UMD Analysis Qualifying Exam/Aug06 Real

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

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

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

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-*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-*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.

Solution 3
Let $$A=||(x^p+\frac{1}{x^p})f||_{L^2(0,\infty)}$$ then we can write

$$ \begin{align} &= ||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
 * 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| \\

\end{align} $$

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

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.