Suppose $X=\lbrace f \in C^1[0,1]:f(0)=f(1)=0 \rbrace$ and $$p(f) = \sup_{t\in[0,1]} |f'(t)|$$ is a norm in $X$. Prove that $p$ is equivalent to the norm $\|\cdot\|_{C^1}$ in $X$. Prove this equivalence norm through Fundamental Theorem of Calculus and the fact that $$\sup_{n\in\mathbb{N}} \int_0^1 f_n(s)ds \leq \int_0^1 \sup_{n\in\mathbb{N}}f_n(s)ds.$$
We are supposed to prove that two norms are equivalent on $C^1([0,1])$ by finding two constants $c,C>0$: $cp(f)\le\| f\|_{C^1}\le Cp(f)$ where $$\| f\|_\infty=\sup_{x\in[0,1]} |f(x)|$$ $$\| f\|_{C^1}=\|f\|_\infty+ \|f'\|_\infty$$ are the norms. I have some difficulties of using the fundamental theorem of calculus and addressing $c$ and $C$.
$\|f'\|_{\infty} \leq \|f\|_{\infty}+\|f'\|_{\infty}$ so we can take $c=1$. Now $f(x)=f(0)+\int_0^{x} f'(t) dt$ so $|f(x)| \leq 0+\int_0^{x} \|f'\|_{\infty}\leq \|f'\|_{\infty}$. Hence $\|f\|_{\infty} \leq \|f'\|_{\infty}$ and $\|f\|_{\infty}+\|f'\|_{\infty} \leq 2\|f'\|_{\infty}$. Hence we can take $C=2$.