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I'm studying the Banach Space of Hölder continuous functions $f:[0,1]\to\mathbb{R}^{+}$ with a parameter $\alpha$. In this space, I consider the usual Hölder norm $\|\cdot\|_\alpha$ and I'm looking for conditions needed to an operator be compact under this norm.

Arzelà-Ascoli Theorem gives conditions to get that for the supremum norm, but I'd like a version of this for the Hölder norm. What results we have about that?

Another possibility can be a result that allow to get compactness on Hölder norm by using the compactness on supremum norm given by Arzelà-Ascoli. Of course, in this case we will need extra hypotesys like to suppose a bound on Hölder norm (maybe plus something nice).

RobPratt
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  • It is unclear what you are asking. Can you separate your questions and also add some mathematical symbols? – Thomas Kojar Apr 12 '23 at 02:02
  • Holder functions with same constant L and exponent alpha, are equicontiniuous and so you can check the other assumptions for Arzela-Ascoli i.e. total boundedness and compact domain. – Thomas Kojar Apr 12 '23 at 02:04
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    Generally one has to be careful with subsequences between uniform and Holder norm, see counterexample here https://math.stackexchange.com/questions/4550870/does-uniform-convergence-imply-h%C3%B6lder-convergence of having uniform but not Holder norm convergence. – Thomas Kojar Apr 12 '23 at 02:04
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    I tried to rewrite the question to be more clear, but I don't know how to express that in a better way since I'm not looking to solve an exercise or something like that. I'm looking for a general result on how to show compactness on Hölder norm (in a similar fashion as Arzelà-Ascoli for supremum norm) – NewGuy23 Apr 12 '23 at 02:48
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    The simplest criterion follows from the compactness of the embedding of $C^\beta ([0,1])$ into $C^\alpha ([0,1])$ when $\beta >\alpha$. – Giorgio Metafune Apr 12 '23 at 08:46
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    Something that's important: Arzela-Ascoli provides a characterization of precompact sets of continuous functions. That is to say, it is an if and only if. Do you actually need that? I ask because @GiorgioMetafune 's comment referring to the compactness of the embedding from higher to lower Holder spaces is the most natural version for PDE specialists, but it is not a characterization. The singleton set ${f}\subset C^{\alpha}$ is compact, but it is possible for $f\in C^\alpha$ but not any $C^\beta$ with $\beta > \alpha$, and so the embedding does not provide a characterization. – Willie Wong Apr 12 '23 at 12:57
  • it is not clear to me what you are asking. You are saying "an operator", but what kind of an operator? Could you give an example of an application of Arzela-Ascoli to a proof of compactness of an operator (as opposed to, a set of fuctions) that you have in mind? – Kostya_I Apr 12 '23 at 13:02
  • Another way to think about this is: Arzela-Ascoli is so prominent in analysis precisely because it is a characterization. To show that a set of continuous functions on a compact space is precompact, it provides the minimal conditions that you need to check. If you are just interested in sufficient conditions, it would help to specify how much "room" we have in terms of an acceptable answer. // Lastly, I want to mention that the $C^\beta \hookrightarrow C^\alpha$ embedding result can be further improved: https://math.stackexchange.com/a/547816/1543 Though it still doesn't treat singletons. – Willie Wong Apr 12 '23 at 13:05
  • @Kostya_I Here's a standard homework exercise in that vein: https://math.stackexchange.com/questions/3087124/compact-operator-by-proving-ascoli-arzel%C3%A0 – Willie Wong Apr 12 '23 at 13:15
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    @WillieWong In the little Holder space $h_\alpha={u \in C^\alpha: [u]\delta \to 0}$, with $[u]\delta=\sup |u(x)-u(y)||x-y|^{-\alpha}, |x-y| \leq \delta}$ the characterization of compactness should be the following. A set $B \subset h_\alpha$ is pre-compact if and only if is bounded and $[u]\delta \to 0$ as $\delta \to 0$, uniformly in $B$. I doubt that there is a simple characterization in $C^\alpha$ since the relationship between $h\alpha, C_\alpha$ is similar to that between $C, L^\infty$. – Giorgio Metafune Apr 12 '23 at 17:10
  • Sounds like you want the Holder analogue of equicontinuity. – Deane Yang Apr 13 '23 at 13:01
  • @DeaneYang in a certain way it is! An analogue of equicontinuity in the sense that, together with a bound, implies compacity on the Holder norm. – NewGuy23 Apr 13 '23 at 16:46
  • In fact, I don't actually need an equivalence like we get from Arzelà-Ascoli theorem. Just sufficient conditions to get compacity would be enough – NewGuy23 Apr 13 '23 at 17:00
  • Also check this: https://mathoverflow.net/questions/372891/arzelà-ascoli-theorem-and-hölder-spaces/372918#372918 – Pietro Majer Jan 09 '24 at 08:54
  • And this: https://mathoverflow.net/questions/93638/dual-of-the-space-of-hölder-continuous-functions/93667#93667 – Pietro Majer Jan 09 '24 at 09:03

