Let $B_1$ be the unit ball in $\mathbb{R}^2$. Define $A=\{u \in W^{1,2}(B_1) : \|u\|_{ W^{1,2} } \leq 1\}$, $B= \{u \in L^{2}(B_1) : \|u\|_{ L^2 } \leq 1\}$, $C=\{u \in C^{0,\alpha}(B_1) : \|u\|_{ C^{0,\alpha} } \leq 1\}$ and $D=\{u \in C^{0,\beta}(B_1) : \|u\|_{ C^{0,\beta} } \leq 1\}$ with $0<\alpha<\beta<1$.
Let $u_n$ be any sequence in $A$, we can find a convergent subsequence such that $u_n$ converges to some limiting function $u_\infty$ weakly in $W^{1,2}$ and strongly in $L^2$. By Fatou Lemma, it is observed that $u_\infty$ also belongs $A$. In this case, can I say $A$ is compact in $B$ (under the topology of $B$)?
Let $v_n$ be any sequence in $D$, we can find a convergent subsequence such that $v_n$ converges to some limiting function $v_\infty$ in $C^{0,\alpha}$ by Arzel`a–Ascoli theorem. By lower semi continuity of Holder norm, it is observed that $v_\infty$ also belongs $D$. In this case, can I say $D$ is compact in $C$ (under the topology of $C$)?
Here comes to the last question. For some sets $X \subset Y$, $X$ is compact in $X$ if and only if $X$ is compact in $Y$. If the above two assertions are true, then $A$ and $D$ are compact, I think it is weired. Can someone helps? Thanks!