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 \begin{equation} \Delta u=0 \text{ on }U, \end{equation} \begin{equation} u=1 \text{ on } \partial U_1, \quad u=0 \text{ on } \partial U_2. \end{equation} 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 \begin{equation} \Delta v=0 \text{ on } U_3 \backslash \bar{U}_1, \end{equation} \begin{equation} v=1 \text{ on } \partial U_1, \quad v=0 \text{ on } \partial U_3. \end{equation} 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.


1 Answer 1


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 $\zeta\in\partial D$.

  • $\begingroup$ 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 $x \to \partial U$? $\endgroup$
    – user101829
    Commented Dec 27, 2016 at 13:12
  • $\begingroup$ Sorry for typos. How can you get $\limsup_{x \to \partial U} \leqslant 0$. $\endgroup$
    – user101829
    Commented Dec 27, 2016 at 13:15
  • $\begingroup$ see the assertion I have added about existence of a solution to the Dirichlet problem. $\endgroup$
    – user111
    Commented Dec 27, 2016 at 13:32
  • $\begingroup$ 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. $\endgroup$
    – user101829
    Commented Dec 28, 2016 at 12:38
  • $\begingroup$ 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. $\endgroup$
    – user111
    Commented Dec 28, 2016 at 17:35

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