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Let $U\subset \Bbb{R}^n\to\Bbb{R}$ be an open set (not necessarily bounded) and $u:U\to\Bbb{R}$ be a bounded continuous function. In Evans's PDE textbook, the author defines a norm $$ \|u\|_{C(\overline{U})}:=\sup_{x\in U}|u(x)|, $$ where $C(\overline{U})$ is defined as $$ C(\overline{U})=\{u\in C(U)\mid u\ \hbox{ is uniformly continuous on bounded subsets of}\ U\} $$ and $$ C(U)=\{u: U\to\Bbb{R} \mid u\ \hbox{continuous}\}. $$ As I see from the answer to a previous question, $u$ can be extended to be a continuous function on $\overline{U}$. Denote this extension (I'm not sure if is unique though) as $\hat u$. Here are my questions:

Is it true that $\sup_{x\in U}|u(x)|=\sup_{x\in\overline{U}}|\hat u(x)|$? Is there a particular reason that Evans uses the supremum in $U$ instead of $\overline{U}$?

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1 Answers1

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A quick remark: the extension is unique, indeed let $u_1$, $u_2$ be two such extensions of $u$ and let $a \in \partial U$. Considering a sequence in $U$ that converges to $a$ we deduce that $u_1(a) = u_2(a)$.

To answer your first question, yes, we can prove that $$\sup_{x \in U}|u(x)| = \sup_{x \in \overline{U}}|\hat{u}(x)| =: S < \infty.$$

Given $\epsilon > 0$, by definition of supremum, there exists $x \in \overline{U}$ such that $|\hat{u}(x)| > S - \epsilon$. By continuity there is $y \in U$ such that $|\hat{u}(x) - u(y)| < \epsilon$, then the triangle inequality implies $$|u(y)| > S - 2\epsilon.$$ Since this holds for every $\epsilon$, we get that $$\sup_{x \in U}|u(x)| \ge S.$$ This proves the result since the other inequality is trivially true.

(I don't see any particular reason to prefer one definition over the other)

Giovanni
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