$\nabla^2(\|\boldsymbol{x}-\boldsymbol{x}_0\|^{-1})=-4\pi\delta(\boldsymbol{x}-\boldsymbol{x}_0)$ with distributions defined on Schwartz space

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.

• Note that Schwartz functions have compact support up to an $\epsilon$. Try to make this precise. – Peter Jan 26 '16 at 11:11

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 $\tilde{C}\cdot \frac{\log R}{R^{N-1}}$.

• Wonderful wonderfully detailed and explained proof. I know the definition of $\mathscr{S}(\mathbb{R}^3)$ as $\{\varphi\in C^\infty(\mathbb{R}^3):\forall\alpha,\beta\in\mathbb{N}_{\ge 0}^3 \quad \sup_{x\in\mathbb{R}^3}|x_1^{\alpha_1}x_2^{\alpha_2}x_3^{\alpha_3}\frac{\partial^{\beta_1+\beta_2+\beta_3}\varphi(x)}{\partial^{\beta_1}x_1\partial^{\beta_2}x_2 \partial^{\beta_3} x_3}|<\infty\}$ ... – Self-teaching worker Jan 26 '16 at 21:00
• ... I suppose that implies that $\forall N\in\mathbb{N}_{\ge 0}\quad\exists C: |\|x\|^N\frac{\partial^{\beta_1+\beta_2+\beta_3}\varphi(x)}{\partial^{\beta_1}x_1\partial^{\beta_2}x_2\partial^{\beta_3}x_3}|\le C$ but I cannot prove it... Thank you so much!!! – Self-teaching worker Jan 26 '16 at 21:00
• Suppose $N = 2k$ (if $N$ is odd, replace it with $N+1$, on the unit ball everything is bounded anyway, and outside the unit ball that makes the left hand side larger). Then $\lVert x\rVert^N = (x_1^2 + x_2^2 + x_3^2)^k$. That is a polynomial, so you can write it in the form $\sum c_\alpha\cdot x^{\alpha}$ (using multi-index notation to save typing). For every monomial in that we have a bound $\lvert x^{\alpha} D^{\beta}\varphi(x)\rvert\leqslant M_{\alpha,\beta}$. Then $C:=\sum\lvert c_{\alpha}\rvert\cdot M_{\alpha,\beta}$ is a bound for $\lVert x\rVert^N\cdot\lvert D^{\alpha}\varphi(x)\rvert$. – Daniel Fischer Jan 26 '16 at 21:21
• If $N$ was odd, you may need to adapt the bound to work also on the unit ball. – Daniel Fischer Jan 26 '16 at 21:22
• Very kind and able to transmit the beauty of mathematics. I $\aleph_1$-ly thank you! – Self-teaching worker Jan 27 '16 at 12:29