# True/False Real Analysis Qualifying Exam Questions

The following is from a practice real analysis qualifying exam, and I had a couple questions about some of them.

$a$) I think this one is true, but I'm not sure how to prove it.

$b$) I know that $\partial^{\alpha}\hat{f} = \hat{[(-2\pi ix)^\alpha f]} (\xi)$. I don't necessarily see any reason that setting this equal to $0$ implies that $f=0$, so that would lead me to believe it might be false, but I'm unable to construct a counterexample.

$c$) We haven't done this yet, so skip.

$d$) True -- HW problem

$e$) Should be false if $d$ is true, but can't construct counterexample

$f$) I believe it's false and that $X=[0,1]$ with $f = \frac{1}{\sqrt x}\chi_{[0,1]}$ works.

$g$) I believe it's true using a comparison test argument

$h$) I believe it's false using $X = [0,1], M = \mathbb{B}_{X}, \mu_1 =$ Lebesgue measure, \mu_2 = $counting measure So I primarily need help with a,b,e and just want to verify that my answers for d,f,g,h are correct. Thank you! • this is really 8 questions not 1 – Mark Joshi May 12 '16 at 0:08 ## 2 Answers •$(a)$is false, for a counterexample I'd suggest to look at an infinite dimensional space, say$\ell^2(\mathbb{N})$. (Recall that the unit ball is not compact) •$(b)$is true: the Fourier Transform maps Schwartz functions into Schwartz functions. The only constant Schwartz function is$0$. This also implies that$f = 0$by Plancherel's equality, for example. •$(e)$Let's disprove the statement: assume the space is only complete wrt one of the two norms. If it is complete wrt$\|\cdot\|_2$then it is complete wrt$\|\cdot\|_1$using the inequality given, a contradiction. Then it must be complete wrt$\|\cdot\|_1$. Assume that it is possible to prove the other inequality, then it would be complete wrt$\|\cdot\|_2$arguing as before. •$(f),(g)$: you are correct. •$(h)$: You can also take a look at this a) false, take Z and make every point one away from every other point b) true, f hat would have to have support at zero so must be zero so f is zero c) this is false, delta distribution • a) but Z is not a complete metric space, nor is every other integer a bounded sequence. – fleablood May 12 '16 at 0:17 • if you make$d(x,y)=1$if$ x \neq y$then it is complete and bounded. More generally any metric space can be made bounded by replacing$d$with$d/(1+d)\$ so the conjecture is definitely false. – Mark Joshi May 12 '16 at 0:20
• Good point. But you should point out Z is now vacuuously complete in that no sequence is cauchy.so all cauchy sequence converge. I will confess I hadn't considered that. – fleablood May 12 '16 at 0:37
• constant sequences are Cauchy so not vacuous – Mark Joshi May 12 '16 at 3:19
• Argh. Good point.... But I'd say constant functions are trivially cauchy. – fleablood May 12 '16 at 16:14