# Thread: Show that... Using some functions

1. ## Show that... Using some functions

Hello everyone!

Thank you all,

this excercise is solved.

Thx

Rapha

2. 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.

3. 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!

4. Hello.

Alright,

thank you so much

Best regards
Rapha