I have che following exercise and some dubts:
Let $M>0$ and $\mathcal{F}=\{ f\in C^{1}([a,b]) \, | \, \| f \|_{C^{1}} \leq M\}$. Prove that $\mathcal{F}$ is relatively compact in $(C^{0}([a,b]), \| \cdot \|_{\infty})$ and $\mathcal{F}$ is not compact in $(C^{1}([a,b]), \| \cdot \|_{C^{1}} )$.
Just to be clear with $\| \cdot \|_{\infty}$ I mean the uniform norm, hence $\| f \|_{\infty}=sup_{x\in [a,b]}|f(x)|$, and with $\| \cdot \|_{C^{1}}$ i mean the $C^{1}([a,b])$ norm hence $\| f\|_{C^{1}}=\sum_{|\alpha|\leq 1}\| D^{\alpha}f \|_{\infty}$.
For the relative compactness in $(C^{0}([a,b]), \| \cdot \|_{\infty})$ I argued as follows: I have to prove that $\overline{\mathcal{F}}$ is compact in $(C^{0}([a,b]), \| \cdot \|_{\infty})$ that is equivalent to show that $\overline{\mathcal{F}}$ is closed and bounded in $(C^{0}([a,b]), \| \cdot \|_{\infty})$. Given the definition of $\mathcal{F}$ I know that exists a positive constant $N\leq M$ such that \begin{equation} \| f \|_{\infty} \leq N\text{ for any $f \in \overline{\mathcal{F}}$}, \end{equation} hence $\overline{\mathcal{F}}$ is bounded. Additionaly $\overline{\mathcal{F}}$ is closed, in fact if I take a sequence $(f_{n})_{n}\subset \mathcal{F}$ I have that $f_{n}\rightarrow f$ where $f \in \overline{\mathcal{F}}$ and the convergence of $f_{n}$ is induced by the norm $\| \cdot \|_{\infty}$. Thus $\overline{\mathcal{F}}$ is closed in $(C^{0}([a,b]), \| \cdot \|_{\infty})$.
For the non-compactness of $\mathcal{F}$ in $(C^{1}([a,b]), \| \cdot \|_{C^{1}})$ I used the following counterexample: I choose $[a,b]=[-1,1]$ and the sequence \begin{equation} f_{n}(x)=|x|^{1+\frac{1}{h}} \end{equation} I have that $(f_{h})_{n} \in C^{1}([-1,1])$ and $f_{n}\rightarrow f$, where $f(x)=|x|$ that does not belong to $C^{1}([-1,1])$. So $\mathcal{F}$ is not compact in $(C^{1}([a,b]), \| \cdot \|_{C^{1}})$.
Is my argument flawed at some point?
The second part is fine. For the first part, use Arzela-Ascoli: a subset $\mathcal F$, of Banach space $X$ is relatively compact if and only if it is bounded and equicontinuous. For convenience, take $M=1$ and $[a,b]=[0,1].$ So we are dealing with the closed unit ball $B=B_{C^1}(f,0).$
$\underline{\mathcal F\ \text{is equicontinuous}}:$
By definition of the $C^1$ norm, for any $f\in B,\ \|f\|_{\infty}+\|f'\|_{\infty}\le 1.$ Therefore, by the mean value theorem, $f$ is Lipschitz bounded on $[0,1]:\ f(x)-f(y)\le 1\cdot |x-y|$ for all $x,y\in [0,1],$ which implies immediately that $\mathcal F$ is equicontinuous on $B$.
$\underline{\mathcal F\ \text{is bounded}}:$
It is enough to note that $B\subseteq B_{C^0}(f,0)$ so $\mathcal F$ is bounded $\textit{in the norm of the larger space} \ C([0,1])$.