weak $C^1$ norm dominates $C^0$ norm on $C^1[0,1]$ (to be used in tutorial).

The weak $C^k$ norm for $f:[a,b] \to \R$ is defined by $$\|f\|_{W^{k,1}} =\sum_{i=0}^k \int_a^b |f^{(i)}(t)|\,dt.$$ This is an exercise (a problem stated in some forum) to show that the weak $C^1$ norm (hence weak $C^k$ norm, this concept comes from Sobolev spaces) dominates the $C^0$ norm (i.e., sup-norm):

Problem. If $f\in C^1[0,1]$, show that $$\sup_{x\ge\in [0,1]}|f(x)|\leq \int_0^1|f(t)|\,dt +\int_0^1 |f'(t)|\,dt.$$  

Solution. Denote $I=[0,1]$. Let $x_0\in I$ be such that $|f(x_0)|=\|f\|_{[0,1]}$. Then for any $\alpha,\beta\in I$ we have $$\int_{x_0}^\beta |f'(t)|\,dt \ge |f(x_0)-f(\beta)|$$ and $$\int_{\alpha}^{x_0} |f'(t)|\,dt \ge |f(x_0)-f(\alpha)|.$$ On summing up, by triangle inequality we have $$\int_0^1|f'(t)|\,dt \ge \int_\alpha^\beta |f'(t)|\,dt \ge |2f(x_0)-f(\alpha)-f(\beta)|.$$ On the other hand, by continuity of $f$ and the differentiability of $g(x):=\int_0^x|f(t)|\,dt$ on $(0,1)$ we have $$\int_0^1|f(t)|\,dt =g(1)-g(0)=g'(c)=|f(c)|,$$ for some $c\in (0,1)$. Therefore by triangle inequality again we have $$\begin{align*} \|f\|_{W^1[0,1]}&:= \int_0^1|f(t)|\,dt+\int_{0}^{1} |f'(t)|\,dt\\ &\ge  |2f(x_0)-f(\alpha)-f(\beta)| + |f(c)|\\ &\ge |2f(x_0)-f(\alpha)-f(\beta)+f(c)|. \end{align*}$$ Now we set $\alpha = x_0$ and $\beta=c$ to get $\|f\|_{W^1[0,1]}\ge |f(x_0)|=\|f\|_{[0,1]}$, as desired.$\qed$