Multivariable Advanced Calculus

270 BROUWER DEGREELemma 10.3.2 Let g ∈ C 2 ( Ω; R n) be an odd map. Then for every ε > 0, thereexists h ∈ C 2 ( Ω; R n) such that h is also an odd map, ||h − g|| ∞< ε, and 0 is a regularvalue of h.Proof: In this argument η > 0 will be a small positive number and C will be aconstant which depends only on the diameter of Ω. Let h 0 (x) = g (x) + ηx where η ischosen such that det Dh 0 (0) ≠ 0. Now let Ω i ≡ {x ∈ Ω : x i ≠ 0}. In other words, leaveout the plane x i = 0 from Ω in order to obtain Ω i . A succession of modifications isabout to take place on Ω 1 , Ω 1 ∪ Ω 2 , etc. Finally a function will be obtained on ∪ n j=1 Ω jwhich is everything except 0.Define h 1 (x) ≡ h 0 (x)−y 1 x 3 1 where ∣ ∣y 1∣ ∣ < η and y 1 = ( y1, 1 · · · , yn) 1 is a regular valueof the function, x → h0(x) for x ∈ Ωx 3 1 . the existence of y 1 follows from Sard’s lemma1because this function is in C 2 ( Ω 1 ; R n ). Thus h 1 (x) = 0 if and only if y 1 = h0(x) .x 3 1Since y 1 is a regular value, it follows that for such x,( ( )h0i,j (x) x 3 1 − ∂)∂xdetjx31 h0i (x)=implying thatdetx 6 1( ( )h0i,j (x) x 3 1 − ∂ )∂x jx31 y1i x 3 1x 6 ≠ 01(det h 0i,j (x) −∂ )( )x3∂x 1 y1i = det (Dh 1 (x)) ≠ 0.jThis shows 0 is a regular value of h 1 on the set Ω 1 and it is clear h 1 is an odd map inC ( 2 Ω; R n) and ||h 1 − g|| ∞≤ Cη where C depends only on the diameter of Ω.Now suppose for some k such that 1 ≤ k < n there exists an odd mapping h kin C ( 2 Ω; R n) such that 0 is a regular value of h k on ∪ k i=1 Ω i and ||h k − g|| ∞≤ Cη.Sard’s theorem implies there exists y k+1 a regular value of the function x → h k (x) /x 3 k+1defined on Ω k+1 such that ∣ ∣ ∣y k+1∣ ∣ ∣ < η and let h k+1 (x) ≡ h k (x)−y k+1 x 3 k+1. As before,h k+1 (x) = 0 if and only if h k (x) /x 3 k+1 = yk+1 , a regular value of x → h k (x) /x 3 k+1 .Consider such x for which h k+1 (x) = 0. First suppose x ∈ Ω k+1 . Then(hki,j (x) x 3 k+1det− ( ) ∂∂x jx3k+1yk+1)i x 3 k+1x 6 ≠ 0k+1which implies that whenever h k+1 (x) = 0 and x ∈ Ω k+1 ,(det h ki,j (x) −∂)( )x3∂x k+1 yk+1i = det (Dh k+1 (x)) ≠ 0. (10.11)jHowever, if x ∈ ∪ k i=1 Ω k but x /∈ Ω k+1 , then x k+1 = 0 and so the left side of 10.11reduces to det (h ki,j (x)) which is not zero because 0 is assumed a regular value of h k .Therefore, 0 is a regular value for h k+1 on ∪ k+1i=1 Ω k. (For x ∈ ∪ k+1i=1 Ω k, either x ∈ Ω k+1 orx /∈ Ω k+1 . If x ∈ Ω k+1 0 is a regular value by the construction above. In the other case,0 is a regular value by the induction hypothesis.) Also h k+1 is odd and in C ( 2 Ω; R n) ,and ||h k+1 − g|| ∞≤ Cη.Let h ≡ h n . Then 0 is a regular value of h for x ∈ ∪ n j=1 Ω j. The point of Ω which isnot in ∪ n j=1 Ω j is 0. If x = 0, then from the construction, Dh (0) = Dh 0 (0) and so 0 isa regular value of h for x ∈ Ω. By choosing η small enough, it follows ||h − g|| ∞< ε.This proves the lemma.