Multivariable Advanced Calculus

264 BROUWER DEGREE= −A.Therefore, H ′ (t) = 0 and so H is a constant.Now let g ∈ U y ∩ C 2 ( Ω; R n) . By Sard’s lemma, Lemma 9.9.9 there exists a regularvalue y 1 of g which is very close to y. This is because, by this lemma, the set of pointswhich are not regular values has measure zero so this set of points must have emptyinterior. Letg 1 (x) ≡ g (x) + y − y 1and let y 1 − y be so small thaty /∈ (1 − t) g 1 + gt (∂Ω) ≡ g 1 + t (g − g 1 ) (∂Ω) for all t ∈ [0, 1] .Then g 1 (x) = y if and only if g (x) = y 1 which is a regular value. Note also D (g (x)) =D (g 1 (x)). Then from what was just shown, letting f = g and g = g 1 in the above andusing g − y 1 = g 1 − y,∫ϕ ε (g (x) − y 1 ) det (D (g (x))) dxΩ∫= ϕ ε (g 1 (x) − y) det (D (g (x))) dxΩ∫= ϕ ε (g (x) − y) det (D (g (x))) dxΩSince y 1 is a regular value of g it follows from the first part of the argument that thefirst integral in the above is eventually constant for small enough ε. It follows the lastintegral is also eventually constant for small enough ε. This proves the claim aboutthe limit existing and in fact being constant for small ε. The last claim follows rightaway from the above. Suppose 10.2 holds. Then choosing ε small enough, it followsd (f, Ω, y) = d (g, Ω, y) because the two integrals defining the degree for small ε areequal. This proves the lemma. Next I will show that if f ∼ g where f, g ∈ U y ∩ C ( 2 Ω; R n) then d (f, Ω, y) =d (g, Ω, y) . In the special case whereh (x,t) = tf (x) + (1 − t) g (x)this has already been done in the above lemma. In the following lemma the two functionsk, l are only assumed to be continuous.Lemma 10.2.6 Suppose k ∼ l. Then there exists a sequence of functions of U y ,{g i } m i=1 ,such that g i ∈ C 2 ( Ω; R n) , and defining g 0 ≡ k and g m+1 ≡ l, there exists δ > 0 suchthat for i = 1, · · · , m + 1,B (y, δ) ∩ (tg i + (1 − t) g i−1 ) (∂Ω) = ∅, for all t ∈ [0, 1] . (10.5)Proof: This lemma is not really very surprising. By Lemma 10.2.3, [k] is an openset and since everything in [k] is homotopic to k, it is also connected because it is pathconnected. The Lemma merely asserts there exists a piecewise linear curve joining kand l which stays within the open connected set, [k] in such a way that the verticesof this curve are in the dense set C 2 ( Ω; R n) . This is the abstract idea. Now here is amore down to earth treatment.