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Schwartz function and norm

  1. Jan 12, 2016 #1
    • Member warned about posting without the homework template
    Hello : CONTEXT : let be E the space of rapidly decreasing functions on
    ##\mathbb{R}^{n}## in ##\mathbb{R}##.

    I define
    $$(||.||_{i})_{i \in \mathbb{R}^{n+1}}$$ with forall ##i = (k, m_{1},\ldots, m_{n}) = (k, m)## we define for ##f## in ##E## ##||f||_{i} = \sup_{x \in \mathbb{R}^{n}} \Big|(1 + ||x||^{k}) \frac{\partial^{m}f}{\partial x_{1}^{m_{1}}\ldots \partial x_{n}^{m_{n}}}(x)\Big|##.
    They are seminorm. Let wright this family of semi norm as ##(||.||_{n})_{n ×\in \mathbb{N}}##.
    My goal is to show that the topologie associate to the family of semi norm is not normable.

    What I do for the moment : I've define a distance on E with ##\forall (x, y) \in E^{2} d(x, y) = \sum_{n \in \mathbb{N}}2^{-n}min(1, ||x - y||_{n})##. It define the same topology(not rally complicate.). I also prevously show that for all ##\epsilon < 1##, ##\{B(0, \frac{\epsilon}{n +1} / n \in \mathbb{N}\}## is a base of neighbourhood of 0. We advice me to use that.
    Wich is denombrable. But I find nothing else on.


    Then I'll try to show an absurdity : by using Riesz theorem by showing E is finite dimension by showing the unity sphere is compacts. I try to use my distance.
    I find nothing.

    Could you help me pelase?

    Thank you in advance and have a nice afternoon:oldbiggrin:.
     
    Last edited by a moderator: Jan 12, 2016
  2. jcsd
  3. Jan 13, 2016 #2

    Samy_A

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    Hint: Arzelà-Ascoli.
     
  4. Jan 13, 2016 #3
    Yeah I made a guide proof with the Alexandrov compactify but it's a little bit long using Ascoli.
    And my curse said it's more simple if we remarc that $$
    \{B(0, \frac{\epsilon}{n +1}) / n \in \mathbb{N}\}$$ is a base of neighbourhood of 0.
    What do you think please?
    Is their any property liking a base and the compacity propertie please?
    It's to use Riesz property.

    Thank you in advance and have a nice afternon:oldbiggrin:.
     
  5. Jan 13, 2016 #4

    Samy_A

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    I frankly don't understand how the following sets are defined:
    $$\{B(0, \frac{\epsilon}{n +1}) / n \in \mathbb{N}\}$$
     
    Last edited: Jan 13, 2016
  6. Jan 14, 2016 #5
    I wroght $$B(0; \frac{\epsilon}{n+1}) = \{x \in E / ||x - 0|| < \frac{\epsilon}{n+1}\}$$/
     
  7. Jan 14, 2016 #6
    This set I show is a fondamental set of neigbourhood of 0. And I wanna deduce an absurdtity of that if E is normable.
     
  8. Jan 14, 2016 #7

    Samy_A

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    So we had to guess that those B's where balls in the norm whose existence you want to disprove. I still don't understand why you need the ##\epsilon## there, as ##\{B(0; \frac{1}{n+1}) ) | n \in \mathbb N \}## is a base too.

    Anyway, to prove that these balls are relatively compact, it suffices to show that any bounded sequence ##(f_n)_n \subset E## has a convergent subsequence (all in the E topology, which you can consider with the seminorms or the norm, as they are supposedly the same).
    That is relatively easy using Arzelà-Ascoli.

    I don't see a different approach out of the box.

    I cheated, googled it, and found a proof not using Arzelà-Ascoli. What it does is first claim that the assumed norm must satisfy ##||f|| \leq Cp_i(f)##, where ##C## is a constant and ##p_i## one of the seminorms. As all seminorms must be continous in the norm, it follows that all seminorms ##p_j## with ##j \geq i## are equivalent. That then leads to a contradiction.
    I leave it to you to fill in the details.
     
