Lip_0(X)$$ space of all Lipschitz functions on $$X$$ vanishing at $$e$$, equipped with the following norm $$\|f\|_{\Lip_0} := \Lip(f),$$ where $$\Lip(f)$$ denotes the Lipschitz constant.

Question

Is there a Hilbert space $$\mathcal H$$ such that $$(\AE(X), \mathcal H, \Lip_0(X))$$ form a Gelfand triple? It would suffice for me to think of $$X$$ as the unit ball in $$\mathbb R^n$$ with base point $$0$$, equipped with the euclidean metric.

Your question is equivalent to asking if there is an injective bounded linear operator from $$AE(X)$$ into a Hilbert space when $$X$$ is a compact metric space. The answer is "yes" because $$AE(X)$$ is separable. It is elementary to construct a nuclear injective linear operator from an arbitrary separable Banach space into a Hilbert space.
• Also note that this works for arbitrary separable $X$, it doesn't have to be compact. Dec 21, 2021 at 13:28
• Thank you all very much! Is there a ‘natural’ identification of such a Hilbert space? In certain cases (as discussed in this question), $AE$ is linearly isomorphic to $\ell^1$ and $Lip_0$ to $\ell^\infty$, so we could take $\mathcal H = \ell^2$, but I would like to identify $\mathcal H$ with some space of functions/distributions on $X$. Dec 21, 2021 at 15:32
• Well, If $X=C[0,1]$, then you can use $L_2(0,1)$ as the Hilbert space. That is pretty "natural". Every separable Banach space embeds isometrically isomorphically into $C[0,1]$, but it is a stretch to claim that there is a "natural" isometric embedding of every separable space into $C[0,1]$. Dec 21, 2021 at 22:04