278 CHAPTER 11. INTEGRATION ON MANIFOLDS

3.) limx→0

∣∣∣y|x|r −rx|x|

∣∣∣2 = limx→0

(|x|2r2 |y|2 + r2−2(x ·y)

)= r2.

4.)|ay+b|=(

a2 |y|2 +2a(y ·b)+ |b|2)1/2

and so

∇y |ay+b|= 12

(|ay+b|2

)−1/2 (2a2y1 +2ab1, ...,2a2yn +2abn

)=

1|ay+b|

(a2y1 +ab1, ...,a2yn +abn

)=

1|ay+b|

(a2y+ab

)= a

ay+b

|ay+b|

5.) Say n > 2. The other case will be similar. From 4.) and the chain rule,

∇y

(|ay+b|−(n−2)

)=−(n−2) |ay+b|−(n−1)

∇y |ay+b|

=−(n−2) |ay+b|−(n−1) aay+b

|ay+b|=−(n−2) |ay+b|−n a(ay+b)

Then ∇y ·∇y

(|ay+b|−(n−2)

)=

n(n−2) |ay+b|−(n+1)∇y |ay+b| ·a(ay+b)−a(n−2) |ay+b|−n

∇y · (ay+b)

= n(n−2) |ay+b|−(n+1) aay+b

|ay+b|·a(ay+b)−a(n−2) |ay+b|−n an

= n(n−2)a2 |ay+b|−(n+1) |ay+b|2

|ay+b|−a2 (n−2)n |ay+b|−n = 0

In case n = 2 it works similarly. Thus if x ̸= 0,∆ψx = 0 = ∆rx. In case x= 0 and bmight not be defined, there is nothing to show because ψ0 is a constant. This shows 5.)since both functions fit into the above situation. ■

Now let B(x,ε)≡ Bε be a small ball inside U and let Vε be the region between them.

U

x Vε

Note that when |y|= r,∣∣∣∣y |x|r− rx|x|

∣∣∣∣2 = r2 |x|2

r2 +r2 |x|2

|x|2−2y ·x= |y−x|2 .

Lemma 11.7.3 Let v(y) ≡ rx (y)−ψx (y). Thus v = 0 on ∂U. Then ∇yv ·n ≡ ∂v∂n =

(n−2)r|x|2−r2

|y−x|n if n > 2 and ∂v∂n = 1

rr2−|x|2

|y−x|2if n = 2. This is on ∂B(0,r).

∣∣∣ ∂ψx

∂n

∣∣∣ is uniformly

bounded on ∂Bε for all ε small enough.