Topics In Analysis

21.1 Sard’s Lemma and Approximation

First is an easy result about approximation of continuous functions with smooth ones.

Proof:  Form g ≡ f ∗ψ_n for a mollifier ψ_n. This will approximate f uniformly on Ω, and will be in C_c^∞

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Using the Weierstrass approximation theorem, you could also get g to equal a function from G described in the development of the Fourier transform for all x ∈ Ω. Simply apply Theorem 7.2.9 to the functions in G.

Applying this result to the components of a vector valued function yields the following corollary.

The following lemma is a wonderful application of the Vitali covering theorem.

Proof: For k ∈ ℕ

T ≡ {x ∈ T : ||Dh (x)|| < k} k

and let ε > 0 be given. Since T_k is a subset of a set of measure zero, it is measurable, but we don’t need to pay much attention to this fact. Now by outer regularity, there exists an open set V , containing T_k which is contained in U such that m_p

< ε. Let x ∈ T_k. Then by differentiability,

h(x + v) = h (x )+ Dh (x)v+ o (v)

and so there exist arbitrarily small r_x < 1 such that B

⊆ V and whenever ≤ 5r_x, < . Thus

From the Vitali covering theorem, there exists a countable disjoint sequence of these balls,

_i=1^∞ such that _i=1^∞ = _i=1^∞ covers T_k Then letting m_p denote the outer measure determined by m_p,

( ( )) ( ( )) mp-(h (Tk)) ≤ mp- h ∪ ∞i=1B^i ≤ mp- ∪∞i=1h ^Bi

∑∞ ---( (^ )) ∞∑ ≤ mp h Bi ≤ mp (B (h (xi),6krxi)) i=1 i=1

Since ε > 0 is arbitrary, this shows m_p = m_p = 0. Now

mp (h (T)) = kli→m∞ mp (h(Tk)) = 0.■

The following is Sard’s lemma. It is important in defining the degree.

Proof: For convenience, assume the balls in the following argument come from

_∞. First note that Z is a Borel set because h is continuous and so the component functions of the Jacobian matrix are each Borel measurable. Hence the determinant is also Borel measurable.

Suppose that U is a bounded open set. Let ε > 0 be given. Also let V ⊇ Z with V ⊆ U open, and

mp (Z)+ ε > mp(V ).

Now let x ∈ Z. Then since h is differentiable at x, there exists δ_x > 0 such that if r < δ_x, then B

⊆ V and also o < η for < r. Thus

h (x+B (0,r)) = h(B (x,r)) ⊆ h (x )+ Dh (x)(B (0,r)) +B (0,rη), η < 1.

Regard Dh

as an n×n matrix, the matrix of the linear transformation Dh with respect to the usual coordinates. Since x ∈ Z, it follows that there exists an invertible matrix A such that ADh is in row reduced echelon form with a row of zeros on the bottom. Therefore,

mp (A (h (B (x,r)))) ≤ mp (ADh (x )(B (0,r))+ AB (0,rη)) (21.1.1)

(21.1.1)

The diameter of ADh

is no larger than 2r and it lies in ℝ^p−1 ×. The diameter of AB is no more than .Therefore, the measure of the right side in 21.1.1 is no more than

That is,

m (A (h(B (x,r)))) ≤ C (||A ||,||Dh (x)||)(2r)p η p

Hence from the change of variables formula for linear maps,

C-(||A||,||Dh--(x)||) mp (h(B (x,r))) ≤ η |det(A)| mp (B (x,r))

Then letting δ_x be still smaller if necessary, corresponding to sufficiently small η,

m (h (B (x,r))) ≤ εm (B (x,r)) p p

The balls of this form constitute a Vitali cover of Z. Hence, by the Vitali covering theorem, Corollary 11.4.6, there exists

_i=1^∞,B_i = B_i, a collection of disjoint balls, each of which is contained in V, such that m_p ≤ εm_p and m_p = 0. Hence from Lemma 21.1.3,

mp (h(Z) ∖∪ih(Bi)) ≤ mp (h(Z ∖∪iBi)) = 0

Therefore,

Since ε is arbitrary, this shows m_p = 0. What if U is not bounded? Then consider Z_n = Z ∩ B. From what was just shown, h has measure 0 and so it follows that h also does, being the countable union of sets of measure zero. ■

Sard’s lemma implies the following fundamental result.

Proof: Using Corollary 21.1.2, let f₀ ∈ C^∞

, _∞ < . Let ₁ be a regular value for f₀ and < . Such ỹ₁ exists because of Sard’s lemma. There are points ỹ₁ not in f₀ arbitrarily close to y₁. Let f₁ ≡ f₀ + y₁ −₁ for all x. Thus y₁ is a regular value of f₁ because Df₁ = Df₀ and if f₁ = y₁, this is the same as having f₀ = ₁ where ₁ is a regular value of f₀. Then also

Suppose now there exists f_k ∈ C^∞ with each of the y_i for i = 1,,k a regular value of f_k and

δ k ( δ) ||f − fk||∞ < -+ -- - . 2 r 3

Then letting S_k denote the singular values of f_k, Sard’s theorem implies there exists

_k+1 such that

δ |^yk+1 − yk+1| < 3r

and

^y ∕∈ S ∪ ∪k (S + y − y). (21.1.2) k+1 k i=1 k k+1 i

(21.1.2)

Let

fk+1 (x ) ≡ fk (x )+ yk+1 − ^yk+1. (21.1.3)

(21.1.3)

If f_k+1

= y_i for some i ≤ k, then

fk(x)+ yk+1 − yi = ^yk+1

and so f_k

is a regular value for f_k since by 21.1.2, _k+1S_k + y_k+1 − y_i and so f_kS_k. Therefore, for i ≤ k, y_i is a regular value of f_k+1 since by 21.1.3, Df_k+1 = Df_k. Now suppose f_k+1 = y_k+1. Then

y = f (x)+ y − ^y k+1 k k+1 k+1

so f_k

= _k+1 implying that f_k = _k+1S_k. Hence detDf_k+1 = detDf_k≠0. Thus y_k+1 is also a regular value of f_k+1. Also,

Let ≡ f_r. Then

∥ ∥ ( ) ∥∥^f − f∥∥ < δ+ δ < δ ∞ 2 3

and each of the y_i is a regular value of

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