I am going through some notes about how to compute that $\nabla^2 G = \nabla^2 (1/4 \pi |x|) = - \delta(x)$ in a sense of distributions, where $x \in \mathbb{R}^3$. Notes are not expected to be completely rigorous, however, I was not able to figure out all of the details to make the arguments precise.

The approach starts by defining function $K : \mathbb{R}^3 \to \mathbb{R}$ such that $K(x) = G(x) = 1/4\pi|x|$ for all $|x| > 1/2$ in a sense of ordinary functions, and continued smoothly on the rest of $\mathbb{R}^3$. I believe the motivation is that the point of singularity of $G(x)$ is $x = 0$, and so we would like to define a function that is the same outside of the singularity, but is smoothly continued near singularity so we can treat it using usual real analysis theorems.

Question 1: How do I know that such function $K$ can be indeed found? Is there a constructive formula for such function? I was thinking about defining $K(x) = G(x)\Theta(|x|-1/2)$, where $\Theta$ is a step function, and then to take smooth approximations $\{ f_n \}_{n \in \mathbb{N}}$ for step function, but I believe for all $n \in \mathbb{N}$, I would get that $K(x) \neq G(x)$ for $|x| > 1/2$.

If such function $K$ indeed exists, then as it is a smooth extension, all derivatives exist in a pointwise real analysis sense. Then, for example, one might compute that for all $|x| > 1/2$, $\nabla^2 K (x) = 0$ pointwise. Also, one might apply divergence theorem to compute the following, where $B_{1/2}(0)$ is an open ball of radius $1/2$ around $0 \in \mathbb{R}^3$.

$$ \int_{\mathbb{R}^3} \nabla^2K (x) \mathrm{d}x = \int_{B_{1/2}(0)} \nabla^2K (x) \mathrm{d}x = -1 $$

Now, one defines one-parameter family of functions (which will create a mollifiers). For each $\varepsilon > 0$, define $K_{\varepsilon} : \mathbb{R}^3 \to \mathbb{R}$ as follows.

$$ K_{\varepsilon}(x) = \frac{1}{\varepsilon} K\left(\frac{x}{\varepsilon}\right) $$

I think I understand the intuition, as for each $\varepsilon > 0$, $K_{\varepsilon}(x) $ is again smooth, and this leads to the following scaling on Laplacian, which later can be used to change variables under the integrals.

$$ \nabla^2 K_{\varepsilon}(x) = \frac{1}{\varepsilon^3} \nabla^2 K \left( \frac{x}{\varepsilon} \right)$$

Question 2: Notes motivate choice of definition of $K_{\varepsilon}$ with the claim that $K_{\varepsilon} * \phi \to G * \phi$, as $\varepsilon \to 0^+$, where $*$ is convolution operator and $\phi$ is suitable test function, say, smooth with a compact support. However, I don't understand in what sense is this true, as I thought that the whole point not working with $G$ is because it is singular at the origin. For example, naively, I would understand $G * \phi$ is the following.

$$ (G * \phi)(x) = \int_{\mathbb{R}^3} G(x-y) \phi(y) \mathrm{d}y $$

However, as $G(x) = 1/4 \pi |x|$ defined on $\mathbb{R}^3 - \{0\}$ is not bounded, I believe (Darboux) integral does not exist. One maybe might define it as follows, but I am not sure if this is the convention here.

$$ (G * \phi)(x) = \lim \limits_{\delta \to 0} \int_{\mathbb{R}^3 - B_{\delta}(0)} G(x-y) \phi(y) \mathrm{d}y $$

Question 3: Fixing interpretation of $G * \phi$, how to prove it is well defined (as an integral or as a limit)?

