Help understanding a step in proving that $\Delta 1/|x| = \delta(x)$, in $\mathcal{D}'(\mathbb{R}^3)$. So, I'm currently taking a module in linear partial differential equations, and in said module we're taught much about the calculus (or theory) of distributions. There's a particular example that we study, in which we prove that $\Delta 1/|x| = \delta(x)$, in  $\mathcal{D}'(\mathbb{R}^3)$ (the set of distributions on the set of test functions on $\mathbb{R}^3$). I've included below a screenshot of some of the working seen in said example, and just below it I've elaborated on a question that I have from said working.



Here is my interpretation of the working shown above. In proving the desired claim, we arrived at the conclusion that
$$
\bigg{\langle}\Delta\frac{1}{|x|},\phi\bigg{\rangle} = \lim_{\varepsilon\to0}\int_{\mathbb{R}^{3}\setminus B_\varepsilon}\frac{\Delta\phi(x)}{|x|}~dx
$$
(for $\phi\in\mathcal{D}(\mathbb{R}^{3})$), and up to this point the working makes perfect sense. From here, we note that since $\phi\in\mathcal{D}(\mathbb{R}^{3})$, $\mathrm{supp}(\phi)$ is a compact set, meaning that we may instead integrate over a bounded subset of $\mathbb{R}^{3}\setminus B_\varepsilon$. In particular, we may let $R>0$ be sufficiently large so that
\begin{equation}
\bigg{\langle}\Delta\frac{1}{|x|},\phi\bigg{\rangle} = \lim_{\varepsilon\to0}\int_{B_R\setminus B_\varepsilon}\frac{\Delta\phi(x)}{|x|}~dx\tag{1}\label{1}
\end{equation}
Again, this makes perfect sense. From here, it is said that we apply Green's formula to arrive at the conclusion that
$$
\bigg{\langle}\Delta\frac{1}{|x|},\phi\bigg{\rangle} = \lim_{\varepsilon\to0}\bigg{(} \int_{B_R\setminus B_\varepsilon}\phi(x)\Delta|x|^{-1}~dx + \int_{\partial(B_R\setminus B_\varepsilon)}\frac{1}{|x|}\partial_n\phi(x) - \phi(x)\partial_n|x|^{-1}~dS\bigg{)}\tag{2}\label{2}
$$
where $n = -x/|x|$. It's at this point that I lose the plot somewhat. I'll try to include in the following list all the questions that I have regarding this working.
Question 1
I'll be completely honest: I'm not even completely sure which Green's formula is being referenced here. I assume it's the formula found in the statement of the conventional Green's theorem (the one of which Stokes' theorem is a generalization), but I have no idea how we might extract the above expression from any variation of the statement of Green's theorem. I understand that in order for us to actually be able to apply Green's theorem, we require that the domain of integration be bounded, hence why the observation that precedes \eqref{1} is important, but beyond that I'm rather lost. I had thought that maybe we're actually meant to apply Gauss' theorem (or the divergence theorem, rather) here, since $\Delta\phi(x) = \mathrm{div}(\nabla\phi(x))$, and this makes particular sense since:

*

*The divergence theorem requires integration over a compact domain, and we know that $B_R\setminus B_\varepsilon$ can be made compact.

*The derivative $\partial_n$, in addition to the fact that $n = -x/|x|$, seems to indicate that we are, in some way or another, evaluating the outward unit normal to the surface $B_\varepsilon$, which is an important quantity to bear in mind when applying the divergence theorem.

