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Let $U$ be an open subset of $\Bbb{R}^d$. In Evans's PDE book, $$ C(U)=\{u: U\to\Bbb{R} \mid u\ \hbox{continuous}\} $$ and $$ C(\bar U)=\{u\in C(U)\mid u\ \hbox{ is uniformly continuous on bounded subsets of}\ U\}. \tag{1} $$

I've seen $C(\bar U)$ defined as $$ C(\bar U)=\{u: \bar U\to\Bbb{R} \mid u\ \hbox{continuous}\} \tag{2} $$ before. Is it the same as Evans's version? What's the point of the definition (*) in Evans's book?

(2) obviously implies (1). To show (1) implies (2), I set it as an exercise:

Suppose $f:U\to \Bbb{R}$ is continuous and for any bounded $A\subset U$, $f$ is uniformly continuous on $A$. Then $f$ has a continuous extension on $\bar U$.

The bounded case is trivial. How can one have the general case that $U$ might be an unbounded open set?

As a nontrivial special case, consider $U=(0,\infty)$ and $f\in C(U)$ such that it satisfies (1). How should I define $f(0)$?

  • When Evans writes $C(\overline{U})$, he usually means functions that are continuous on $U$ and which extend continuously to the boundary $\partial U$. For example, let $f(x) = \frac{1}{x}$, then $f \in C( (0,1))$, but $f \not\in C ([0,1])$. These are the kinds of examples that we would like to exclude. – msteve Sep 07 '15 at 18:28
  • @msteve: Thanks for your attention. So what is the difference between (1) and (2)? –  Sep 07 '15 at 18:38
  • Obviously, there's no difference if $U$ is bounded, as is always the case in that book. I would say that there's no difference on unbounded domains also, but those things are often slippery. (The way I see it, a function space on $\overline{U}$ should be composed of restrictions to $U$ of functions defined on $\mathbb{R}^d$. This is always the case when $U$ has a regular enough boundary, and indeed many theorems have that requirement as an assumption. Without this assumption, one typically enters the field of geometric measure theory, and that's another world.) – Giuseppe Negro Sep 07 '15 at 18:45
  • (N.B.: Those were my 2 cents). – Giuseppe Negro Sep 07 '15 at 18:46
  • Why there is no difference if $U$ is bounded? –  Sep 08 '15 at 14:01
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    Because in that case any function in $C(\overline{U})$ (with definition (1)) is uniformly continuous on $U$ and so it can be continuously extended to $\overline{U}$. Conversely, a function in $C(\overline{U})$ (with definition (2)) is uniformly continuous by Heine-Cantor's theorem. – Giuseppe Negro Sep 08 '15 at 16:29
  • @GiuseppeNegro: Fair enough. Thanks for your attention. I still don't get the unbounded case though. –  Sep 08 '15 at 17:57

2 Answers2

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The two definitions are equivalent, whether $U$ is bounded or not.

Consider a point $x \in \partial U$. The set $A = U \cap B_1(x)$ is a bounded subset of $U$, so $f\lvert_A$ is uniformly continuous, and therefore has a continuous extension $g_A$ to $\overline{A}$. Since $f$ is continuous, we have $g_A \equiv f$ on $\overline{A}\cap B_1(x)\cap U$, so, naming $C = \partial U \cap B_1(x)$,

$$F_C \colon y \mapsto \begin{cases} f(y) &, y \in U \\ g_A(y) &, y \in C\end{cases}$$

is a continuous extension of $f$ to $U \cup C = U \cup (\overline{A}\cap B_1(x))$. If $M,N$ are two subsets of $\partial U$ and $F_M,F_N$ are continuous extensions of $f$ to $U\cup M$ resp. $U \cup N$, then since $U$ is dense in $\overline{U}$, we have $F_M\lvert_{M\cap N} \equiv F_N\lvert_{M\cap N}$, so any two continuous extensions of $f$ to parts of the boundary coincide on their common domain, whence the local continuous extensions of $f$ to parts of the boundary fit together to give a global continuous extension to $\overline{U}$.

In the case $U = (0,+\infty)$, the uniform continuity of $f$ on $(0,1)$ yields the existence of $L := \lim\limits_{x\to 0} f(x)$ and the continuous extension of $f$ to $[0,+\infty)$ is obtained by setting $\tilde{f}(0) = L$.

Daniel Fischer
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  • What are $F_M$ and $F_N$? –  Sep 08 '15 at 18:33
  • They are continuous extensions of $f$ to $U\cup M$ resp. $U\cup N$. – Daniel Fischer Sep 08 '15 at 18:34
  • Ah, thanks. Since $M,N$ are two subsets of $\partial U$ while $A$ is not, I'm just confused with the notation $f_A$ with $f_M$ and $f_N$. –  Sep 08 '15 at 18:51
  • Indeed, suboptimal naming on my part. Should have named say $C = \partial U \cap B_1(x)$ and called it $F_C$ rather than $F_A$ for consistency. – Daniel Fischer Sep 08 '15 at 18:55
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    @ThomasShelby: that follows from the assumption that $f$ is uniformly continuous on any bounded subset of $U$. –  Aug 15 '19 at 03:37
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(1) and (2) are not the same in the sense that functions are defined in different domains unless $U=\mathbb{R}^d$. The notation $C(\overline{U})$ is seldom defined as in (2) in PDE books (I have never seen one), although the space $C(X)$ (which denotes the set of continuous complex or real functions) where $X$ is a compact Hausdorff space is well discussed in functional analysis.

Instead $C(\overline{U})$ is usually defined as a subset of $C(U)$ such that (1) is satisfied. A common alternative (see e.g. Introduction to Partial Differential Equations by Folland) is defining $C(\overline{U})$ as the space of all $u\in C(U)$ such that $u$ extends continuously to the closure $\overline{U}$.

Here is a related old question:
Extension of partial derivatives and the definition of $C^k(\overline{\Omega})$

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