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  "body": "\n## Proof\n\nWe solve the constraints for $m$ of the variables, and reduce to an extremum without constraints.\n\n1. Let $U$, $m \u003c n$, $f$, $g$, $M$ and $a$ be as in the statement. The functions $f$ and $g$ are of [[class C^1|0TCBZMEYGN]], so they are [[differentiable|0KXSK8TEEB]] on $U$.\n2. Let $J = Dg(a)$, the $m \\times n$ [[Jacobian matrix|0YF0VXKXGB]]. Its rows are the [[gradients|0NS2QYVC4Y]] $\\nabla g_1(a), \\dots, \\nabla g_m(a)$, which are [[linearly independent|0EVHJFTG9Z]]. So the row space of $J$ has dimension $m$. Since [[row rank equals column rank|0TZAG60ZYT]], the columns of $J$ span $\\R^m$, and some $m$ of them are linearly independent.\n3. Reordering the coordinates of $\\R^n$ reorders the entries of every gradient in the same way. So it does not change the statement. We may therefore assume that the last $m$ columns of $J$ are linearly independent.\n4. Write $x = (u, w)$ with $u \\in \\R^{n-m}$ and $w \\in \\R^m$, and $a = (a', b)$. Write $D_u$ and $D_w$ for the matrices of partial derivatives with respect to $u$ and $w$. Then $D_w g(a)$ is the $m \\times m$ matrix of the last $m$ columns of $J$, so it is invertible.\n5. Since $g(a) = 0$, the [[implicit function theorem|0BXBADR94C]] gives open sets $U' \\ni a'$ and $V' \\ni b$, and a $C^1$ function $\\varphi \\colon U' \\to V'$, with this property. For $(u, w) \\in U' \\times V'$, $g(u, w) = 0$ if and only if $w = \\varphi(u)$. Here $U' \\times V' \\subseteq U$, since $g$ is evaluated on this set.\n6. The theorem also gives $D\\varphi(a') = -\\big( D_w g(a) \\big)^{-1} D_u g(a)$. Since $(a', b) \\in U' \\times V'$ and $g(a', b) = 0$, we have $\\varphi(a') = b$.\n7. Define $\\psi(u) = (u, \\varphi(u))$ for $u \\in U'$. Then $\\psi(u) \\in M$ for every $u \\in U'$, and $\\psi(a') = a$.\n8. Assume that $f$ restricted to $M$ has a [[local maximum|04CPG9K1C4]] at $a$. So there is $\\varepsilon \u003e 0$ with $f(x) \\le f(a)$ for $x \\in M$ with $\\norm{x - a} \u003c \\varepsilon$.\n9. The map $\\psi$ is [[continuous|0S9EK5MQPE]] at $a'$. So there is $\\delta \u003e 0$ with $\\norm{\\psi(u) - a} \u003c \\varepsilon$ for $u \\in U'$ with $\\norm{u - a'} \u003c \\delta$. For such $u$, step 8 gives $f(\\psi(u)) \\le f(a) = f(\\psi(a'))$.\n10. Thus $f \\circ \\psi$ has a local maximum at $a'$, on the open set $U'$. The case of a local minimum is the same, with the inequalities reversed.\n11. By the [[chain rule|0EAAS2EHDR]], $f \\circ \\psi$ is differentiable at $a'$, with $D(f \\circ \\psi)(a') = D_u f(a) + D_w f(a) \\, D\\varphi(a')$.\n12. By the theorem that [[the derivative vanishes at a local extremum|0F405477F8]], this derivative is $0$. With step 6, $D_u f(a) = D_w f(a) \\big( D_w g(a) \\big)^{-1} D_u g(a)$.\n13. Define the row vector $\\lambda = D_w f(a) \\big( D_w g(a) \\big)^{-1}$, with entries $\\lambda_1, \\dots, \\lambda_m$. Then $D_w f(a) = \\lambda \\, D_w g(a)$, and step 12 says $D_u f(a) = \\lambda \\, D_u g(a)$.\n14. Together, $Df(a) = \\lambda \\, Dg(a)$ as $1 \\times n$ matrices. The row $\\lambda \\, Dg(a)$ is $\\sum_i \\lambda_i \\nabla g_i(a)$. So $\\nabla f(a) = \\lambda_1 \\nabla g_1(a) + \\dots + \\lambda_m \\nabla g_m(a)$.\n15. If $\\mu \\in \\R^m$ also satisfies this equation, then $\\sum_i (\\lambda_i - \\mu_i) \\nabla g_i(a) = 0$. The gradients are linearly independent, so $\\mu = \\lambda$. This proves uniqueness.\n\n## Idea\n\nNear $a$, the constraint set is the graph $w = \\varphi(u)$, by the implicit function theorem. On this graph, $f(u, \\varphi(u))$ is a function of $u$ alone with a free local extremum, so its derivative is zero. The formula for $D\\varphi$ then shows that $Df(a)$ is a combination of the rows of $Dg(a)$.\n",
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