Here $C^{k,\alpha}(\mathbb R^n)$ refers to the usual Banach space of functions that are $k$ times continuously differentiable, have bounded derivatives, and whose $k$th derivatives have finite Hoelder norm, with Hoelder exponent $\alpha$.

Motivation: I am taking an introductory functional analysis class, and I wonder if the Banach-Alaoglu theorem can be used to prove the Arzela-Ascoli theorem.


1 Answer 1


No, it's not reflexive. Rule of thumb: if the definition of norm involves a supremum or a limit of some sort, the space is not reflexive. If it involves a sum or an integral, it may be.

To prove nonreflexivity, it suffices to show that the space contains a subspace isomorphic to $\ell^\infty$ (since the latter is nonreflexive). Let $Q_k$ be a sequence of disjoint closed cubes that are far apart from one another. For each of them pick a function $f_k$ supported on $Q_k$ and such that $\|f_k\|_{C^{k,\alpha}}=1$. Map every sequence $(c_k)\in \ell^\infty$ to the sum $\sum c_k f_k$. Since $Q_k$ do not interact with each other, the sum makes sense and defines a $C^{k,\alpha}$ function of norm $\sup |c_k|$ (if you separated the cubes sufficiently; otherwise it may be a little large but that is not necessarily a problem). This is the desired embedding.

The same works for $C^{k,\alpha}$ on a bounded domain; the cubes will have to shrink, but they can still be placed relatively far apart, etc.

  • 2
    $\begingroup$ +1 for the rule of thumb. Also noted that if the norm on the dual space involves sup, the space is not reflexive either, for example $L^1$. $\endgroup$
    – Shuhao Cao
    May 11, 2013 at 4:17

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.