Section 31

Derivatives on the Line*

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Problem 31.1

Extend Examples 31.1 and 31.3: Let p0,…,pr−1p_{0}, \ldots , p_{r-1} be nonnegative numbers adding to 1, where r≥2r \geq 2; suppose there is no ii such that pi=1p_{i}=1. Let X1X_{1}, X2,…X_{2}, \ldots be independent, identically distributed random variables such that P(Xn=i)=pi,0≤i<r\mathbb {P}\left(X_{n}=i\right)=p_{i}, 0 \leq i<r, and put X=∑n=1∞Xnr−nX=\sum_{n=1}^{\infty } X_{n} r^{-n}. Let FF be the distribution function of XX. Show that FF is continuous. Show that FF is strictly increasing over the unit interval if and only if all the pip_{i} are strictly positive. Show that F(x)≡xF(x) \equiv x for 0≤x≤10 \leq x \leq 1 if pi≡r−1p_{i} \equiv r^{-1} and that otherwise FF is singular; prove singularity by extending the arguments both of Example 31.1 and of Exampie 31.3. What is the analogue of (31.17)?

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Problem 31.2

↑\uparrow In Problem 31.1 take r=3r=3 and p0=p2=12,p1=0p_{0}=p_{2}=\frac{1}{2}, p_{1}=0. The corresponding FF is called the Cantor function. The complement in [0,1][0,1] of the Cantor set (see Problems 1.5 and 3.16) consists of the middle third (13,23)\left(\frac{1}{3}, \frac{2}{3}\right), the middle thirds (19,29)\left(\frac{1}{9}, \frac{2}{9}\right) and (79,89)\left(\frac{7}{9}, \frac{8}{9}\right), and so on. Show that FF is 12\frac{1}{2} on the first of these intervals, 14\frac{1}{4} on the second, 34\frac{3}{4} on the third, and so on. Show by direct argument that F′=0F^{\prime }=0 except on a set of Lebesgue measure 0.

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Problem 31.3

A real function ff of a real variable is a Lebesgue function if [x:f(x)≤α][x: f(x) \leq \alpha ] is a Lebesgue set for each α\alpha.

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(a)

Show that, if f1f_{1} is a Borel function and f2f_{2} is a Lebesgue function, then the composition f1f2f_{1} f_{2} is a Lebesgue function.

(b)

Show that there exists a Lebesgue function f1f_{1} and a Lebesgue (even Borel, even continuous) function f2f_{2} such that f1f2f_{1} f_{2} is not a Lebesgue function. Hint: Use Example 31.4.

†{ }^{\dagger } This uses the fact that ∥F∥Δ\left\| F\right\|_{\Delta } cannot decrease under passage to a finer partition.

Problem 31.4

↑\uparrow An arbitrary function ff on ( 0,1]] can be represented as a composition of a Lebesgue function f1f_{1} and a Borel function f2f_{2}. For xx in ( 0,1 ], let dn(x)d_{n}(x) be the nnth digit in its nonterminating dyadic expansion, and define f2(x)=∑n=1∞2dn(x)/3nf_{2}(x)= \sum_{n=1}^{\infty } 2 d_{n}(x) / 3^{n}. Show that f2f_{2} is increasing and that f2(0,1]f_{2}(0,1] is contained in the Cantor set. Take f1(x)f_{1}(x) to be f(f2−1(x))f\left(f_{2}^{-1}(x)\right) if x∈f2(0,1]x \in f_{2}(0,1] and 0 if x∈(0,1]−f2(0,1]x \in (0,1]-f_{2}(0,1]. Now show that f=f1f2f=f_{1} f_{2}.

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Problem 31.5

Let r1,r2,…r_{1}, r_{2}, \ldots be an enumeration of the rationals in ( 0,1 ) and put F(x)=Σkrk≤x2−kF(x)= \Sigma_{k r_{k} \leq x} 2^{-k}. Define φ\varphi by (14.5) and prove that it is continuous and singular.

