I'm trying to solve the below problem to prepare for the mid-term exam.
Let $\mathcal C^1([0,1],\mathbb R)$ be endowed with the norm $\|\cdot\|$ where $$\|f\| = |f(0)| + \int_{0}^{1} |f'(x)| \, \mathrm{d}x, \quad f\in \mathcal C^1([0,1],\mathbb R)$$ Prove that $\|f\|_\infty \le\|f\|$.
Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!
My attempt:
By Fundamental Theorem of Calculus, $f(x) = \int_{0}^{x} f'(t) \, \mathrm{d}t$ for all $x \in [0,1]$. It follows that $$\begin{aligned} \|f\|_\infty &= \max_{x \in [0,1]} |f(x)|\\ &= \max_{x \in [0,1]} \left |\int_{0}^{x} f'(t) \, \mathrm{d}t \right |\\&\le \max_{x \in [0,1]} \int_{0}^{x} |f'(t)| \, \mathrm{d}t \\ &=\int_{0}^{1} |f'(t)| \, \mathrm{d}t \\&\le |f(0)| + \int_{0}^{1} |f'(t)| \, \mathrm{d}t \\ &=\|f\|\end{aligned}$$
Best Answer
Your proof is correct except for one mistake. It is not true that $f(x)=\int_0^{x} f'(t)dt$. What is true is $f(x)=f(0)+\int_0^{x} f'(t)dt$. You can now correct your argument.