Suppose $0<\alpha<\beta<1$, and $\Omega$ is a bounded subset of $\mathbb{R}^n$. Then the Holder space $C^{\beta}(\Omega)$ is compactly embedded into $C^{\alpha}(\Omega)$. But if $\Omega=\mathbb{R}^n$, then the compact embedding is not true.

However, if we consider the weaker weighted Holder space $C^{\alpha, -\delta}(\mathbb{R}^n)$ (for any $\delta>0$) instead of $C^{\alpha}(\mathbb{R}^n)$. Then is $C^{\beta}(\mathbb{R}^n)$ compactly embedded to $C^{\alpha, -\delta}(\mathbb{R}^n)$?

Here $$ \|f\|_{C^{\alpha, -\delta}}=\|(1+|\cdot|^2)^{-\frac{\delta}{2}}f\|_{C^{\alpha}}. $$

I could not find a precise reference from some books on functional analysis. Any comment is welcome.


1 Answer 1


Since the norm is quite specific, I am not sure if you can find it in any book. However, you can prove compactness of the embedding directly. Given a sequence $f_k\in C^{\beta}(\mathbb{R}^n)$, the compactness for bounded domains and a standard diagonal argument shows that you can find a subsequence $f_{k_\ell}$ that converges to some $f$ in $C^\alpha$ on any ball $\mathbb{B}^n(0,R)$. Then it is easy to prove that this subsequence converges to $f$ in $C^{\alpha,-\delta}$ because roughly speaking: on a large ball the $C^\alpha$ norm of $f-f_{k_\ell}$ is small and on the complement on a large ball the $C^{\alpha,-\delta}$ norm is small.


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