Uniqueness Proof for solution to $\nabla^2 G(\textbf{r}) = \delta(\textbf{r})$ with $G \rightarrow 0$ when $|\textbf{r}| \rightarrow \infty$ I'm having difficulty understanding the derivation of solution to this equation: $\nabla^2 G (\textbf{r}) = \delta(\textbf{r})$ with $G \rightarrow 0$ when $|\textbf{r}| -> \infty$ in $R^n$ where the boundary condition is defined at infinity. 
I suppose that a reasonable approach is to first guess a solution and prove the uniqueness of it, so here's what I've tried to prove the uniqueness:
Assume that $G_1, G_2$ are different solutions, take $G' = G_1 - G_2$ then $\nabla^2 G' = 0$. My purpose is to prove $G' = 0$.
Consider $\int (\nabla G')^2 d^n{\textbf{r}}$ where $d^n{\textbf{r}} \stackrel{\Delta}{=} dV$ is the volume element. In a FINITE space $\Omega$ containing the origin with boundary condition $G'(\textbf{r}) = 0$ when $\textbf{r} \in \partial \Omega$ I can have 
$\int_{\Omega} (\nabla G')^2 \cdot dV = \int_{\Omega} \nabla \cdot (G' \cdot \nabla G') \cdot dV - \int_{\Omega} G' \cdot \nabla^2 G' \cdot dV = \int_{\Omega} \nabla \cdot (G' \cdot \nabla G') \cdot dV$
and by divergence theorem
$\int_{\Omega} \nabla \cdot (G' \cdot \nabla G') \cdot dV = \int_{\partial \Omega} (G' \cdot \nabla G') \cdot \textbf{n}(\textbf{r}) \cdot dS = 0$ 
where $\textbf{n}(\textbf{r})$ is the norm vector of surface element on $\partial \Omega$. Thus $\nabla G' = 0$ in $\Omega$ and $G' = 0$ in $\Omega$.
However when the boundary condition involves $|\textbf{r}| \rightarrow \infty$ I'm not sure whether the step at which I applied divergence theorem is still valid.
Any help is appreciated :)
Edited:
$\delta(\textbf{r})$ here is the Dirac Delta function.
 A: Maybe you could use the following argumentation:
Let $B_R =\{x\in \mathbb{R}^n | ||x||\leq R \}$ be a ball of size $R$. Let $\phi_R\in C_0^\infty(B_R)$ with $\phi_R(0)\not = 0$ be arbitrary.
Then from $G(r)=\delta(r)$, we get
\begin{align}
&\langle \nabla^2 G,\phi_R\rangle_{B_R} = \langle \delta,\phi_R\rangle_{B_R}=\phi_R(0)\\
\Leftrightarrow&\langle \nabla G,\nabla \phi_R\rangle_{B_R}=-\phi_R(0) \quad \quad \quad (*)
\end{align}
Now $G'=G_1-G_2$ are all three supposed to be solutions and $G_1\not \equiv G_2$. Further note, that $G'\equiv 0$ is not a solution. Then $(*)$ is true for $G'$, i.e.
