Of course the Dirac delta is not a function. Despite, I think the concept of a measure is much easier than that of a distribution. Therefore, I was wondering: In what sense is the concept of a Dirac distribution equivalent to the Dirac measure? Are you (in principle) able to prove all the properties of the distribution if you are using the concept of a measure? Or is the only thing that the Dirac delta measure is good for to say:

$$\int_{\mathbb{R}} f(x)\delta(x-x_0) d\mu(x):= \int_{\mathbb{R}} f(x)d\delta_{x_0}=f(x_0)?$$

Or differently: Would this definition be an appropriate definition? Or do we have to refer to the theory of distributions to prove all the properties that the Dirac-delta has?

  • 6
    $\begingroup$ The Dirac distribution is the Dirac measure. Every sufficiently nice measure is (defines) a distribution. You can't prove all properties of $\delta$ within the framework of measures only, unless I'm much mistaken, since the derivatives of $\delta$ aren't measures, for example. But you can do a lot looking at it as a measure already. $\endgroup$ – Daniel Fischer Jun 23 '14 at 13:16
  • $\begingroup$ @DanielFischer in that case I have a question: In physics you fairly often write down the distribution without the integral, like if you want to say I have a particle carrying a charge q that is at position $r$ and moving with speed $v$. Then you can identify its current with: $j(r')= qv\delta(r'-r)$. Now, in what sense is this rigorous then? I mean, according to your comment, the distribution is only defined as the measure we use for integration. $\endgroup$ – user66906 Jun 29 '14 at 10:16
  • $\begingroup$ @DanielFischer if this physics interpretation is uncomfortable to you, then I could also refer en.wikipedia.org/wiki/Green%27s_function#Definition_and_uses this example. Is the delta distribution without an integral meaningful? In the part' motivation ' it appears inside an integral. Is this the only way to interpret this line? $\endgroup$ – user66906 Jun 29 '14 at 12:43

The Dirac distribution is not the Dirac measure (point measure), but induced by the Dirac measure. Every Radon measure $\mu$ on $\mathbf{R}$ induces a distribution by $$\phi\mapsto \int_{\mathbf{R}}\phi \ d\mu.$$ Yes we can define the distribution $\delta_0\in\mathscr{D}'(\mathbf{R})$ to be the one induced by the Dirac measure, or simply by $\delta_0(f)=f(0)$. These are obviously the same distribution. I don't quite get your second question, if you consider the Dirac distribution as a distribution (and as such it is properly viewed) you cannot get around distribution theory. Apart from that distribution theory is not that complicated (apart from, say, the topology on $\mathscr{D}$ but according to Hörmander you don't need to understand it) and this is one reason why it is so successful.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy