2
$\begingroup$

The following is a definition from Evans' PDE book:

Let $U \subset \mathbb{R}^n$ be open and bounded, $k = \{1, 2, \ldots, \}$. We say that a boundary $\partial U$ is $C^k$ if for each point $x^0 \in \partial U$ there exist $r > 0$ and a $C^k$ function $\gamma: \mathbb{R}^{n-1} \rightarrow \mathbb{R}$ such that - upon relabeling and reorienting the coordinate axes if necessary - we have $$U \cap B(x^0, r) = \{x \in B(x^0, r) | x_n > \gamma(x_1, \ldots, x_{n-1})\}.$$ Likewise, $\partial U$ is $C^\infty$ if $\partial U$ is $C^k$ for $k = 1, 2, \ldots$ and $\partial U$ is analytic if the mapping $\gamma$ is analytic.

This definition seems to break down when $n = 1$, since then the domain of $\gamma$ is defined as $\mathbb{R}^0$ which is often interpreted as a single point. The inspiration for this question is that a lot of nice theorems in PDE depend on the regularity of the boundary, and I would like to know how these can be translated to the case $n = 1$. Is there an intuitive way to visualize a $C^k$ boundary in $\mathbb{R}$?

$\endgroup$

1 Answer 1

4
$\begingroup$

Usually it is assumed that $U$ is open and connected.

For $n=1$ the only open and connected sets are intervals. In the sense of the above definition, intervals have $C^k$ boundary for all $k$. The boundary regularity is needed to prove extension theorems for Sobolev functions or regularity of solutions of pdes. All these theorems do work for the $n=1$ case. Although in that case the proofs are much easier.

$\endgroup$
2
  • $\begingroup$ Thank you! So I guess we can take the $C^k$ function to be any constant function $y = c$ such that $c \leq \min{U}$? $\endgroup$
    – CBBAM
    Commented May 24 at 6:05
  • 1
    $\begingroup$ yes, the inequalities would be $x>a$ and $x<b$ close to the ends of the interval $U=(a,b)$. $\endgroup$
    – daw
    Commented May 24 at 8:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .