Evans' definition of $C(\bar{U})$ and the trace theorem in his PDE book I'm not sure if it is appropriate to ask this question here; after all, not everyone has Evans' PDE book. Anyway, I'm kind of confused by his notation. In Appendix A, $U$ denotes an open subset of $\mathbb{R}^n$, $\bar{U}:=U\cup\partial U$, and Evans defines
$$C(U)=\{u:U\to\mathbb{R}\mid\text{$u$ is continuous}\}$$
and
$$C(\bar{U})=\{u\in\color{red}{C(U)}\mid\text{$u$ is uniformly continuous on bounded subsets of $U$}\}.$$
His definition of $C(\bar{U})$ seems to disagree with our common knowledge, but this is fine if he can stick to it throughout the book. However, in the opening of the section on the trace theorem, he says

... Now if $u\in C(\bar{U})$, then clearly $u$ has values on $\partial U$ in the usual sense. ...

It just doesn't make sense to me that ... that could be possible, because $u\in C(\bar{U})$ merely guarantees the definition of $u$ on $U$. Did Evans change his mind here and tacitly consent to the fact that $u$ is defined on all of $\bar{U}$? I don't get it, and this really stops me from understanding the motivation behind the trace theorem. Can anyone get me out? Thank you.
Update(2022.7.17): I think Evans is assuming that $U$ is bounded. Then $u$ is uniformly continuous on its domain, which, according to @Glitch, enables us to uniquely extend $u$ to $\bar{U}$. Now it makes sense to talk about the values of $u$ along $\partial U$.
 A: The key fact that he's using without clearly stating it is that a uniformly continuous function on a dense subset of a metric space uniquely extends to a uniformly continuous function on the full metric space.  If Evans assumed that $u$ were uniformly continuous on all of $U$, then the density of $U$ in $\bar{U}$ would give us the extension of $u$ to $\partial U$ immediately.  However, he only assumes the local uniform continuity condition, so we need to do a bit more work.  The idea is that if we fix a point $p \in \partial U$ and consider the ball $B(p,r)$, then $U \cap B(p,r)$ is dense in $\bar{U} \cap B(p,r)$, and our function $u$ is uniformly continuous in the former set by assumption.  So, we can extend it uniquely to a continuous function on $\bar{U} \cap B(p,r)$, which in particular allows us to uniquely assign values to $u$ on $\partial U \cap B(p,r)$.  This works for all $p \in \partial U$, and so we can uniquely extend $u$ to $\partial U$ using this trickery, and the resulting extension is continuous.
On the other hand, you can ask about the space
$$
\dot{C}(\bar{U}) = \{u : \bar{U} \to \mathbb{R} \;\vert\; u \text{ is continuous on } \bar{U}\}
$$
(I've added the dot to distinguish this from his space).  We've just seen that any $u \in C(\bar{U})$ (his space) uniquely extends to define an element of $\dot{C}(\bar{U})$.  On the other hand, any $u \in \dot{C}(\bar{U})$ restricted to $U$ defines an element of $C(\bar{U})$ since $u$ will be uniformly continuous on compact subsets of $\bar{U}$, and any bounded subset of $U$ will be contained in such a compact subset.
Ultimately, what we've done is built a linear extension isomorphism (you can check easily that it's linear) $E : C(\bar{U}) \to \dot{C}(\bar{U})$, and a linear restriction isomorphism $R: \dot{C}(\bar{U}) \to C(\bar{U})$ such that $ER = I$ and $RE = I$.  In other words, these spaces are naturally isomorphic via these extension and restriction processes.  So, what's the point?  In some sense the space $C(\bar{U})$ is preferable because it only requires the functions involved to be defined on $U$.  This means that statements like $W^{k,p}(U) \hookrightarrow C(\bar{U})$ (for appropriate Sobolev parameters $k,p$) are slightly more natural because the functions in the Sobolev space are only defined on $U$ a priori.  If we were to write $W^{k,p}(U) \hookrightarrow \dot{C}(\bar{U})$, then there would be a slight abuse of notation in that the functions in the first space are only defined a priori in the smaller space $U$.  This isn't a big deal, as we've seen above with $E$, and many texts simply gloss over this issue and take the statement $W^{k,p}(U) \hookrightarrow \dot{C}(\bar{U})$ to mean
$$
W^{k,p}(U) \hookrightarrow C(\bar{U}) \xrightarrow{E} \dot{C}(\bar{U}).
$$
EDIT: Here is a proof of the claim that $U \cap B(p,r)$ is dense in $\bar{U} \cap B(p,r)$ in any metric space.  Let $\varepsilon >0$ and pick a point $y \in \bar{U} \cap B(p,r)$.  We know that $\bar{U} = U \cup \partial U$.  If $y \in U \cap B(p,r)$, then we simply pick $x = y \in U \cap B(p,r)$ and note that $d(x,y) = 0 < \varepsilon$.  Suppose, then, that $y \in \partial U \cap B(p,r)$.  By definition of the boundary, we can pick $x \in U \cap B(y,s)$ for $s = \min\{\varepsilon, r-d(y,p) \}$.  Then, certainly, $d(x,y) < s \le  \varepsilon$, but also $d(x,p) \le d(x,y) + d(y,p) < s + d(y,p) \le r$, so $x \in U \cap B(p,r)$.  We have now shown that for every $y \in \bar{U} \cap B(p,r)$ and $\varepsilon >0$ there exists $x \in U \cap B(p,r)$ such that $d(x,y) < \varepsilon$, so $U \cap B(p,r)$ is dense in $\bar{U} \cap B(p,r)$.
