-2

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.

juaninf
  • 1,264
  • Please write your question in an intelligible human language. – Braindead Nov 09 '12 at 01:48
  • sorry now you understand? – juaninf Nov 09 '12 at 01:53
  • What did you try? –  Nov 09 '12 at 02:03
  • 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. – glebovg Nov 09 '12 at 02:46
  • Isn't the inner product bilinear? – Pragabhava Nov 09 '12 at 02:59
  • The notation in this question is very confusing. As far as I can tell, $<\varphi,\delta>$ means $\int_{ - \infty }^\infty {\varphi (x)\delta (x)dx}$. – glebovg Nov 09 '12 at 03:02
  • 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. –  Nov 09 '12 at 03:04
  • Why the confusion? in other thread I wrote a similar question without problems "http://math.stackexchange.com/questions/208906/delta-dirac-function" – juaninf Nov 09 '12 at 03:05
  • Your notation could mean many different things. – glebovg Nov 09 '12 at 03:06
  • @glebovg There is a confusion indeed, what is being defined is, as Ross pointed out, the delta functional in the following way $$\delta_0(\phi) := \langle \delta(x),\phi(x)\rangle$$ – Pragabhava Nov 09 '12 at 03:06
  • 1
    @Pragabhava, I think it's quite clear that $\delta_0 (\phi) := \phi(0)$? i.e. Delta function is considered as a distribution here. –  Nov 09 '12 at 03:07
  • It says Dirac delta in the title. – glebovg Nov 09 '12 at 03:08
  • @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? – Pragabhava Nov 09 '12 at 03:10
  • 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. –  Nov 09 '12 at 03:11
  • @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. – Pragabhava Nov 09 '12 at 03:18
  • $\delta_0 = \delta(0)$, I am working in Sobolev spaces – juaninf Nov 09 '12 at 03:18
  • @Juan What is the definition of $\phi_1+\phi_2$? – Phira Nov 09 '12 at 06:46

2 Answers2

4

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

Ross Millikan
  • 383,099
  • 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. – Ross Millikan Nov 09 '12 at 03:14
  • $\delta_0 = \delta(0)$, I am working in Sobolev spaces, see that $\phi_1$ and $\phi_2$ $\in W^{m,p}$. – juaninf Nov 09 '12 at 03:27
  • @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? –  Nov 09 '12 at 03:30
  • $\phi$ is a functional that belong to $W^{m,p}$, Where I wrote value at a point? – juaninf Nov 09 '12 at 03:39
  • @Juan, I mean when you write $\phi(0)$, what does that mean? $L^p$ functions cannot be evaluated at a point? –  Nov 09 '12 at 03:49
  • @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 $\delta_0+b$? – Ross Millikan Nov 09 '12 at 03:57
  • $W^{m,p}(\Omega)=\left{{u\in L^p(\Omega); D^{\alpha}u \in L^p(\Omega) \forall |\alpha| \leq m}\right}$ – juaninf Nov 09 '12 at 03:59
  • @RossMillikan I don't know that happen – juaninf Nov 09 '12 at 04:05
  • @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. – Ross Millikan Nov 09 '12 at 04:13
  • @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? –  Nov 09 '12 at 09:04
  • This is ill-defined $\delta(\phi) = \phi(0)$ – juaninf Nov 09 '12 at 13:11
  • $\left<{\delta_0,\phi}\right> = \left<{\delta(0),\phi}\right> = \phi(0)$ – juaninf Nov 09 '12 at 13:12
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.

glebovg
  • 10,452