Consider Banach space $C^{1}[0,1]$ of continuously differentiable functions provided with the norm $$ ||f|| = \sup_{x \in [0,1]} |f(x)| + \sup_{x \in [0,1]} |f'(x)|.$$
Similarly we define space $C^{n}[0,1]$ of $n$ times continuously differentiable functions.
$\textbf{My question}$: Does $C^{1}[0,1]$ contains a subspace which is isometrically isomorphic to the Banach space $c_0$ of all sequences converging to zero? What about $C^{n}[0,1]$?
$\textbf{My thoughts}$: I know that Banach space C[0,1] of all continuous functions equipped with norm $||f|| = \sup_{x \in [0,1]} |f(x)|$ contains isometric copy of $c_0$. To produce such copy one chooses any sequence $(f_n) \subset C[0,1]$ of disjointly supported functions of norm one and considers space $X = \overline{span}(f_n: n \in \mathbf{N})$. Then $\sum \alpha_n f_n$ converges iff $\alpha_n \rightarrow 0$ or, in other words, $(\alpha_n) \in c_0$. Consequently, mapping $$c_0 \ni (\alpha_n) \mapsto \sum \alpha_n f_n \in X $$ is isometric isomorphism and $X$ is copy of $c_0$ sitting inside $C[0,1]$.
Example of sequence $(f_n) \subset C[0,1]$ satisfying the above properties is easy to define: put $f_n(t) = 0$ for $t \in [0,1] \setminus \Bigl(\frac{1}{2^{n+1}},\frac{1}{2^{n}}\Bigr)$ and $f_n\Bigl( \frac{3}{2^{n+2}} \Bigr) = 1$ and let $f_n$ be linear elsewhere.
Let us pass to the space $C^{1}[0,1]$ with the norm defined at the beginning. I think the above procedure will also work in this case i.e. in order to find copy of $c_0$ inside this space one only needs to find a sequence $(f_n) \subset C^{1}[0,1]$ of smooth disjointly supported functions such that $\|f_n\| = 1$ for all $n$. However I'm having troubles to construct explicitly such a sequence.