Let $S\subset\mathbb{R}^n$ be compact, $\alpha,\beta\in(0,1)$, $\alpha>\beta$ and $X$ a Banach space. Under which assumptions on $X$ is the embedding $$C^\alpha(S;X)\subset C^\beta(S;X)$$ compact?

The For $X=\mathbb{R}^N$ compactness holds and is a consequence of Ascoli-Arzela's theorem. The above question seems to boil down to whether there's a Banach space valued version of Ascoli-Arzela's theorem. References are welcome. Thanks.


2 Answers 2


Associating to every $x\in X$ the constant function with value $x$ is an isometric embedding of $X$ into $C^\alpha(S,X)$ equipped with the norm $\|f\|_\alpha=\sup\lbrace \|f(s)\|_X:s\in S\rbrace +\sup\lbrace\frac{\|f(s)-f(t)\|_X}{|s-t|^\alpha}: s\neq t\rbrace$. If the inclusion is compact this yields the compactness of the identity on $X$ which hods only in the finite dimensional case. Of course, there is a version of Arzela-Ascoli for Banach space valued continuous functions: A subset $A$ of $C(S,X)$ is relatively compact if and only if $A(s)=\lbrace f(s): f \in A \rbrace$ is relatively compact in $X$ for every $s\in S$ and $A$ is equicontinuous.


Adding to the answer by Jochen Wengenroth, the theorem is proven in "Real and Functional Analysis" (Ch. III §3) by Serge Lang. There are certainly more sources, but I remember having to look through surprisingly many standard textbooks before finding a reference.

  • $\begingroup$ Erm. Good reference. But Lang does not study Hölder spaces at all. He only covers the classical Arzela-Ascoli theorem that characterizes the relatively compact sets in $C^0(S;X)$... $\endgroup$ Commented Aug 14, 2019 at 15:53
  • $\begingroup$ Yes, "the theorem" in my answer refers to the Banach-space valued Arzelà-Ascoli theorem. The actual answer to the question is the one Jochen Wengenroth gave above. I have added the answer with the reference since both the accepted answer and the OP question refer to/ask for it. $\endgroup$
    – Hannes
    Commented Aug 14, 2019 at 19:16

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