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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.

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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.

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    $\begingroup$ +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

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