# 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 1b

 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.

## Problem 3

 Suppose $(x^p+\frac{1}{x^p})f\in L^2(0,\infty)$, where $p>1/2$. Show that $f\in L^1(0,\infty)$.

### Solution 3

Let $A=||(x^p+\frac{1}{x^p})f||_{L^2(0,\infty)}$ then we can write

\begin{align} ||f||_{L^1(0,\infty)}=\int_0^\infty |f| &= \int_0^1 |f \frac{1}{x^p}| |x^p| +\int_1^\infty |\frac{1}{x^p}| |f x^p| \\ &= ||f \frac{1}{x^p}||_{L^2(0,1)} ||x^p||_{L^2(0,1)} +||\frac{1}{x^p}||_{L^2(1,\infty)} ||f x^p||_{L^2(1,\infty)} \\ &\leq A ||x^p||_{L^2(0,1)} +A ||f x^p||_{L^2(1,\infty)} < \infty \end{align}

Hence $f\in L^1(0,\infty)$.

## Problem 5

 Let $f\in L^1(\mathbb{R})$, $F(x)=\int_\mathbb{R} f(t) \frac{\sin xt}{t}\, dt.$ (a) Show that $F$ is differentiable a.e. and find $F'(x)$. (b) Is $F$ absolutely continuous on closed bounded intervals $[a,b]$?

### Solution 5

Look at the difference quotient:

$F'(x)=\lim_{h\to 0} \frac{1}{h} (F(x+h)-F(x)) = \lim_{h\to 0} \int f(t) \frac{[\sin ((x+h)t) - \sin(xt)]}{ht}\, dt$

We can justify bringing the limit inside the integral. This is because for every $x$, $|\sin(xt)/t| <1$. Hence, our integrand is bounded by $2f(t)$ and hence is $L^1$ for all $n$. Then by Lebesgue Dominated Convergence, we can take the pointwise limit of the integrand. to get

$F'(x)= \int f(t)\cos(xt) \, dt.$

It is easy to show that $F'(x)$ is bounded (specifically by $||f||_L^1$) which implies that $F$ is Lipschitz continuous which implies that it is absolutely continuous.