Originally Posted by
Rapha 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 $
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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.