Let $C=C[0,1]$ be the space of all continuous functions on $[0,1].$ $$K_n(a)=\{x.\in C:|x_0|\leq 2^n,|x_t-x_s|\leq N(a)|t-s|^a \enspace\forall |t-s|\leq 2^{-n}\},$$
where $N(a)=\frac{2^{2a+1}}{2^a-1},\ t,s\in[0,1],\ n\geq0,$ and $a>0.$ Show that $K_n(a)$ are compact sets in $C$.
I am currently reading a book in Random Processes and this was given as one of the exercises. A theorem for compactness was given before this as:
Let $K$ be a closed subset of $C$. It is compact if and only if the family of functions $x.\in K$ is uniformly bounded and equicontinuous, i.e. if and only if
(i) there is a constant $N$ such that $$\sup_{t}|x_t|\leq N \enspace\enspace \forall x.\in K$$ (ii) for each $\epsilon>0$ there exists a $\delta>0$ such that $|x_t-x_s|\leq\epsilon$ whenever $x.\in K$ and $|t-s|\leq\delta,\enspace t,s\in[0,1]$.
So I guess I first need to show that $K_n(a)$ are closed and then they are uniformly bounded and equicontinuous.
For closure, we want to show that every Cauchy sequence in $K_n(a)$ converges within the set.
Let $\{x_.(k)\}$ be a Cauchy sequence with $x_.(k)\in K_n(a)$, all $k$. Then by a known theorem, there exists a $k\in\mathbb{N}$ such that $\{x_.(k)\}$ uniformly converges to $x.^*$ ( $\lim_{k\rightarrow\infty}\sup_{t\in [0,1]}\{|x_t^*-x_t(k)|\}=0$), then $x.^*$ is continuous on $[0,1]$.
It remains to show that $x.^*$ satisfies the conditions to be in $K_n(a)$. Let $\epsilon > 0.$ Then \begin{align} |x_t^*-x_s^*|=&|x_t^*-x_t(k)+x_t(k)-x_s(k)+x_s(k)-x_s^*|\\ \leq& |x_t^*-x_t(k)|+|x_t(k)-x_s(k)|+|x_s(k)-x_s^*|\\ \leq& \ \frac{\epsilon}{2}+N(a)|t-s|^a+\frac{\epsilon}{2}\\ =& \ \epsilon+N(a)|t-s|^a,\end{align}
for all $t,s\in[0,1]$, such that $|t-s|\leq 2^{-n}.$ Hence, we have that $|x_t^*-x_s^*|\leq N(a)|t-s|^a \enspace \forall |t-s|\leq 2^{-n}.$
Morevover, \begin{align} |x_0^*|=&|x_0^*-x_0(k)+x_0(k)| \\ \leq& |x_0^*-x_0(k)|+|x_0(k)| \\ \leq& \epsilon+2^n \end{align}
and hence $ x.^*\in K_n(a)$ and we have closure.
For uniform boundedness it is enough to notice $|x_t| \leq 2^n + 2^n N(a)2^{-an}$.
As for equicontinuity, let $\delta = \min(2^{-n},\sqrt[\leftroot{-2}\uproot{2}a]{\frac{\epsilon}{N(a)}})$, since then $|x_t-x_s|\leq N(a)|t-s|^a \leq \epsilon$, whenever $|t-s|\leq \delta\implies |t-s|^a\leq\delta^a$.