Results 1 to 4 of 4

Math Help - Show that... Using some functions

  1. #1
    Senior Member
    Joined
    Nov 2008
    Posts
    461

    Show that... Using some functions

    Hello everyone!

    Thank you all,

    this excercise is solved.

    Thx


    Rapha
    Last edited by Rapha; May 20th 2009 at 11:04 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by Rapha View Post
    Hello everyone!

    1)
    Let l_2(N) = \{ \phi : N \to \mathbb{C} : \sum_{n \in \mathbb{Z}} |\phi(a_n)|^2 < \infty \} and N := \{a_n, n \in \mathbb{Z} \}


    g(0) = 1

    g = 0, iff  x \in ( ] - \infty, - \delta] \cup [\delta, \infty[ )

     g \in C_c^\infty(\mathbb{R} ) (actually this means g :  \mathbb{R} \to \mathbb{R} is continious, all derivatives of g exists and the support of g is compact)

     \delta:= \inf_n (a_{n+1}-a_n) > 0
    ___________________
    in this case it is
    2)
     g(0) = 1

     g = 0, iff  x \in ]-\infty, -a+2] \cup [ a+2, \infty[

     g \in C_c^\infty(\mathbb{R})

     \delta:= \inf_n (a_{n+1}-a_n) > 0

    3)

     f := \sum_{n \in \mathbb{Z}} \phi(a_n)g(x-a_n)

    EXCERCISE

    Show that f = \phi on N

    Does anyone have an idea for this problem, because I don't have any clue. So any help would be much appreciated.

    Thank you,
    Rapha
    There is a lot of notation, but the answer is almost obvious: you are asked to prove that f(a_n)=\phi(a_n) for every n\in\mathbb{Z}. However,

    f(a_n)=\sum_{p\in\mathbb{Z}}\phi(a_p)g(a_n-a_p)

    and g(a_n-x) is 0 as soon as |a_n-x|>\delta by definition. However, \delta is defined in such a way that it is smaller than the space between two values of the sequence (a_n)_n (which is increasing), so that \delta<|a_n-a_p| for any p\neq n, and finally g(a_n-a_p)=0 for any p\neq n. We conclude that the only non-zero term in the previous sum is \phi(a_n)g(0). Remember g(0)=1 and you are done.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    In case 1), there seems to be an implicit assumption here that (a_n) is an increasing sequence of real numbers. If so, then the condition implies that |a_m-a_n|\geqslant\delta whenever m\ne n. But you are given that g(t) = 0 whenever |t|\geqslant\delta, and therefore g(a_m-a_n)=0 whenever m\ne n. Hence f(a_m) := \textstyle\sum_{n\in\mathbb{Z}}\phi(a_n)g(a_m-a_n) = \phi(a_m)g(0) = \phi(a_m) (because the only nonzero term in the sum is the one where m=n). So f=\phi on N.

    The data for case 2) looks wrong. In fact, the two conditions g(0)=1 and g(x) = 0 if x\in (-\infty,-a+2] will contradict each other if a<2. And there is nothing to indicate what a is supposed to be.

    Edit. Beaten to it by Laurent!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Nov 2008
    Posts
    461
    Hello.

    Alright,

    thank you so much

    Best regards
    Rapha
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: April 21st 2011, 06:43 PM
  2. Replies: 4
    Last Post: April 19th 2011, 06:39 PM
  3. Replies: 4
    Last Post: February 21st 2011, 03:37 PM
  4. Replies: 1
    Last Post: February 8th 2011, 11:20 AM
  5. How to show these functions are continuous?
    Posted in the Calculus Forum
    Replies: 6
    Last Post: February 2nd 2011, 08:39 PM

Search Tags


/mathhelpforum @mathhelpforum