Suppose $$ X=\left\{x \in C^2(\Bbb R,\Bbb R):x(t+T)=x(t)\; \text{for all}\;t \in \Bbb R \right\}, $$ $$ Y=\left \{h \in C(\Bbb R,\Bbb R):h(t+T)=h(t)\;\text{for all}\;t \in \Bbb R \right \} ,$$ and define the operator $ A:X\rightarrow Y$ by $$Aw=w''+3w'$$ I want to prove that $A$ is compact and think it needs "Arzela-Ascoli theorem" .
proof. $ \text{Since}\; X \;\text{and}\; Y\; \text{are bounded by continuity and periodicity, for arbitrary sequence}\{w_n\}\subset X, |Aw_n(t)|\text{ is uniformly bounded, and by Mean value theorem and } |Aw_n(t)-Aw_n(s)|\le|w_n''(t)-w_n''(s)|+3|w_n'(t)-w_n'(s)|,\{w_n\} \text{ is equicontinuous.}$ Thus, $\text{there is a convergent subsequence of } \{Aw_n\}$ .
Is the proof correct?
I suppose that the norms of $X$ and $Y$ are the ordinary $C^2$ and $C^0$ norm respectively. Let $T = 2\pi$ for simplicity. Consider the functions $f_n\in X$,
$$f_n(x) = \frac{1}{n^2}\sin nx,\ \ \ \ n\in \mathbb N.$$
Then $f_n$ has bounded $C^2$ norm. But
$$Af_n = -\sin nx + \frac 1n \cos nx$$
has no convergent subsequence. Hence $A$ is not compact.