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.$\qed$