# A concrete description of the projective tensor product of Lipschitz spaces

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

Background
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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Question
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 $$D \times D$$?

Thoughts
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.

• To identify the projective tensor product with the space of nuclear operators requires the approximation property. Commented Jan 24, 2022 at 18:37
• This is a good question, I'd like to know the answer too! Commented Jan 24, 2022 at 20:43
• @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 Commented Jan 25, 2022 at 13:52
• @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? Commented Feb 11, 2022 at 16:39
• $\ell^{\infty}\hat{\otimes}_{\pi}\ell^{\infty}$ 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. Commented Feb 12, 2022 at 4:46