Prove that $\bar{B}_1(0)=\{f \in \mathcal{C}^{1}([0,1]) | \: \lVert f \rVert_{1,\infty} \leq 1\}$ is relatively compact but not compact

Question

Define on $\mathcal{C}^{1}([0,1])$ the norm $\lVert f \rVert_{1,\infty} := \lVert f\rVert_{\infty} + \lVert f' \rVert_{\infty}$.
Prove that the closed unit ball $\bar{B}_1(0) \subset \mathcal{C}^{1}([0,1])$ with respect to this norm is relatively compact in ($\mathcal{C}([0,1]), \lVert \cdot \rVert_{\infty})$ but not compact.

Answer 1

The first conclusion follows directly from Arzelà–Ascoli theorem. Namely:

• If $f\in \bar B_1(0)$ then $\|f'\|_\infty\leq 1$, so $f$ is 1-Lipschitz. Therefore $\bar B_1(0)$ is a uniformly equicontinuous family of functions.
• If $f\in \bar B_1(0)$ then $\|f\|_\infty\leq 1$, so $|f(x)|\leq 1$ for all $x$. Therefore $\bar B_1(0)$ is a uniformly bounded family of functions.

From theorem, any sequence $f_n\in \bar B_1(0)$ has a uniformly convergent subsequence (that is convergent in $\|\cdot\|_\infty$).

To show that this set isn't compact it sufficies to show that it isn't closed. One can find a sequence of functions $f_n\in\bar B_1(0)$ such that $f_n\to f$ uniformly and $f\notin C^1([0,1])$. One can exploit the functions $$g_n(x)=\sqrt{x^2+\frac 1n},\ g(x)=|x|.$$ It's easy to show that $g_n\to g$ uniformly. By scaling and shifting these functions one can find a proper sequence $f_n$. More precisely, consider $f_n:=c\cdot g_n(x-0.5)$, where $c$ is a constant. Then $$\|f_n\|_\infty = f_n(0)=c\cdot \sqrt{1/4+1/n}<2c,\quad \|f_n'\|\leq c.$$ We see that for $c\leq 1/3$ we have $f_n\in \bar B_1(0)$ and that $f_n\to f$, where $f(x)=c\cdot|x-0.5|$.