Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This question is posted and unanswered from math.stackexchange.

Suppose $0 < \alpha < \beta$ and $\Omega$ is bounded. Then, the Hölder space $C^\beta(\Omega)$ is compactly imbedded to $C^\alpha(\Omega)$. See the wikipedia page:

http://en.wikipedia.org/wiki/H%C3%B6lder_condition

More precisely, I want to know the exact reference of the theory related to the following statement: [Claim] Given $\{f_n\}$ is a sequence of functions with $\|f_n\|_\beta <1$ for all $n$. Then, there exists a subsequence $\{n_k\}$ and $f\in C^\alpha$ such that, $\lim_{k\to \infty} \|f_{n_k} - f\|_\alpha = 0$. In the above, $\|\cdot\|_\alpha$ is Hölder-$\alpha$ norm.

However, I could not find a precise reference from some books on functional analysis.

1) Can anybody indicate a precise reference for this theorem?

2) If possible, I would like to know a reference on the similar result on parabolic Hölder space.

Thanks.

share|improve this question
    
The proof for (1) is already given in the Wiki article you linked to. –  Willie Wong Jul 24 '13 at 11:52
    
@WillieWong Yes, I guess for (2), the proof is similar, but more complicated. I just want to save my work by by citing a reference, but wiki is not acceptable formally. Nevertheless, it's not for parabolic H\"older space. –  kenneth Jul 24 '13 at 14:21
1  
I think that the Arzela-Ascoli theorem does the trick. –  Liviu Nicolaescu Jul 24 '13 at 15:12
add comment

2 Answers 2

For non-integer values of $\alpha$, the space $C^\alpha$ has a nice characterisation in terms of wavelet coefficients, see "Wavelets and Operators" by Yves Meyer. With that characterisation (essentially a weighted $\ell^\infty$ bound on the wavelet coefficients), the compactness statement boils down to the (trivial) statement that the set of sequences bounded by some fixed sequence $\{a_n\}$ converging to $0$ is compact in $\ell^\infty$. For parabolic Hölder spaces (or any non-Euclidean scaling for that matter), Meyer's characterisation and therefore the compactness of the embedding still works, provided that one considers a suitably scaled tensor product wavelet basis.

share|improve this answer
add comment

I may be wrong, but I think this fits the requirements of what you are asking. Consider Holder continuous functions defined on $\mathbb{R}$. Specifically, $f_n:\mathbb{R}\to\mathbb{R}$ given by $$ f_n(x) = \frac{1}{2}(\max\{ x-n,0\})^\beta \ \ \ \forall x\in\mathbb{R},\ n\in\mathbb{N} .$$ It follows that $f_n\in C^\beta (\mathbb{R})$ and $||f_n||_\beta = \tfrac{1}{2}<1$ for all $n\in\mathbb{N}$. It also follows that $f_n(x)\to 0$ for all $x\in\mathbb{R}$ as $n\to\infty$ (albeit not uniformly). However, on any compact interval (where $f_n$ does converges uniformly to 0 as $n\to\infty$) any subsequence of $f_n$ must also converge to 0 as $n\to\infty$. Therefore, there is only one function corresponding to $f$ in this scenario, namely $f\equiv 0$. It also follows that for any $\alpha \in (0,\beta )$, we have $f_n\not\in C^\alpha (\mathbb{R})$ (as it is only locally Holder continuous of degree $\alpha$, not globally). This eliminates the possibilty of taking the difference $||f_n-f||_\alpha$ completely.

I also feel I should ask why you have supposed the condition $||f_n||_\beta <1$. It doesn't seem that there is too much special about 1. It does seem that the intention here is to consider $f_n$ defined on a compact/bounded set, however in your claim, it was not explicitly stated.

share|improve this answer
    
perhaps one should take $||f_{n_k}- f||_\alpha$ with the Holder $\alpha$ norm on an arbitrary compact subset of the original set. –  JCM Jul 24 '13 at 11:46
2  
Compact embedding is only true on bounded domains. In the Wiki article that the OP referred to the assumption is stated, but the OP failed to copy it into the question. –  Willie Wong Jul 24 '13 at 11:54
    
@JCM Thanks, your answer is correct. But, as you mentioned, I forgot to mention its domain is a bounded set. This is already added in the above question now. By the way, one can just put $\|f_n\|_\beta < K$ instead of $<1$. –  kenneth Jul 24 '13 at 14:42
    
On second look, it seems that I was using the Holder semi-norm and not the Holder norm. The argument I employed was not quite correct. –  JCM Jul 24 '13 at 14:51
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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