Intereting Posts

$\operatorname{Aut}(\mathbb Z_n)$ is isomorphic to $U_n$.
Solving closed form proximal operators of L2(not squared) without using moreau decomposition?
Showing equality of sequence of holomorphic functions to limit function if converges uniformly locally
Differential Geometry without General Topology
Introductory Treatment of Differential Geometry
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How prove this $S_{\Delta ABC}\ge\frac{3\sqrt{3}}{4\pi}$
Shall remainder always be positive?
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Combinatorial Interpretation of a Certain Product of Factorials
Correct process for proof in graph theory.
Simplifying $\sum_{k=0}^{\lfloor{\frac{n}{2}}\rfloor}\binom{n}{2k}2^{2k}$
Are there $a,b>1$ with $a^4\equiv 1 \pmod{b^2}$ and $b^4\equiv1 \pmod{a^2}$?
derivative of x^x^x… to infinity?

I know, from a recent enlightening answers received here, that, if we define the distribution represented by Dirac’s $\delta$ on the space $K$ of test functions of class $C^\infty$ whose support is contained in a compact subset of $\mathbb{R}^3$, then$$\nabla^2\left(\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}_0\|}\right)=-4\pi\delta(\boldsymbol{x}-\boldsymbol{x}_0)$$where the Laplacian obviously is to be intended in the sense of the derivatives of distributions. More explicitly, that means that $$\forall\varphi\in K\quad \int_{\mathbb{R}^3}f\,\nabla^2\varphi\,d\mu=-4\pi\varphi(\boldsymbol{x}_0)=:-4\pi\int_{\mathbb{R}^3}\delta(\boldsymbol{x}-\boldsymbol{x}_0)\varphi(\boldsymbol{x})$$where $f:\boldsymbol{x}\mapsto\|\boldsymbol{x}-\boldsymbol{x}_0\|^{-1}$ and the first integral is intended as a Lebesgue integral.

I suspect that the identity may well also hold with $\varphi$ as a more generical function belonging to the Schwartz space, but I cannot generalise this excellent proof, to whose author I am immensely grateful, for the $\varphi\in K$. If the identity really holds with the distribution defined on the Schwartz space, how can it be proved? I $\infty$-ly thank any answerer.

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- Dirac's delta in 3 dimensions: proof of $\nabla^2(\|\boldsymbol{x}-\boldsymbol{x}_0\|^{-1})=-4\pi\delta(\boldsymbol{x}-\boldsymbol{x}_0)$
- Derivation of Dirac-Delta with complicated argument $\delta(f(x))$
- How to prove this Dirac delta identity involving $\int_{-\infty}^\infty \delta(f(x)) \,s(x) \,dx$?

- Why is the topology of compactly supported smooth function in $\mathbb R^d$ not first countable?
- How to solve integral of formula consisting of derivative of the delta function.
- Questions concerning a proof that $\mathcal{D}$ is dense in $\mathcal{S}$.
- Dirac delta and non-test functions
- Iterated Limits Schizophrenia
- A sufficient condition for a function to be of class $C^2$ in the weak sense.
- Dirac Delta or Dirac delta function?
- Dirac Delta function inverse Fourier transform
- Continuous inclusions in locally convex spaces
- How to prove that $\lim_{k\to+\infty}\frac{\sin(kx)}{\pi x}=\delta(x)$

The proof is essentially the same, the only difference is that with $\varphi \in \mathscr{S}(\mathbb{R}^3)$ we can’t use one fixed outer radius $R$ of the spherical shell over which we integrate, since Schwartz functions generally don’t have compact support. But Schwartz functions decay rapidly as $\lVert x\rVert \to \infty$, and thus if we let the radius of the outer sphere in Green’s formula tend to $\infty$, that part of the boundary integral tends to $0$.

Since $\varphi \in \mathscr{S}(\mathbb{R}^3) \implies \nabla^2 \varphi \in \mathscr{S}(\mathbb{R}^3)$ and $f(x) = \lVert x-x_0\rVert^{-1}$ is a locally integrable function of polynomial growth – in fact, $\lim\limits_{\lVert x\rVert\to\infty} f(x) = 0$ – we have

$$\int_{\mathbb{R}^3} f(x) \nabla^2\varphi(x)\,d\mu = \lim_{\substack{\varepsilon \to 0 \\ R \to \infty}} \int_{A(\varepsilon,R;x_0)} f(x)\nabla^2\varphi(x)\,d\mu,\tag{1}$$

where $A(\varepsilon,R;x_0) = \{ x\in \mathbb{R}^3 : \varepsilon < \lVert x-x_0\rVert < R\}$ and $0 < \varepsilon < R < \infty$. In $(1)$, we can take iterated limits in either order, or a simultaneous limit. Since $f\cdot \nabla^2\varphi$ is integrable, all these limits are well-defined and coincide.

