# UMD Analysis Qualifying Exam/Aug06 Real

## Problem 1a

 Prove the following version of the Riemann-Lebesque Lemma: Let ${\displaystyle f\in L^{2}[-\pi ,\pi ]\!\,}$. Prove in detail that ${\displaystyle {\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x)e^{-inx}dx\rightarrow 0\!\,}$ as ${\displaystyle n\rightarrow \infty \!\,}$ Here ${\displaystyle n\!\,}$ 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 ${\displaystyle e^{-inx}=\cos(nx)-i\sin(nx)\!\,}$.

Hence we can equivalently show

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

#### Claim

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

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

#### Proof

{\displaystyle {\begin{aligned}\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{aligned}}\!\,}

#### Step functions approximate L^1 functions well

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

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

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

{\displaystyle {\begin{aligned}\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{aligned}}}

### Problem 1b

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

### Solution 1b

For the sake of brevity, let S be the set of such x. If S has positive measure, then it contains a subset of positive measure on which liminf[sin(nkx)] is bounded below by some positive constant; ie, the integral of liminf[sin(nkx)] over S will be positive. If S has zero measure, then the same integral will be zero. Thus, we must only compute an appropriate integral to show that m(S)=0.

Since we do not know that S has finite measure, take a sequence of functions fk(x)=2-|x|*sin(nkx), each of which is clearly integrable. By Fatou's lemma, the integral of liminf[fk(x)] over S <= the liminf of the integrals of fk(x) over S. However, each fk is the L1 function 2-|x|*sin(nkx). By the Riemann-Lebesgue Lemma, this goes to 0 as nk goes to infinity.

Hence our original integral of a strictly positive function over S is bounded above by 0, so m(S)=0.

## Problem 3

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

### Solution 3

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

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

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

## Problem 5

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

### Solution 5

Look at the difference quotient:

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

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

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

It is easy to show that ${\displaystyle F'(x)}$ is bounded (specifically by ${\displaystyle ||f||_{L}^{1}}$) which implies that ${\displaystyle F}$ is Lipschitz continuous which implies that it is absolutely continuous.