Question 4: I tried performing the argument on $K_{\varepsilon}$ as the one that is used to show that $\nabla^2 (K_{\varepsilon} * \phi) \to - \phi(x)$ as $\varepsilon \to 0^+$. However, I do not understand what goes wrong. I provide (intuitive and, I think, wrong) argument below.

$$ \lim \limits_{\varepsilon \to 0^+} (K_{\varepsilon} * \phi)(x) = \lim \limits_{\varepsilon \to 0^+} \int \limits_{\mathbb{R}^3} K_{\varepsilon}(x-y)\phi(y)\mathrm{d}y = \lim \limits_{\varepsilon \to 0^+} \int \limits_{\mathbb{R}^3} K_{\varepsilon}(y)\phi(x-y)\mathrm{d}y = \lim \limits_{\varepsilon \to 0^+} \int \limits_{\mathbb{R}^3} \frac{1}{\varepsilon} K\left( \frac{y}{\varepsilon} \right)\phi(x-y)\mathrm{d}y $$ $$= \lim \limits_{\varepsilon \to 0^+} \int \limits_{\mathbb{R}^3} \varepsilon^2 K\left( y \right)\phi(x-\varepsilon y)\mathrm{d}y$$

So, I would think that the following happens.

$$ \lim \limits_{\varepsilon \to 0^+} \int \limits_{\mathbb{R}^3} K\left( y \right)\phi(x-\varepsilon y)\mathrm{d}y = \phi(x) \int \limits_{\mathbb{R}^3} K\left( y \right) \mathrm{d}y $$

Here, I think the latter integral does not exist, but I feel that there might a different choice of $K(x)$ such that that integral exists (for example, by also taking large $|x|$ cutoff). In that case, I feel that this limit would be finite but because of the $\varepsilon^2$ factor, $K_{\varepsilon} * \phi \to 0$ as $\varepsilon \to 0^+$.


1 Answer 1


I suggest you first take a look at Dirac's $\delta$ smooth approximation, (the answer I provide there will be helpful for your Question 4.) We can actually deal with the $n$-dimensional case directly, assuming $n\geq 3$. Let $G:\Bbb{R}^n\setminus\{0\}\to\Bbb{R}$, $G(x)=\frac{\gamma}{|x|^{n-2}}$, where $\gamma$ is some constant (in fact by choosing $\gamma:=\frac{1}{(n-2)A_{n-1}}$, where $A_{n-1}$ is the surface area of $S^{n-1}$, it will follow that $\nabla^2G=-\delta$. In the case $n=3$, $\gamma=\frac{1}{4\pi}$).

Question 1.

You should take a look at the existence of bump functions. This means we can find a smooth function $\psi:\Bbb{R}^n\to\Bbb{R}$, such that $\psi=1$ on $B_{1/4}(0)$ and $\psi=0$ outside say $B_{1/3}(0)$. Now, set $K:= (1-\psi)\cdot G$; then $K=0$ on $B_{1/4}(0)$, and outside $B_{1/3}(0)$, we have $K=G$, and this will be smooth.

I hope you don't mind, but in order to stay consistent with the notation in my linked answer above, let me define with a superscript $K^{\epsilon}(x):=\frac{1}{\epsilon^{n-2}}K\left(\frac{x}{\epsilon}\right)$. Then, $\nabla^2(K^{\epsilon})(x)=\frac{1}{\epsilon^n}(\nabla^2K)\left(\frac{x}{\epsilon}\right)\equiv (\nabla^2K)_{\epsilon}(x)$, where the subscript $\epsilon$ has the same meaning as in the linked answer. Now, for $|x|\geq 1$, $\nabla^2K(x)=\nabla^2G(x)=0$, while for $|x|\leq 1$, $K$ is smooth, so $\nabla^2K$ is also smooth, and thus $\nabla^2K$ is in $C^{\infty}_c$ (and thus is Lebesgue integrable), with \begin{align} \int_{\Bbb{R}^n}\nabla^2K(x)\,dx&=\int_{B_1(0)}\nabla^2K(x)\,dx\\ &=\int_{S^{n-1}} \frac{\partial G}{\partial r}\,d\sigma\\ &=\int_{S^{n-1}}(2-n)\gamma\,d\sigma\\ &=-1, \end{align} where we have used the divergence theorem in the middle and that $K=G$ in an open neighborhood of $S^{n-1}$. The reason I wrote all of this out is because now, following the notation of the linked answer with $\zeta:=\nabla^2K$ and $c=-1$, we have as $\epsilon\to 0^+$, that

\begin{align} (\nabla^2K)_{\epsilon}:=\zeta_{\epsilon}\to c\delta=-\delta, \end{align} where the convergence is in the space of distributions $\mathcal{D}'(\Bbb{R}^n)$.

