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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* $\\mathcal{U}_f$ consists of the functions $u$ on $\\Omega$, [[superharmonic|X4V2S56YQQ]] or identically $+\\infty$ and bounded below, with $\\liminf_{x \\to \\xi} u(x) \\ge f(\\xi)$ for all $\\xi \\in \\partial\\Omega$.\n2. The *lower class* $\\mathcal{L}_f$ consists of the functions $v$ on $\\Omega$, [[subharmonic|Z1MHMGFNFW]] or identically $-\\infty$ and bounded above, with $\\limsup_{x \\to \\xi} v(x) \\le f(\\xi)$ for all $\\xi \\in \\partial\\Omega$.\n3. The *upper solution* is $\\overline{H}_f = \\inf_{u \\in \\mathcal{U}_f} u$ and the *lower solution* is $\\underline{H}_f = \\sup_{v \\in \\mathcal{L}_f} v$, using the [[infimum|04KFTA34C5]] and [[supremum|0BTMQ04556]] pointwise.\n\nThe function $f$ is *resolutive* if $\\overline{H}_f = \\underline{H}_f$ and this function is harmonic on $\\Omega$.\nIn that case the *Perron–Wiener–Brelot solution* of $f$ is $H_f = \\overline{H}_f = \\underline{H}_f$.\n\n## Remarks\n\nAlways $\\underline{H}_f \\le \\overline{H}_f$, by the maximum principle applied to $u - v$ for $u \\in \\mathcal{U}_f$ and $v \\in \\mathcal{L}_f$.\nEach of $\\overline{H}_f$ and $\\underline{H}_f$ is harmonic, identically $+\\infty$, or identically $-\\infty$ on each connected component of $\\Omega$.\nExample: let $f$ be continuous on $\\partial\\Omega$, and let $u$ be continuous on $\\overline{\\Omega}$, harmonic on $\\Omega$ and equal to $f$ on $\\partial\\Omega$.\nThen $u$ lies in both classes, so $\\overline{H}_f \\le u \\le \\underline{H}_f$ and $H_f = u$.\nNon-example: let $f \\equiv +\\infty$ on $\\partial\\Omega$.\nBy the minimum principle, a superharmonic function in $\\mathcal{U}_f$ would be at least every constant $M$.\nSo $\\mathcal{U}_f$ contains only the constant $+\\infty$, and $\\overline{H}_f \\equiv +\\infty$ is not harmonic; hence $f$ is not resolutive.\nFor $\\Omega = \\set{0 \u003c \\abs{x} \u003c 1}$ in $\\R^2$, let $f = 0$ on $\\abs{x} = 1$ and $f(0) = 1$.\nThen $f$ is resolutive with $H_f = 0$, which does not attain the value $f(0)$.\nSo the PWB solution need not solve the classical Dirichlet problem.\nBy Wiener's theorem every continuous $f$ is resolutive.\n",
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