Perron–Wiener–Brelot solution

Remarks

Always $\underline{H}_f \le \overline{H}_f$, by the maximum principle applied to $u - v$ for $u \in \mathcal{U}_f$ and $v \in \mathcal{L}_f$. Each of $\overline{H}_f$ and $\underline{H}_f$ is harmonic, identically $+\infty$, or identically $-\infty$ on each connected component of $\Omega$. Example: let $f$ be continuous on $\partial\Omega$, and let $u$ be continuous on $\overline{\Omega}$, harmonic on $\Omega$ and equal to $f$ on $\partial\Omega$. Then $u$ lies in both classes, so $\overline{H}_f \le u \le \underline{H}_f$ and $H_f = u$. Non-example: let $f \equiv +\infty$ on $\partial\Omega$. By the minimum principle, a superharmonic function in $\mathcal{U}_f$ would be at least every constant $M$. So $\mathcal{U}_f$ contains only the constant $+\infty$, and $\overline{H}_f \equiv +\infty$ is not harmonic; hence $f$ is not resolutive. For $\Omega = \set{0 < \abs{x} < 1}$ in $\R^2$, let $f = 0$ on $\abs{x} = 1$ and $f(0) = 1$. Then $f$ is resolutive with $H_f = 0$, which does not attain the value $f(0)$. So the PWB solution need not solve the classical Dirichlet problem. By Wiener's theorem every continuous $f$ is resolutive.