Now we use $\nabla^2 f(x) = 0$ on $\mathbb{R}^3 \setminus \{x_0\}$ and Green’s formula to rewrite

\begin{align}

\int_{A(\varepsilon,R;x_0)} f(x)\nabla^2\varphi(x)\,d\mu

&= \int_{A(\varepsilon, R; x_0)} f(x)\nabla^2\varphi(x) – \varphi(x)\nabla^2 f(x)\,d\mu \\

&= \int_{\partial A(\varepsilon, R; x_0)} f(x) \frac{\partial \varphi}{\partial \nu}(x) – \varphi(x)\frac{\partial f}{\partial \nu}(x)\,dS,

\end{align}

where $\frac{\partial}{\partial \nu}$ is the normal derivative (in direction of the outer normal).

$\partial A(\varepsilon, R; x_0)$ consists of two pieces, the outer sphere $S_R = \{ x : \lVert x-x_0\rVert = R\}$ and the inner sphere $S_{\varepsilon} = \{ x : \lVert x-x_0\rVert = \varepsilon\}$. For the integral over $S_{\varepsilon}$, the argument that

$$\lim_{\varepsilon \to 0} \int_{S_{\varepsilon}} f(x) \frac{\partial \varphi}{\partial \nu}(x) – \varphi(x)\frac{\partial f}{\partial \nu}(x)\,dS = -4\pi \varphi(x_0)$$

is completely identical to the case of compactly supported $\varphi$. It remains to see that

$$\lim_{R \to \infty} \int_{S_R} f(x) \frac{\partial \varphi}{\partial \nu}(x) – \varphi(x)\frac{\partial f}{\partial \nu}(x)\,dS = 0.\tag{2}$$

But that follows from the fast decay of $\varphi$ and its derivatives. By definition of $\mathscr{S}(\mathbb{R}^3)$, for all $N \in \mathbb{N}$ there is a constant $C \in (0,+\infty)$ such that $\lvert \lVert x\rVert^N\varphi(x)\rvert \leqslant C$ and $\bigl\lvert \lVert x\rVert^N\frac{\partial \varphi}{\partial x_k}(x)\bigr\rvert \leqslant C$ for all $x\in \mathbb{R}^3$ and $1\leqslant k \leqslant 3$. For $R > 2\lVert x_0$ we have

$$\lVert x\rVert = \lVert (x-x_0) + x_0\rVert \geqslant \lVert x-x_0\rVert – \lVert x_0\rVert = R – \lVert x_0\rVert \geqslant \frac{R}{2}$$

on $S_R$, and so we can estimate

$$\lvert\varphi(x)\rvert \leqslant \frac{C}{\lVert x\rVert^N} \leqslant \frac{2^NC}{R^N},\quad \biggl\lvert \frac{\partial \varphi}{\partial\nu}(x)\biggr\rvert \leqslant \frac{2^N\sqrt{3}\,C}{R^N},$$

which yields

$$\Biggl\lvert \int_{S_R} f(x) \frac{\partial \varphi}{\partial \nu}(x) – \varphi(x)\frac{\partial f}{\partial \nu}(x)\,dS \Biggr\rvert \leqslant \biggl(\frac{1}{R}\cdot \frac{2^N\sqrt{3}\,C}{R^N} + \frac{2^NC}{R^N}\cdot \frac{1}{R^2}\biggr)\cdot 4\pi R^2 \leqslant \frac{\tilde{C}}{R^{N-1}},$$

and we see that taking any $N > 1$ gives us $(2)$.

It may be worth noting that the argument isn’t specific to dimension $3$. For any dimension $d \neq 2$, the function $f_d(x) = \lVert x-x_0\rVert^{2-d}$ is harmonic on $\mathbb{R}^d\setminus \{x_0\}$, and the same computation shows $\nabla^2 f_d = (2 – d)\omega_{d-1}\cdot \delta_{x_0}$, where $\omega_{d-1}$ is the $d-1$-dimensional volume of the unit sphere in $\mathbb{R}^d$. The last estimate becomes

$$\Biggl\lvert \int_{S_R} f(x) \frac{\partial \varphi}{\partial \nu}(x) – \varphi(x)\frac{\partial f}{\partial \nu}(x)\,dS \Biggr\rvert \leqslant \biggl(\frac{1}{R^{d-2}}\cdot \frac{2^N\sqrt{d}\,C}{R^N} + \frac{2^NC}{R^N}\cdot \frac{\lvert d-2\rvert}{R^{d-1}}\biggr)\cdot \omega_{d-1} R^{d-1} \leqslant \frac{\tilde{C}}{R^{N-1}}.$$

For dimension $d = 2$, the corresponding function is $f_2(x) = -\log \lVert x-x_0\rVert$ with $\nabla^2 f_2 = 2\pi \delta_{x_0}$ and for the integral over the outer boundary circle we eventually get an estimate $\tilde{C}\cdot \frac{\log R}{R^{N-1}}$.

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