Is the inclusion $C^1[0,1]\subset C[0,1]$ compact? I am working on this problem but i couldn't succeed . 
Consider the space $C^1[0,1]$ with the norm $$\|f\|=\max \{\|f\|_{C[0,1]}, \|f'\|_{C[0,1]}\},$$  
I don't know if the inclusion map is compact, but my friend said it is . How do i show that the inclusion map from $$(C^1[0,1],\| .\|) \to (C[0,1],\| .\|)$$ is compact  ? 
Thanks . 
 A: I will try to write an answer as @Martin suggested, Please correct me if i am wrong : 
Here we use Arzela Ascoli : which says that  $M$ , a subset of Banach space $X=C[0,1]$ is relatively compact iff $M$ is bounded and equicontinuous .  
Here , Let us take unit ball of $C^1[0,1]$ then $\|f\|_{C^1} \le 1$ for all $f \in B_1(0_{C^1[0,1]}) \subset  C^1[0,1]$ , hence for  such that $\frac {f(x)-f(y)}{x-y} \le 1 $ ,   for all $f \in B_1(0_{C^1[0,1]})$ 
That is because for $f \in M=B_1{(0_{C_1[0,1]})}$ we have
$1\ge ∥f∥_ {C_1[0,1]}=\{\max{∥f∥_{C^0[0,1]},∥f′∥_{C^0[0,1]}} \}$
which implies that both $∥f∥_{C^0}≤1$ and $∥f′∥_{C^0}≤1$
Note the above follows from mean value theorem ie there exists $\zeta \in (x,y)$ , because $f$ is continuous , which $\implies$  $f(x)- f(y) = f'(\zeta) (x-y)$ 
But as we know that $|f'(\zeta) | \le 1$ from the definition of $C^1$ norm . 
So we have a common lipschitz constant for all $f\in B_1(0_{C^1[0,1]} )$ , and we follow that for $|x-y| < \epsilon $ we get 
which implies that $|f(x)-f(y)| < \epsilon $ for all $f$ in the unit ball of $C^1[0,1]$ 
This gives me equicontinuity . ie restated "if a set of functions is bounded by common Lipschitz constant then the set is equicontinuous  " . 
Now our task is to prove that unit ball of $C^1[0,1]$ is bounded in unit ball of $C[0,1]$ , that means $B_1(0_{C^1[0,1]}) \subset B_1{(0_{C[0,1]})}$
which seems clear while the $\|.\|_{C^1[0,1]} \ge \|.\|_C[0,1]$
It is useful to note that $T(f)(s) =\int_0^t f(s) ds$ is a compact operator , it follows directly from the above observation because $T$ maps from $C[0,1]$ to $C^1[0 1]$ . $T$ is continuous because $\|T\| \le \|f\|_{C[0,1]} \le \|f\|_{C^1[0,1]}$ , now composing $T$ with the inclusion map name it $i : C^1[0,1] \to C[0,1]$ which is compact . As we know that subspace of compact operators from Banach space to Banach space form a Ideal in the space of continuous operators . It follows that $T$ is compact. 
If the proof is right then credit goes to Martin. 
Which gives us what we want . ( hopefully)  
