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  "body": "\n## Definition\n\nLet $n \\ge 2$, let $\\Omega \\subseteq \\R^n$ be a bounded [[open set|02BH3MCBQB]] with [[boundary|05D5EGMFYV]] $\\partial\\Omega$, and let $f \\colon \\partial\\Omega \\to [-\\infty, +\\infty]$ be a function.\n\n1. The *upper class* $\\mathcal{U}_f$ is the set of functions $u$ on $\\Omega$ with these properties.\nOn each connected component of $\\Omega$, $u$ is [[superharmonic|X4V2S56YQQ]] or identically $+\\infty$.\nThe function $u$ is bounded below, and $\\liminf_{x \\to \\zeta,\\, x \\in \\Omega} u(x) \\ge f(\\zeta)$ for every $\\zeta \\in \\partial\\Omega$.\n2. The *lower class* $\\mathcal{L}_f$ is the set of functions $v$ on $\\Omega$ with these properties.\nOn each connected component of $\\Omega$, $v$ is [[subharmonic|Z1MHMGFNFW]] or identically $-\\infty$.\nThe function $v$ is bounded above, and $\\limsup_{x \\to \\zeta,\\, x \\in \\Omega} v(x) \\le f(\\zeta)$ for every $\\zeta \\in \\partial\\Omega$.\n3. The *upper solution* is $\\overline{H}_f(x) = \\inf\\set{u(x) : u \\in \\mathcal{U}_f}$ and the *lower solution* is $\\underline{H}_f(x) = \\sup\\set{v(x) : v \\in \\mathcal{L}_f}$, for $x \\in \\Omega$.\n\nThe function $f$ is *resolutive* if $\\overline{H}_f = \\underline{H}_f$ on $\\Omega$ and this common function is [[harmonic|0DEB0JMN0W]] on $\\Omega$, in particular finite.\nThe common function is then written $H_f$.\n\n## Remarks\n\nThe inequality $\\underline{H}_f \\le \\overline{H}_f$ always holds, by the maximum principle applied to $v - u$ for $v \\in \\mathcal{L}_f$ and $u \\in \\mathcal{U}_f$.\nWiener's theorem says that every continuous $f$ is resolutive, so $H_f$ is a generalised solution of the [[Dirichlet problem|8D56GQ4AG8]].\nFor a bounded domain, Brelot showed that $f$ is resolutive if and only if $f$ is integrable with respect to [[harmonic measure|CFD7BPM51W]].\nThen $H_f(x)$ is the integral of $f$ against the harmonic measure for $x$.\n\nExample: let $\\Omega$ be the unit disc in $\\R^2$ and $f$ continuous on the unit circle.\nThe Poisson integral $P[f]$ is harmonic in $\\Omega$ and tends to $f(\\zeta)$ at each boundary point $\\zeta$.\nSo $P[f]$ lies in both $\\mathcal{U}_f$ and $\\mathcal{L}_f$, which gives $\\overline{H}_f \\le P[f] \\le \\underline{H}_f$.\nWith $\\underline{H}_f \\le \\overline{H}_f$ this forces $\\overline{H}_f = \\underline{H}_f = P[f]$, which is harmonic, so $f$ is resolutive.\n\nNon-example: let $f \\equiv +\\infty$ on $\\partial\\Omega$.\nThe only member of $\\mathcal{U}_f$ is the constant $+\\infty$, so $\\overline{H}_f \\equiv +\\infty$.\nEvery constant $c$ lies in $\\mathcal{L}_f$, so $\\underline{H}_f \\equiv +\\infty$ too.\nThe two solutions agree but are not harmonic, so $f$ is not resolutive.\n",
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