Let $E=\{ f: \,f\in C^1([0,1]), f(0)=0\}$ and the norm defined on $E$ be $$ \left\|f\right\|_E = \sup_{x\in [0,1]} \,|f'(x)|,\,\, f\in E, $$ and $F=C\,([0,1])$ with the norm $$ \left\|f\right\|_F = \sup_{x\in [0,1]} \,|f(x)|,\,\, f\in F.$$
I need to prove that $\forall f\in E, $ $$ \left\|f\right\|_F \leq \left\|f\right\|_E.$$
Since the interval $[0,1]$ is compact, $f\in E, f'\in F$ are continuous, the maxima of these functions are attained in the unit intarval. Thus, $\exists x_1, x_2\in [0,1]$ such that we can write: $$ \left\|f\right\|_F= \max_{x\in [0,1]}|f(x)|=|f(x_1)| \,\,\text{and}\,\,\left\|f\right\|_E= \max_{x\in [0,1]}|f'(x)|=|f'(x_2)|.$$ Also, $f(x+h)=f(x)+f'(x)h +o(h)$. After taking the norm and using the triangle inequality I can not reach the point which would suggest me the solution.
Can you provide a solution proposal or give me a hint. Thanks.
Best Answer
You were on the right track. Let $x_1 \in [0,1]$ such that $|f(x_1)| = \|f\|_F$.
If $x_1 = 0$ then $f \equiv 0$ so the statement holds.
If $x_1 > 0$, the mean value theorem implies that $\exists c \in \langle 0, x_1\rangle$ such that $$f(x_1) - f(0) = f'(c)(x_1 - 0)$$ so $$\|f\|_F = |f(x_1)| = |f'(c)||x_1| \le |f'(c)| \le \|f\|_E$$