A Banach space $\mathcal{X}$ is said to have the approximation property (AP) if, for every compact set $K \subset \mathcal{X}$, there is a sequence of finite rank operators $\{U_n : \mathcal{X} \to \mathcal{X}\}_{n=1}^\infty$ such that $$\sup_{f \in K}\|f - U_n f\|_\mathcal{X} \to 0 \quad \text{as} \quad n \to \infty.$$ Let $\Omega \subset \mathbb{R}^d$ be a bounded, open set. I am wondering how much is known about standard function spaces possessing AP. I am particularly interest in the Lebesgue spaces $L^p(\Omega)$ for $1 \leq p \leq \infty$, the Sobolev spaces $W^{k,p}(\Omega)$ with $k \in \mathbb{N}$ and $1 \leq p \leq \infty$, the uniformly continous functions $C^k(\bar{\Omega})$ for $k \in \mathbb{N}_0$, and maybe even the Besov and Holder spaces, but the previous three are the main ones. The best I could find is this which says that $W^{k,p}(\Omega)$ with $1 < p < \infty$ and $\partial \Omega$ is class $C^\infty$ has AP. Or this which explicitly constructs a Schauder basis for $C^k ([0,1]^d)$ and $W^{k,p}([0,1]^d)$ with $k \in \mathbb{N}_0$ and $1 \leq p < \infty$. What results are known for more general domains?

  • $\begingroup$ Your definition of AP is not quite correct; you should say that the identity is in the closure of the finite rank operators for the topology of uniform convergence on compact sets. That means that there is a net $(U_n)$ such that for each $K$ etc. Note that all $L_p$ spaces have the AP, even the MAP. $\endgroup$ Jul 7, 2021 at 19:09
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    $\begingroup$ All $L^p$ spaces have the MAP, as Dirk said. You can deduce this from what is in Lindenstrauss-Tzafriri. Every separable $L^p$ space, $1\le p < \infty$ is isometrically isomorphic to $X \oplus_p Y$, where $X$ is either the zero space or $L^p(0,1)$ and $Y$ is $\ell^p(n)$ with $n = 0,1,...$ or $n=\aleph_0$. All $L^\infty$ spaces are $C(K)$ spaces, so they also have the MAP. $\endgroup$ Jul 7, 2021 at 20:23
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    $\begingroup$ Your definition of AP is OK but looks strange. I had to look twice to see that it is equivalent to one of the usual definitions (what Dirk wrote or what you wrote in your comment). $\endgroup$ Jul 7, 2021 at 20:27
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    $\begingroup$ As for $C(K)$: The space $L_\infty$ is an abstract $M$-space with unit, hence by Kakutani's representation theorem isometrically isomorphic to a space $C(K)$ with a nonmetrisable compact $K$. (Or use the commutative Gelfand-Naimark theorem). -- Many Besov spaces are isomorphic to $\ell_p$-sums of $\ell_q$'s (and hence have the AP); I don't know what happens when the boundary is non-smooth... $\endgroup$ Jul 7, 2021 at 22:16
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    $\begingroup$ Here's a comment on $C^1 [0,1]^2$. Grothendieck, in his 1953 Canad. J. paper claimed that this space is complemented in a $C(K)$-space. A couple of years later, in a short paper in Ann. Inst. Fourier (Zbl 0072.12003) he listed all his errors in his work on functional analysis (I think there were four of them), including the above claim, now saying that $C^1 [0,1]^2$ is not complemented in a $C(K)$. If it were, it would inherit the bounded AP from $C(K)$... $\endgroup$ Jul 9, 2021 at 20:05


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