TITLE: Poincare Inequality
QUESTION [2 upvotes]: I've been reading Evan's PDE and I came upon the following statement:

Assume $U$ is a bounded, open subset of $\mathbb{R}^n$. Suppose $u\in W_0^{1,p} (U)$ for some $1\le p <n$. Then we have the estimate
$$\|u\|_{L^q(U)} \le C \|Du\|_{L^p(U)}$$
for each $q\in [1,p^*]$, the constant $C$ depending only on $p,q,n$ and $U$.
In particular, for all $1\le p \le \infty$,
$$\|u\|_{L^p(U)} \le C \|Du\|_{L^p(U)}$$

I'm not sure about how is the Poincare inequality (by this I'm referring to the inequality after "in particular", i.e. $\| u \|_{L^p(U)} \le C \|\partial u\|_{L^p(U)}$) a particular case of this general theorem (everything before "in particular").
I understand that the Sobolev conjugate $p^* > p$, so we can choose $q = p$, but that's only for $1\le p <n$. I also understand that if $n \le p < \infty$, then we can choose $1 \le q < n$ sufficiently close to $n$ so that $q^*>p$ and then apply the general theorem to the particular case. However, what if $p = \infty$ or if $n = 1$ (in which case we can't choose $1\le q <n$)?

REPLY [2 votes]: Regarding the Poincare inequality, I suppose it's a question of terminology.  What do you take as your definition of Poincare's inequality?  For what it's worth, I'm looking at the book and Evans writes "This estimate is sometimes called Poincare's inequality."  (Page 282 in the second edition.)  See also the Wiki article or Wolfram Mathworld, which have somewhat divergent opinions on what should or shouldn't be called a Poincare inequality.
I'm sad to say I didn't notice this issue with $p = \infty$ when I first read the book.  Later on Evans proves that, at least if $U$ is $C^{1}$, then $W^{1,\infty}(U)$ is embedded in $C^{0,1}(U)$ (i.e. every $W^{1,\infty}$ function has a Lipschitz continuous representative and $\|u\|_{C^{0,1}(U)} \leq C \|u\|_{W^{1,\infty}(U)}$).  This makes the solution to your question quite straightforward.  If $u \in W_{0}^{1,\infty}(U)$ (and we identify it with its Lipschitz representative), then $u$ is uniformly Lipschitz on $\overline{U}$.  By section 5.8.2b of Evans, $\text{Lip}(u) = \|Du\|_{L^{\infty}(U)}$.  Thus, if $x \in U$ and $y \in \partial U$, 
\begin{equation*}
|u(x)| = |u(x) - u(y)| \leq \|Du\|_{L^{\infty}(U)} |x - y|,
\end{equation*}
which implies that $\|u\|_{L^{\infty}(U)} \leq D \|Du\|_{L^{\infty}(U)}$ if $D = \text{diam}(\overline{U})$.  
When $n = 1$, there's very little to do.  Suppose $U = I = (a,b)$.  If $u \in W^{1,p}_{0}(I)$, then $u$ is absolutely continuous.  In particular, since $u(a) = 0$, it follows that $u(x) = \int_{a}^{x} u'(x) \, dx$ almost everywhere.  Thus, by Minkowski's inequality (if $p < \infty$), 
\begin{equation*}
\|u\|_{L^{p}(I)} \leq (b - a)^{\frac{1}{p}} \|u'\|_{L^{p}(I)}.
\end{equation*}
When $p = \infty$, we immediately have $|u(x)| \leq (b - a) \|u'\|_{L^{\infty}(I)}$ a.e. (or we could obtain the same by sending $p \to \infty$ above).
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