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  "body": "\n## Definition\n\nLet $U \\subseteq \\R^n$ be open.\nA function $u \\colon U \\to \\R$ is *locally integrable* if it is measurable and $\\int_K \\abs{u} \\, dx \u003c \\infty$ for every compact set $K \\subset U$.\nLet $C_c^\\infty(U)$ be the space of smooth functions on $U$ with compact support in $U$, the test functions.\nFor a multi-index $\\alpha = (\\alpha_1, \\ldots, \\alpha_n)$ write $\\abs{\\alpha} = \\alpha_1 + \\cdots + \\alpha_n$ and $\\partial^\\alpha = \\partial_1^{\\alpha_1} \\cdots \\partial_n^{\\alpha_n}$.\nLet $u$ and $v$ be locally integrable on $U$.\nThe function $v$ is the *$\\alpha$-th weak derivative* of $u$, written $D^\\alpha u = v$, if\n$$\\int_U u \\, \\partial^\\alpha \\varphi \\, dx = (-1)^{\\abs{\\alpha}} \\int_U v \\, \\varphi \\, dx \\quad \\text{for all } \\varphi \\in C_c^\\infty(U).$$\n\n## Remarks\n\nThe weak derivative, when it exists, is unique up to a set of measure zero.\nIf $u$ is of class $C^{\\abs{\\alpha}}$ on $U$, then its classical [[partial derivative|09EGC6F8PT]] $\\partial^\\alpha u$ is also its weak derivative, by integration by parts.\n\nExample: on $U = (-1, 1)$, the function $u(x) = \\abs{x}$ has weak derivative $v = \\sgn x$.\nFor $\\varphi \\in C_c^\\infty(-1, 1)$, integration by parts on $(-1, 0)$ and on $(0, 1)$ gives $\\int_{-1}^{1} \\abs{x} \\varphi'(x) \\, dx = \\int_{-1}^{0} \\varphi \\, dx - \\int_{0}^{1} \\varphi \\, dx = -\\int_{-1}^{1} \\sgn(x) \\varphi(x) \\, dx$, since the boundary terms at $0$ vanish.\n\nNon-example: on $(-1, 1)$, the function $\\sgn x$ has no weak derivative.\nFor any $\\varphi \\in C_c^\\infty(-1, 1)$ one finds $\\int \\sgn(x) \\varphi'(x) \\, dx = -2 \\varphi(0)$.\nA locally integrable $v$ with $\\int v \\varphi \\, dx = 2 \\varphi(0)$ for all $\\varphi$ would vanish almost everywhere on $(-1, 0)$ and on $(0, 1)$, and then the integral would be $0$, not $2 \\varphi(0)$.\nIn the language of distributions, the derivative of $\\sgn x$ is $2\\delta$, which is not a function.\n",
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