Fine properties of monotone maps arising in optimal transport for non-quadratic costs

Theorem 3.1

If T : R n → P ( R n ) is a multivalued map that is h-monotone, with h ∈ C 2 homogeneous of degree p for some p ≥ 2 and satisfying (2.3), then T ( x ) is a singleton for all points x ∈ dom ( T ) except on a set of measure zero.

Proof

Let

we will prove that S has Lebesgue measure zero. For each k ≥ 1 integer, let

We have S = ∪ k = 1 ∞ S k . Also for each k ≥ 1 integer, let { B j } j = 1 ∞ be a family of Euclidean balls in R n each one with radius ε ⁄ k with ε > 0 small to be chosen later after inequality (3.10), depending only Λ ⁄ λ , the constants in (2.3), and such that R n = ∪ j = 1 ∞ B j . Let

We have ∪ j = 1 ∞ B j * = dom ( T ) , and so S = ∪ k , j = 1 ∞ S k j . We shall prove that ∣ S k j ∣ = 0 for all k and j . To do this we recall that a point x ∈ R n is a density point for the set E ⊂ R n , with E not necessarily Lebesgue measurable, if

where ∣ ⋅ ∣ * denotes the Lebesgue outer measure and B r ( x ) is the Euclidean ball with center x and radius r . We shall prove in Lemma 3.7, that if E ⊂ R n is any set, then almost all points in E are density points for E . In view of this, if we prove that each x 0 ∈ S k j is not a density point for S k j , then it follows that ∣ S k j ∣ = 0 . We remark that a priori it is not known that the sets S k j are Lebesgue measurable, and therefore, we cannot apply the Lebesgue differentiation theorem directly. This is why we use the Lebesgue outer measure and prove Lemma 3.7.

Let us fix x 0 ∈ S k j . Then diam T ( x 0 ) > 1 ⁄ k and T ( x 0 ) ∩ B j ≠ ∅ , so we can pick y 1 ∈ T ( x 0 ) ∩ B j . There exists y 2 ∈ T ( x 0 ) such that ∣ y 1 − y 2 ∣ ≥ 1 ⁄ 2 k , because otherwise T ( x 0 ) ⊂ B 1 ⁄ 2 k ( y 1 ) , which would imply that diam T ( x 0 ) ≤ 1 ⁄ k . Also notice that y 2 ∉ B j if ε < 1 ⁄ 4 , since B j has radius ε ⁄ k . In addition, if x j is the center of B j , it follows that ∣ x j − y 2 ∣ ≥ ∣ y 2 − y 1 ∣ − ∣ y 1 − x j ∣ ≥ 1 2 k − ε k ( > 0 ) if ε < 1 ⁄ 2 . And so for each ξ ∈ B j , ∣ ξ − y 2 ∣ ≥ ∣ x j − y 2 ∣ − ∣ ξ − x j ∣ ≥ 1 2 k − 2 ε k ≥ 2 ε k if ε < 1 ⁄ 8 .

Set

Given x ≠ x 0 , x ∈ dom ( T ) , let z = x − x 0 − ( x − x 0 ) ⋅ e ∣ e ∣ e ∣ e ∣ . Then z ⋅ e ∣ e ∣ = 0 , ( x − x 0 ) ⋅ z ∣ z ∣ = ∣ z ∣ , and

If δ is the angle between the unit vectors x − x 0 ∣ x − x 0 ∣ and e ∣ e ∣ , then we have 0 < δ < π and

Let ξ ∈ T ( x ) and consider the matrix A ( x , x 0 ; ξ , y 2 ) defined by (2.2). From (2.4),

and since x ≠ x 0 , Φ ( x , x 0 ; ξ , y 2 ) > 0 . To simplify the notation, we write A ( x , x 0 ; ξ , y 2 ) = A and Φ ( x , x 0 ; ξ , y 2 ) = Φ .

