Sets where lip is infinite Thomas Z urcher July 2017 lip = 1 / - PowerPoint PPT Presentation

Sets where lip is infinite Thomas Z¨ urcher July 2017 lip = ∞ 1 / 18 �

Two story lines ◮ The story of lip. lip = ∞ 2 / 18 �

Two story lines ◮ The story of lip. ◮ Characterizing sets of non-differentiability. lip = ∞ 2 / 18 �

Rademacher and Stepanov Theorem (Rademacher) A Lipschitz function f : R n → R is differentiable almost everywhere. lip = ∞ 3 / 18 �

Rademacher and Stepanov Theorem (Rademacher) A Lipschitz function f : R n → R is differentiable almost everywhere. Theorem (Stepanov) If f : R n → R satisfies | f ( y ) − f ( x ) | Lip f ( x ) := lim sup sup r r → 0 y ∈ B ( x , r ) | f ( y ) − f ( x ) | = lim sup < ∞ � x − y � y → x almost everywhere, then f is almost everywhere differentiable. lip = ∞ 3 / 18 �

Cheeger Cheeger proves an analogue to Rademacher’s theorem in (quite general) metric measure spaces. This includes giving sense to the notion of differentiability almost everywhere in metric measure spaces. lip = ∞ 4 / 18 �

Keith Keith gives a weaker condition for Cheeger’s result to hold. It involves the Lip − lip-inequality: | f ( y ) − f ( x ) | Lip f ( x ) ≤ K lip f ( x ) := K lim inf sup r r → 0 y ∈ B ( x , r ) for all Lipschitz functions f : X → R and almost all points x . lip = ∞ 5 / 18 �

Schioppa In the correct, quite general setting: if a Lip − lip-inequality holds, then it holds for K = 1. lip = ∞ 6 / 18 �

Balogh–Rogovin–Z¨ urcher / Wildrick–Z¨ urcher Stepanov’s theorem holds in metric measure spaces as well (where Rademacher holds). With Kevin Wildrick, we have a version for RNP-mappings as well (remember Sean Li’s talk). lip = ∞ 7 / 18 �

Balogh–Cs¨ ornyei In general, Stepanov’s theorem fails for lip. Theorem (Balogh–Cs¨ ornyei) There exist a set A ⊂ [0 , 1] of positive Lebesgue measure and a continuous function f : [0 , 1] → R such that lip f ( x ) < ∞ for all x ∈ [0 , 1] , lip f ∈ L s [0 , 1] for all s < 1 and f is not differentiable at any point of A. lip = ∞ 8 / 18 �

Balogh–Cs¨ ornyei In general, Stepanov’s theorem fails for lip. Theorem (Balogh–Cs¨ ornyei) There exist a set A ⊂ [0 , 1] of positive Lebesgue measure and a continuous function f : [0 , 1] → R such that lip f ( x ) < ∞ for all x ∈ [0 , 1] , lip f ∈ L s [0 , 1] for all s < 1 and f is not differentiable at any point of A. Theorem (Balogh–Cs¨ ornyei) There exists a nowhere differentiable, continuous function f : [0 , 1] → R such that lip f ( x ) = 0 for L 1 -a.e. x ∈ [0 , 1] . lip = ∞ 8 / 18 �

Theorem (Balogh–Cs¨ ornyei) Let Ω ⊂ R n be a domain and f : Ω → R a continuous function. Assume that lip f ( x ) < ∞ for x ∈ Ω \ E, where the exceptional set E has σ -finite ( n − 1) -dimensional Hausdorff measure. Assume loc (Ω) for some 1 ≤ p ≤ ∞ . Then f ∈ W 1 , p that lip f ∈ L p loc (Ω) . If, in addition p > n, we have for L n -a.e. x ∈ Ω . lip f ( x ) = Lip f ( x ) = �∇ f ( x ) � lip = ∞ 9 / 18 �

Balogh–Cs¨ ornyei, Heuristics Sobolev functions behave well on almost every line. Show that only very few lines hit the exceptional set in an essential portion. The function is Sobolev on almost all lines, hence on the whole space (taking the correct integrability into account). In metric spaces: modulus of curves. lip = ∞ 9 / 18 �

Lorentz space version With Kevin Wildrick we have a Lorentz space-version for metric measure spaces. Question: What is the correct assumption on the Poincar´ e inequality? We cannot use openendedness (see Luk´ aˇ s Mal´ y’s talk). lip = ∞ 10 / 18 �

Zahorski A set E ⊂ R is the set of non-differentiability for a Lipschitz function if and only if E is a G δσ set of measure zero. lip = ∞ 11 / 18 �

A related question Given a set E ⊂ R of measure zero, is there a Lipschitz function that is not differentiable at any point of E ? lip = ∞ 12 / 18 �

A related question Given a set E ⊂ R of measure zero, is there a Lipschitz function that is not differentiable at any point of E ? Yes. Each set of measure zero is contained in a G δ set of measure zero. lip = ∞ 12 / 18 �

Higher dimensions are different Theorem (Preiss–Speight) Suppose n > 1 . Then there exists a Lebesgue null set N ⊂ R n containing a point of differentiability for every Lipschitz function f : R n → R n − 1 . Also results by Michael Dymond And Olga Maleva. Compare to the talk by Andrea Pinamonti and Gareth Speight. lip = ∞ 13 / 18 �

Question Can we characterize the sets where lip is infinite? By the way, the corresponding sets for Lip are the G δ sets. There are results concerning the sets where derivatives are infinite (for example by Jarn´ ık). lip = ∞ 14 / 18 �

Status With Zolt´ an Buczolich, Bruce Hanson, and Martin Rmoutil: F σ -case almost done. The difficulty is in the construction of the function: ◮ countable sets ◮ perfect, nowhere dense sets ◮ union of nowhere dense, perfect sets ◮ union of nowhere dense sets ◮ F σ -case With David Preiss and Martin Rmoutil: Strategy for F σδ is there, we need to work out the details. Difficulty: steps have to be carried out “simultaneously”. lip = ∞ 15 / 18 �

Typical results Theorem (Banach) The subset of all function in C ([0 , 1] , �·� ∞ ) that are such that in no point t � = 1 both right-sided derived numbers are finite is of the second category. lip = ∞ 16 / 18 �

Typical results Theorem The typical function f ∈ C ([0 , 1] d ) satisfies lip f ( x ) = 0 at points of a residual set of full measure in [0 , 1] d . lip = ∞ 17 / 18 �

Bibliography Zolt´ an Buczolich, Bruce Hanson, Martin Rmoutil, and Thomas Z¨ urcher. On sets where lip f is finite. Work in progress. David Preiss, Martin Rmoutil, and Thomas Z¨ urcher. On sets where lip f is finite. Work even more in progress. lip = ∞ 18 / 18 