limit of harmonic functions Gilbarg and Trudinger page 27 Here is a passage from Gilbarg and Trudinger page 27
Let $\Sigma$ be a bounded domain in $\mathbb{R}^n$ ($n\geq 3$) with smooth boundary $\partial\Omega$, and let $u$ be the harmonic function (often called conductor potential) defined in the complement of $\bar{\Omega}$ and satisfying the boundary conditions $u=1$ on $\partial\Omega$ and $u=0$ at infinity. The existence of $u$ is easily established as the (unique) limit of harmonic functions $u'$ in an expanding sequence of bounded domains having $\partial\Omega$ as an inner boundary (on which $u'=1$) and with outer boundaries (on which $u'=0$) tending to infinity. If $\Sigma$ denotes $\partial\Omega$ or any smooth closed surface enclosing $\Omega$, then the quantity
$$\text{cap }\Omega = -\int_{\underset{\bar{}}{v}}\frac{\partial u}{\partial v}\text{d}s = \int_{\mathbb{R}^n-\Omega} |Du|^2\text{d}x\quad v= \text{outer normal}$$
is defined to be the capacity of $\Omega$.
1) I do not understand why the part in bold is obviously true?
2) In the definition of capacity, I do not really understand what the first integral really mean and why does it equals to the second integral. My guess is integration by parts of some sort, but it is not obvious to me.
 A: Concerning the first question, let us denote the domains by $G_n$, and the outer boundary of $G_n$ by $S_n$, and let $u_n$ be the harmonic function on $G_n$ with boundary conditions $u_n = 1$ on $\partial \Omega$ and $u_n = 0$ on $S_n$.
Then by the maximum principle, we have $0 \leqslant u_n \leqslant 1$ on $G_n$. Consequently, $u_{n+1}(x) \geqslant 0$ for all $x \in S_n$. Thus $u_{n+1}\lvert_{G_n} - u_n$ is a harmonic function on $G_n$ with boundary values $0$ on $\partial \Omega$ and $\geqslant 0$ on $S_n$. Therefore $u_{n+1}\lvert_{G_n} - u_n \geqslant 0$ on $G_n$.
Thus for every $x \in G_n$, the sequence $(u_m(x))_{m \geqslant n}$ is monotonically increasing (not necessarily strictly), and bounded ($u_m(x) \leqslant 1$).
So $u(x) = \lim u_n(x)$ is well-defined for all $x \in \mathbb{R}^n\setminus \overline{\Omega}$. Use your favourite argument to conclude that $u$ is harmonic. Harnack's inequality shows that a monotonic sequence of harmonic functions converges locally uniformly, for example. Or use that the limit has the mean value property by the monotone convergence theorem.
It is clear that $u$ has boundary values $1$ on $\partial \Omega$. It remains to be seen that $\lim\limits_{\lVert x\rVert\to\infty} u(x) = 0$.
The credit for the argument for that goes to Tomás.
We choose an $r > 0$ such that $\Omega \subset B_r(0)$, and let $v(x) = \dfrac{r^{n-2}}{\lVert x\rVert^{n-2}}$. $v$ is harmonic and satisfies $v(x) = 1$ on $\partial B_r(0)$ and $v(\infty) = 0$. For $n$ large enough, $\overline{B_r(0)} \subset G_n$, and then $v - u_n$ is harmonic on $G_n\setminus \overline{B_r(0)}$ with boundary values $1 - u_n(x) \geqslant 0$ on $\partial B_r(0)$ and $v(x) - 0 \geqslant 0$ on $S_n$, hence $u_n \leqslant v$ on $G_n\setminus \overline{B_r(0)}$. It follows that $0 \leqslant u(x) \leqslant v(x)$ on $\mathbb{R}^n\setminus \overline{B_r(0)}$, and hence $\lim\limits_{\lVert x\rVert\to\infty} u(x) = 0$.
A: I have now obtained the answer for the second part from a friend.
$\int_V \nabla u.\nabla u\text{d}V= \int_V \nabla (u\nabla u)-u\nabla^2 u \text{d}V = \int_S u\nabla u.n\text{d}s-0=-\int_v \frac{\partial u}{\partial v}\text{d}s$
where $\nabla^2 u =0$ by harmonicity and the last line follws because the normals $n$ is pointing in the opposite direction to the definition.
