Properties of Dirac delta

Question

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?

Answer 1

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 $R\to\infty.$

Answer 2

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.