To show that x 0 is not a density point for S k j , we analyze the sizes of the angles between various vectors using the h -monotonicity. For the sake of clarity, we divide the proof into three steps.

Step 1. We shall estimate F ( δ ) ≔ angle A 1 ⁄ 2 x − x 0 ∣ x − x 0 ∣ , A 1 ⁄ 2 e ∣ e ∣ , proving (3.5).

We have cos F ( δ ) = A 1 ⁄ 2 x − x 0 ∣ x − x 0 ∣ , A 1 ⁄ 2 e ∣ e ∣ A 1 ⁄ 2 x − x 0 ∣ x − x 0 ∣ A 1 ⁄ 2 e ∣ e ∣ , and from (3.1),

Hence,

From (3.2),

If

then by Cauchy-Schwarz,

Setting

we obtain from (3.3) that

Notice that since ∣ B ∣ ≤ C , we have 1 + C s 2 + 2 B s ≥ 0 for all s ∈ R . If ∣ B ∣ < C , then 1 + C s 2 + 2 B s > 0 for all s ∈ R ; and if ∣ B ∣ = C , then 1 + C s 2 + 2 B s = 1 + B 2 s 2 + 2 B s = ( 1 + B s ) 2 , which is strictly positive except when s = − 1 ⁄ B . For s > 0 , let us now estimate

Since − C ≤ B ≤ C , it follows that

for − ∞ < s < ∞ . Therefore, if ∣ s ∣ ≤ 1 ⁄ ( 2 C ) , then we obtain 1 − C ∣ s ∣ ≥ 1 ⁄ 2 . Also 1 + B s ≥ 1 − C s if s > 0 , and 1 + B s ≥ 1 + C s for s < 0 . So 1 + B s ≥ 1 − C ∣ s ∣ . Hence,

In particular, since λ ⁄ Λ ≤ C ≤ Λ ⁄ λ and ∣ B ∣ ≤ C , we obtain the bound

This implies that

for δ such that 0 < tan δ ≤ ( 1 ⁄ 2 ) λ ⁄ Λ . Since tan δ ∼ δ for δ sufficiently small, we then obtain the estimate

for all 0 < δ ≤ δ 0 , with δ 0 a positive number depending only on λ ⁄ Λ . Hence,

On the other hand, arccos ( 1 − x 2 ) x → 2 as x → 0 + , it follows that

Therefore, from the definitions of δ and F ( δ ) , we obtain

for all ξ ∈ T ( x ) if angle ( x − x 0 , e ) ≤ δ 0 , x ≠ x 0 , x ∈ dom ( T ) . On the other hand, from the monotonicity (2.1),

that is, ⟨ A ( x , x 0 ; ξ , y 2 ) 1 ⁄ 2 ( x − x 0 ) , A ( x , x 0 ; ξ , y 2 ) 1 ⁄ 2 ( ξ − y 2 ) ⟩ ≥ 0 for all ξ ∈ T ( x ) , and therefore,

Hence, from (3.6),

when angle ( x − x 0 , e ) ≤ δ 0 , x ≠ x 0 , x ∈ dom ( T ) .

Step 2. Let Γ = { x : angle ( x − x 0 , e ) ≤ δ 0 } . We shall prove that

for ε sufficiently small depending only on Λ ⁄ λ ; B j = B ε ⁄ k ( x j ) .

Recall that x 0 ∈ S k j , y 1 ∈ T x 0 ∩ B j , and y 2 ∈ T x 0 with ∣ y 2 − y 1 ∣ ≥ 1 ⁄ 2 k . Let C be the cone with vertex y 2 , axis e = y 2 − y 1 , and opening π + θ 0 . We have C ∩ B j = ∅ when θ 0 > 0 is small, for all ε < 1 . Suppose by contradiction that (3.8) does not hold, that is, there is ξ ∈ T ( x ) ∩ B j for some x ∈ Γ , and let θ be the angle between ξ − y 2 and e . Since ξ ∈ B j , we then have θ ≥ ( π + θ 0 ) ⁄ 2 for all ε < 1 . To obtain a contradiction, we first right down the left-hand side of (3.7) in terms of the angle θ and will show that it is close to π when θ ∼ π , for ε sufficiently small, contradicting (3.7).

