Let $B\subset \mathbb{R}^n$ be a open ball. Let $\{f_i\}$ be a sequence of functions bounded in the Hölder norm $C^{k,\alpha}(B)$ for a given integer $k\geq 0$ and $\alpha\in (0,1)$.

Does there exist a subsequence which converges to a function $f$ (necessarily $f\in C^{k,\alpha}(B)$) in the norm $C^{k,\alpha/2}(\bar B')$ for any closed ball $\bar B'\subset B$?

A reference would be very helpful.

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    $\begingroup$ You don't need smaller balls.A simpler and more general statement is: for any bounded open $\Omega\subset\mathbb{R}^n$, $k\in\mathbb{N}$, and $0<\beta<\alpha\le1$, there is a compact embedding $C^{k,\alpha}(\Omega)\to C^{k,\beta}(\Omega)$. $\endgroup$ – Pietro Majer Sep 29 at 20:03

At first, if partial derivatives of order at most $k$ of $f_{n_i}$ converge to those of $f$, than automatically $f\in C^{k,\alpha}(B)$, since $$|(D^k f)(x)-(D^k f)(y)|\leqslant \limsup_i |(D^k f_{n_i})(x)-(D^k f_{n_i})(y)|\leqslant c\cdot |x-y|^\alpha$$ unformly over $x,y\in B$ (here $D^k$ denotes the vector of all partial derivatives of order at most $k$).

Choose closed balls $B_1\subset B_2\subset B_3\ldots$ such that $B=\cup B_i$. It suffices to solve the problem for each $B_i$ separately, then use diagonalization.

For fixed $B_m$ using Arzela -- Ascoli we may suppose that $f_i$ converge to $f$ in $C^k$. Denote $g_i=D^k (f_i-f)$. Then $g_i$ converge to $0$ in $C(B_m)$, and we need to prove that it holds in $C^{\alpha/2}(B_m)$ too. Assume the contrary, then again passing to subsequence we may suppose that $|g_i(x_i)-g_i(y_i)|\geqslant \kappa\cdot |x_i-y_i|^{\alpha/2}$ for fixed $\kappa$ and certain $x_i,y_i\in B_m$. Without loss of generality $x_i\to x_0$, $y_i\to y_0$. Consider two cases.

  1. $x_0\ne y_0$. But then $|g_i(x_0)|+|g_i(y_0)|\geqslant |g_i(x_i)-g_i(y_i)|$, liminf of the last expression is at least $\kappa\cdot |x_0-y_0|^{\alpha/2}>0$, a contradiction.

  2. $x_0=y_0$. Then $$\|g_i\|_{C^\alpha}\geqslant \frac{|g_i(x_i)-g_i(y_i)|}{|x_i-y_i|^\alpha}\geqslant \kappa\cdot |x_i-y_i|^{-\alpha/2}\to \infty,$$ a contradiction.

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For completeness, let's mention a simpler and more general statement: For $\Omega\subset\mathbb{R}^n$ a bounded open set, $k\in\mathbb{N}$ and $0<\beta<\alpha\le1$ there is a compact embedding $$ C^{k,\alpha}(\Omega) \to C^{k,\beta}(\Omega) . $$

Some details:

1. For a compact metric space $(E,d)$, and for $0<\beta<\alpha\le1$ we have a compact embedding of the space of the $\alpha$-Hölder functions into the space of $\beta$-Hölder functions: $$\big( C^\alpha(E),\|\cdot\|_{\alpha,E}\big)\to\big( C^\beta(E),\|\cdot\|_{\beta,E}\big).$$ Here $\|u\|_{\alpha,E}:= \|u \|_\infty+|u|_{\alpha,E}$ and $$|u|_{\alpha,E}:=\sup_{x\neq y\in E} \frac{|u(x)-u(y)|}{d(x,y)^\alpha} . $$

Indeed, let $(u_k)_{k\in\mathbb{N}}\subset C^\alpha(E)$ be a $\|\cdot\|_{\alpha,E}$-bounded sequence, that is, it is uniformly bounded and equicontinuous w.r.to a common modulus of continuity $Ct^\alpha$. By Ascoli-Arzelà, some subsequence $(u_{k_j})$ converges uniformly to some $u$ with the same modulus of continuity, so that $u\in C^\alpha(E)$. We may assume w.l.o.g. that $u$ is the null function (for we just replace $(u_{k_j} )$ with $(u_{k_j}-u)$). The thesis then follows since for $j\to\infty$ we have $\|u_{k_j}\|_\infty= o(1)$ and $$\left|\frac{u_{k_j}(x)-u_{k_j}(y)}{d(x,y)^\beta}\right|=\left| \frac{u_{k_j}(x)-u_{k_j}(y)}{d(x,y)^\alpha}\right|^{\frac{\beta}{\alpha}} \left|u_{k_j}(x)-u_{k_j}(y)\right|^{1-\frac{\beta}{\alpha}} $$whence also $$ |u_{k_j}|_\beta\le |u_{k_j}|_\alpha^{\frac{\beta}{\alpha}}\left(2\|u_{k_j}\|_\infty \right)^{1-\frac{\beta}{\alpha}}=o(1).$$

2. The same compact embedding holds true if $(E,d)$ is only assumed totally bounded: its completion $(\tilde E,\tilde d)$ is compact, and the map "extension by density of uniformly continuous functions" gives an isometry (whose inverse map is the restriction to $E$) $$C^\alpha(E)\to C^\alpha(\tilde E).$$

3. For $\Omega\subset\mathbb{R}^n$ a bounded open set, $k\in\mathbb{N}$ and $0<\beta<\alpha\le1$ the analogous compact embedding $$\big( C^{k,\alpha}(\Omega),\|\cdot\|_{k,\alpha}\big)\to\big( C^{k,\beta}(\Omega),\|\cdot\|_{k,\beta}\big) $$ follows from the case $k=0$, because of the usual closed-range embedding $$ C^{k,\alpha}(\Omega) \to C^{0,\alpha}(\Omega)^N$$ given by $u\mapsto \big( \partial^\nu u \big)_{\nu\in\mathbb{N^n},|\nu|\le k}$, for $N:=\big({k+n-1\atop k}\big)$.

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  • $\begingroup$ Can I ask why you write Hölder rather than Hölder? $\endgroup$ – Olivier Bégassat Sep 29 at 21:19
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    $\begingroup$ well, times ago I made a great mess here on MO, because I tried to correct all occurrences of the wrong spelling Holder to Hölder, and edited dozens of posts :) So now I wrote ö just to recall that funny moment. $\endgroup$ – Pietro Majer Sep 29 at 21:24

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