Although ultimately, I was once again unsure of how I was meant to proceed. In truth, it's primarily the first of the terms listed in \eqref{2} that's throwing me off, because both the application of Green's theorem, and the application of the divergence theorem, requires some-or-other manipulation of the domain of integration, and yet the first term listed in \eqref{2} demonstrates no manipulation in the domain of integration at all, so I remain rather lost in this regard.
Question 2
I'm not sure how we're meant to apply the derivative operator $\partial_n$. Particularly since $n = -x/|x|$, and $x\in\mathbb{R}^{3}$. It's mentioned in a subsequent claim that $n = -x/|x|\implies \partial_n = -\partial_r$, which makes intuitive sense, but I'm sure how to either show that this is rigorously true, or how to proceed, given this information.
These are the aspects of this working that confuse me at the moment. I've tried to cross-reference the working I've mentioned above with the working found in the mark scheme of past examination papers containing similar questions, and I've managed to posit that, in general, given an appropriate function, $f:\mathbb{R}^{3}
    \to\mathbb{R}$
$$
\lim_{\varepsilon\to0}\int_{\mathbb{R}^{3}\setminus B_\varepsilon}f(x)\Delta\phi(x)~dx = \lim_{\varepsilon\to0}\bigg{(} \int_{B_R\setminus B_\varepsilon}\phi(x)\Delta f(x)~dx + \int_{\partial(B_R\setminus B_\varepsilon)}f(x)\partial_n\phi(x) - \phi(x)\partial_nf(x)~dS \bigg{)}
$$
$\forall\phi\in\mathcal{D}(\mathbb{R}^{3})$, although, firstly, I don't know what constitutes an "appropriate" function in this case, and secondly, I don't imagine it'll be sufficient to simply commit this fact to memory in order to excel in the exam (I'm currently studying this problem as part of preparation for an exam).
Any advice in this regard is greatly appreciated. Also please do not hesitate to ask for any clarification as far as notation is concerned, I understand that this field is rather notation-centric, so I have no problem with clarifying what is meant by the symbols seen in this question.
 A: Question 1:
See https://en.wikipedia.org/wiki/Green%27s_identities (Green's Second Identity - $\varepsilon = 1$ case). To derive this Green's Second Identity, this is nothing but a divergence theorem + product rule (which is basically what we call "integration by parts in $\mathbb{R}^n$") applied twice.
When a book says Green's formula, it could really mean a lot of things. For instance, from the same Wikipedia page, you can find Green's first and second identities (which could mean Green's formula), and Green's theorem (like you have mentioned), or even divergence theorem!
By applying Green's second identity directly, you should see $\partial_n$ and the terms in $\partial(B_R \setminus B_\varepsilon)$ almost immediately. (See that (2), by considering terms on the left and right, are a total of 4 terms, which you can see from Green's second identity.)
Question 2:
The intuitive picture that you have is absolutely correct, since the normal direction along the inner ball $B_\varepsilon$ should be pointing towards the origin (since the actual region is the annulus of radius $\varepsilon$ to $R$, as represented by $B_R \setminus B_\varepsilon$).
To be more rigorous, $\partial_n u$ is defined as the directional derivative (https://en.wikipedia.org/wiki/Directional_derivative) of the function $u$ in the $\mathbf{n}$ direction, where $\mathbf{n}$ is a unit vector. The formula is given by
$$\frac{\partial u}{\partial n} = \mathbf{n}\cdot \nabla u $$
for any $u$ (with the appropriate regularity).
Thus, we can look at this equation in the operator sense, ie
$$\frac{\partial}{\partial n} = \mathbf{n} \cdot \nabla. $$
You can then compute this, with $\mathbf{n} = -\frac{x}{|x|} \cdot \nabla $
Note that this is $\frac{x}{|x|} = \hat{\mathbf{r}}$. Furthermore, we can write the gradient in spherical coordinates as (see https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates)
$$\hat{\mathbf{r}}\frac{\partial}{\partial r} +  \hat{\boldsymbol{\theta}}(...) + \hat{\boldsymbol{\phi}}(...)$$
where $(...)$ are some expressions that do not really matter since we know that at each point, the basis vectors $\mathbf{r}, \hat{\boldsymbol{\theta}},$ and $\hat{\boldsymbol{\phi}}$ are orthonormal.
Thus, we have
$$-\hat{\mathbf{r}} \cdot [\hat{\mathbf{r}}\frac{\partial}{\partial r} +  \hat{\boldsymbol{\theta}}(...) + \hat{\boldsymbol{\phi}}(...)] = -\frac{\partial}{\partial r}.$$
Note that the part on spherical coordinates and orthonormality of these unit vectors can be computed directly through really tedious computations. (They are intuitively true; $\hat{\mathbf{r}}$ refers to the unit vector pointing in the direction of increasing $\mathbf{r}$, $\hat{\boldsymbol{\theta}}$ in increasing $\theta$, and so on)
Disclaimer: There might be an easier way to derive this, but this is the way I understand it.
Hope this helps!
