515 – Homework 2

This set is due Feb. 29 at the beginning of lecture. Let me know if more time is needed or anything like that. Problem 4 was incorrect as stated; I have fixed it now. Thanks to Tara Sheehan for bringing the problem to my attention.

1. We have used repeatedly that if f:[a,b]\to{\mathbb R} is Riemann integrable, then it is bounded. We briefly discussed this in class, but I did not give a proof. Please prove it.

2. Recall that Cantor’s middle third set is defined as C=\bigcap_n I_n where I_0=[0,1] and, for each n, I_{n+1} is the result of removing from each of the closed intervals making up I_n their (open) middle third, so

I_1=[0,1/3]\cup[2/3,1],

I_2=[0,1/9]\cup[2/9,1/3]\cup[2/3,7/9]\cup[8/9,1],

etc. Show that C is a Jordan measurable set of measure 0.

3. Solve exercise 1.1.12 from the book. To be specific, if A is Jordan null and B\subseteq A, you need to prove that B is also measurable, and its measure is 0. (In particular, every subset of the Cantor set C is Jordan measurable.)

4. Modify the construction of the Cantor set as follows: Let r be a constant, 0<r<1. Define C_r as \bigcap_n I_{n,r}, where I_{0,r}=[0,1] and, for each n, I_{n+1,r} is the result of removing from each of the closed intervals making up I_{n,r} their (open) middle rth portion, so each of these intervals {}[a,b] is replaced with {}[a,c]\cup[d,b] where \displaystyle c=\frac{a+b}2-\frac{(b-a)r}2 and \displaystyle d=\frac{a+b}2+\frac{(b-a)r}2. (In particular, C=C_{1/3}.)

Also, for 0<r<1, define C_r^\prime so that at stage n we remove from each interval its middle portion of length r^n (and if the interval has length less than r^n, then we remove it).

Check that all C_r are Jordan measurable of measure 0, and find (with proof) all the values of r for which C_r^\prime is Jordan measurable. (If proceeding in this generality is not possible, at least prove that C_{1/5}^\prime is not Jordan measurable.)

On the other hand, all the sets C_r^\prime are Lebesgue measurable, as we will see in lecture.

5. Suppose that f_n:[a,b]\to{\mathbb R} are continuous functions, and that f_n\to f uniformly. Suppose that g:[a,b]\to{\mathbb R} is continuous. Prove that

\displaystyle \int_a^b g(x)f_n(x)dx\to\int_a^b g(x)f(x) dx.

6. Suppose that f:[0,1]\to{\mathbb R} is continuous, and that \displaystyle \int_0^1 f(x)x^n dx=0 for all n=0,1,2,\dots Prove that f is the constant zero function. This is a theorem of Hausdorff.

What if we are only given that the integrals vanish for integers n>0?

The quantities \displaystyle \int_0^1 f(x) x^n dx are called moments. They are particularly useful in probability theory and in physics. (For example, in classical mechanics, if f(x)\ge0 for all x\in[0,1], one thinks of f as a mass density, and of the zero-th moment M=\displaystyle \int_0^1 f(x)dx as its associated mass; the quantity \displaystyle \frac1M\int_0^1 f(x)x dx is the center of gravity, and \int_0^1 f(x) x^2 dx is the moment of inertia about 0.)

7. Hausdorff’s moment problem asks whether the moments of a continuous f uniquely determine f. If f is defined on a bounded interval, the answer is yes, by the result from the previous problem. But this is no longer the case if we consider functions f defined on unbounded intervals (where, of course, the integrals are now interpreted as improper integrals). For a specific example, define f:[0,\infty)\to{\mathbb R} by

\displaystyle f(x)=e^{-x^{1/4}}\sin(x^{1/4}).

Show that \displaystyle \int_0^\infty f(x)x^n dx=0 for all n=0,1,\dots

Hint: Let \zeta=e^{i\pi/4}. Show that

\displaystyle \int_0^\infty t^{n+1} e^{-\zeta t}dt =\frac {n+1}{\zeta}\int_0^\infty t^n e^{-\zeta t}dt.

Conclude that \displaystyle \int_0^\infty t^{4n+3}e^{-\zeta t}dt is real. Consider its imaginary part and, by an appropriate change of variables, deduce that all the moments

\displaystyle \int_0^\infty e^{-x^{1/4}}\sin(x^{1/4}) x^n dx

are 0.

8. A trigonometric series is a series \displaystyle \sum_{n\in{\mathbb Z}} c_n e^{inx}, where the series is understood as

\displaystyle \lim_{N\to\infty}\sum_{n=-N}^N c_n e^{inx}.

Such a series is a Fourier series iff there is a function f periodic of period 2\pi and integrable on the interval {}[-\pi,\pi] such that \displaystyle c_n=\hat f(n)=\frac1{2\pi}\int_{-\pi}^\pi f(x)e^{-inx}dx for all integers n. Not all trigonometric series are Fourier series; for example, \displaystyle \sum_{n=2}^\infty\frac{\sin nx}{\ln n} is not a Fourier series. (You do not need to prove this.)

Suppose that f is continuous and periodic of period 2\pi. Show that \hat f(n)\to0 as |n|\to\infty. This is a particular case of the Riemann-Lebesgue theorem.

(Cantor used this result in 1870 to prove that if \displaystyle \sum_{n\in{\mathbb Z}} c_n e^{inx}=0 for all x, then c_n=0 for all n. Set theory was born of his research in generalizations of this result.)

Extra Credit Problem.

Suppose that S(x)=\sum_n c_n e^{inx} is a trigonometric series, and that the c_n are bounded. The result of formally integrating S twice is the Riemann function F_S of S, defined by

\displaystyle F_S(x)=\frac{c_0x^2}2-{\sum_{n\in{\mathbb Z}}}'\frac1{n^2}c_ne^{inx},

where the \sum' means that n=0 is excluded.  Show that F_S is continuous (though in general, it is no longer periodic). One would expect that F_S'' exists and equals S, but this is in general not true. However, one can prove a variant of this:

Define the second symmetric derivative of a function F by

\displaystyle D^2 F(x)=\lim_{h\to 0}\frac{\Delta^2 F(x,h)}{h^2},

where \Delta^2F(x,h)=F(x+h)-2F(x)+F(x-h).

Prove that if F''(x) exists, then so does D^2 F(x) and F''(x)=D^2 F(x). Give an example where D^2 F(x) exists but F''(x) does not.

Prove Riemann’s first lemma: If S(x) is a trigonometric series with bounded coefficients (that converges at the point x_0) then D^2 F_S(x_0) exists and equals S(x_0).

For much more on this topic, see for example

  • Alexander Kechris, Set theory and uniqueness for trigonometric series, 1997,

available at Kechris’s webpage.

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