# Arzelà-Ascoli theorem and Hölder spaces

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.

• 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 $C^{k,\alpha}(\Omega)\to C^{k,\beta}(\Omega)$. – 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.

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 $$N:=\big({k+n-1\atop k}\big)$$.

• Can I ask why you write Hölder rather than Hölder? – Olivier Bégassat Sep 29 at 21:19
• 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. – Pietro Majer Sep 29 at 21:24