UMD Analysis Qualifying Exam/Aug06 Real

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Problem 1a[edit]

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[edit]

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 \!\,


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

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


\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

Step functions approximate L^1 functions well[edit]

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\!\,

\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

Problem 1b[edit]

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[edit]

Problem 3[edit]

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[edit]

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

||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


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

Problem 5[edit]

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[edit]

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.