Q) Given $0 < \alpha < \beta \leq 1$. Show that the inclusion map $i : C^{0,\beta}[0, 1] \rightarrow C^{0,\alpha}[0, 1] $ is compact.
Ans) Let {$u_n$}$_1^\infty$ ⊂ $C^{0,β}[0,1]$ such that $\|u_n\|_{C^{0,β}} \leq 1$ , i.e. $\|u_n\|_\infty \leq 1$ and $$|u_n(x) − u_n(y)| \leq |x − y|^\beta \ \text{ for all } \ x, y ∈ [0,1]$$ By Arzela-Ascoli, there exists a subsequence {$\tilde u_n$}$_1^\infty$ of {$u_n$}$_1^\infty$ and $u ∈ C[0,1]$ such that $\tilde u_n \rightarrow u$ in $C$. Since $$|u(x) − u(y)| = \lim_{n→∞} |\tilde u_n(x) − \tilde u_n(y)| ≤ |x − y| β$$ $u ∈ C^{0,β}$ as well. Define $g_n := u − \tilde u_n ∈ C^{0,β}$, then $$[g_n]_{0,β} + \|g_n\|_C = \|g_n\|_{C^{0,β}} ≤ 2$$ and $g_n → 0$ in $C$. To finish the proof we must show that $g_n → 0$ in $C^{0,α}$. Given $δ > 0$, $$[g_n]_{0,α} = \sup_{\substack{x,y\in [0,1] \\ x \neq y}} \frac{|g_n(x)-g_n(y)|}{|x-y|^{\alpha}} ≤ A_n + B_n$$ where $$A_n = \sup \Biggr \{ \frac {|g_n(x) − g_n(y)|}{|x − y|^α} : x \neq y \ \text{ and } \ |x − y| ≤ δ \Biggr \}$$ $$= \sup \Biggr \{ \frac {|g_n(x) − g_n(y)|}{|x − y|^β}·|x − y|^{β−α} : x \neq y \ \text{ and } \ |x − y| ≤ δ \Biggr \}$$ $$≤ δ^{β−α}·[g_n]_{0,β} ≤ 2δ^{β−α}$$ and $$B_n = \sup \Biggr \{ \frac {|g_n(x) − g_n(y)|}{|x − y|^α} : |x − y| > δ \Biggr \}$$ $$≤ 2^{δ−α}\|g_n\|_C → 0 \ \ \ \ \ \text{ as } \ n → ∞$$ Therefore, $$\limsup_{n→∞} [g_n]_{0,α} ≤ \limsup_{n→∞} A_n + \limsup_{n→∞} B_n ≤ 2δ^{β−α} + 0 → 0 \ \ \ \text{ as } \ δ ↓ 0$$
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Now the Definition of compactness for the map being used is
Let $X$ and $Y$ be normed spaces and $T : X → Y$ a linear operator. Then $T$ is compact if for any bounded sequence $(x_n)_{n \in \mathbb {N}}$ in X, the sequence $(Tx_n)_{n \in \mathbb {N}}$ contains a converging subsequence
Assuming the proof has been done correctly, I am stuck in this one step which seems minor, but is holding up the rest of it. How do I prove that inclusion map $i$ is linear, for this defintion to be applicable and thus the proof to be valid?
$C^{0,\beta}[0,1]$ is a subset of $C^{0,\alpha}[0,1]$ and $i$ is nothing but the inclusion map so that $i(f) = f$ as a function (the only difference is that $i(f)$ is an element of a larger vector space). Hence, for linearity of $i$ you only need to know that the vector space operations on $C^{0,\beta}[0,1]$ are the restrictions of the ones on $C^{0,\alpha}[0,1]$ to $C^{0,\beta}[0,1]$. This is trivial since both spaces define addition by $$(f+g)(x) = f(x) + g(x)$$ and scalar multiplication by $$(\lambda f)(x) = \lambda f(x).$$