Poincaré 's inequality? I claim that 
$$ 
\| u - \int_{(0,1)} u(x) \, dx \|_{L^{\infty} (0,1)} \leq \| u' \|_{L^1 (0,1)} 
$$
for every $u \in W^{1,1}(0,1)$ (Here, $(0,1) \subseteq \mathbb{R}$.)
I have checked that 
\begin{equation} 
\| u - \int_{(0,1)} u(x) \, dx \|_{L^{\infty} (0,1)} \leq C \| u' \|_{L^1 (0,1)} 
\end{equation}
for some constant $C$ by using the Poincaré-Wirtinger inequality. 
How can I show that this inequality holds for $C = 1$? 
Any help will be appreciated! 
 A: We first get the following result formally.
$$ 
\| u - \int_{(0,1)} u(x) \, dx \|_{L^{\infty} (0,1)} 
=\|\int_{0}^1  u(y)-u(x) dx\|_\infty
=\|\int_{0}^1 \int_x^y u'(t) dt dx\|_\infty,   
$$
which implies 
$$
\| u - \int_{(0,1)} u(x) \, dx \|_{L^{\infty} (0,1)} \leq \int_0^1\int_0^1 
  |u'(t)|dt dx=\|u'\|_{L^1(0,1)}
$$
The formal computation above also woks for functions in $W^{1,1}$. First, we can use the density argument: We first prove the inequality in $C^1[0,1]$ (by assuming $u\in C^1$) and obtain the general inequality by passing limit, since $C^1[0,1]$ is dense in $W^{1,1}(0,1)$. That is, we choose $\{u_n\}\subset C^1[0,1]$ such that $u_n\to u$ in $W^{1,1}$, and prove that 
$$
\|u_n-\int_0^1 u'_n(y)dy\|_\infty\leq \|u'_n\|_{L^1(0,1)},
$$
and then passing limit as $n\to\infty$ in both sides, which is valid because $W^{1,1}$ is embedded into $L^\infty$. This argument is standard in the theory of Sobolev space.  
The second method is more elementary.  We observe that
$u\in W^{1,1}(0,1)$ if and only if $u$ is AC and $\frac{d u}{dt}= u'\in L^1$ a.e. More precisely, we have the following result:

By using this result, we first prove that 
$$
\|\widetilde{u}-\int_0^1 \widetilde{u}(x)dx\|_\infty\leq \|u'\|_{L^{1}}.
$$
Then we replace $\widetilde{u}$ by $u$ above, since they are identical almost everywhere. 
