Let be $\Omega=]-1,1[\times ]-1,1[$. How I will be able to show Whats is the smaller space where $\delta_0$ (delta Dirac) belong?. $\delta_0$ is defined than $\left<{\delta_0,\phi}\right> = \phi(0)$ $\forall\phi\in C^{\infty}_0(\Omega)$
Smaller space $\delta_0$
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0If you don't provide context, it is impossible to know what "smaller" could mean. – 2012-10-22
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0@Martin: 'the smaller' intends to mean 'the smallest'. But some context would be good, yes. – 2012-10-22
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0@Berci: indeed. Still, we don't know whether it's the smallest vector space, Banach space, Hilbert space. And even if we fix one of those categories, it might not be obvious what "smaller" means. My answer would be that the smallest space that contains $\delta_0$ is $\mathbb C\delta_0$, but I doubt this is what the asker wants. – 2012-10-23
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0It looks to me like you are trying to define a distribution (or even a measure) $\delta_0$ on $\Omega$, and asking what the support of the distribution is. I have one concern about your definition, though. What does "$\phi(0)$" mean, when there is no point $0$ in $\Omega$? – 2012-10-23
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0My teacher wrote this excercise, ... when the vector space is Banach for example – 2012-10-23
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0I asked about space and yes is Banach Space – 2012-10-23
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0$\phi(0)$ is a $\phi$ evaluated in the point zero – 2012-10-23
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0@Juan: Notice that $\Omega$ does not contain the point $0$. If $\phi\in C_0^\infty(\Omega)$, then it follows that $0$ does not lie in the support of $\phi$. Unless I am misinterpreting your question, the only sensible way to define $\phi$ at $0$ is by setting $\phi(0) = 0$. But then $\langle \delta_0,\phi\rangle = 0$ for each $\phi$, i.e., $\delta_0 = 0$. – 2012-10-23
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0@froggie understand mmm, I thinking ... – 2012-10-24
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0@froggie then Whats is a smaller space that contains a zero function ? – 2012-10-24
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0I edit a question my teacher tell that wrote with mistake – 2012-10-24
1 Answers
It is still unclear what you mean by "the smaller" space $\delta_0$ belongs to. I think there are two possible good answers to this question.
Answer 1: Right now you have defined $\delta_0$ to be a linear map $C^\infty_0(\Omega)\to \mathbb{R}$ (or $\mathbb{C}$, depending on whether you are working over the real or complex numbers; I'll work over $\mathbb{R}$). That is, $\delta_0$ is an element of the dual space $C^\infty_0(\Omega)^*$. However, notice that the definition of $\langle\delta_0, \phi\rangle$ does not depend on any derivatives of $\phi$, it only cares what the value of $\phi$ is at $0$. In fact, $\delta_0$ defines a continuous linear map $C_0(\Omega)\to \mathbb{R}$. Thus you can think of $\delta_0$ as an element of the dual space $C_0(\Omega)^*$, which is smaller than $C_0^\infty(\Omega)^*$.
Answer 2: Another thing you may be asking is, what is the support of $\delta_0$? You should think of the support of $\delta_0$ as the smallest closed subset of $\Omega$ on which $\delta_0$ takes nonzero values. More precisely, the complement of the support of $\delta_0$ is the largest open set $U\subseteq\Omega$ such that $\langle \delta_0,\phi\rangle = 0$ for all $\phi\in C_0^\infty(U)$. In this problem, notice $\langle \delta_0, \phi\rangle = 0$ for every $\phi\in C_0^\infty(\Omega\smallsetminus\{0\})$, so the support of $\delta_0$ is the point $\{0\}$.
My guess is answer 1 is what you are looking for, but hopefully one of these helped.
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0Are you assuming that $\delta_0$ is a Dirac delta function? – 2012-10-25
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0@Juan: I'm not *assuming* $\delta_0$ is a Dirac delta function, that's how you've *defined* it. – 2012-10-25
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0tha function is a Dirac Detla, $\delta_0$, defined than $\left<{\delta_0,\phi}\right> = \phi(0)$ $\forall\phi\in C^{\infty}_0(\Omega)$ – 2012-10-25