Section 2.5 Regularity of domains
We also need to talk about the regularity of the set \(\Omega\) and its boundary. (Recall from Notation 2.8 that \(\Omega \subseteq \R^N\) is a non-empty open set.) Thankfully, we can make do with just a few definitions.
Definition 2.23. Open balls.
A ball in \(\R^N\) is a set of the form
\begin{equation*}
B_r(x_0) = \{ x \in \R^N : \abs{x-x_0} \lt r \},
\end{equation*}
where here the real number \(r \gt 0\) is called the radius of the ball and the point \(x_0\) is its centre.
Definition 2.24. Boundary.
The boundary \(\partial A\) of a subset \(A \subset \R^N\) is the set of \(x \in
\R^N\) such that, for all \(r \gt 0\text{,}\) both \(B_r(x) \cap A\) and \(B_r(x) \without A\) are nonempty.
Definition 2.25. Interior and exterior ball properties.
A open set \(\Omega\) has the interior ball property at a point \(p \in \partial \Omega\) if there exists a ball \(B \subseteq \Omega\) such that \(p \in \partial B\text{.}\) If this holds for all \(p \in \partial \Omega\text{,}\) we simply say that \(\Omega\) has the interior ball property. The exterior ball property is defined analogously, except that we now require that the ball \(B \subseteq \R^N \without \overline \Omega\text{.}\)
Definition 2.26. Connected set.
An open set \(\Omega \subseteq \R^N\) is called connected if it cannot be written as the disjoint union of two nonempty open sets \(\Omega_1,
\Omega_2 \subseteq \R^N\text{.}\)
Exercises Exercises
1. (PS3) A domain with a corner.
(See comment below about the examinability of this question.) Consider the open set \(\Omega = \{(x,y) \in \R^2 : y \lt \abs x\}\) and the point \(p=(0,0) \in \partial\Omega\text{.}\)
(a)
Show that \(\Omega\) satisfies the interior ball property at \(p\text{.}\)
Hint.
Look for an interior ball. Putting the centre along the \(y\)-axis will make things easier.
Solution.
To see that \(\Omega\) has the interior ball property at the origin, let \(B=B_1(-e_2)\text{.}\) Then we easily check that \(0 \in \partial B\text{.}\) Moreover, all points \((x,y) \in B\) have \(y \lt 0\text{,}\) which in particular implies that \((x,y) \in \Omega\text{.}\)
(b)
Show that \(\Omega\) does not satisfy the exterior ball property at \(p\text{.}\)
Hint 1.
Draw some pictures to convince yourself that this is true! Proving it requires a bit more ingenuity than the previous part.
Hint 2.
One relatively simple proof given by a student in previous years can be outlined as follows. Suppose that \(B_r(x_0,y_0)\) were an exterior ball at \(0 \in \partial\Omega\text{,}\) and assume without loss of generality that \(x_0 \ge
0\text{.}\) Then argue that for sufficiently small \(x \gt 0\text{,}\) the point \((x,x) \in \partial\Omega\) satisfies \(\abs{(x,x)-(x_0,y_0)}^2 \lt r^2\text{.}\)
Solution.
The following is adapted from a student’s solution in a previous year. Let \(B=B_r(x_0,y_0)\) with \(0 \in \partial B_r(x_0,y_0)\) and hence \(\abs{(x_0,y_0)}=r\text{.}\) Supposing for contradiction that \(B \subseteq \R^2
\without \overline \Omega\text{,}\) we can assume without loss of generality that \(x_0 \ge 0\text{,}\) in which case \(y_0 \gt \abs{x_0} = x_0 \ge 0\text{.}\) We will show that points \((x,x)\) with \(0 \lt x \ll 1\) lie in \(\overline \Omega
\cap B\text{,}\) which is the desired contradiction. Clearly all such points lie in \(\partial \Omega \subset \overline \Omega\text{.}\) To see that they lie in \(B\text{,}\) we simply expand
\begin{align}
\abs{(x,x)-(x_0,y_0)}^2
\amp =
(x-x_0)^2+(x-y_0)^2 \notag\\
\amp =
r^2 - 2(x_0+y_0)x + 2x^2\text{.}\tag{✶}
\end{align}
Comment 1.
On the exam, you may need to determine whether a given domain \(\Omega\) satisfies the interior or exterior ball properties, as these are hypotheses of some of our main results. But doing this ‘by eye’ without giving a detailed rigorous proof will be completely acceptable.
Comment 2.
There were several other nice solutions this year, several of which did not involve any calculus.
2. (PS3) Locally constant functions.
Let \(\Omega \subseteq \R^N\) be an open set and \(u \in C^0(\Omega)\text{.}\) Suppose that \(u\) is locally constant in the sense that, for all \(y \in
\Omega\text{,}\) there exists \(\delta \gt 0\) such that \(u \equiv u(y)\) on the ball \(B_\delta(y)\text{.}\)
(a)
Prove that, if \(\Omega\) is connected, then \(u\) is globally constant.
Hint.
Fix \(z \in u(\Omega)\text{.}\) Supposing that \(u \not \equiv z\text{,}\) prove that the sets \(\{ x \in \Omega : u(x)=z \}\) and \(\{x \in \Omega : u(x) \ne z\}\) are both open and nonempty.
Solution.
Fix a value \(z \in u(\Omega)\text{,}\) and suppose for the sake of contradiction that \(u \not \equiv z\) on \(\Omega\text{.}\) Then \(\Omega = A \cup B\) where the sets \(A = \{ x \in \Omega : u(x) = z \}\) and \(B = \{x \in \Omega : u(x) \ne z\}\) are disjoint and nonempty. Moreover, since \(u\) is continuous and \(\R
\without \{z\}\) is open, a standard argument shows that \(B\) is open. Since \(u\) is locally constant, we easily check that \(A\) is open as well. But then we have written \(\Omega\) as the disjoint union of two nonempty open sets, contradicting the assumption that \(\Omega\) was connected.
Comment 1.
If \(u \maps \Omega \to \R\) is locally constant, then it is very easy to check that it is automatically continuous. So the requirement \(u \in C^0(\Omega)\) is somewhat redundant.
Comment 2.
Often students attempt a more hands-on argument involving chaining together balls on which \(u\) is constant. While this can be made to work, it requires some serious effort because the definition of ‘locally constant’ gives us no control (from below) on the radii of these balls, and so if we are not careful they could rapidly shrink to zero.
(b)
Give a counterexample to show that the conclusion in the previous part can fail if \(\Omega\) is not connected.
Solution.
Let \(A,B \subset \R^N\) be any two nonempty disjoint open sets, \(\Omega = A
\cup B\text{,}\) and define a continuous function \(u\) by \(u \equiv 0\) in \(A\) and \(u \equiv 1\) in \(B\text{.}\) Then \(u\) is locally constant in the sense above, but clearly it is not globally constant.
