$\newcommand{\projtenprod}[2]{#1 \; \hat\otimes_\pi #2}$ $\DeclareMathOperator\Lip{Lip}\DeclareMathOperator\AE{AE}$ $\newcommand{\norm}[1]{\| #1\|}$ $\newcommand{\abs}[1]{| #1|}$

Projective tensor product. Let $X$ and $Y$ be Banach spaces. Let $$ X \otimes Y := \bigcup_{n\in\mathbb N} \bigcup_{\substack{x_i \in X \\y_i \in Y}}\sum_{i=1}^n {x_i \otimes y_i} $$ denote their algebraic tensor product. The projective tensor product $\projtenprod{X}{Y}$ is the completion of the algebraic tensor product with respect to the projective norm $$ \pi(u) := \inf \left\{\sum_{i=1}^n \norm{x_i} \, \norm{y_i} \colon u = \sum_{i=1}^n {x_i \otimes y_i} \right\}, $$ where the infimum is taken over all representations of the tensor $u = \sum_{i=1}^n {x_i \otimes y_i}$.

The projective tensor product is given by (Ryan, "Introduction to Tensor Products of Banach spaces", Prop. 2.8) $$ \projtenprod{X}{Y} = \bigcup_{\substack{x_i \in X \\y_i \in Y}}\sum_{i=1}^\infty {x_i \otimes y_i}, $$ and $$ \pi(u) = \inf \left\{\sum_{i=1}^\infty \norm{x_i} \, \norm{y_i} \colon u = \sum_{i=1}^\infty {x_i \otimes y_i} \right\}, $$ where the infimum is taken over all representations of the tensor $u = \sum_{i=1}^\infty {x_i \otimes y_i}$.

If $X$ itself is a dual space with a predual $X^\diamond$, and either $X$ or $Y$ has the approximation property, then the projective tensor product $\projtenprod{X}{Y}$ can be identified with the space of nuclear operators $\mathcal N(X^\diamond,Y)$.

Space of Lipschitz functions. Let $D \subset \mathbb R^n$ be compact and $e \in D$. Denote by $\Lip_0(D)$ the space of all Lipschitz functions on $D$ (with respect to the Euclidean metric $d(p,q) := \norm{p-q}$, $p,q \in D$) vanishing at $e$, equipped with the following norm $$ \|f\|_{\Lip_0} := \sup_{\substack{p,q \in D \\ p \neq q}} \frac{\abs{f(p)-f(q)}}{d(p,q)}. $$

For $0<\alpha<1$, denote by $D^\alpha$ the metric space $(D,d^\alpha)$, where $d^\alpha(p,q) := \norm{p-q}^\alpha$. By $\Lip_0(D^\alpha)$ we will denote the space of Lipschitz functions with respect to the metric $d^\alpha$.

The space $\Lip_0(D)$ (resp. $\Lip_0(D^\alpha)$) has a unique predual, called the Arens-Eells space $\AE(D)$ (resp. $\AE(D^\alpha)$), also known as the Lipschitz-free space. It is known (Nik Weaver, "Lipschitz Algebras", 2nd ed., Thm. 8.49) that $\AE(D^\alpha)$ is linearly homeomorphic to $\ell^1$, and $\Lip_0(D^\alpha)$ to $\ell^\infty$ if $0<\alpha<1$. In this case, $\Lip_0(D^\alpha)$ also has a second predual $\mathrm{lip}_0(D^\alpha)$ (so-called little Lipschitz functions), which is linearly homeomorphic to $c_0$.

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Is there a concrete description of the projective tensor product $\projtenprod{\Lip_0(D)}{\Lip_0(D)}$ or $\projtenprod{\Lip_0(D^\alpha)}{\Lip_0(D^\alpha)}$ in terms of Lipschitz functions on $D \times D$?

There is a canonical identification of $\Lip_0(D \times D)$ with the space of vector-valued Lipschitz functions $\Lip_0(D,\Lip_0(D))$, which in its turn can be identified with the space of bounded linear operators $\mathcal L(\AE(D),\Lip_0(\Omega))$. With the inclusion $\mathcal N(\AE(D),\Lip_0(\Omega)) \subset \mathcal L(\AE(D),\Lip_0(\Omega))$, we have $$ \Lip_0(D \times D) = \Lip_0(D,\Lip(D)) = \mathcal L(\AE(D),\Lip_0(\Omega)) \\\supset \mathcal N(\AE(D),\Lip_0(D)) = \projtenprod{\Lip_0(D)}{\Lip(D)}. $$ If $g \in \Lip_0(D \times D)$ is a Lipschitz function, is there an intuitive condition on $g$ so that $g \in \projtenprod{\Lip_0(D)}{\Lip_0(D)}$ holds?

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Any help will be much appreciated.

  • 3
    $\begingroup$ To identify the projective tensor product with the space of nuclear operators requires the approximation property. $\endgroup$ Commented Jan 24, 2022 at 18:37
  • 1
    $\begingroup$ This is a good question, I'd like to know the answer too! $\endgroup$
    – Nik Weaver
    Commented Jan 24, 2022 at 20:43
  • 1
    $\begingroup$ @DirkWerner I'm also not sure what $L^\infty([0,1]) \, \hat \otimes_\pi \, L^\infty([0,1])$ would look like. Or even $\ell^\infty \, \hat \otimes_\pi \ell^\infty$, for that matter $\endgroup$ Commented Jan 25, 2022 at 13:52
  • 1
    $\begingroup$ @OnurOktay Many thanks for the reference, Onur. For $\alpha<1$, it is indeed easy to show that $B_{1,1}^\alpha(D \times D) \subset Lip_0(D^\alpha) \hat \otimes_\pi Lip_(D^\alpha)$. There doesn't seem to be an equality, though. My intuition is that it could be the space of functions that are in $B_{1,1}^\gamma(D \times D)$ for all $\gamma<\alpha$, i.e. a conjecture could be that $Lip_0(D^\alpha) \hat \otimes_\pi Lip_(D^\alpha) = \bigcap_{0<\gamma<\alpha} B_{1,1}^\gamma(D \times D)$. Are you aware of any results in this direction? $\endgroup$ Commented Feb 11, 2022 at 16:39
  • 1
    $\begingroup$ $\ell^{\infty}\hat{\otimes}_{\pi}\ell^{\infty}$ contains a complemented copy of $\ell^2$, whereas $B^{\alpha}_{1,1}$ is isomorphic to $\ell^1$ as a Banach space. I don't think your conjecture holds. $\endgroup$
    – Onur Oktay
    Commented Feb 12, 2022 at 4:46


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