# Is maximum principle valid in the case of non-smooth boundaries?

Let $U_1$ and $U_2$ be two bounded domains in $\mathbb{R}^n$ such that $U_1 \Subset U_2$. Note that we don't assume $\partial U_i$ to be smooth or Lipschitz, they may be very bad.

Denote $U=U_2 \backslash \bar{U_1}$, consider the Dirichlet problem $$\Delta u=0 \text{ on }U,$$ $$u=1 \text{ on } \partial U_1, \quad u=0 \text{ on } \partial U_2.$$ Let $$U^r_1=\{x \in \mathbb{R}^n: d(x,U_1)\leqslant r\},$$ $$g_r=\max\{ 1-\frac{d(x,U_1)}{r}, 0\},$$ an extension from $1,0$ function to a Lipschitz function defined on $\bar{U}$. By the standard argument, we can prove that there exists a function $u \in W^{1,2}(U)$ satisfying the above boundary value problem and $u-g_r \in W^{1,2}_0(U)$. For different $r$ and different extensions, the solutions remain unchanged, i.e. the solution is unique.

Now suppose $U_2 \Subset U_3$, consider the Dirichlet problem $$\Delta v=0 \text{ on } U_3 \backslash \bar{U}_1,$$ $$v=1 \text{ on } \partial U_1, \quad v=0 \text{ on } \partial U_3.$$ Do we have $v\geqslant u$ on $U$?

If $\partial U_i$ are smooth for $i=1,2,3$, then $u\in C(\bar{U})$, $v\in C(\bar{U_3} \backslash U_1)$. $u, v|_{\partial U_1}=1$, $u|_{\partial U_2}=0$ and $v|_{\partial U_3}=0$. Then $v-u \geqslant 0$ on $\partial U_1$ and $\partial U_2$, by the maximum principle, we have $v\geqslant u$. However, the boundaries we considered here are not smooth.

Yes, indeed, the maximum principle for subharmonic functions (hence for harmonic functions) is valid for domains, independently of the smoothness of their boundaries :

Let $u$ be a subharmonic function on a domain (i.e. open connected set) $D$ in $\mathbb{C}$ (or $\mathbb{R}^n$), if $\limsup_{z\to\zeta}u(z)\leq 0$ for all $\zeta\in\partial D$, then $u\leq 0$ on $D$,

See e.g. Theorem 2.3.1 in the book by T. Ransford, Potential theory in the complex plane.

Concerning existence (and uniqueness) of a solution to the Dirichlet problem, the following holds true (see Corollary 4.2.6 of the above-mentioned book):

Let $D$ be a domain with non-polar (intuitively not too small) boundary, and let $\phi:\partial D\to\mathbb{R}$ be a bounded function, continuous on $\partial D$. Then there exists a unique bounded harmonic function $h$ on $D$ such that $\lim_{z\to\zeta}h(z)=\phi(\zeta)$ for nearly every $\zeta\in\partial D$ (nearly meaning outside a set of capacity zero). If the domain is regular (with respect to the Dirichlet problem) the limit holds for all $\zeta\in\partial D$.

• Let $w=v-u$ on $U$. Since the boundaries are bad, we don't have $u(x)\to 1$ when $x \to \partial U_1$ and $u(x) \to 0$ when $x \to \partial U_2$. The same is for $v$. How can you get $\lim sup w(x) \ls 0$ when $x \to \partial U$? Commented Dec 27, 2016 at 13:12
• Sorry for typos. How can you get $\limsup_{x \to \partial U} \leqslant 0$. Commented Dec 27, 2016 at 13:15
• see the assertion I have added about existence of a solution to the Dirichlet problem. Commented Dec 27, 2016 at 13:32
• Let $w=u-v$, I want to prove that $w\leqslant 0$ on $U$. We know that $\lim_{z\to x} u=1$ and $\lim_{z\to x} v=1$ for nearly every $x\in \partial U_1$. $\lim_{z\to x} u=0$ for nearly every $x\in \partial U_2$ and $v\geqslant 0$ on $\partial U_2$. Then $\limsup_{z\to x} w \leqslant 0$ for \textbf{nearly} every $x\in \partial U$. It's not enough for the condition of maximum principle you mentioned. Commented Dec 28, 2016 at 12:38
• Actually, the fact that $\limsup_{z\to\zeta}w(z)\leq 0$ for nearly every $\zeta\in\partial D$ is sufficient to deduce that $w\leq 0$ on $D$, see the extended maximum principle, Theorem 3.6.9 of the book by Ransford. Commented Dec 28, 2016 at 17:35