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Problem 31.6

Suppose that μ\mu and FF are related by (31.18). If FF is not absolutely continuous, then μ(A)>0\mu (A)>0 for some set AA of Lebesgue measure 0. It is an interesting fact, however, that almost all translates of AA must have μ\mu-measure 0. From Fubini's theorem and the fact that λ\lambda is invariant under translation and reflection through 0, show that, if λ(A)=0\lambda (A)=0 and μ\mu is σ\sigma-finite, then μ(A+x)=0\mu (A+x)=0 for xx outside a set of Lebesgue measure 0.

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Problem 31.7

17.431.6↑17.431 .6 \uparrow Show that FF is absolutely continuous if and only if for each Borel set A,μ(A+x)A, \mu (A+x) is continuous in xx.

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Problem 31.8

Let F∗(x)=lim⁡δ→0inf⁡(F(v)−F(u))/(v−u)F_{*}(x)=\lim_{\delta \rightarrow 0} \inf (F(v)-F(u)) /(v-u), where the infimum extends over uu and vv such that u<x<vu<x<v and v−u<δv-u<\delta. Define F∗(x)F^{*}(x) as this limit with the infimum replaced by a supremum. Show that in Theorem 31.4, F′F^{\prime } can be replaced by F∗F^{*} in part (i) and by F∗F_{*} in part (ii). Show that in Theorem 31.6, FD{ }_{F} D can be replaced by F∗F_{*} (note that F∗(x)≤FD(x)F_{*}(x) \leq { }_{F} D(x) ).

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Problem 31.9

Lebesgue's density theorem. A point xx is a density point of a Borel set AA if λ((u,v]∩A)/(v−u)→1\lambda ((u, v] \cap A) /(v-u) \rightarrow 1 as u↑xu \uparrow x and v↓xv \downarrow x. From Theorems 31.2 and 31.4 deduce that almost all points of AA are density points. Similarly, λ((u,v]∩A)/(v−u)→0\lambda ((u, v] \cap A) /(v-u) \rightarrow 0 almost everywhere on AcA^{c}.

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Problem 31.10

Let f:[a,b]→Rkf:[a, b] \rightarrow R^{k} be an aic; f(t)=(f1(t),…,fk(t))f(t)=\left(f_{1}(t), \ldots , f_{k}(t)\right). Show that the arc is rectifiable if and only if each fif_{i} is of bounded variation over [a,b][a, b].

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Problem 31.11

↑\uparrow Suppose that FF is continuous and nondecreasing and that F(0)=0F(0)=0, F(1)=1F(1)=1. Then f(x)=(x,F(x))f(x)=(x, F(x)) defines an arc f:[0,1]→R2f:[0,1] \rightarrow R^{2}. It is easy to see by monotonicity that the arc is rectifiable and that, in fact, its length satisfies L(f)≤2L(f) \leq 2. It is also easy, given ϵ\epsilon, to produce functions FF for which L(f)>2−ϵL(f)>2-\epsilon. Show by the arguments in the proof of Theorem 31.4 that L(f)=2L(f)=2 if FF is singular.

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Problem 31.12

Suppose that the characteristic function of FF satisfies lim sup⁡t→∞∣φ(t)∣=1\limsup_{t \rightarrow \infty }\left|\varphi (t)\right|=1. Show that FF is singular. Compare the lattice case (Problem 26.1). Hint: Use the Lebesgue decomposition and the Riemann-Lebesgue theorem.

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Problem 31.13

Suppose that X1,X2,…X_{1}, X_{2}, \ldots are independent and assume the values ±1\pm 1 with probability 12\frac{1}{2} each, and let X=∑n=1∞Xn/2nX=\sum_{n=1}^{\infty } X_{n} / 2^{n}. Show that XX is uniformly distributed over [−1,+1][-1,+1]. Calculate the characteristic functions of XX and XnX_{n} and deduce (1.40). Conversely, establish (1.40) by trigonometry and conclude that XX is uniformly distributed over [−1,+1][-1,+1].

