The length of the graph of an increasing function with an almost everywhere zero derivative

Description

The length of the graph of an increasing function with an almost everywhere zero derivative. 1 Answer to problem 1. Lemma 1 If A is a set on which f has a null derivative, then f(A) is a null set. Proof of lemma: Let y = f(x), with x ? A. The inverse function of f has an infinite derivative at y = f(x). This inverse function is, as f , an increasing function. So it has almost everywhere a finite derivative by Lebesgue's theorem. This forces y to stay inside a null set. Lemma 2 Let f be continous, increasing defined on [a, b]. Then the length of f 's graph is bounded below by b? a and f(b)? f(a), and bounded above by b? a + f(b)? f(a). Proof of lemma: This length is the upper-bound of n∑ k=1 √ (xi ? xi?1)2 + (f(xi)? f(xi?1))2 with a = x0 < x1 < · · · < xn = b. Each term of the sum is greater than xi ? xi?1, greater than f(xi)? f(xi?1), and at most xi ? xi?1 + f(xi)? f(xi?1).

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Sujets

• Length

Informations

• Publié par: profil-urra-2012
• Nombre de lectures: 13
• Langue: English

Extrait

The length of the graph of an increasing function with an almost everywhere zero derivative.

1 Answerto problem 1. Lemma 1IfAis a set on whichfhas a null derivative, thenf(A)is a null set.

Proof of lemma: Lety=f(x), withx∈Ainverse function of. Thefhas an infinite derivative aty=f(x). Thisinverse function is, asf, an increasing function.So it has almost everywhere a finite derivative by Lebesgue’s theorem.This forcesyto stay inside a null set. Lemma 2Letfbe continous, increasing defined on[a, b]the length of. Thenf’s graph is bounded below byb−aandf(b)−f(a), and bounded above byb−a+f(b)−f(a).

Proof of lemmalength is the upper-bound of: This n X p 2 2 (x i −x i−1 ) +(f(x i )−f(x i−1 )) k=1 witha=x 0 < x 1 <∙ ∙ ∙< x n =b. Eachterm of the sum is greater thanx i −x i−1 , greater than f(x i )−f(x i−1 ), and at mostx i −x i−1 +f(x i )−f(x i−1 ). Solution of problem 1note: Wec=f(a) andd=f(blemma (2), the length of). Byf’s graph is not greater thanb−a+d−cWe construct a partitionwe are looking for a lower bound.. Now of [a, b], with a finite number of intervals, on which the length of the corresponding arc offis essentially composed of the increase of thex–coordinate, or of the increase of they–coordinate. This is possible because there is a subset of [a, b] with measureb−awhose image is null. Let (I n ) be a sequence of disjoint intervals of total length< , outside of whichfhas a zero derivative. Letus noteA= [a, b]\ ∪ n I n , andJ n =f(I n ). [c, d] is the union off(A) and the f(I n ); by lemma (1),f(ASo the sum of the lengths of the) is null.J n isd−c. Afterrenaming we P p can assume thanlength(J k )≥d−c−[. Now,a, b]\ ∪ 1≤k≤p I k is a finite union of intervals k=1 0 (V n ) 1≤n≤q , whose total length is greater thenb−a−. Thelength offs graph is the sum of the lengths offrestricted to theI k , and offrestricted to theV k (2),. Usingfrestricted to V k has length greater than length(V k ), andfrestricted toI k has length greater than length(J k ). Summing on all intervals, we get the lower bound p q X X length(J k ) +length(V j )≥d−c−+b−a−. k=1 j=1

2 Answerto problem 2. k Lemma 3Ifβ=we have the following upper bound for the binomial coefficient: n   n1 n ≤δ(β)withδ(β) = β1−β k β(1−β) 1 andδis a strictly increasing function ofβforβ≤, andδ(1/2) = 2. 2

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