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

${\begin{aligned}\int _{-\pi }^{\pi }\psi (x)\cos(nx)&=\sum _{i=1}^{m}c_{i}\int _{\xi _{i-1}}^{\xi _{i}}\cos(nx)dx\\&=\sum _{i=1}^{m}c_{i}{\frac {1}{n}}\underbrace {(\left.\sin(nx)\right|_{\xi _{i-1}}^{\xi _{i}})} _{<2}\\&\leq 2\max _{i}c_{i}{\frac {1}{n}}\\&\rightarrow 0{\mbox{ as }}n\rightarrow \infty \end{aligned}}\!\,$

Step functions approximate L^1 functions well

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

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

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

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

Problem 1b

Let $n_{k}\!\,$ 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.

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(n_kx)] is bounded below by some positive constant; ie, the integral of liminf[sin(n_kx)] 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 f_k(x)=2^-|x|*sin(n_kx), each of which is clearly integrable. By Fatou's lemma, the integral of liminf[f_k(x)] over S <= the liminf of the integrals of f_k(x) over S. However, each f_k is the L¹ function 2^-|x|*sin(n_kx). By the Riemann-Lebesgue Lemma, this goes to 0 as n_k 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 $(x^{p}+{\frac {1}{x^{p}}})f\in L^{2}(0,\infty )$, 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

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

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

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.