\begin{align}
&\langle \nabla G',\nabla \phi_R\rangle_{B_R}=-\phi_R(0) \\
\Leftrightarrow& \langle \nabla G_1,\nabla \phi_R\rangle_{B_R}
-\langle \nabla G_1,\nabla \phi_R\rangle_{B_R}=-\phi_R(0)
\end{align}
But since $G_1$ is a solution and $(*)$ has to hold,
\begin{align}
&\langle \nabla G_2,\nabla \phi_R\rangle_{B_R}=0 \\
\Leftrightarrow & \langle \nabla^2 G_2, \phi_R\rangle_{B_R}=0
\end{align}
Since $\phi_R$ was arbitrary, by the fundaental theorem of variational calculus
\begin{align}
\nabla^2G_2 \equiv 0
\end{align}
has to hold. This means, that $G_2$ is linear in $r$ independent of $B_R$. But since $G\rightarrow 0$ for $|r|\rightarrow \infty$ we have $G_2\equiv 0$ which is a contradiction because the solution is not identically zero. $\blacksquare$
EDIT:
1)
$C^\infty_0(B_R)$ is the set of all functions that are differentiable of arbitrary order (that's what $\infty$ refers to) and have compact support on the set $B_R$, i.e. for $\phi \in C^\infty_0(B_R)$, we have $\phi(x)=0,\forall x \not \in B_R$
2) First, I assumed $\delta(r)$ is the Dirac delta distribution, which has the property, that $\int f(x)\delta(x)\,dx = f(0)$ which explains the right side of the equation for $(*)$. The notation $\langle f,g \rangle_{B_R}$ simply means the scalar product $\int_{B_R}f(x)g(x)\, dx$. Thus \begin{align}
\langle \nabla^2g,\phi \rangle_{B}&=\int_B \nabla^2 g(x)\phi(x)\,dx
\\&{}^*=[\nabla g \phi]_{\partial B}-\int_B \nabla g(x)\nabla\phi(x)\,dx
\end{align}
where we use integration by parts at $*$. The first term $[\nabla g \phi]_{\partial B}=0$ because $\phi(x)=0 \forall x\in \partial B$
2nd EDIT:
Let me give this one last try. I'll keep the fist part, i.e. $\langle \nabla G,\nabla \phi\rangle = -\phi(0)$
Now if $G$ is not a unique solution, then there exists another solution $G'\not \equiv G$. Then $G'=G+H$ with $H\not \equiv 0$. However, $H$ does not have to be a solution. Since $G'$ and $G$ are solutions, we have
\begin{align}
 -\phi(0)&= \langle \nabla G',\nabla \phi\rangle \\
&= \langle \nabla G,\nabla \phi\rangle+ \langle \nabla H,\nabla \phi\rangle \\
&=-\phi(0)+ \langle \nabla H,\nabla \phi\rangle
\end{align}
So we get
\begin{align}
\langle \nabla H,\nabla \phi\rangle=0
\end{align}
Again with integration by party and the fundamental lemma of variational calculus, $H$ has to be linear in $r$. Assume now $H$ is not zero , i.e. $H\not \equiv 0$. Then $H$ is either constant $\not = 0$ or linear with slope $\not = 0$
Since we know, that $G(r)=0$ for $r\rightarrow \infty$, we know that $G(r)+H(r)=H(r)$ for $r \rightarrow \infty$. But $H(r)\not =0 $ for $r \rightarrow \infty$ since it is either constant $\not = 0$ or linear. But then $G'(r)=G(r)+H(r)\not = 0$ for $r \rightarrow \infty$. Which means that $G'$ can't be a solution. This is a contradiction, so $H\equiv 0$ and thus $G'\equiv G$, thus the solution is unique.
A: @sonystarmap, I think I get the idea now: you denote $G'=G_1 - G_2$, a sphere $B_R$ with radius $R$ and functions 
$$\phi_R(\textbf{r}) = \{ {\begin{array}{cc} arbitrary & |\textbf{r}| < R \\ 0 & \textbf{r} \ge R \end{array}} $$
Then with $\langle \nabla G', \nabla \phi_R \rangle_{B_R}=\langle \nabla G_1, \nabla \phi_R \rangle_{B_R}=-\phi_R(\textbf{0})$ got (ignoring the intermediate reasoning) $\nabla^2 G_2 \equiv 0$, is it? 
In my calculation $\langle \nabla G', \nabla \phi_R\rangle_{B_R}=0$, because $\nabla^2 G'=\nabla^2 G_1 - \nabla^2 G_2 = 0$ and $\langle \nabla G', \nabla \phi_R\rangle_{B_R} = \int_{\partial B_R} \nabla G' \cdot \phi_R \cdot d\textbf{S} - \int_{B_R} \nabla^2 G' \cdot \phi_R \cdot d\Omega = 0$. 
So $\langle \nabla G', \nabla \phi_R \rangle_{B_R} = -\phi_R(\textbf{0})$ doesn't seem right to me.
