1
$\begingroup$

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

enter image description here

$a$) 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!

$\endgroup$
  • $\begingroup$ this is really 8 questions not 1 $\endgroup$ – Mark Joshi May 12 '16 at 0:08
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ a) but Z is not a complete metric space, nor is every other integer a bounded sequence. $\endgroup$ – fleablood May 12 '16 at 0:17
  • $\begingroup$ 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. $\endgroup$ – Mark Joshi May 12 '16 at 0:20
  • $\begingroup$ 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. $\endgroup$ – fleablood May 12 '16 at 0:37
  • $\begingroup$ constant sequences are Cauchy so not vacuous $\endgroup$ – Mark Joshi May 12 '16 at 3:19
  • $\begingroup$ Argh. Good point.... But I'd say constant functions are trivially cauchy. $\endgroup$ – fleablood May 12 '16 at 16:14
0
$\begingroup$
  • $(a)$ 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
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.