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For a commutative Banach algebra $A$ and for any $0<\alpha<1$, let $\text{Lip}_\alpha(K,A)$ consist of all $A$-valued functions $f$ on a metric space $(K,\text d)$ with the property that $\rho_\alpha(f)<\infty$, where $$\rho_\alpha(f):=\sup\left\{\frac{\|f(x)-f(y)\|}{\text d(x,y)^\alpha}: x,y\in X, x\neq y\right\}.$$ As in the well-known scalar-valued case, the vector-valued Lipschitz algebra $\text{Lip}_\alpha(K,A)$ of order $\alpha$ becomes a complex Banach algebra with respect to pointwise operations, when endowed with the norm $\|\cdot\|_\alpha$ given by $\|f\|_\alpha:=\|f\|_\infty+\rho_\alpha(f)$ for all $f\in\text{Lip}_\alpha(K,A)$.

$\text{lip}_\alpha(K,A)$ consists of all $f\in\text{Lip}_\alpha(K,A)$ for which $$\lim_{\text d(x,y)\rightarrow0}\frac{\|f(x)-f(y)\|}{\text d(x,y)^\alpha}=0.$$ It is easily seen that $\text{lip}_\alpha(K,A)$ is a closed subalgebra of $\text{Lip}_\alpha(K,A)$. My questions are:

  1. Denote by $\text{Lip}_\alpha(K)\tilde\otimes A$ the closure of $\text{Lip}_\alpha(K)\otimes A$ in $\text{Lip}_\alpha(K,A)$. Is $\text{Lip}_\alpha(K)\tilde\otimes A$ an ideal of $\text{Lip}_\alpha(K,A)$?
  2. Similarly let $\text{lip}_\alpha(K)\tilde\otimes A$ be the closure of $\text{lip}_\alpha(K)\otimes A$ in $\text{lip}_\alpha(K,A)$. In this paper it was shown that for an interval $I=[a,b]$, $\text{lip}_\alpha(I)\tilde\otimes A=\text{lip}_\alpha(I,A)$. Does the same equality hold for an arbitrary comapct metric space $K$ rather than $I$?
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  • $\begingroup$ 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. $\endgroup$
    – Nik Weaver
    Commented Jan 21, 2023 at 22:29
  • $\begingroup$ 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). $\endgroup$
    – Nik Weaver
    Commented Jan 21, 2023 at 22:31
  • $\begingroup$ @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. $\endgroup$
    – MSMalekan
    Commented Jan 22, 2023 at 12:42
  • $\begingroup$ @NikWeaver in number 1, I'm looking for equality, is this unexpected? $\endgroup$
    – MSMalekan
    Commented Jan 22, 2023 at 13:13
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    $\begingroup$ 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. $\endgroup$
    – Nik Weaver
    Commented Jan 23, 2023 at 12:23

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