Proof that the uniqueness theorem is valid for zero potential at infinity Suppose a boundary condition is that the potential tends to 0 at infinity.  Can I use this as a boundary condition for uniqueness theorem? How do you prove this?
 A: Given your other similar question I assume you're talking about Laplace's equation. So, does Laplace's equation together with the boundary condition that the function vanishes at infinity have a unique solution? If not, what further conditions would guarantee that it does?
It is true if the domain is the whole of $\mathbb{R}^n$, but trivially so. More precisely for $u:\mathbb{R}^n\to\mathbb{R}$ harmonic,
$$\lim_{|x|\to\infty}\frac{u(x)}{|x|}=0\implies u \,\,\,\mathrm{constant}$$
so in particular if $u(x)\to 0$, then $u$ must be constant and therefore $0$. See this Math.SE question for a sketch of the proof.
If the domain is not the whole space, then generally it does not hold. Consider for instance Laplace's equation in $\mathbb{R}^3\setminus\{0\}$. Then both $1/r$ and $0$ are solutions which decay to $0$ at infinity.
You can make the solution unique if you specify further boundary conditions, but the theorem is more technical. Roughly speaking for Laplace's equation on $\mathbb{R}^3\setminus\Omega$ the solution is unique if you specify the value of the solution on $\partial\Omega$ and require that it vanishes at infinity. See here for a precise statement and a proof.
A: Infinity is only one boundary. You can use this as one of the boundary conditions, but there are (obviously) multiple electrostatic potentials that satisfy the condition that $\phi(\infty)=0$ -- for example, the potential due to any localized, finite charge distribution.
Note that it also may be necessary to specify that $\phi(\infty)=0$ for there to be a unique solution. The usual statement that the potential of a point charge is $\phi(r)=q/r$ (rather than $\phi(r)=q/r+\phi_0$) reflects that the boundary condition $\phi_0=\phi(\infty)=0$ has been assumed.
