Metrics of curves in shape optimization and analysis - Andrea Carlo ...

We remark that• a measurable representation always exists; for example, letbe the inverse ofarc : S 1 → [0, 2π)α ↦→ (cos(α), sin(α))when α ∈ [0, 2π). arc() is a Borel function; then ((arc ◦ ξ ′ )(s) + a) ∈ S, fora choice of a ∈ lR.• The measurable representation is never unique: for example, given anymeasurable A, B ⊂ [0, 2π] with |A| = |B|,will as well represent ξ.α(s) + 2π1 A (s) − 2π1 B (s)This implies that the family A ξ of measurable α ∈ S that represent the samecurve ξ is infinite. It may be then advisable to define a quotient distance ˆdas follows:ˆd(ξ 1 , ξ 2 ) := inf d(α 1 , α 2 ) (8.3)α 1∈A ξ1 ,α 2∈A ξ2where d(α 1 , α 2 ) = ‖α 1 − α 2 ‖ L 2, or alternatively d = d g is the geodesic distanceon S.8.1.2 Continuous vs measurable lifting — no windingIf ξ ∈ C 1 , we have an unique 13 continuous representation α ∈ S; but notethat, even if ξ 1 , ξ 2 ∈ C 1 , the infimum (8.3) may not be given by the continuousrepresentations α 1 , α 2 of ξ 1 , ξ 2 . Moreover there are rectifiable curves ξ that donot admit a continuous representation α, as for example the polygons.A problem similar to the above is expressed by this proposition.Proposition 8.7 For any h ∈ Z, the set of closed smooth curves ξ with rotationindex h, when represented in S using the continuous lifting, is dense in S \ Z.It implies that we cannot properly extend the concept of rotation index to S.The proof is based on this lemma.Lemma 8.8 Suppose that ξ is not flat, let τ be one of the measurable liftings ofξ. There exists a smooth projection π : V → S defined in a neighborhood V ⊂ L 2of τ such that, if f ∈ L 2 ∩ C ∞ then π(f) is in L 2 ∩ C ∞ .This is the proof of both the lemma and the proposition.13 Indeed, the continuous lifting is unique up to addition of a constant to α(s), which isequivalent to a rotation of ξ; and the constant is decided by Φ 1 (α) = 2π 255