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While the OP is pondering which interpretation of the question is the right one, here are some relevant references. The characterization of compactness in the little Hölder space given by Giorgio Metafune is Thm.3.2 in (Johnson, 1970).

As for the big Hölder and Lipschitz spaces, a 2019 monograph dedicated to these function spaces does not seem to contain an analog of an Arzelà-Ascoli-like characterization of compact subsets. So that could still be an open problem.


Update: Some incidental searching of the literature uncovered this nice (and essentially hot of the presses) article:

Gulgowski, Jacek; Kasprzak, Piotr; Maćkowiak, Piotr, Compactness in Lipschitz spaces and around, ZBL07762688, arXiv:2205.04543.

Let me quote from their Introduction, which confirms that there was a gap in the literature concerning compactness characterizations in Lipschitz and Hölder spaces:

Because of their applications—[...]—Lipschitz and Hölder continuous functions constantly enjoy great interest among researchers. Each year numerous scientific articles and chapters are devoted just to studying their properties. Even whole books on Lipschitz and Hölder functions are written. [...] What they lack, however, is a (strong) compactness criterion in such spaces. No matter how hard we tried, we could not find one in the literature, although, as it turned out, in the case of real-valued functions it had stayed hidden in a plain sight for more than half a century. In a paper of Cobzaş from 2001 it is even stated: “[...] apparently there is no compactness criterion in spaces of Hölder functions, and some criteria given in the literature turned to be false” (see [8, p. 9]).

In the body of the article they fill this void with a mixture of old and recent ideas. The recent ideas are really only needed to state the results in the generality that they want, which is for functions valued in Banach spaces. For scalar valued functions old ideas are are actually sufficient, which is kind of obvious in retrospect.

The key idea relies on the following observation. Let $(X,d)$ be a metric space and denote by $\def\Lip{\operatorname{Lip}} \Lip_\alpha(X)$ the space of $\alpha$-Lipschitz/Hölder continuous scalar valued functions on $X$ ($0 < \alpha \le 1$, with $\alpha=1$ corresponding to the Lipschitz case), as well as $\Delta_X = \{(x,x) \mid x \in X\}$, $X^{(2)} = X \times X \setminus \Delta_X$.
Lemma: The ( so-called de Leeuw's ) map $$ \Phi\colon \Lip_\alpha(X) \to C(X^{(2)}), \quad \Phi(f)(x,y) = \frac{f(x) - f(y)}{d(x,y)^\alpha} $$ has only constants in its kernel, has a closed range and is a homeomorphism of $\Lip_\alpha(X) / \{\text{const}\}$ onto its image.

