prove an equivalence norm

Question

Consider the space $U=\lbrace f\in C^1[0,1]:f(0)=f(1)=0\rbrace$ and $p(f)=sup_{t\in[0,1]}|f'(t)|$ is a norm in $U$. Prove that $p$ is equivalent to the norm $\|·\|_{C^1}$ in U. I don't know how to prove this equivalence norm through fundamental theorem of Calculus. Could anyone help me? Thanks in advance!

Answer 1

Note that for $f \in U$, and $x \in [0,1]$ we have (by the fundamental theorem of calculus) $$\lvert f(x) \rvert = \left \lvert \int^x_0 f'(t) dt \right \rvert \le \int^x_0 \lvert f'(t) \rvert dt \le x p(f) \le p(f).$$ Since this holds for all $x$, you can pass to the supremum to conclude that $$\|f\|_{C^{0}} \le p(f)$$ for all $f \in U$. By definition, $$\|f\|_{C^1} = \|f\|_{C^0} + p(f).$$ Thus we see $$p(f) \le \|f\|_{C^1} \le 2 p(f).$$