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Problem 31.14
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(a)

Suppose that X1,X2,…X_{1}, X_{2}, \ldots are independent and assume the values 0 and 1 with probability 12\frac{1}{2} each. Let FF and GG be the distribution functions of ∑n=1∞X2n−1/22n−t\sum_{n=1}^{\infty } X_{2 n-1} / 2^{2 n-t} and ∑n=1∞X2n/22n\sum_{n=1}^{\infty } X_{2 n} / 2^{2 n}. Show that FF and GG are singular but that F∗GF * G is absolutely continuous.

(b)

Show that the convolution of an absolutely continuous distribution function with an arbitrary distribution function is absolutely continuous.

Problem 31.15

31.2 ↑ Show that the Cantor function is the distribution function of ∑n=1∞Xn/3n\sum_{n=1}^{\infty } X_{n} / 3^{n}, where the XnX_{n} are independent and assume the values 0 and 2 with probability 12\frac{1}{2} each. Express its characteristic function as an infinite product.

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Problem 31.16

Show for the FF of Example 31.1 that FD(1)=∞{ }_{F} D(1)=\infty and DF(0)=0D^{F}(0)=0 if p0<12p_{0}<\frac{1}{2}. From (31.17) deduce that FD(x)=∞{ }_{F} D(x)=\infty and DF(x)=0D^{F}(x)=0 for all dyadic rationals xx. Analyze the case p0>12p_{0}>\frac{1}{2} and sketch the graph

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Problem 31.17

6.14↑6.14 \uparrow Let FF be as in Example 31.1, and let μ\mu be the corresponding probability measure on the unit interval. Let dn(x)d_{n}(x) be the nnth digit in the nonterminating binary expansion of xx, and let sn(x)=∑k=1ndk(x)s_{n}(x)=\sum_{k=1}^{n} d_{k}(x). If In(x)I_{n}(x) is the dyadic interval of order nn containing xx, then

−1nlog⁡μ(In(x))=−(1−sn(x)n)log⁡p0−sn(x)nlog⁡p1.(31.33) -\frac{1}{n} \log \mu \left(I_{n}(x)\right)=-\left(1-\frac{s_{n}(x)}{n}\right) \log p_{0}-\frac{s_{n}(x)}{n} \log p_{1}. \tag {31.33}
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(a)

Show that (31.33) converges on a set of μ\mu-measure 1 to the entropy h=−p0log⁡p0−p1log⁡p1h=-p_{0} \log p_{0}-p_{1} \log p_{1}. From the fact that this entropy is less than log⁡2\log 2 if p0≠12p_{0} \neq \frac{1}{2}, deduce in this case that on a set of μ\mu-measure 1,F1, F does not have a finite derivative.

(b)

Show that (31.33) converges to −12log⁡p0−12log⁡p1-\frac{1}{2} \log p_{0}-\frac{1}{2} \log p_{1} on a set of Lebesgue measure 1. If p0≠12p_{0} \neq \frac{1}{2} this limit exceeds log⁡2\log 2 (arithmetic versus geometric means), and so μ(In(x))/2−n→0\mu \left(I_{n}(x)\right) / 2^{-n} \rightarrow 0 except on a set of Lebesgue measure 0. This does not prove that F′(x)F^{\prime }(x) exists almost everywhere, but it does show that, except for xx in a set of Lebesgue measure 0. if F′(x)F^{\prime }(x) does exist, then it is 0.

(c)

Show that, if (31.33) converges to ll, then

lim⁡nμ(In(x))(2−n)α={∞ if α>l/log⁡2,0 if α<l/log⁡2.(31.34) \lim _{n} \frac{\mu \left(I_{n}(x)\right)}{\left(2^{-n}\right)^{\alpha }}= \begin{cases} \infty & \text{ if } \alpha >l / \log 2, \tag {31.34} \\ 0 & \text{ if } \alpha <l / \log 2. \end{cases}

If (31.34) holds, then (roughly) FF satisfies a Lipschitz condition †{ }^{\dagger } of (exact) order l/log⁡2l / \log 2. Thus FF satisfies a Lipschitz condition of order h/log⁡2h / \log 2 on a set of μ\mu-measure 1 and a Lipschitz condition of order (−12log⁡p0−12log⁡p1)/log⁡2\left(-\frac{1}{2} \log p_{0}-\frac{1}{2} \log p_{1}\right) / \log 2 on a set of Lebesgue measure 1.

