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Show that $\delta_0$, Dirac function, defined than $\left<{\delta_0,\phi}\right> = \phi(0)$ is linear.

I trying: Let be $\phi_1,\phi_2$ $\in W^{m,p}(\Omega)$ then $\delta_0(\phi_1+\phi_2)=(\phi_1+\phi_2)(0)$, but I need more steps.

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Please write your question in an intelligible human language. – Braindead Nov 9 '12 at 1:48
sorry now you understand? – juaninf Nov 9 '12 at 1:53
I think the notation means a function instead, i.e. $\delta_0 (\phi)$ is defined to be $\phi(0)$, where $\phi$ is some function he's considering. – user27126 Nov 9 '12 at 3:04
@Pragabhava, I think it's quite clear that $\delta_0 (\phi) := \phi(0)$? i.e. Delta function is considered as a distribution here. – user27126 Nov 9 '12 at 3:07
@Pragabhava, I don't understand what you are saying. Is there inner product involved at all? This is just a distribution written in a pairing form. – user27126 Nov 9 '12 at 3:11

Hint: What is the definition of a linear functional? Just plug this into the definition and see if it works. Under trying, $(\phi_1 + \phi_2)(0)=\phi_1 (0) + \phi_2 (0)$

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Downvoter: please explain. When I was in college long ago, I found these problems valuable to see if I understood the definitions. My approach was always to review the definition and see if I could satisfy it. I am trying to encourage OP to do that and explain where the problem is. – Ross Millikan Nov 9 '12 at 3:14
$\delta_0 = \delta(0)$, I am working in Sobolev spaces, see that $\phi_1$ and $\phi_2$ $\in W^{m,p}$. – juaninf Nov 9 '12 at 3:27
@Juan, now I feel confused. What is $\phi$? My understanding of Sobolev space is that they are really $L^p$ functions. So what do you mean by value at a point? – user27126 Nov 9 '12 at 3:30
$\phi$ is a functional that belong to $W^{m,p}$, Where I wrote value at a point? – juaninf Nov 9 '12 at 3:39
@Juan, I mean when you write $\phi(0)$, what does that mean? $L^p$ functions cannot be evaluated at a point? – user27126 Nov 9 '12 at 3:49

I am a little confused about your question, so please do not downvote my answer. If $<>$ represents the inner product, use the fact that the Heaviside step function is the antiderivative of the Dirac delta as follows: $$< \varphi ,\delta > = \int_{ - \infty }^\infty {\varphi (x)\delta (x)dx = \left[ {\varphi (x)H(x)} \right]_{ - \infty }^\infty } - \int_{ - \infty }^\infty {\varphi '(x)} H(x)dx$$ which simplifies to $$- \int_0^\infty {\varphi '(x)dx = \left[ { - \varphi (x)} \right]_0^\infty = \varphi (0)}.$$ Note that I used the definition of the Dirac delta.

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Why downvote? The question is confusing. At least I tried to help. – glebovg Nov 9 '12 at 3:05
I make upvote I don't understand the confusing – juaninf Nov 9 '12 at 3:06
@Juan: if you understand now, please upvote an answer. If you don't know which, this one-I have more rep than I need. If not, please comment on what the problem is and the clarification may get you the answer you are looking for. – Ross Millikan Nov 9 '12 at 3:37
I don't understand yet only I get upvote because the claim of @glebovg is fair. – juaninf Nov 9 '12 at 3:41
@RossMillikan Thanks Ross. – glebovg Nov 9 '12 at 4:29

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