Proof: Mostly self-evident, taking into account the form of the Hölder norm and the $\sup$ norm on $C(X^{(2)})$. It is worth noting that the range is closed, since it can be realized as the kernel of a continuous linear map, for instance $\Psi \colon C(X^{(2)}) \to C(X^{(2)}) \times C(X^{(3)})$, where $X^{(3)} = X \times X \times X \setminus (\Delta_x \times X \cup X \times \Delta_X)$ and $$\Psi(g) = (g(x,y)+g(y,x), ~~ d(x,y)^\alpha g(x,y) + d(y,z)^\alpha g(y,z) - d(x,z)^\alpha g(x,z)). \tag*{$\Box$} $$

The statement of the Lemma is deliberately vague. There are different versions of Lipschitz spaces (local, global, global and uniformly bounded), which can be accommodated by adjusting $C(X^{(2)})$ and the topology on it. The simplest case is just to take uniformly bounded continuous functions on $X^{(2)}$ under the uniform norm, corresponding to uniformly bounded Lipschitz functions on $X$. For locally Lipschitz functions, one should take a filter of neighborhoods of the missing $\Delta_X$ rather than all of $X^{(2)}$.

Finally, the desired characterization is provided by the now obvious
Proposition: Compactness of $A \subset \Lip_\alpha/\{\text{const}\}$ is characterized requiring boundedness of $|f(x)|$ uniformly over $f\in A$ (for any one point $x\in X$) and by applying the appropriate version of the Arzelà-Ascoli theorem in $C(X^{(2)})$ to the image $\Phi(A)$ of de Leeuw's map.


In addition: When $X \subseteq \mathbb{R}^n$ is (at least) a sufficiently nice subset, there is another approach. Namely, via a combination of Rademacher's theorem and Morrey's inequality, we have (e.g., Thm.5.8.4 in Evans PDEs, or Thm.1.41 of Weaver Lipschitz functions) a linear isometry $\Lip_1(X) \cong W^{1,\infty}(X)$, which is the first order Sobolev space consisting of bounded $f$ such that $df \in L^\infty(X)^n$ with respect to Lebesgue measure. By definition of the Sobolev space, the exterior derivative is a linear isometry $d \colon W^{1,\infty}(X) \cong L^\infty(X)^n$.

So, as earlier, compactness of $A\subset \Lip_1(X)$ can be characterized by equiboundedness at a point and compactness of $d(A)$, while compactness in $L^\infty$ can be characterized using Thm.IV.8.18 (or adapting Thm.IV.5.6) of Dunford & Schwartz, vol.I.

I've not seen anything similar for Hölder spaces, though.

Igor Khavkine
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  • Theorem IV. 5.6, pag 260 in the first volume of Dunford and Sshwartz characterizes compactness in $L^\infty$ through small oscillation on measurable sets rather than neighborhoods. Maybe a similar results hold for $C^\alpha$ (and maybe it is useless). – Giorgio Metafune Apr 13 '23 at 09:10
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    @GiorgioMetafune Googling around, one can find a 2009 bachelor's thesis that tried to implement exactly that idea (at least for the Lipschitz case). Unfortunately, the reports on the thesis (in Czech only!) indicate that the attempted proof contains an unrecoverable error. So that path is open, but apparently not trivial. – Igor Khavkine Apr 13 '23 at 10:00
  • Oh, nice references! This will help me a lot, thank you! Definetely looks like it is an open problem. I need to dig more and try to find at least one side of the result: good (for me) sufficient conditions to get compacity. Should be easier than an equivalence. – NewGuy23 Apr 13 '23 at 16:57
  • And what do you mean by “Little” and “Big” Hölder spaces? – Pietro Majer Jan 09 '24 at 09:11
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    @PietroMajer "Big" Hölder is the usual definition. In the "little" space the $\alpha$-Hölder constant is zero, but could be nonzero for some $\beta > \alpha$. The definition was mentioned here by Giorgio Metafune, in his linked comment. – Igor Khavkine Jan 09 '24 at 10:00
  • OK thanks, I understand, $h_\alpha$ is the subset of $C^{0,\alpha}(X)$ that embeds isometrically into $C^0(X\times X)$, so that if $X$ is compact (or just totally bounded) then AA applies – Pietro Majer Jan 09 '24 at 11:00