-2
$\begingroup$

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.

  • 0
    Please write your question in an intelligible human language.2012-11-09
  • 0
    sorry now you understand?2012-11-09
  • 0
    What did you try?2012-11-09
  • 0
    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
  • 0
    Isn't the inner product _bilinear_?2012-11-09
  • 0
    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
  • 1
    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
  • 0
    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
  • 0
    Your notation could mean many different things.2012-11-09
  • 0
    @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
  • 1
    @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
  • 0
    It says Dirac delta in the title.2012-11-09
  • 0
    @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
  • 1
    @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
  • 0
    @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
  • 0
    $\delta_0 = \delta(0)$, I am working in Sobolev spaces2012-11-09
  • 0
    @Juan What is the definition of $\phi_1+\phi_2$?2012-11-09

2 Answers 2

4

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

  • 1
    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.2012-11-09
  • 0
    $\delta_0 = \delta(0)$, I am working in Sobolev spaces, see that $\phi_1$ and $\phi_2$ $\in W^{m,p}$.2012-11-09
  • 0
    @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?2012-11-09
  • 0
    $\phi$ is a functional that belong to $W^{m,p}$, Where I wrote value at a point?2012-11-09
  • 0
    @Juan, I mean when you write $\phi(0)$, what does that mean? $L^p$ functions cannot be evaluated at a point?2012-11-09
  • 0
    @Juan: you use $W^{m,p}(\Omega)$ without definition. Maybe others find it standard, but I don't. What happens when you define $a\delta_0$ and $\delta_0+b$?2012-11-09
  • 0
    $W^{m,p}(\Omega)=\left\{{u\in L^p(\Omega); D^{\alpha}u \in L^p(\Omega) \forall |\alpha| \leq m}\right\}$2012-11-09
  • 0
    @RossMillikan I don't know that happen2012-11-09
  • 0
    @Juan: these are terms that occur in the definition of a linear functional. If you look at $(a\delta)_0=a(\delta_0)$ and $(\delta_0+b)x=\delta_0(x)+b$ you should get there.2012-11-09
  • 0
    @RossMillikan, I assume you know what the question is asking, so do you mind explaining to me what $\delta(\phi) = \phi(0)$ means in this context?2012-11-09
  • 0
    This is ill-defined $\delta(\phi) = \phi(0)$2012-11-09
  • 0
    $\left<{\delta_0,\phi}\right> = \left<{\delta(0),\phi}\right> = \phi(0)$2012-11-09
2

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.

  • 1
    Why downvote? The question is confusing. At least I tried to help.2012-11-09
  • 0
    I make upvote I don't understand the confusing2012-11-09
  • 0
    @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.2012-11-09
  • 0
    I don't understand yet only I get upvote because the claim of @glebovg is fair.2012-11-09
  • 0
    @RossMillikan Thanks Ross.2012-11-09