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Solutions for Homework 1 & New Homework (MATH 1540 Advanced Calculus-undergraduate)

Homework 1 was a difficult one. Many of the problems were given at Qualifying Examinations from universities around the US. Others appeared as part of the Putnam Competition.

The problem that sparked the most attention was the following:

Problem 7. If f, g are continuous functions on \mathbb{R} of period 1, then

\displaystyle\lim_{n\to\infty}\int_0^1f(x)g(nx)dx=\int_0^1f(x)dx\int_0^1g(x)dx.

Although elementary, this problem has deep connections with Fourier analysis and it is often also regarded as the Riemann-Lebesgue lemma. The classical Riemann-Lebesgue lemma states that for a Riemann integrable function f:[a, b]\to\mathbb{R},

\displaystyle\lim_{n\to\infty}\int_a^bf(x)\sin nxdx=\lim_{n\to\infty}\int_a^bf(x)\cos nxdx=0.

This implies one of the most important facts in classical Fourier analysis, if f:\mathbb{R}\to\mathbb{R} is a periodic function of period 2\pi, then under suitable conditions,

\displaystyle S_{n}(x)=a_{0}+\sum_{k=1}^n(a_{k}\cos kx+b_{k}\sin kx)

converges to f(x) as n\to\infty, where a_{n}, b_{n} are the Fourier coefficients,

\displaystyle a_{0}=\frac{1}{2\pi}\int_0^{2\pi}f(x)dx, \displaystyle a_{n}=\frac{1}{\pi}\int_0^{2\pi}f(x)\cos nxdx, and \displaystyle b_{n}=\frac{1}{\pi}\int_0^{2\pi}f(x)\sin nxdx.

The following result generalizes the classical Riemann-Lebesgue lemma:

Theorem 1. Let f:[a, b]\to\mathbb{R} be a continuous function, 0\leq a<b. Suppose that g:[0, \infty)\to\mathbb{R} is a periodic function of period T>0. Then we have

\displaystyle\lim_{n\to\infty}\int_a^bf(x)g(nx)dx=\frac{1}{T}\int_0^Tg(x)dx\int_a^bf(x)dx.

However, in 1962, Luxemburg proved the following more general result:

Theorem 2. ([1]) Let \phi be a bounded real or complex valued measurable function defined on \mathbb{R}, and assume that \phi is periodic with period p>0. If I is an arbitrary interval and f a real or complex-valued measurable function defined on I and integrable over I, then

\displaystyle\lim_{|\lambda|\to\infty}\int_{I}f(x)\phi(\lambda x)dx=\left(\frac{1}{p}\int_0^p\phi(x)dx\right)\left(\int_{I}f(x)dx\right).

The reference is given below

[1] W. A. J. Luxemburg, A property of the Fourier coefficients of an integrable function, The American Mathematical Monthly, 69 (1962), 94-98.

The solution of the above theorem as well as many other homework problems can be found here. Also, homework 3 is here.

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