Real Analysis Qualifying Exam Practice Questions

I’m doing some practice questions for a real analysis qualifying exam coming up in a few weeks. I have a couple questions, namely on the “if the statement is true, prove it. Otherwise, give a counterexample” questions.

a) In an infinite-dimensional Hilbert space $H$, for any weakly convergent sequence $\left\lbrace x_n \right\rbrace$, there exists a subsequence that is convergent with respect to the norm

b) Since two iterated integrals exist and $$\int_{(0,1)}\int_{(0,1)} \frac{x^2-y^2}{(x^2+y^2)^2} dm(x)dm(y) = \int_{(0,1)}\int_{(0,1)} \frac{x^2-y^2}{(x^2+y^2)^2} dm(y)dm(x)$$ we can conclude, via the Tonelli-Fubini theorem, that the double integral exists.

c) There exists a function $f \geq 0$ on $(0, \infty))$ such that $f \in L^p((0,\infty)$ if and only if $p=1$

I have no idea how to solve part a.

For b I think the answer is true since I can switch the order of integration if the inside integral is finite, and since the integral exists by assumption, that solves b.

For c, I think the answer should be true and the function should be some modified version of $\frac{1}{x}$. The space on the paper for this answer is rather short, so there should be simple counterexample of this form I would think, but I can’t construct it properly.

Any help would be tremendously appreciated!-

Solutions Collecting From Web of "Real Analysis Qualifying Exam Practice Questions"

a. $\{\sin(n\cdot)\}$ converges weakly to $0$ in $L^2([0,2\pi])$ but $$\int_0^{2\pi}\sin^2(nx)\,dx = \pi – \frac{\sin(4\pi n)}{4n} \ge 1.$$ To show that this provides a counterexample to the statement recall that if a subsequence converges strongly then it converges strongly to the weak limit, which in this case is $0$. Hence we would have $$1 \le \lim_{k\to \infty}\|\sin(n_k\cdot)\|_{L^2} = 0.$$

b. The iterated integrals are not equal.

c. Let
f =
\frac{1}{(\log x)^2 x} & \text{if}\ x \in (0,0.5) \\
\frac{1}{x^2} &\ \text{otherwise.}

Then $f$ is positive and integrable, but $f \notin L^p$ for any $p > 1$ since the singularity at $0$ is not integrable. Indeed let $p = 1 + 2\epsilon$, with $\epsilon > 0$. Then, for $x$ small, $$|f(x)|^p = \frac{1}{x^{1 + \epsilon}}\frac{1}{|\log x|^{2p}x^{\epsilon}} \ge \frac{1}{x^{1 + \epsilon}} \notin L^1.$$
The last inequality follows from the fact that $$\lim_{x \to 0} (\log x)^{2p}x^{\epsilon} = 0,$$ so that we can find $\delta < \frac 12$ such that if $x < \delta$ then $$\frac{1}{|\log x|^{2p}x^{\epsilon}} \ge 1.$$