  9. Jan 14, 2016 #8
    I have the ennonce of the proof in a cruse but in french. I can send it.
     
  10. Jan 14, 2016 #9

    Samy_A

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    Please do, I'm from Belgium, my French is quite good. :)
     
  11. Jan 14, 2016 #10
  12. Jan 15, 2016 #11
    So what do you think of it please?
     
  13. Jan 15, 2016 #12

    Samy_A

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    I learned from your pdf that the French style is still very much inspired by Bourbaki. :oldsmile:

    I don't see an easy way to finish the proof (doesn't mean there is none, just that I don't see it).
    From the definition of the topology (on page 18) or from the definition of the metric, one can indeed deduce that ##||f|| \leq Cp_i(f)## for some ##i## as I wrote in a previous post.
    Also, for every seminorm ##p_j(f) \leq C_j||f||## for some constant ##C_j##.
    Now you could construct a function that converges to 0 in the seminorms up to ##p_i##, but not in some "higher" seminorm, thereby getting a contradiction.

    I assume your prof has something simpler in mind, but I don't see it.
     
    Last edited: Jan 15, 2016
  14. Jan 15, 2016 #13
    OK. Really strange.
     
  15. Jan 18, 2016 #14
    But I maid a mistake in my enouncee. I said something wrong.
    I have to show first that if my topologie come from a norm $$N$$, so $$\{\frac{1}{n+1}B_{d_{\phi}}(0, \epsilon) / n \in \mathbb{N}\}$$ is a base of neighbourhood of 0 and not the first base I wroght.
    Where $$d_{\phi}$$ is a distance define page 19. This distance made the same topologie than $$N$$.
     
  16. Jan 18, 2016 #15
    In fact I've got another problem : To show it let introduce $$\epsilon \in ]0; 1[$$. Let be $$U \in V(0)$$.
    So $$\exists \epsilon' > 0 / B_{N}(0, \epsilon') \subset U$$ so as $$\{N(x) / x \in B_{d_{\phi}}(0; \epsilon)\}$$ is borne* let's said by a certain M > 0. Let be n integer with $$\frac{1}{n+1} < \frac{\epsilon'}{M}$$

    If, $$y \in \frac{1}{n+1}B_{d_{\phi}}(0; \epsilon), \exists x \in B_{d_{\phi}}(0; \epsilon) / y = \frac{1}{n+1}x$$
    We've got $$N(y) = \frac{1}{n+1}N(x) \leq \epsilon'$$. So $$\frac{1}{n + 1}B_{d_{\phi}}(0; \epsilon) \subset U$$.
    But don't even use my $$\epsilon$$ so I think their's a mistake. It should be there*.

    What do you think please?

    Thank you in advance and have a nice afternoon:oldbiggrin:.
     
  17. Jan 18, 2016 #16

    Samy_A

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    It's difficult to parse your proof. I think you are making this too complicated.
    You can assume that the topologies defined by the (supposed) norm and by the metric are the same.
    So you know that for some ##\epsilon>0##, ##B_d(0, \epsilon) \subset B_N(0,1)##.
    But you also know that the "norm-balls" form a basis for the topology.
    All you have to show now is that for every "norm ball" ##B_N(0,\delta)## (where ##\delta>0##), there is an ##n \in \mathbb N## such that ##\frac{1}{n+1}B_d(0,\epsilon) \subset B_N(0,\delta)##.
     
  18. Jan 18, 2016 #17
    Ah quite smart I didn't see it sorry. Anytime.
    I purhaps have a proof with my neighbourhood.
     
  19. Jan 20, 2016 #18

    Samy_A

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    I'm actually also interested in the solution your professor has in mind.

    As said I can solve the exercise using Arzelà-Ascoli.

    I sketched another solution in posts 7 and 12.
    To make it simple, let's take n=1.
    Define the sequence of functions ##(f_m )_m## by ##\displaystyle f_m(x)=\frac{e^{-x²}\sin(mx)}{m^{k+1}}##, where k is sufficiently large that for some constant ##C##, ##||f|| \leq Cp_k(f)## for all ##f \in \mathcal S (\mathbb R)##. (##p_k## is the seminorm ##p_k(f)=\sup_{x \in \mathbb{R}}{(1+|x|^k)|f^{(k)}(x)|}##).
    It is elementary (but tedious) to prove that the sequence converges to 0 in ##p_k## when ##m \to \infty##, and thus in the norm, but that it doesn't converge in all the seminorms ##p_j## where ##j>k+1##. That's a contradiction.

    This doesn't use the second part of the exercise about the balls, though.

    When you know the intended solution, please post it. I'm curious.
     
    Last edited: Jan 21, 2016
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