Question 2.

First let me remark (as I've already mentioned in the comments), that yes, although $G$ is unbounded near the origin, it is still integrable there, i.e the singularity of $G$ is not so bad for integration. To see why, note that \begin{align} \int_{B_1(0)}|G(x)|\,dx&=\int_{0}^1\int_{S^{n-1}}|G(r\xi)|\,r^{n-1}\,d\sigma(\xi)\,dr\\ &=\int_0^1\int_{S^{n-1}}\frac{\gamma}{r^{n-2}}\cdot r^{n-1}\,d\sigma\,dr\\ &=\int_0^1\gamma A_{n-1}r\,dr\\ &<\infty \end{align} Note that the finiteness of this integral is not a surprising thing, because for example in one dimension, $\int_0^1\frac{dt}{\sqrt{t}}=2$ is finite even though $\frac{1}{\sqrt{t}}\to \infty$ as $t\to 0^+$. So, this just goes to show unboundedness alone says nothing about finiteness of integrals.

Because of this, all integrals over $\Bbb{R}^n$ involving $G$ and a compactly supported function will exist in the proper sense as a Lebesgue integral (near the singularity of $G$, it is integrable, and for very large arguments, the compact support of the function makes it zero, so there's no issues at $\infty$ nor at the singularity of $G$).

Having said this, I can't speak for in what sense they intended the convergence. Though I do not see why convergence of the convolutions would be useful. More useful would be to show that $K^{\epsilon}\to G$ in the topology of distributions $\mathcal{D}'(\Bbb{R}^n)$. What this means is we have to show for every $\phi\in\mathcal{D}(\Bbb{R}^n)$ (i.e smooth compactly supported function) $\langle K^{\epsilon},\phi\rangle\to\langle G,\phi \rangle$. This is easily verified: \begin{align} \langle K^{\epsilon},\phi\rangle&:=\int_{\Bbb{R}^n}K^{\epsilon}(x)\,\phi(x)\,dx\\ &=\int_{|x|\leq \epsilon}\frac{1}{\epsilon^{n-2}}K\left(\frac{x}{\epsilon}\right)\phi(x)\,dx+\int_{\epsilon<|x|}\frac{1}{\epsilon^{n-2}}K\left(\frac{x}{\epsilon}\right)\phi(x)\,dx\\ &=\int_{|y|\leq 1}\epsilon^2K(y)\phi(\epsilon y)\,dy +\int_{\epsilon<|x|}\frac{1}{\epsilon^{n-2}}G\left(\frac{x}{\epsilon}\right)\phi(x)\,dx\tag{$*$}\\ &=\int_{|y|\leq 1}\epsilon^2K(y)\phi(\epsilon y)\,dy +\int_{\epsilon<|x|}G(x)\phi(x)\,dx \end{align} In $(*)$, for the first term, we used the change of variables $x=\epsilon y$ (note that this makes $dx=\epsilon^n\,dy$ which cancels with $\frac{1}{\epsilon^{n-2}}$ to give $\epsilon^2$), and in the second term, we note that $\epsilon<|x|$ means $1<\frac{|x|}{\epsilon}$, which means the arugments are large enough for $K$ to equal $G$. Now, the first integral is bounded by $\epsilon^2 \|\phi\|_{\infty}\cdot \int_{|y|\leq 1}|K(y)|\,dy$, which clearly vanishes as $\epsilon\to 0^+$. For the second integral, note that $G\cdot \phi\in L^1(\Bbb{R}^n)$ (because as I mentioned above, the singularity of $G$ is integrable and because $\phi$ has compact support). Thus as $\epsilon\to 0^+$, the second term becomes $\int_{\Bbb{R}^n}G(x)\phi(x)\,dx$. Thus, we have shown \begin{align} \lim_{\epsilon\to 0^+}\langle K^{\epsilon},\phi \rangle&=\langle G,\phi\rangle, \end{align} and since $\phi$ was arbitrary, this proves convergence $K^{\epsilon}\to G$, as $\epsilon\to 0^+$, in the space of distributions $\mathcal{D}'(\Bbb{R}^n)$.

