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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.2012-11-09
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    sorry now you understand?2012-11-09
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    What did you try?2012-11-09
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    Is $<>$ the Bra-ket notation or is it the inner product in Hilbert space? If it is the latter, use the fact that the Heaviside step function is the antiderivative of the Dirac delta.2012-11-09
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    Isn't the inner product _bilinear_?2012-11-09
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    The notation in this question is very confusing. As far as I can tell, $<\varphi,\delta>$ means $\int_{ - \infty }^\infty {\varphi (x)\delta (x)dx}$.2012-11-09
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    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.2012-11-09
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    Why the confusion? in other thread I wrote a similar question without problems "http://math.stackexchange.com/questions/208906/delta-dirac-function"2012-11-09
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    Your notation could mean many different things.2012-11-09
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    @glebovg There is a confusion indeed, what is being defined is, as Ross pointed out, the _delta functional_ in the following way $$\delta_0(\phi) := \langle \delta(x),\phi(x)\rangle$$2012-11-09
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    @Pragabhava, I think it's quite clear that $\delta_0 (\phi) := \phi(0)$? i.e. Delta function is considered as a distribution here.2012-11-09
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    It says Dirac delta in the title.2012-11-09
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    @Sanchez Exactly, and the inner product is $$\langle f,g \rangle = \int_\Omega f g d\Omega$$ where $\Omega \in \mathbb{R}^n$ I pressume?2012-11-09
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    @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.2012-11-09
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    @Sanchez It's the same thing. The way I see it: Let $$ \delta_0(\phi) = \int_\Omega \delta(X) \phi(X) dX \equiv \langle \delta, \phi \rangle.$$ Prove that $\delta_0$ is linear (_in_ $\phi$). It's the [Riesz representation theorem](http://en.wikipedia.org/wiki/Riesz_representation_theorem).2012-11-09
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    $\delta_0 = \delta(0)$, I am working in Sobolev spaces2012-11-09
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    @Juan What is the definition of $\phi_1+\phi_2$?2012-11-09

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