To do this, we proceed as before as in the writing of F ( δ ) letting now ζ = ξ − y 2 − ( ξ − y 2 ) ⋅ e ∣ e ∣ e ∣ e ∣ . Then ζ ⋅ e ∣ e ∣ = 0 , ( ξ − y 2 ) ⋅ ζ ∣ ζ ∣ = ∣ ζ ∣ , and

recalling that θ is the angle between ξ − y 2 and e . Then the left-hand side of (3.7) can be written as follows:

and since θ > π ⁄ 2 , it follows as in (3.3) that

with

Let us now analyze the function G ( θ ) when θ is close to π . For s < 0 and with g defined as in (3.4) but with B ¯ and C ¯ instead of B and C , we have

Applying the estimates for the last denominator obtained in Step 1 yields

In particular, since λ ⁄ Λ ≤ C ¯ ≤ Λ ⁄ λ , and ∣ B ¯ ∣ ≤ C ¯ , we obtain the bound

This implies that

for θ such that − ( 1 ⁄ 2 ) λ ⁄ Λ < tan θ < 0 . Now tan θ ∼ θ − π when θ → π − , and so

for π − θ 1 < θ < π with θ 1 > 0 small depending only on Λ ⁄ λ , which implies the following lower bound for the left-hand side of (3.7)

for all π − θ 1 < θ < π . We now choose ε > 0 so that for each ξ ∈ B j = B ε ⁄ k ( x j ) , the angle θ = angle ( ξ − y 2 , e ) satisfies π − θ 1 < θ < π . Recall once again that x 0 ∈ S k j , y 1 ∈ T x 0 ∩ B j , and y 2 ∈ T x 0 with ∣ y 2 − y 1 ∣ ≥ 1 ⁄ 2 k . Then ∣ y 2 − x j ∣ ≥ ∣ y 2 − y 1 ∣ − ∣ x j − y 1 ∣ ≥ 1 2 k − ε k = 1 2 ε − 1 ε k ; that is, y 2 ∉ B 1 2 ε − 1 ε k ( x j ) . For each δ ′ large, there is ε sufficiently small such that 1 2 ε − 1 = 1 + δ ′ . Let α be the angle between the vectors y 2 ξ → and − e = y 2 y 1 → , and consider the convex hull of B j and y 2 , i.e., the ice-cream cone containing B j . If β is the angle between the vector y 2 x j → and the edge of the convex hull, we have α ≤ 2 β , see Figure 1. So sin α ≤ 2 sin β = 2 ε ⁄ k ∣ y 2 − x j ∣ ≤ 2 ε ⁄ k 1 2 ε − 1 ε ⁄ k = 2 ⁄ ( 1 + δ ′ ) , a number than can be made arbitrarily small taking ε small, in particular, it can be made smaller than sin θ 1 . Therefore, with this choice of ε , the ball B j is determined and the inequality (3.10) can be applied when ξ ∈ B j . Then combining (3.7), (3.9), and (3.10), we obtain that

for π − θ 1 < θ < π . Since x ∈ Γ , angle ( x − x 0 , e ) ≤ δ 0 . Thus, if θ → π − , this yields a contradiction since arccos ( − 1 ) = π . The proof of (3.8) is then complete.

Step 3. We are now in a position to prove that x 0 ∈ S k j cannot be a point of density for S k j . Recall S k j = S k ∩ B j * , where B j * = { x ∈ dom ( T ) : T x ∩ B j ≠ ∅ } . From (3.8), it follows that B j * ∩ Γ = ∅ with Γ the cone in Step 2. Let B r ( x 0 ) be the Euclidean ball centered at x 0 with radius r , then