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  "body": "\n## Definition\n\nLet $U \\subset \\R^n$ be a bounded open set with $C^1$ boundary $\\partial U$, carrying its surface measure, and let $1 \\le p \u003c \\infty$.\nThe *trace operator* is the unique bounded linear map\n$$T \\colon W^{1,p}(U) \\to L^p(\\partial U)$$\nsuch that $T u = u|_{\\partial U}$ for every $u \\in W^{1,p}(U)$ that is continuous on $\\overline{U}$.\nHere $W^{1,p}(U)$ is the [[Sobolev space|M4CEZ5K9HP]].\nThe function $T u$ is the *trace* of $u$ on $\\partial U$.\n\n## Remarks\n\nThe operator exists and is unique.\nFor $u$ smooth up to the boundary, the divergence theorem gives $\\norm{u|_{\\partial U}}_{L^p(\\partial U)} \\le C \\norm{u}_{W^{1,p}(U)}$ with $C$ depending on $p$ and $U$.\nSuch functions are dense in $W^{1,p}(U)$, so restriction extends uniquely to a [[continuous linear map|03TM135G0N]] on all of $W^{1,p}(U)$.\n\nExample: on $U = (0, 1)$ the boundary is $\\set{0, 1}$, and every $u \\in W^{1,p}(0, 1)$ has a continuous representative.\nIts trace is the pair of values $(u(0), u(1))$ of that representative.\n\nNon-example: on $U = (0, 1)$, restriction to the boundary is not bounded for the $L^p$ norm alone.\nThe functions $u_k(x) = (1 - x)^k$ satisfy $u_k(0) = 1$ while $\\norm{u_k}_{L^p(0,1)} \\to 0$, so no bounded map $L^p(0, 1) \\to \\R^2$ extends restriction.\nThis is why the definition uses $W^{1,p}(U)$.\n\nThe kernel of $T$ is $W^{1,p}_0(U)$.\nFor $1 \u003c p \u003c \\infty$, the image of $T$ is the [[fractional Sobolev space|N6NX1VT837]] $W^{1 - 1/p, p}(\\partial U)$, a result of Gagliardo.\n",
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