$\DeclareMathOperator\Lip{Lip}\DeclareMathOperator\AE{AE}$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.
Best Answer
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.