394 CHAPTER 14. INTEGRATION ON MANIFOLDS

Proof: Suppose x is not a boundary point with respect to the chart (U,R) but is aboundary point with respect to (V,S). Then U ∩V is open in Ω so Rx ∈ B ⊆ R(U ∩V )where R(U ∩V ) is open in HR and B is an open ball contained in R(U ∩V ). But then, byTheorem 8.10.5, S◦R−1 (B) is open in Rp and contains Sx so x is not a boundary point withrespect to (V,S) after all. ■

Definition 14.1.4 Let V ⊆ Rq. Ck(V ;Rp

)is the set of functions which are restric-

tions to V of some function defined on Rq which has k continuous derivatives which hasvalues in Rp . When k = 0, it means the restriction to V of continuous functions. A functionis in D

(V ;Rp

)if it is the restriction to V of a differentiable function defined on Rq. A

Lipschitz function f is one which satisfies ∥f (x)−f (y)∥ ≤ K ∥x−y∥.

Thus, if f ∈Ck(V ;Rq

)or D

(V ;Rp

), we can consider it defined on V and not just on

V . This is the way one can generalize a one sided derivative of a function defined on aclosed interval.

Lemma 14.1.5 Suppose A is a m×n matrix in which m > n and A is one to one. Then∥v∥ ≡ |Av| is a norm on Rn equivalent to the usual norm.

Proof: All the algebraic properties of the norm are obvious. If ∥v∥ = 0 then |Av| = 0and since A is one to one, it follows v = 0 also. Now recall that all norms on Rn areequivalent. ■

We have in mind, from now on that our manifold will be a compact subset of Rq forsome q≥ p.

Proposition 14.1.6 Suppose in the atlas for a manifold with boundary Ω it is also the

case that each chart (U,R) hasR−1 ∈C1(R(U)

)and DR−1 (x) is one to one onR(U).

Then for two charts (U,R) and (V,S) , it will be the case thatS◦R−1 :R(U ∩V )→S (V )

will be also C1(R(U ∩V )

).

Proof: Then

DR−1 (x)h+o(h) = R−1 (x+h)−R−1 (x)

= S−1 (S (R−1 (x+h)))−S−1 (S (R−1 (x)

))(14.1)

= DS−1 (S (R−1 (x)))(S(R−1 (x+h)

)−S

(R−1 (x)

))+o(S(R−1 (x+h)

)−S

(R−1 (x)

))(14.2)

By continuity ofR−1,S, if h is small enough, which will always be assumed,∣∣o(S (R−1 (x+h))−S

(R−1 (x)

))∣∣≤ α

2

∣∣S (R−1 (x+h))−S

(R−1 (x)

)∣∣where here there is α > 0 such that∣∣DS−1 (S (R−1 (x)

))(S(R−1 (x+h)

)−S

(R−1 (x)

))∣∣≥ α

∣∣(S (R−1 (x+h))−S

(R−1 (x)

))∣∣