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Jan 23, 2023 at 12:23 comment added Nik Weaver Oh, you're right. It's false even if $K$ is compact. Take $K = [0,1]$ and $A = l^\infty$ and define $f: K \to A$ by setting $f(2^{-n}) = 2^{-n}e_n$ and extending linearly between the points $2^{-n}$. That shouldn't be approximable in Lipschitz norm by anything in the tensor product.
Jan 22, 2023 at 16:51 comment added MSMalekan @NikWeaver: I did that, but this proves that $Lip(K)\otimes A$ is uniformly dense in $Lip(K,A)$, not w.r.t norm $\|f\|_\alpha=\|f\|_\infty+\rho_\alpha(f)$.
Jan 22, 2023 at 16:45 comment added Nik Weaver Find an $\epsilon$-net and use a Lipschitz partition of unity.
Jan 22, 2023 at 15:56 comment added MSMalekan @NikWeaver that if $K$ is compact then we have the equality in both questions 1 and 2. I tried but I could not. This is my first encounter with Lipschitz functions, so I don't know how to start.
Jan 22, 2023 at 15:08 comment added Nik Weaver Which statement? Anyway, I think these are good exercises for you.
Jan 22, 2023 at 15:03 comment added MSMalekan @NikWeaver Great! Could you please provide me a survey which does prove this statement.
Jan 22, 2023 at 14:26 comment added Nik Weaver You should be okay if $X$ is compact, but for noncompact $X$ there are easy counterexamples. E.g., take $X=\mathbb{N}$ with standard metric and $A = l^\infty$. The function $n \mapsto e_n$ is then Lipschitz but can't be uniformly approximated by finite linear combinations appearing in the tensor product.
Jan 22, 2023 at 13:13 comment added MSMalekan @NikWeaver in number 1, I'm looking for equality, is this unexpected?
Jan 22, 2023 at 12:42 comment added MSMalekan @NikWeaver thank you for the comments. By the closure I mean the closure with the norm introduced in my explanation. But it is of interest for me to know the predual of this space if $A$ is a dual Banach algebra.
Jan 21, 2023 at 22:31 comment added Nik Weaver You say "the closure of ${\rm Lip}_\alpha(K) \otimes A$" --- closure in what, the norm topology I assume? Here the general advice is that the norm topology is rarely useful in Lipschitz spaces. You probably want to use the weak* topology, which will exist if $A$ is a dual space (and you choose a better norm).
Jan 21, 2023 at 22:29 comment added Nik Weaver For number 1, notice that if $A$ us unital then so is ${\rm Lip}_\alpha(K) \overline{\otimes} A$, so it's going to be hard for this to be an ideal.
Jan 21, 2023 at 20:14 history asked MSMalekan CC BY-SA 4.0