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 $\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)e^{-inx}dx\to 0$ as $n\to \mathrm{\infty }$ 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

$\underset{-\pi }{\overset{\pi }{\int }}f(x)\cos (nx)\to 0$ as $n\to \mathrm{\infty }$

Claim

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

$\underset{-\pi }{\overset{\pi }{\int }}\psi (x)\cos (nx)\to 0$ as $n\to \mathrm{\infty }$

Proof

$\begin{array}{cc}\underset{-\pi }{\overset{\pi }{\int }}\psi (x)\cos (nx)& =\sum _{i=1}^m{c}_i\underset{{\xi }_{i-1}}{\overset{{\xi }_i}{\int }}\cos (nx)dx\\ & =\sum _{i=1}^m{c}_i\frac{1}{n}\underset{<2}{\underbrace{({\sin (nx)|}_{{\xi }_{i-1}}^{{\xi }_i})}}\\ & \le 2\underset{i}{\max }{c}_i\frac{1}{n}\\ & \to 0\text{ as }n\to \mathrm{\infty }\end{array}$

Step functions approximate L^1 functions well

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

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

$\underset{-\pi }{\overset{\pi }{\int }}|f(x)-\psi (x)|dx<ϵ$

$\begin{array}{cc}\underset{-\pi }{\overset{\pi }{\int }}f(x)\cos (nx)& =\underset{-\pi }{\overset{\pi }{\int }}([f(x)-\psi (x)+\psi (x)]\cos (nx))dx\\ & \le \underset{-\pi }{\overset{\pi }{\int }}|f(x)-\psi (x)|\cos (nx)+\underset{-\pi }{\overset{\pi }{\int }}|\psi (x)|\cos (nx)\\ & \le ϵ\cdot 2\pi +ϵ\end{array}$

Problem 1b

Let $n_k$ be an increasing sequence of positive integers. Show that $\{x|\lim \underset{k\to \mathrm{\infty }}{\inf }\sin (n_{k}x)>0\}$ 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 $(x^p+\frac{1}{x^p})f\in L^2(0,\mathrm{\infty })$, where $p>1/2$. Show that $f\in L^1(0,\mathrm{\infty })$.

Solution 3

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

$\begin{array}{cc}||f|{|}_{L^1(0,\mathrm{\infty })}=\underset{0}{\overset{\mathrm{\infty }}{\int }}|f|& =\underset{0}{\overset{1}{\int }}|f\frac{1}{x^p}||x^p|+\underset{1}{\overset{\mathrm{\infty }}{\int }}|\frac{1}{x^p}||f{x}^p|\\ & =||f\frac{1}{x^p}|{|}_{L^2(0,1)}||x^p|{|}_{L^2(0,1)}+||\frac{1}{x^p}|{|}_{L^2(1,\mathrm{\infty })}||f{x}^p|{|}_{L^2(1,\mathrm{\infty })}\\ & \le A||x^p|{|}_{L^2(0,1)}+A||f{x}^p|{|}_{L^2(1,\mathrm{\infty })}<\mathrm{\infty }\end{array}$

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

Problem 5

Let $f\in L^1(ℝ)$, $F(x)=\underset{ℝ}{\int }f(t)\frac{\sin xt}{t}dt.$ (a) Show that $F$ is differentiable a.e. and find $F^{\prime }(x)$. (b) Is $F$ absolutely continuous on closed bounded intervals $[a,b]$?

Solution 5

Look at the difference quotient:

$F^{\prime }(x)=\underset{h\to 0}{\lim }\frac{1}{h}(F(x+h)-F(x))=\underset{h\to 0}{\lim }\int f(t)\frac{[\sin ((x+h)t)-\sin (xt)]}{ht}dt$

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

$F^{\prime }(x)=\int f(t)\cos (xt)dt.$

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