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Feb 14, 2022 at 9:40 comment added Yury Korolev @OnurOktay ok, thanks.
Feb 12, 2022 at 4:46 comment added Onur Oktay $\ell^{\infty}\hat{\otimes}_{\pi}\ell^{\infty}$ 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.
Feb 11, 2022 at 16:48 history edited Yury Korolev CC BY-SA 4.0
corrected a typo
Feb 11, 2022 at 16:39 comment added Yury Korolev @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?
Jan 26, 2022 at 13:45 comment added Yury Korolev @OnurOktay Thank you! This actually could be something, since a certain type of wavelets form a basis in $lip_0(D^\alpha)$, hence a possible connection to Besov spaces (see here as well as Section 8.5 in Weaver's book). Thank you for the reference, I will look into this.
Jan 25, 2022 at 16:55 comment added Onur Oktay Did you consider to adapt the atomic decompositions of the classical Hölder spaces $C^{\alpha}$ to $Lip(D\times D)$ for common (e.g., Lipschitz domain) $D$? Please see Triebel's monographs, for instance Chapter 2 in "The theory of Function spaces III". In the same book, you can find atomic decompositions for more general sets (see Ch. 8). Perhaps this method won't characterize the projective tensor product, but provide various (perhaps non-closed) subspaces that can be characterized via their atoms, for example, the Besov spaces $B_{1,1}^{s}$ for $s\geq\alpha$.
Jan 25, 2022 at 14:12 comment added Onur Oktay $\ell^{\infty}\hat{\otimes}_{\epsilon}\ell^{\infty}$ and $\ell^{\infty}$ are not linearly isomorphic. Thus, $Lip_0(D)\hat{\otimes}_{\epsilon} Lip_0(D) \neq Lip_0(D\times D)$ whenever both $Lip_0(D)$ and $Lip_0(D\times D)$ are both isomorphic to $\ell^{\infty}$ for some $D$. This doesn't make the question above any harder or easier though.
Jan 25, 2022 at 13:54 comment added Nik Weaver No idea. I've never thought about projective tensor products of Lipschitz spaces --- I would expect injective tensor products to be more natural (but I haven't thought about them either).
Jan 25, 2022 at 13:52 comment added Yury Korolev @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
Jan 25, 2022 at 13:50 comment added Yury Korolev @NikWeaver Is there at least a way to check for a given function $g \in Lip_0(D \times D)$ that $g \in Lip_0(D) \, \hat \otimes_\pi \, Lip_0(D)$?
Jan 25, 2022 at 13:28 comment added Yury Korolev @OnurOktay On pp. 49-50 in his book, Ryan shows that $C(D,X)$ can be idetified with $C(D) \, \hat \otimes_\varepsilon \, X$ for any Banach space $X$, but this involves a density argument which may fail for $Lip_0$ since it is non-separable. I haven't checked, though
Jan 25, 2022 at 13:21 comment added Onur Oktay Can we also identify $Lip_0(D\times D)$ with $Lip_0(D)\hat{\otimes}_{\epsilon} Lip_0(D)$? If so, does the analogy with the $C(D)$ answer the question above up to some extent?
Jan 25, 2022 at 13:15 comment added Yury Korolev @DirkWerner Thank you for noticing, I have edited the post now
Jan 25, 2022 at 13:15 history edited Yury Korolev CC BY-SA 4.0
corrcted inaccuracy regarding nuclear operators and projective tensor product
Jan 25, 2022 at 13:06 comment added Yury Korolev @OnurOktay It is known that $C(D \times D)$ can be identified with the injective tensor product $C(D) \, \hat \otimes_\varepsilon \, C(D)$. I am not sure about the projective product
S Jan 25, 2022 at 6:46 history suggested Dirk Werner CC BY-SA 4.0
typos and LaTeX
Jan 24, 2022 at 20:59 comment added Onur Oktay I wonder if the same question has an answer for $C(D)$ in place of $Lip_0(D)$. Let's also assume that $D$ is hyperstonean so that $C(D)$ is a dual Banach algebra.
Jan 24, 2022 at 20:43 comment added Nik Weaver This is a good question, I'd like to know the answer too!
Jan 24, 2022 at 18:39 comment added Dirk Werner Have you looked at $D=[0,1]$? Here $\mathrm{Lip}_0(D)$ is isometric to $L_\infty(D)$.
Jan 24, 2022 at 18:37 comment added Dirk Werner To identify the projective tensor product with the space of nuclear operators requires the approximation property.
Jan 24, 2022 at 18:34 review Suggested edits
S Jan 25, 2022 at 6:46
Jan 24, 2022 at 17:17 history asked Yury Korolev CC BY-SA 4.0