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Let $K \in \mathbb{R}^d$ be a compact set and consider the space of Hölder continuous functions $C^{0,\gamma}(K)$ with norm $||f||_{C^{0,\gamma}}:=||f||_{\infty}+\sup_{x,y \in K,x \neq y}\frac{|f(x)-f(y)|}{|x-y|^{\gamma}}$. Assume we have a bounded sequence $\{f_n\} \subset C^{0,\gamma}(K)$, i.e. $\exists C>0$ s.t. $\sup_{n}||f_n||_{C^{0,\gamma}} \leq C$. Under what conditions can we say the sequence $\{f_n\}$ is pre-compact in $C^{0,\gamma}(K)$? In other words, what are the sufficient conditions that guarantee there exist a subsequence $f_{n_k}\subset f_n$ and a $f \in C^{0,\gamma}(K)$ such that $f_{n_k} \xrightarrow{C^{0,\gamma}}f$ ?

Thank you very much!

Davide Giraudo
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  • Equicontinuity of $f_n$ would be enough. This pretty much follows from Arzela-Ascoli, with a little work. –  Nov 01 '13 at 04:30
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    @Sanchez It would not. Every sequence $f_n$ that is bounded in $C^{0,\gamma}$ is equicontinuous. But since the space is not finite-dimensional, its unit ball is not precompact. (For an explicit counterexample, take $f(x)=|x-1/n|^\gamma$ in one dimension). – user103402 Nov 01 '13 at 10:38
  • @user103402, ah you are right. I was thinking that once we show the existence of limit of sa ubsequence, we only need to make sure it still lies in the Holder space, but I forgot that I need to check convergence in this norm too. –  Nov 01 '13 at 15:38

1 Answers1

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A sufficient condition is: the sequence $f_n$ is bounded in $C^{0,\gamma+\epsilon}$ for some $\epsilon>0$. For a proof, see Is there a reference for compact imbedding of Hölder space?

A slightly weaker sufficient condition: there exists a modulus of continuity $\omega:(0,\infty)\to(0,\infty)$ such that $\omega(\delta)=o(\delta^\gamma) $ as $\delta\to 0$ and $$|f_n(a)-f_n(b)|\le \omega(|a-b|) \tag{1}$$ for all $a,b$ in $K$, and for all $n$.

The proof is the same as in thread linked above. Namely, extract a uniformly convergent subsequence by Ascoli-Arzelà and use (1) in the form $$\frac{|f_n(a)-f_n(b)|}{|a-b|^\gamma} \le \tau(|f_n(a) -f_n(b) |) $$ where $\tau$ is such that $\tau(t)\to 0$ as $t\to 0$. The existence of $\tau$ follows by rewriting (1) as $$\omega^{-1}(|f_n(a)-f_n(b)|)\le |a-b| $$ where $\omega^{-1}$ is the inverse of $\omega$, hence satisfies $\omega^{-1}(t)/t^{1/\gamma}\to\infty$ as $t\to0$.

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