Let $B = \{x \in \mathbb{R}^2; |x| < 1\}$. Let $\{f_k\}$ be a sequence of continuous functions on $B$ satisfying
$$
\sup_{x \in B} |f_k(x)| + \sup_{\substack{x \ne y \\ x, y \in B}} \frac{ | f_k(x) – f_k(y)| }{|x – y|^{1/2}} \le C
$$
for some positive constant $C$. Show that there is a subsequence of $\{f_k\}$ convergent uniformly on $B$ to a function $g$, and $g$ is continuous on $B$.
My attempts:
I think that I have to apply the Ascoli-Arzela theorem on $C([-1,1])$. But now, I have a problem with the definition of uniformly convergent. We say that the sequence $\{f_k\}$ is uniformly convergent if for every $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that for all $n \in \mathbb{N}$ and $x \in X$, $\lvert f_k(x)-g(x) \rvert \leq \varepsilon$.
Now, let $\{f_{k_i}\}$ be a subsequence of $\{f_k\}$. So, we have
$$\lvert f_{k_i}(x)-g(x) \rvert= \lvert f_{k_i}(x)-f_k(x)+f_k(x)-g(x) \rvert \leq \lvert f_{k_i}(x)-f_k(x)\rvert+\lvert f_k(x)-g(x) \rvert $$
But I don't know how to use the assumptions of the question and also how to use the mentioned theorem.
Best Answer
First step is to extend the definition of $f_k$'s to $\{x \in \mathbb R^{2}: |x| \leq 1\}$. Note that $|f_k(x)-f_k(y)| \leq C|x-y|^{1/2}$. If $|x| \leq 1$ then there is a sequence $(x^{j})$ in $B$ convergeing to $x$. Now $|f_k(x^{j})-f_k(x^{l})|\leq C|x^{j}-x^{l}|^{1/2} \to 0$. Hence $f_k(x)=\lim_{j \to \infty} f_k(x^{j})$ exists. I will let you check that this defines a continuous function $F_k$ on the closure of $B$ which extends $f_k$.
Now apply Arzela-Ascoli Theorem to $(F_k)$. The inequality $|F_k(x)-F_k(y)| \leq C|x-y|^{1/2}$ holds for $|x|,|y| \leq 1$ and this immediately gives equi-continuity of $(F_k)$. ALso $|F_k(x)| \leq C$ and this gives uniform boundedness. Hence, there is a subsequence of $(F_k)$ which converges uniformly on $\{x: |x| \leq 1\}$. Of course, this implies that there is a subsequence of $(f_k)$ which converges uniformly on $B$.