Question 3.

I think it is unnecessary to address this question at this point.

Question 4.

Again, it is not clear to me why we're considering the convolution. Here's how I'd finish the argument. We have shown in step 2 that $K^{\epsilon}\to G$ in the space of distributions. Thus, any distributional derivative of $K^{\epsilon}$ (which is the same as the ordinary derivative) also converges to those of $G$. In particular, for the Laplacian. So we have (all derivatives are in distributional sense, and all limits are in the space of distributions $\mathcal{D}'(\Bbb{R}^n)$) \begin{align} \nabla^2G&=\lim_{\epsilon\to 0^+}\nabla^2(K^{\epsilon}) =\lim_{\epsilon \to 0^+}(\nabla^2K)_{\epsilon} =-\delta. \end{align} As I just mentioned, the first equality is because $K^{\epsilon}\to G$, and the second and third equalities have been explained in step 1.

Alternative Approaches.

There are of course several other approaches to proving this fact. The classical approach is as I mentioned in the comments.

Classic Approach.

Remove a small ball around the origin, integrate by parts and take limits to conclude. I leave it to you to fill in the details of this argument: let $\phi\in\mathcal{D}(\Bbb{R}^n)$, and say it has support lying inside a ball of radius $R>0$. Then, \begin{align} \langle \nabla^2G, \phi\rangle&:=\langle G,\nabla^2\phi\rangle\tag{since $G\in L^1_{loc}(\Bbb{R}^n)$}\\ &=\int_{\Bbb{R}^n}G(x) \nabla^2\phi(x)\,dx\\ &=\int_{|x|\leq R}G(x)\nabla^2\phi(x)\,dx\\ &=\lim\limits_{\epsilon\to 0^+}\int_{\epsilon\leq |x|\leq R}G(x)\nabla^2\phi(x)\,dx\tag{DCT}\\ &=\lim_{\epsilon \to 0^+}\left[-\int_{|x|=\epsilon}\phi\frac{\partial G}{\partial r}\,d\sigma+\int_{|x|=\epsilon}G\frac{\partial \phi}{\partial r}\,d\sigma\right]\tag{$\ddot{\smile}$}\\ &=-\phi(0)+0\\ &=\langle -\delta, \phi\rangle \end{align} Thus, $\nabla^2G=-\delta$. In $(\ddot{\smile})$ we used Greens identity (really an integration by parts using the divergence theorem), I leave the details to you, and in the next step, I leave it to you to estimate the integrals to see why they converge respectively to $-\phi(0)$ and $0$ (the second integral is actually $\mathcal{O}(\epsilon)$).

Yet another approach is outlined in Folland's Real Analysis book Chapter 9, Exercise 14. The proof of this exercise can be found in his other book Introduction to Partial Differential Equations, page 75, Theorem 2.17. Of course, if you're interested, you should try to follow the problem outline first and attempt it yourself before looking at the proof (btw the above classical approach was left as an exercise in Folland's PDE book).

Edit in response to comment.

So, obviously the second term in the integral for question 2 doesn't converge to $0$ (because I've already provided a correct proof that it converges to $\int G\phi$). But let us see what would actually happen if we made the substitution: \begin{align} \int_{\epsilon<|x|}\frac{1}{\epsilon^{n-2}}G\left(\frac{x}{\epsilon}\right)\phi(x)\,dx &=\int_{1<|y|}\epsilon^2G(y)\phi(\epsilon y)\,dy \end{align} Note that here, if I made an analogous estimate as for the first integral then \begin{align} \left|\int_{1<|y|}\epsilon^2G(y)\phi(\epsilon y)\,dy\right| \leq \epsilon^2\|\phi\|_{\infty}\int_{1<|y|}|G(y)|\,dy = \infty, \end{align} because the integral is infinity (since $G$ doesn't decay rapidly enough... in fact if you change to polar coordinates, you'll see that it is $\epsilon^2\|\phi\|_{\infty}\int_1^{\infty}\frac{1}{r^{n-2}}\,A_{n-1}r^{n-1}\,dr =\epsilon^2\|\phi\|_{\infty}A_{n-1} \int_1^{\infty}r\,dr=\infty$). Therefore, we have bounded our original integral by $\infty$, which is a completely true, but useless estimate.

