UMD Analysis Qualifying Exam/Aug06 Real

Problem 1a

Prove the following version of the Riemann-Lebesque Lemma: Let $f \in L^2[-\pi,\pi] \!\,$. Prove in detail that $\frac{1}{2\pi}\int_{-\pi}^{\pi} f(x) e^{-inx}dx \rightarrow 0 \!\,$ as $n \rightarrow \infty\!\,$ Here $n \!\,$ denotes a positive integer. You may use any of a variety of techniques, but you cannot simply cite another version of the Riemann-Lebesque Lemma.

Solution 1a

Note that $e^{-inx}=\cos(nx) - i \sin (nx) \!\,$.

Hence we can equivalently show

$\int_{-\pi}^{\pi}f(x) \cos(nx) \rightarrow 0 \!\,$ as $n \rightarrow \infty \!\,$

Claim

Let $\psi(x) \!\,$ be a step function.

$\int_{-\pi}^{\pi}\psi(x) \cos(nx) \rightarrow 0\!\,$ as $n \rightarrow \infty \!\,$

Proof

$\begin{align} \int_{-\pi}^{\pi} \psi(x) \cos(nx) &= \sum_{i=1}^m c_i \int_{\xi_{i-1}}^{\xi_i}\cos(nx) dx \\ &= \sum_{i=1}^m c_i \frac{1}{n} \underbrace{( \left. \sin(nx) \right|_{\xi_{i-1}}^{\xi_i} ) }_{<2} \\ &\leq 2 \max_i c_i \frac{1}{n} \\ &\rightarrow 0 \mbox{ as } n \rightarrow \infty \end{align} \!\,$

Step functions approximate L^1 functions well

Since $f \in L^2[-\pi, \pi] \!\,$, then $f \in L^1 [\pi,\pi] \!\,$

Hence, given $\epsilon >0 \!\,$, there exists $\psi(x) \!\,$ such that

$\int_{-pi}^{\pi} |f(x)-\psi(x)| dx < \epsilon\!\,$

$\begin{align} \int_{-pi}^\pi f(x) \cos(nx) &= \int_{-pi}^{\pi} ([f(x)-\psi(x)+\psi(x)]\cos(nx))dx \\ &\leq \int_{-\pi}^{\pi} |f(x)-\psi(x)| \cos(nx) + \int_{-pi}^\pi |\psi(x)| \cos(nx) \\ &\leq \epsilon \cdot 2 \pi + \epsilon \end{align}$

Problem 1a

Let $n_k \!\,$ be an increasing sequence of positive integers. Show that $\{ x | \lim \inf_{k \rightarrow \infty} \sin(n_k x) > 0 \} \!\,$ has measure 0. Notes: You may take it as granted that the above set is measurable.