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The most important propety of the 2-dimensional Dirac delta is

$$ \int_{\mathbb{R}^2} d^2 y\, \delta^{(2)}(x-y)\, f(y) = f(x) $$

where $f(y)$ is some test-function.

Let's take a ball $B_r$ centered in $x$ with radius $r$ such that $\mathbb{R}^2 = \lim_{r\rightarrow +\infty} B_r$.

I want to know if the following statements are true

  • 1) $\lim_{r\rightarrow 0} \mathbb{R}^2 \setminus B_r = \mathbb{R}^2$ where this limit is meant in terms of domain of integration. The analogous of what I mean in a one-dimensional space is $B_{r} =(x-r/2,x+r/2)$ and $ \mathbb{R}\setminus B_r = (-\infty,x-r/2] \cup [x+r/2,+\infty)$ and $\lim_{r\rightarrow 0} \mathbb{R} \setminus B_r = (-\infty,+\infty)$

  • 2)

$$ \lim_{r\rightarrow 0^+} \int_{\mathbb{R}^\infty\setminus B_r} d^2 y\, \delta^{(2)}(x-y) = 1. $$

If this is not true, is there a way to regularize this integral in order to be this limit well-defined?

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    $\begingroup$ I think the space you are referring to is more commonly referred to as $\mathbb{R}^2$, not $\mathbb{R}^\infty$. $\endgroup$ Commented Dec 4, 2017 at 19:28
  • $\begingroup$ If you mention it, you can define $\langle \delta(x-a),f \rangle = f(a)$ for $f$ continuous around $a$. Then you are asking about $\lim_{r \to \infty} \langle \delta(x-a),1_{|x-a| < r} \rangle$ and $\lim_{r \to \infty} \langle \delta(x-a),1_{|x-a| > r} \rangle$. $\endgroup$ Commented Dec 4, 2017 at 19:28
  • $\begingroup$ @reuns sorry, there was a typo on the question I corrected $\endgroup$ Commented Dec 4, 2017 at 19:30
  • $\begingroup$ But the Dirac delta isn't very interesting, what is interesting it to find if for some sequence of distributions, it holds that $\lim_{n \to \infty} \langle T_n, f \rangle = \langle \delta(x-a), f \rangle = f(a)$. In that case, $f$ continuous, piecewise continuous, or $C^\infty$ or $C^\infty_c$ makes a huge difference, and the result isn't always the same. $\endgroup$ Commented Dec 4, 2017 at 19:32
  • $\begingroup$ @reuns thanks... can you be a little bit more specific? $\endgroup$ Commented Dec 4, 2017 at 19:35

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  1. When we are talking about integration of functions then one can say that $\lim_{r\rightarrow 0} \mathbb{R}^2 \setminus B_r = \mathbb{R}^2$ is valid. However, the Dirac delta "function" is not a function; it's a distribution with support $\{(0,0)\}$ (in the 2-dimensional case). For that the limit is not valid.

  2. Since $\delta$ has support $\{(0,0)\}$ which is not in the domain of integration $\mathbb R^2 \setminus B_r$ we have $$\int_{\mathbb R^2 \setminus B_r} \delta(x-y) \, dy = 0$$ and therefore also $$\lim_{r\rightarrow 0^+} \int_{\mathbb R^2 \setminus B_r} \delta(x-y) \, dy = 0$$ But we can regularize it if we instead consider the integral $$\int \delta(x-y) \, \phi_R(y) \, dy,$$ where $\phi_R(y) = e^{-|y|^2/R}$ and then use the fact that $\delta = \triangle G,$ where $\triangle$ is the Laplace operator and $G : \mathbb R^2 \to \mathbb R$ is defined by $G(x_1,x_2) = \frac{1}{2\pi} \ln \sqrt{x_1^2+x_2^2},$ to rewrite this integral as $$\int G(x-y) \, \triangle\phi_R(y) \, dy.$$ Since $G$ is a function we can now limit the domain of integration and take limits: $$\lim_{r\rightarrow 0^+} \int_{\mathbb R^2 \setminus B_r} G(x-y) \, \triangle\phi_R(y) \, dy$$ Finally we can let $\phi_R$ tend to the function that is constant $\equiv 1$ by letting $R\to\infty.$

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  • $\begingroup$ So, If I use a function as a representation of the dirac delta, am I allowed to do the limit? For example, in d=1 I can say that $\lim_{\Lambda\rightarrow 0} (\int_{-\infty}^{-\Lambda} + \int_{\Lambda}^{+\infty})\frac{1}{\pi} \frac{n}{n^2+x^2} = 1$. Can I do the same for d=2? $\endgroup$ Commented Dec 4, 2017 at 20:42
  • $\begingroup$ Yes, you can do similar for $d=2$. $\endgroup$ Commented Dec 4, 2017 at 21:57
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First off, the thing which you call $\mathbb{R}^\infty$ is simply the Euclidean space itself, $\mathbb{R}^2$.

Secondly, for any arbitrary $\epsilon>0$,

$$\int_{\mathbb{R}^2\backslash B_\epsilon} d^2y\delta^{(2)}(x-y) = 0,$$

so the limit is also equal to 0.

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  • $\begingroup$ The domain $\mathbb{R}^2\setminus B_\epsilon$ does not include the point $x$. How can that quantity be finite for arbitrary $\epsilon$? $\endgroup$ Commented Dec 4, 2017 at 19:32
  • $\begingroup$ Whoops, made a type-o. Edited $\endgroup$ Commented Dec 4, 2017 at 19:33
  • $\begingroup$ @apt45 From what you've said, I get the impression that you think the quantity isn't finite. 0 is finite, and the limit is well defined. $\endgroup$ Commented Dec 4, 2017 at 19:37
  • $\begingroup$ Sorry, for me finite means non-vanishing :) but you are right, 0 is finite but I want to define a limit like the one I wrote such that it is 1. $\endgroup$ Commented Dec 4, 2017 at 19:39
  • $\begingroup$ @apt45 I don't see how a meaningful definition of limits would produce 1 for that limit. The whole point of a limit is that it gives an indication of the behaviour of a function in the neighborhood of some point, independently of the value at that point. The Dirac delta distribution is very well behaved everywhere except at the origin, and limits should reflect this. If you want a value of 1, why take the limit at all? $\endgroup$ Commented Dec 4, 2017 at 19:42

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