Compact linear operator on $C^1[0,1]$

Question

Let $T:(C^1[0,1],\|.\|)\to (C[0,1],\|.\|_{\infty})$ be defined by $Tx=x$ for all $x\in C^1[0,1]$, where $\|x\|=\|x\|_{\infty}+\|x^{\prime}\|_{\infty}$. I have to show that $T$ is compact.

Let $(x_n)$ be a bounded sequence in $(C^1[0,1],\|.\|)$. Thus $(x_n)$ is bounded in $(C[0,1],\|.\|_{\infty})$. But can we show that $(x_n)$ has a convergent subsequence? I am stuck here.

Answer 1

By a standard diagonal argument, you may extract a sequence $\{f_n\}_{n \in \mathbb{N}}$ (this is a subsequence of your original bounded sequence) that converges pointwise on $\mathbb{Q} \cap [0,1]$.

Let's show that $\|f_n - f_m\| \to 0$. Let $\varepsilon >0$ be given. Well, fix an $x$ and choose $q_k$ such that $|x-q_k|<\varepsilon$. Let $M$ be the bound on the $f_n$ in the $\| \cdot \|$ norm.

Then $|f_n(x) - f_m(x)| \le |f_n(x) - f_n(q_k)| + |f_n(q_k) - f_m(q_k)| + |f_m(q_k) - f_m(x)|.$

Now, $|f_n(x) - f_n(q_k)|\le \displaystyle\left|\int_{q_k}^x f_n'(y)dy\right|\le M\varepsilon$, and likewise, $|f_m(q_k) - f_m(x)|\le M\varepsilon$.

Finally, $|f_n(q_k) - f_m(q_k)|$ gets arbitarily small by the diagonal argument.

This concludes the proof. Use the fact that we're working with Banach spaces to obtain a limit function.

Edit: It is no harder to show that the identity operator on $C^k([0,1])$ to $C^{k-1}([0,1])$, with norms defined akin to your $C^1([0,1])$ norm, is a compact operator as well. You might find it instructive to do it.

Answer 2

$\|x_n\|$ is bounded in second space, so $\|x_n'\|$ is bounded in first space.
So subsequence of $\|x_n\|$ exists where $\|x_n'\|$ converges in first space.
And trivially, subsequence of above subsequence exists where $\|x_n\|$ converges in first space.
That is the convergent subsequence that converge in second space.
Thank you for reading.