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.


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.


1 Answer 1


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.

  • $\begingroup$ Is it similarly “obvious” hilbert space embeds into the lipschitz space? $\endgroup$
    – Ryan
    Commented Dec 21, 2021 at 5:07
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    $\begingroup$ @Ryan: Yes, because you may assume that the operator Bill describes has dense range in the Hilbert space (by replacing the Hilbert space by the norm closure of the range of the operator if necessary; this subspace is also a Hilbert space). Under the assumption that the operator described by Bill has dense range, the adjoint of this operator is a continuous, injective linear operator from the dual of the Hilbert space into the Lipschitz space. Identifying the Hilbert space with its dual space gets you the rest of the way. $\endgroup$ Commented Dec 21, 2021 at 12:05
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    $\begingroup$ Also note that this works for arbitrary separable $X$, it doesn't have to be compact. $\endgroup$
    – Nik Weaver
    Commented Dec 21, 2021 at 13:28
  • $\begingroup$ 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 $X$. $\endgroup$ Commented Dec 21, 2021 at 15:32
  • $\begingroup$ 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 $C[0,1]$. $\endgroup$ Commented Dec 21, 2021 at 22:04

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