Analysis

16.3 Definitions And Elementary Properties

First is a simple lemma which links an integral to something which is an integer. It will always be the case that ϕ_ε will denote a mollifier as in Definition 16.2.3. Also, there are various conditions which hold if something is small enough. Numbers like 1/5 and 1/10 are used to express suitable smallness assumptions. There is nothing particularly special about these numbers. I just chose some which were sufficiently small to work out. I am not trying to give anything precise in expressing these estimates. The thing that is interesting is the degree for continuous functions and anything which will get you this is what is wanted.

Proof: First note that the sum is finite. Indeed, if there were a sequence of x_k such that x_k ∈ g⁻¹

, then it would have a subsequence converging to some x ∈Ω . Thus g = y and by assumption, y∂Ω. Since y is a regular value, Dg is invertible and so by the inverse function theorem, there is an open set containing x on which g is one to one, contrary to x being a limit of a sequence of points in g⁻¹. It only remains to verify the equation.

I need to show the left side of this equation is constant for ε small enough and equals the right side. By what was just shown, there are finitely many points,

_i=1^m = g⁻¹. By the inverse function theorem, there exist disjoint open sets U_i with x_i ∈ U_i, such that g is one to one on U_i with det having constant sign on U_i and g is an open set containing y. Then let ε be small enough that B ⊆∩_i=1^mg and let V _i ≡ g⁻¹ ∩ U_i.

Therefore, for any ε this small,

∫ ∑m ∫ Ω ϕε(g(x)− y)detDg (x)dx = V ϕε(g (x) − y )detDg (x )dx i=1 i

The reason for this is as follows. The integrand on the left is nonzero only if g

−y ∈ B which occurs only if g ∈ B which is the same as x ∈ g⁻¹. Therefore, the integrand is nonzero only if x is contained in exactly one of the disjoint sets, V _i. Now using the change of variables theorem, (z = g − y,g⁻¹ = x.)

m∑ ∫ = ϕε(z)detDg (g−1(y + z)) ||detDg −1(y + z)||dz i=1 g(Vi)− y

By the chain rule, I = Dg

Dg⁻¹ and so

( )| | detDg g− 1(y + z) |detDg −1 (y + z)|

( ( )) | ( )|| | = sgn detDg g−1 (y + z) |detDg g−1 (y + z)||detDg −1(y+ z)|

( ( )) = sgn detDg g−1(y +z)

= sgn (detDg (x)) = sgn (detDg (xi)).

Therefore, this reduces to

∑m ∫ sgn(detDg (xi)) ϕ ε(z) dz = i=1 g(Vi)−y

∑m ∫ ∑m sgn (det Dg (xi)) ϕε(z)dz = sgn (detDg (xi)). i=1 B(0,ε) i=1

In case g⁻¹

= ∅, there exists ε > 0 such that g∩ B = ∅ and so for ε this small,

∫ ϕ (g (x) − y )detDg (x)dx = 0.■ Ω ε

Now with this, it is possible to define d

whenever f ∈ C.

Proof: First consider the claim that the degree is well defined. Let y

f so dist > 0 and suppose that g,ĝ are two functions as described in the definition. Then for x ∈ ∂Ω,t ∈,

Then for each ε > 0, and ϕ_ε the mollifier of Definition 16.2.3, it follows from Lemma 16.2.4, that if

ε < t∈in[f0,1]dist(y,g(∂Ω )+ t(ˆg (∂Ω)− g (∂Ω )))

then

∫ t → ϕε(g (x )+ t(ˆg (x)− g(x))− y)det(D (g+ t(ˆg− g))(x))dx Ω

is constant. Therefore,

∫ ∫ ϕ (g(x) − y )det(Dg (x))dx = ϕ (ˆg (x)− y)det(Dˆg (x ))dx Ω ε Ω ε

However, for ε sufficiently small, Lemma 16.3.1 implies the two equal integrals reduce respectively to

Thus the definition is well defined.

Consider the last claim. By Lemma 16.1.5 there is ĥ ∈ C^∞

such that

∥ ∥ 1 ∥∥h − ˆh∥∥ < -- min(dist(h (∂Ω),y),dist(f (∂Ω ),y)) ∞ 10

and y is a regular value for ĥ. For x ∈ ∂Ω

Also