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.



Reference for compact embedding between (weighted) Holder space on $\mathbb{R}^n$
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.


