# Does it make sense to talk about regularity of boundaries in 1D?

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 $$\mathbb{R}$$?

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