1 Lévy Processes and Infinite Divisibility - Department of ...

Exercise 8 Recall that D[0,1] is the space of functions f : [0,1] → R whichare right continuous with left limits.(i) Define the norm ||f|| = sup x∈[0,1] |f(x)|. Use the triangle inequality todeduce that D[0,1] is closed under uniform convergence with respect tothe norm ||·||. That is to say, show that if {f n : n ≥ 1} is a sequence inD[0,1] and f : [0,1] → R such that lim n↑∞ ||f n −f|| = 0 then f ∈ D[0,1].(ii) Suppose that f ∈ D[0,1] and let ∆ = {t ∈ [0,1] : |f(t)−f(t−)| ̸= 0} (theset of discontinuity points). Show that ∆ is countable if ∆ c is countablefor all c > 0 where ∆ c = {t ∈ [0,1] : |f(t)−f(t−)| > c}. Next fix c > 0.By supposing for contradiction that ∆ c has an accumulation point, say x,show that the existence of either a left or right limit at x fails as it wouldimply that there is no left or right limit of f at x. Deduce that ∆ c andhence ∆ is countable.Exercise 9 The explicit construction of a Lévy process given in the Lévy–Itôdecomposition begs the question as to whether one may construct examples ofdeterministic functions which have similar properties to those of the paths ofLévyprocesses. The objective of this exercise is to do preciselythat. The readeris warned however, that this is purely an analytical exercise and one should notnecessarily think of the paths of Lévy processes as being entirely similar to thefunctions constructed below in all respects.(i) Let us recall the definition of the Cantor function which we shall use toconstruct a deterministic function which has bounded variation and thatis right continuous with left limits and whose points of discontinuity aredense in its domain. Take the interval C 0 := [0,1] and perform the followingiteration. For n ≥ 0 define C n as the union of intervals whichremain when removing the middle third of each of the intervals whichmake up C n−1 . The Cantor set C is the limiting object, ⋂ n≥0 C n and canbe described byC = {x ∈ [0,1] : x = ∑ k≥1α k3 k such that α k ∈ {0,2} for each k ≥ 1}.One sees then that the Cantor set is simply the points in [0,1] whichomits numbers whose tertiary expansion contain the digit 1. To describethe Cantor function, for each x ∈ [0,1] let j(x) be the smallest j for whichα j = 1 in the tertiary expansion of ∑ k≥1 α k/3 k of x. If x ∈ C thenj(x) = ∞ and otherwise if x ∈ [0,1]\C then 1 ≤ j(x) < ∞. The Cantorfunction is defined as followsf(x) = 1j(x)−12 + ∑ α ifor x ∈ [0,1].j(x) 2i+1 i=1Now consider the function g : [0,1] → [0,1] given by g(x) = f −1 (x)−axfor a ∈ R. Here we understand f −1 (x) = inf{θ : f(θ) > x}. Note that g is30