206 CHAPTER 8. POSITIVE LINEAR FUNCTIONALS

Lemma 8.7.12 Let f : B(p,r)→ Rn where the ball is also in Rn. Let f be one to one, fcontinuous. Then there exists δ > 0 such that

f(

B(p,r))⊇ B(f(p) ,δ ) .

In other words, f(p) is an interior point of f(

B(p,r))

.

Proof: Since f(

B(p,r))

is compact, it follows that f−1 : f(

B(p,r))→ B(p,r) is con-

tinuous. By Lemma 8.7.11, there exists a polynomial g : f(

B(p,r))→ Rn such that∥∥g− f−1

∥∥f(B(p,r)) < εr,ε < 1,

Dg(f(p))−1 exists, and g(f(p)) = f−1 (f(p)) = p

From the first inequality in the above,

|g(f(x))−x|=∣∣g(f(x))− f−1 (f(x))

∣∣≤ ∥∥g− f−1∥∥f(B(p,r)) < εr

By Lemma 8.7.10,

g◦ f(

B(p,r))⊇ B(p,(1− ε)r) = B(g(f(p)) ,(1− ε)r)

Since Dg(f(p))−1 exists, it follows from the inverse function theorem that g−1 also existsand that g,g−1 are open maps on small open sets containing f(p) and p respectively. Thusthere exists η < (1− ε)r such that g−1 is an open map on B(p,η)⊆ B(p,(1− ε)r). Thus

g◦ f(

B(p,r))⊇ B(p,(1− ε)r)⊇ B(p,η)

So do g−1‘ to both ends. Then you have g−1 (p) = f(p) is in the open set g−1 (B(p,η)) .Thus

f(

B(p,r))⊇ g−1 (B(p,η))⊇ B

(g−1 (p) ,δ

)= B(f(p) ,δ ) ■

pq◦ f

(B(p,r)

)B(p,(1− ε)r))

p = q(f(p))

With this lemma, the invariance of domain theorem comes right away. This remark-able theorem states that if f : U → Rn for U an open set in Rn and if f is one to one andcontinuous, then f(U) is also an open set in Rn.

Theorem 8.7.13 Let U be an open set in Rn and let f : U → Rn be one to one andcontinuous. Then f(U) is also an open subset in Rn.

Proof: It suffices to show that if p ∈ U then f(p) is an interior point of f(U). LetB(p,r) ⊆ U. By Lemma 8.7.12, f(U) ⊇ f

(B(p,r)

)⊇ B(f(p) ,δ ) so f(p) is indeed an

interior point of f(U). ■

206 CHAPTER 8. POSITIVE LINEAR FUNCTIONALSLemma 8.7.12 Let f: B(p,r) > R" where the ball is also in R". Let f be one to one, fcontinuous. Then there exists 6 > 0 such thatf(B(p.7)) 2 B(f(p).6).In other words, f (p) is an interior point of f G (p.")).Proof: Since t(B (p.”)) is compact, it follows that f~! : t(B (p.)) — B(p,r) is con-tinuous. By Lemma 8.7.11, there exists a polynomial g : t(B (p.")) — R” such that1IIg—f (Been) <erneé<l,Dg(f(p)) ' exists, and g(f(p)) =f | (f(p)) =pFrom the first inequality in the above,lg(f(x)) —x| = fe (f(x) —" (£())| < |lB—f"laepny < £7By Lemma 8.7.10,got (B(p.r)) 2 B(p,(1—e)r) =B(g(f(p)),(1—e)r)Since Dg(f(p))~' exists, it follows from the inverse function theorem that g~! also existsand that g,g~! are open maps on small open sets containing f(p) and p respectively. Thusthere exists 1 < (1 —€)r such that g~! is an open map on B(p, 7) C B(p, (1—€)r). Thusgot (B(p,r)) 2B(p,(1—e)r) 2B(p.m)So do g~! to both ends. Then you have g~! (p) = f(p) is in the open set g~! (B(p,n)).Thusf(B(p.7)) 2e'(B(P,n)) 2B(g" (p),8) = BP), 3)B(p,(1—€)r))With this lemma, the invariance of domain theorem comes right away. This remark-able theorem states that if f: U — R” for U an open set in R” and if f is one to one andcontinuous, then f(U) is also an open set in R”.Theorem 8.7.13 Let U be an open set in IR" and let f : U + R" be one to one andcontinuous. Then f(U) is also an open subset in R".Proof: It suffices to show that if p € U then f(p) is an interior point of f(U). LetB(p,r) CU. By Lemma 8.7.12, f(U) D t(B(p.7)) > B(f(p),5) so f(p) is indeed aninterior point of f(U). i