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Section 2.5 Regularity of domains

We also need to talk about the regularity of the set Ω and its boundary. (Recall from Notation 2.8 that ΩRN is a non-empty open set.) Thankfully, we can make do with just a few definitions.

Definition 2.23. Open balls.

A ball in RN is a set of the form
Br(x0)={xRN:|xx0|<r},
where here the real number r>0 is called the radius of the ball and the point x0 is its centre.

Definition 2.24. Boundary.

The boundary A of a subset ARN is the set of xRN such that, for all r>0, both Br(x)A and Br(x)A are nonempty.
Figure 2.2. Interior and exterior balls at a point pΩ.

Definition 2.25. Interior and exterior ball properties.

A open set Ω has the interior ball property at a point pΩ if there exists a ball BΩ such that pB. If this holds for all pΩ, we simply say that Ω has the interior ball property. The exterior ball property is defined analogously, except that we now require that the ball BRNΩ.

Definition 2.26. Connected set.

An open set ΩRN is called connected if it cannot be written as the disjoint union of two nonempty open sets Ω1,Ω2RN.

Exercises Exercises

1. (PS3) A domain with a corner.

(See comment below about the examinability of this question.) Consider the open set Ω={(x,y)R2:y<|x|} and the point p=(0,0)Ω.
(a)
Show that Ω satisfies the interior ball property at p.
Hint.
Look for an interior ball. Putting the centre along the y-axis will make things easier.
Solution.
To see that Ω has the interior ball property at the origin, let B=B1(e2). Then we easily check that 0B. Moreover, all points (x,y)B have y<0, which in particular implies that (x,y)Ω.
(b)
Show that Ω does not satisfy the exterior ball property at p.
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 Br(x0,y0) were an exterior ball at 0Ω, and assume without loss of generality that x00. Then argue that for sufficiently small x>0, the point (x,x)Ω satisfies |(x,x)(x0,y0)|2<r2.
Solution.
The following is adapted from a student’s solution in a previous year. Let B=Br(x0,y0) with 0Br(x0,y0) and hence |(x0,y0)|=r. Supposing for contradiction that BR2Ω, we can assume without loss of generality that x00, in which case y0>|x0|=x00. We will show that points (x,x) with 0<x1 lie in ΩB, which is the desired contradiction. Clearly all such points lie in ΩΩ. To see that they lie in B, we simply expand
|(x,x)(x0,y0)|2=(xx0)2+(xy0)2(✶)=r22(x0+y0)x+2x2.
Since x00 and y0>0, the coefficient of x in (✶) above is strictly negative. Thus the right hand side of (✶) is <r2 for x>0 sufficiently small, implying that (x,x)B.
Comment 1.
On the exam, you may need to determine whether a given domain Ω 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 ΩRN be an open set and uC0(Ω). Suppose that u is locally constant in the sense that, for all yΩ, there exists δ>0 such that uu(y) on the ball Bδ(y).
(a)
Prove that, if Ω is connected, then u is globally constant.
Hint.
Fix zu(Ω). Supposing that uz, prove that the sets {xΩ:u(x)=z} and {xΩ:u(x)z} are both open and nonempty.
Solution.
Fix a value zu(Ω), and suppose for the sake of contradiction that uz on Ω. Then Ω=AB where the sets A={xΩ:u(x)=z} and B={xΩ:u(x)z} are disjoint and nonempty. Moreover, since u is continuous and R{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 Ω as the disjoint union of two nonempty open sets, contradicting the assumption that Ω was connected.
Comment 1.
If u:ΩR is locally constant, then it is very easy to check that it is automatically continuous. So the requirement uC0(Ω) 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 Ω is not connected.
Solution.
Let A,BRN be any two nonempty disjoint open sets, Ω=AB, and define a continuous function u by u0 in A and u1 in B. Then u is locally constant in the sense above, but clearly it is not globally constant.