7
$\begingroup$

Let $(X,d,\mu)$ be a separable metric measure space on which every ball has positive but finite measure.

I've come across the definition of a homogeneous Fractional Hajlasz-Sobolev spaces $M^{s,p}(X,d,\mu)$ which are defined as the collection of functions $f \in L^p_{loc}(X)$ for which there exists some $g \in L^p(X)$ satisfying $$ |f(x)-f(y)|\leq d(x,y)^s[g(x)+g(y)], $$ outside a set of $\mu$-measure $0$. These spaces are Banach spaces with the norm $$ \|f\|\triangleq \inf_{g \in D^s(f)} \|g\|_{L^p(X)}, $$ where $D^s(f)$ is the class of functions $g$ satisfying the first inequality.

If now $(Y,\rho,\nu)$ is another separable metric measure space satisfying the same conditions as $(X,d,\mu)$ then is there a separable Frechet space of Borel-measurable functions from $X$ to $Y$ generalizing the fractional Hajlasz-Sobolev space?

$\endgroup$
  • $\begingroup$ I hope I answered your question and thank you for the bounty, but next time you offer a bounty, you should wait until the end of the bounty period since there might be other, better answers. $\endgroup$ – Piotr Hajlasz Jul 22 at 13:16
  • 1
    $\begingroup$ The answer is fantastic. I'll take that advice also, but since these spaces are named after you I wouldn't expect anyone to give a better answer :) $\endgroup$ – MrMMS Jul 22 at 13:19
7
+50
$\begingroup$

The question is not stated in a very clear manner, but nevertheless, the answer is: no.

Separability. The space $M^{1,p}(X,d,\mu)$ is not separable even if $X$ is the standard ternary Cantor set, $d$ is the Euclidean metric $d(x,y)=|x-y|$ and $\mu$ is the natural Hausdorff measure. This was proved in [R]. Therefore, there is no reason to expect separability in a more general case of mappings between metric spaces.

The space of mappings from $X$ to $Y$ is, in general, not linear so it cannot be a Banach or Frechet space. As I understand MrMMS wants to investigate the Frechet manifold structure, but it is even not clear how to define a metric and hence the topology in the space of $M^{s,p}$ mappings from $X$ to $Y$. A natural way to define a metric is to assume that $Y$ is isometrically embedded in a Banach space, say $Y\subset \ell^\infty$ and then we define the space $M^{s,p}(X,Y)$ as the space of mappings $u\in M^{s,p}(X,\ell^\infty)$ such that $u(x)\in Y$ for $\mu$ almost all $x\in X$. Since $M^{s,p}(X,\ell^\infty)$ is a Banach space, we have a natural metric in $M^{s,p}(X,Y)$ inherited from the norm of $M^{s,p}(X,\ell^\infty)$.

The problem is that very little is known about the metric and the topological structure of the space $M^{s,p}(X,Y)$, even in a very classical case when we have Sobolev mappings from a comapact manifold into a compact metric space $W^{1,p}(M,Y)$.

Why is it so difficult? The proof that the space of smooth mappings between manifolds has a (infinitely dimensional) manifold structure steams from the fact that mappings can be localized in coordinate charts on manifolds. Since Sobolev mappings are not necessarily continuous, this approach is not available and basically, the only way to define the metric (and topology) in $W^{1,p}(M,Y)$ is through the isometric embedding of $Y$ into $\ell^\infty$ (or another Banach space). This approach does not provide any natural way to equip the space $W^{1,p}(M,Y)$ with a structure of a Banach or a Frechet manifold.

Let me just mention one result that show how sensitive the definition is.

Let $Y\subset \mathbb{R}^{n+2}$ be compact. There are two different ways to define $W^{1,p}(M,Y)$, where $M$ is a compact Reiamnnian manifold. One way is as $$ W^{1,p}(M,Y)=\{u\in W^{1,p}(M,\mathbb{R}^{n+2}:\, u(x)\in Y\, a.e.\} $$ and another is through the (Kuratowski) isometric embedding $\kappa:Y\to\ell^\infty$ as described above. Denote the space defined through the Kuratowski embedding by $W^{1,p}_\kappa(M,Y)$. Let $S^n$ be the standard sphere. The next result was proved in [H].

Theorem. There is a compact and connected set $Y\subset \mathbb{R}^{n+2}$ such that Lipschitz mappings $\operatorname{Lip}(S^n,Y)$ are dense in $W^{1,n}(S^n,Y)$, while if $\kappa:Y\to\ell^\infty$ is the Kuratowski embedding, then Lipschitz mappings $\operatorname{Lip}(S^n,Y)$ are not dense in $W^{1,n}_\kappa(S^n,Y)$.

This result shows that the metric and topology of the space of Sobolev mappings into metric spaces heavily depends on what isometric embedding we choose: with one choice Lipschitz mappings are dense, but with another one they are not dense.

[H] P. Hajlasz, Sobolev mappings: Lipschitz density is not an isometric invariant of the target. Int. Math. Res. Not. IMRN Vol. 2011, no.12, 2794-2809.

[R] J. Rissanen, Wavelets on self-similar sets and the structure of the spaces $M(E, \mu)$, Ann. Acad. Sci. Fenn. Math. Diss. 125 (2002).

$\endgroup$
  • $\begingroup$ Wow, this is much more delicate than i realized. I'll have to comb through this answer in more detail and the related papers you posted; but this is good news (I was hoping that it wasn't (/wasn't clear if it is) a Frechet manifold). $\endgroup$ – MrMMS Jul 22 at 13:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.