So, ok we can't be so crude with our approximation. What else might we try? Well, the integrand is $\epsilon^2G(y)\phi(\epsilon y)$. For each $y\in\Bbb{R}^n$, this goes to $0$ as $\epsilon\to 0^+$; i.e the integrand converges to $0$ pointwise. Does it then follow that \begin{align} \lim_{\epsilon\to 0^+}\int_{1<|y|}\epsilon^2G(y)\phi(\epsilon y)\,dy=\int_{1<|y|}\lim_{\epsilon \to 0^+}\epsilon^2G(y)\phi(\epsilon y)\,dy = 0 ? \end{align} So, the question is whether that interchange of limits and integrals is valid. If you've only been studying Riemann integrals then MAJOR GIGANTIC alarm bells should be going off in your brain. If you're dealing with Lebesgue integrals, smaller alarm bells should be going off; we could try to invoke dominated convergence, but then there is no obvious dominating function (in fact there isn't because as I've shown in Q2, the limit is not necessarily $0$). Therefore, the interchange of limits with the integral is not valid. By making the substitution, there is just no way to conclude the limit is $0$, nor can we readily observe what the limit is. In Q2, I didn't make the substitution precisely because $G$ was homogeneous of degree $2-n$, which nicely cancels out the powers of $\epsilon$. Note that while $G\notin L^1$, the product $G\phi\in L^1$, which is why I was able to use dominated convergence to say $\int_{\epsilon<|x|}G\phi\to \int_{\Bbb{R}^n}G\phi$.

  • 2
    $\begingroup$ and I think the classical approach is the classic approach for a reason. It's the most intuitive thing to try to do (remove a small $\epsilon$ ball around the singularity, and then let $\epsilon\to 0^+$), and requires the least amount of technical real analysis lemmas (only Greens identity, which is simply an integration by parts/divergence theorem, followed by a simple limit evaluation at the end). $\endgroup$
    – peek-a-boo
    Aug 13, 2021 at 22:52
  • $\begingroup$ Thanks so much for your time and effort! I really appreciate it! I'll need some time to digest everything in your answer. I have a few quick questions, though. On question 1, you are using integral in the Lebesgue sense, but I know (generalized) Stokes' theorem only for Darboux/Riemann integrals. Is your use of divergence theorem justified? Can you give me some references? On question 2, you split up your integral into two parts, and use substitution $x = \varepsilon y$ for only one of them. I have trouble understanding why can't you use the same substitution for the other integral. (...) $\endgroup$ Aug 15, 2021 at 1:16
  • $\begingroup$ (...) By the same line of reasoning, the second term seems to tend to $0$ as $\varepsilon \to 0^+$. I guess that can't be right as that would mean that $K^{\varepsilon} \to 0$ in a sense of distributions. Can you please explain why the analogous argument fails for the second integral? Again, thanks for your time! $\endgroup$ Aug 15, 2021 at 1:16
  • $\begingroup$ @DanielsKrimans yes, I'm using the Lebesgue integral (as one should when doing any serious analysis stuff, eg involving distributions since Lebesgue integral behaves nicely with limits due to monotone and dominated convergence theorems). The Lebesgue integral is a strict generalization of the Riemann integral, so anything you can do with Riemann, you can do with Lebesgue (divergence, Stokes etc). Anyway, if you want a book which treats the generalized stokes theorem (and much more) from a Lebesgue perspective, see Amann and Eschers Analysis III. $\endgroup$
    – peek-a-boo
    Aug 15, 2021 at 2:22

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