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

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$a$) 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!

Brenton
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2 Answers2

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  • $(a)$ 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
Community
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Giovanni
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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

Mark Joshi
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  • a) but Z is not a complete metric space, nor is every other integer a bounded sequence. – fleablood May 12 '16 at 00:17
  • 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 00:20
  • 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 00:37
  • constant sequences are Cauchy so not vacuous – Mark Joshi May 12 '16 at 03:19
  • Argh. Good point.... But I'd say constant functions are trivially cauchy. – fleablood May 12 '16 at 16:14