**Q)** Show that the inclusion map $i : C^{0,alpha}(0, 1) rightarrow C^{0,beta}(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

$δ > 0$,

$$(g_n)_{0,α} = sup_{substack{x,yin (0,1) \ x neq y}} frac{|g_n(x)-g_n(y)|}{|x-y|^{alpha}} ≤ A_n + B_n$$

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$$

$ $

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?**