Is it a compact operator？

Question

Let $$C^{1}_{2\pi}=\{u\in C^{1}((0,2\pi),\mathbf{R}^n):u(0)=u(2\pi)\}$$ $$C_{2\pi}=\{u\in C^{0}((0,2\pi),\mathbf{R}^n):u(0)=u(2\pi)\}.$$ $C_{2\pi}$ is equipped with the norm $$\|u\|_0=max|u(s)|$$ where $|\cdot|$ denotes the usual Euclidean norm. And $C^{1}_{2\pi}$ is equipped with norm $$\|u\|_1=\|u\|_0+\|\frac{du}{ds}\|_0.$$ Whether the inclusion map $$i:C^{1}_{2\pi}\rightarrow C_{2\pi}$$ is a compact operator or not?

Answer 1

You need to show that $B=i(B(0,1))$ is relatively compact.

Suppose $y_n \in B$. Then since $\|y_n\|_0 + \|y_n'\|_0 < 1$, we see that $\|y_n\|_0 < 1$ for all $n$, hence $y_n$ are uniformly bounded. Furthermore, since $\|y_n'\|_0 <1$, we see that the $y_n$ are Lipschitz with rank at most one, hence equicontinuous.

Applying the Arzelà–Ascoli theorem, we see that $y_n$ has a uniformly convergent subsequence. It follows that $B$ is relatively compact.

Here is a suggestion for an alternative approach which is perhaps more intuitive, but has a few more technicalities:

The idea is to construct a sequence of finite rank operators that converge to $i$.

Define $s_N(f)$ to be the sum of the Fourier components $s_N(f)(t) = \sum_{|n|\le N} \hat{f_n} e^{i n t}$, where $\hat{f_n} = {1 \over 2 \pi} \int_0^{2 \pi} f(t) e^{-int}dt$. It is straightforward to show that $s_N$ is continuous and has finite rank.

If $\|f\|_1 <1$, we have $\|f'\|_0 < 1$ and so $f$ is Lipschitz with rank at most one. Hence Jackson's theorem (http://en.wikipedia.org/wiki/Convergence_of_Fourier_series#Uniform_convergence) shows that $\|f-s_n(f)\|_0 \le K {\ln N \over N}$. In particular, this shows that $\|i-s_N\| = \sup_{\|f\|_1 <1} \|i(f)-s_N(f)\|_0 \to 0$, and so $i$ is compact.