What does $W^{1,p}(I)\subset C\big(\overline{I}\big)$ mean? Let $I\subset\mathbb{R}$ be a bounded open interval. Brézis book states that the injection $W^{1,p}(I)\subset C(\overline{I})$ is compact for all $1<p\leq\infty$.
Elements of $W^{1,p}(I)$ are functions whose domain is $I$ and elements of $C(\overline{I})$ are functions whose domain is $\overline{I}$. So, I would like an explanation on the meaning of the inclusion $W^{1,p}(I)\subset C(\overline{I})$.
In the begining of the chapter is proved that for each $u\in W^{1,p}(I)$ there exists a $\tilde{u}\in C(\overline{I})$ such that $u=\tilde{u}$ a.e. on $I$. When the author says "the injection $W^{1,p}(I)\subset C(\overline{I})$" is he talking about the application
$u\mapsto\tilde{u}$?
Thanks.
 A: This inclusion means that every element of $W^{1,p}(I)$ can be identified with an element of $C\big(\bar I\big)$.
Let me explain what this identification means:
The most subtle issue here is the nature of the elements of $W^{1,p}(I)$. Let's remember that this space is defined to be the completion of $C^\infty(I)$ with respect to the $W^{1,p}$ norm. 
So, we first take
$X=\{u\in C^\infty(I): \|u\|_{1,p}<\infty\}$, where
$$
\|u\|_{1,p}=\left(\int_0^1 |u(x)|^p+|u'(x)|^p\right)^{1/p},
$$
then observe that this space is not complete, and hence it is not a Banach space. And then define its completion as $W^{1,p}(I)$. 
Now, what do the elements of the completion look like? They are not functions yet! They are equivalent classes of equi-convergent Cauchy sequences, i.e., classes of Cauchy sequence, where the difference of two sequences in same the class tends to zero, in the $\|\cdot\|_{1,p}$-norm.
And the critical question: 
Why are we allowed to identify these equivalence classes with elements of $C\big(\bar I\big)$? 
Because, if $\{u_n\}\subset X$ is a $\|\cdot\|_{1,p}$-Cauchy sequence, then it is also a Cauchy sequence with respect to the the supremum norm, as the $\|\cdot\|_{1,p}$-norm is stronger than the supremum norm. And as $\{u_n\}$ is a Cauchy sequence in the supremum norm, it has a  (unique) limit $u\in C\big(\bar I\big)$, since the last space is complete. This allows us to identify the equivalence class with this $u\in C\big(\bar I\big)$, and enables us to think of $W^{1,p}(I)$ as a subset of $C\big(\bar I\big)$.