Problem 31.18

van der Waerden's continuous, nowhere differentiable function is f(x)=∑k=0∞ak(x)f(x)= \sum_{k=0}^{\infty } a_{k}(x), where a0(x)a_{0}(x) is the distance from xx to the nearest integer and ak(x)=2−ka0(2kx)a_{k}(x)=2^{-k} a_{0}\left(2^{k} x\right). Show by the Weierstrass MM-test that ff is continuous. Use (31.8) and the ideas in Example 31.1 to show that ff is nowhere differentiable. †{ }^{\dagger } A Lipschitz condition of order α\alpha holds at xx if F(x+h)−F(x)=O(∣h∣α)F(x+h)-F(x)=O\left(|h|^{\alpha }\right) as h→0h \rightarrow 0, for α>1\alpha >1 this implies F′(x)=0F^{\prime }(x)=0, and for 0<α<10<\alpha <1 it is a smoothness condition stronger than continuity and weaker than differentiability.

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Problem 31.19

Show (see (31.31)) that (apart from addition of constants) a function can have only one representation F1+F2F_{1}+F_{2} with F1F_{1} absolutely continuous and F2F_{2} singular.

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Problem 31.20

Show that the FsF_{\mathrm{s}} in the Lebesgue decomposition can be further split into Fd+FcsF_{\mathrm{d}}+F_{\mathrm{cs}}, where FcsF_{\mathrm{cs}} is continuous and singular and FdF_{\mathrm{d}} increases only in jumps in the sense that the corresponding measure is discrete. The complete decomposition is then F=Fac+Fcs+FdF=F_{\mathrm{ac}}+F_{\mathrm{cs}}+F_{\mathrm{d}}.

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Problem 31.21
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(a)

Suppose that x1<x2<⋯x_{1}<x_{2}<\cdots and ∑n∣F(xn)∣=∞\sum_{n}\left|F\left(x_{n}\right)\right|=\infty. Show that, if FF assumes the value 0 in each interval (xn,xn+1)(x_{n}, x_{n+1}), then it is of unbounded variation.

(b)

Define FF over [0,1][0,1] by F(0)=0F(0)=0 and F(x)=xαsin⁡x−1F(x)=x^{\alpha } \sin x^{-1} for x>0x>0. For which values of α\alpha is FF of bounded variation?

Problem 31.22

14.4↑14.4 \uparrow If ff is nonnegative and Lebesgue integrable, then by Theorem 31.3 and (31.8), except for xx in a set of Lebesgue measure 0,

1v−u∫uvf(t)dt→f(x)(31.35) \frac{1}{v-u} \int _{u}^{v} f(t) d t \rightarrow f(x) \tag {31.35}

if u≤x≤v,u<vu \leq x \leq v, u<v, and u,v→xu, v \rightarrow x. There is an analogue in which Lebesgue measure is replaced by a general probability measure μ\mu : If ff is nonnegative and integrable with respect to μ\mu, then as h↓0h \downarrow 0,

1μ(x−h,x+h]∫(x−h,x+h]f(t)μ(dt)→f(x)(31.36) \frac{1}{\mu (x-h, x+h]} \int _{(x-h, x+h]} f(t) \mu (d t) \rightarrow f(x) \tag {31.36}

on a set of μ\mu-measure 1. Let FF be the distribution function corresponding to μ\mu, and put φ(u)=inf⁡[x:u≤F(x)]\varphi (u)=\inf [x: u \leq F(x)] for 0<u<10<u<1 (see (14.5)). Deduce (31.36) from (31.35) by change of variable and Problem 14.4.

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