The question is: For $K$ and $\alpha$ fixed, show that $\{f\in \operatorname{Lip}_k \alpha : f(0) = 0\}$ is a compact subset of $C[0,1]$.
I was going to attempt this by using by Arzela-Ascoli theorem but I don't know where to start because I don't know what the set actually is. I was guessing that it's saying that $f$ is Lipschitz? But I wasn't too sure.
Any suggestions how to attack the problem?
Let $F = \{f\in \operatorname{Lip}_k \alpha : f(0) = 0\}$ on $[0, 1]$ with fixed constants $K$ and $\alpha$. To begin, we show that $F$ is closed, uniformly bounded and equicontinous.
To show that $F$ is closed. Let $f_n \in F$ and $f_n \rightarrow f$ uniformly. Therefore we have that $$\frac{|f_n (x) - f_n (y)|}{|x - y|^\alpha} \leq K$$ for some fixed $K$ and $\alpha$.
If we take the limit as $n \rightarrow \infty$ and use that $f_n \rightarrow f$ uniformly, we get that: $$\frac{|f(x) - f(y)|}{|x - y|^\alpha} \leq K$$
This shows that $f$ is Lipschitz continuous for some fixed constant $K$ and $\alpha$ so $f \in F$ thus $F$ is closed.
To show $F$ is uniformly bounded, we have that $f(0) = 0$. Using this, we get:
$|f(x) - f(0)| = |f(x)| \leq K*|x - x_0|^\alpha = K*|x|^\alpha \leq K$
Hence $F$ is uniformly bounded
To show that $F$ is equicontinous, let $\epsilon >0$. Therefore:
$|f_n (x) - f_n (y)| \leq K*|x - y|^\alpha < \delta$
From this, we let $\delta = (\frac{\epsilon}{K})^\frac{1}{\alpha}$
So we have that $|f_n (x) - f_n (y)| \leq K*|x - y|^\alpha < \epsilon$
Hence $F$ is equicontinous and applying the Arzela-Ascoli Theorem, we have that $F$ is compact in $C[0,1]$.