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I just encountered this question in my real analysis class involving distribution theory it is question 25 chapter 9 from Folland's real analysis second edition, which reads as follows:

Suppose that P is a polynomial in n variables such that only zero of $ P(\zeta)$ in $ R^n $ is $ \zeta = 0 $ and let P(D) be the operator D substituted in the polynomial P with $ D^{\alpha} = (2 \pi i)^{-|\alpha|} \partial^{\alpha} $ for any multi-index $\alpha$

a. Every tempered distribution F which satisfies P(D)F=0 is a polynomial

b. Every bounded function f that satisfies P(D)f=0 is a constant.

For part a. there is a hint to use two auxiliaries:

  1. The Fourier transforms of linear combinations of $ \delta $ (Dirac's Delta) and its derivatives is precisely the polynomials.
  2. The following exercise which I previously asked here and got help on it: Specific problem on distribution theory.. Basically it speaks about a distribution whose support is a singleton.

I am stuck on both parts and cannot incorporate the hints given here or the fact that the distribution is Tempered. We just started Tempered distributions now and my teacher warned us this might be a difficult exercise. I really do need the help on it.

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Tempered distributions have Fourier transforms. If $P(D)\,F=0$, take the Fourier transform to get $P(\xi)\,\hat F(\xi)=0$ for all $\xi\in\mathbb{R}$. Since $P(\xi)\ne0$ if $\xi\ne0$ the support of $\hat F$ is $\{0\}$. Can you take it fromhere?

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  • $\begingroup$ Thanks got part a thanks to you but part b is still a mystery to me. Could you please help me with part b? $\endgroup$
    – Croc2Alpha
    Jul 31, 2015 at 15:31
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    $\begingroup$ @zbigniew2015 Part b follows from part a! A bounded function is a tempered distribution. A bounded polynomial is constant. $\endgroup$ Jul 31, 2015 at 15:33
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    $\begingroup$ A bounded function is a tempered distribution. How many bounded polynomials do you know? $\endgroup$ Jul 31, 2015 at 15:33
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    $\begingroup$ @DavidC.Ullrich What are the probabilities of writing exactly the same sentence? $\endgroup$ Jul 31, 2015 at 15:35
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    $\begingroup$ @JuliánAguirre Positive, evidently. $\endgroup$ Jul 31, 2015 at 15:36

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