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Thread: Show that... Using some functions

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by Rapha View Post
    Hello everyone!

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


    g(0) = 1

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

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

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

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

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

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

    3)

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

    EXCERCISE

    Show that $\displaystyle 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 $\displaystyle f(a_n)=\phi(a_n)$ for every $\displaystyle n\in\mathbb{Z}$. However,

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

    and $\displaystyle g(a_n-x)$ is 0 as soon as $\displaystyle |a_n-x|>\delta$ by definition. However, $\displaystyle \delta$ is defined in such a way that it is smaller than the space between two values of the sequence $\displaystyle (a_n)_n$ (which is increasing), so that $\displaystyle \delta<|a_n-a_p|$ for any $\displaystyle p\neq n$, and finally $\displaystyle g(a_n-a_p)=0$ for any $\displaystyle p\neq n$. We conclude that the only non-zero term in the previous sum is $\displaystyle \phi(a_n)g(0)$. Remember $\displaystyle g(0)=1$ and you are done.
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  3. #3
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    In case 1), there seems to be an implicit assumption here that $\displaystyle (a_n)$ is an increasing sequence of real numbers. If so, then the condition implies that $\displaystyle |a_m-a_n|\geqslant\delta$ whenever $\displaystyle m\ne n$. But you are given that $\displaystyle g(t) = 0$ whenever $\displaystyle |t|\geqslant\delta$, and therefore $\displaystyle g(a_m-a_n)=0$ whenever $\displaystyle m\ne n$. Hence $\displaystyle f(a_m) := \textstyle\sum_{n\in\mathbb{Z}}\phi(a_n)g(a_m-a_n) = \phi(a_m)g(0) = \phi(a_m)$ (because the only nonzero term in the sum is the one where m=n). So $\displaystyle f=\phi$ on N.

    The data for case 2) looks wrong. In fact, the two conditions $\displaystyle g(0)=1$ and $\displaystyle g(x) = 0$ if $\displaystyle x\in (-\infty,-a+2]$ 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!
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  4. #4
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    Hello.

    Alright,

    thank you so much

    Best regards
    Rapha
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