Evans Partial Differential Equations - Derivation of Green function

Question

I am reading through PDE book by Evans to get a better understanding of what a Green function is. In my understanding the Green function enables an explicit representation of the solution of certain PDE (I think mainly linear ones).

Looking at the derivation is, as always, insightful.

The author specifically derives Green function for the Laplace equation

$$ \left\{ \begin{array}{ll} -\Delta u = f & \text{in $U$} \\ u = g & \text{on $\partial U$} \end{array} \right. $$

I quote the derivation (which isn't long) and I'll put my questions in the middle.

Suppose first of all $u \in C^2(\overline{U})$ is an arbitrary function. Fix $x \in U$ choose $\epsilon > 0$ so small that $B(x,\epsilon) \in U$, and apply the Green's formula from C.2 on the region $V_{\epsilon} = U - B(x,\epsilon)$ to $u(y)$ and $\Psi(y - x)$. We thereby compute $$ \int_{V_{\epsilon}} u(y) \Delta \Psi(y - x) - \Psi(y - x) \Delta u(y) dy = \int_{\partial V_{\epsilon}} u(y) \frac{\partial \Psi}{\partial \nu}(y - x) - \Psi(y - x) \frac{\partial u}{\partial \nu}(y) dS(y), $$ $\nu$ denoting the outer unit normal vector in $\partial V_{\epsilon}$. Recall next $\Delta \Psi (x - y) = 0$ for $x \neq y$. We observe also $$ \left|\int_{\partial B(x,\epsilon)} \Psi(y - x) \frac{\partial u}{\partial \nu}(y) dS(y) \right| \leq C\epsilon^{n-1} \max_{\partial B(0,\epsilon)} \left| \Psi \right| = o(1) $$ as $\epsilon \to 0$. Futhermore the calculations in the proof of Theorem 1 show $$ \int_{\partial B(x,\epsilon)} u(y) \frac{\partial \Psi}{\partial \nu} dS(y) = \frac{1}{\left| \partial B(x,\epsilon) \right|} \int u(y)dS(y) = u(x) $$ as $\epsilon \to 0$. Hence sending $\epsilon \to 0$ yields the formula $$ u(x) = \int_{\partial U} \Psi(y - x)\frac{\partial u}{\partial \nu}(y) - u(y) \frac{\partial \Psi}{\partial \nu}(y - x) dS(y) - \int_{U} \Psi(y - x) \Delta u(y) dy \hspace{10mm} (25) $$

My observation here is that most of the calculation being done so far they're independent from the use of the ball $B(x,\epsilon)$. My question here is: Do we need such a ball so we can apply the mean value property? Apart from that specific calculation I don't see any other use of the ball.

The observation made by the author is that formula (25) would fully define $u$ if we knew the normal derivative at the boundary (which we don't, we only know the function at the boundary but not it's normal derivative). Therefore a corrective term $\phi^x(y)$ for fixed $x$ is added. Such a corrective term is defined as the solution of the following boundary problem

$$ \left\{ \begin{array}{ll} \Delta \phi^x = 0 & \text{in $U$} \\ \phi^x = \Psi(y - x) & \text{on $\partial U$} \end{array} \right. $$

And applying the Green's formula once more yields $$ -\int_U \phi^x(y) \Delta u(y) dy = \int_{\partial U} u(y) \frac{\partial \phi^x}{\partial \nu}(y) - \Psi(y - x) \frac{\partial u}{\partial \nu}(y) dS(y) \hspace{10mm} (27) $$

The Green function is then defined as $$ G(x,y) := \Psi(y - x) - \phi^x(y) \hspace{10mm} (x, y \in U, x \neq y) $$ Adding (25) and (27) together gives

$$ u(x) = - \int_{\partial U} u(y) \frac{\partial G}{\partial \nu}(x,y) dS(y) - \int_{U} G(x,y) \Delta u(y) dy $$

Finally a remark at the end

Remark. Fix $u \in U$. Then regarding $G$ as a function of $y$, we may simbolically write $$ \left\{ \begin{array}{ll} -\Delta G = \delta_x & y \in U \\ G = 0 & y \in \partial U \end{array} \right. $$

Question : Why is the remark true? why can we simbolically write a PDE with the Dirac delta as input?

Question : This derivation requires the use of a boundary term. However I wonder what changes would we need to make if the domain is unbounded and we search for solutions (for example) in $C^2(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$, does the boundary term just disappear? So we only endup with:

$$ u(x) = -\int_{\mathbb{R}^n} G(x,y) \Delta u(y) dy $$