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  "body": "\n## Proof\n\nWe apply the mean value theorem twice to a double difference.\n\n1. Let $U \\subseteq \\R^n$ be open, let $f \\colon U \\to \\R$ be of [[class C^2|0TCBZMEYGN]], and let $a \\in U$. Let $i, j \\in \\set{1, \\dots, n}$.\n2. If $i = j$, the two sides of the claim are the same [[second partial derivative|02RSXVM1ZN]], so there is nothing to prove. Assume $i \\ne j$.\n3. Write $\\partial_l$ for the [[partial derivative|05AXG3QRKR]] in the direction $x_l$, and $e_l$ for the $l$-th standard basis vector. So $\\frac{\\partial^2 f}{\\partial x_j \\, \\partial x_i} = \\partial_j \\partial_i f$.\n4. Since $U$ is open, there is $r \u003e 0$ such that the open ball of radius $2r$ around $a$ lies in $U$.\n5. For real $s, t$ with $\\abs{s} \u003c r$ and $\\abs{t} \u003c r$, the point $a + \\sigma e_i + \\tau e_j$ lies in that ball whenever $\\abs{\\sigma} \\le \\abs{s}$ and $\\abs{\\tau} \\le \\abs{t}$. This holds because its distance to $a$ is at most $\\abs{\\sigma} + \\abs{\\tau} \u003c 2r$.\n6. For such $s, t$ define the double difference $\\Delta(s, t) = f(a + s e_i + t e_j) - f(a + s e_i) - f(a + t e_j) + f(a)$.\n7. Fix $s, t$ with $0 \u003c \\abs{s} \u003c r$ and $0 \u003c \\abs{t} \u003c r$. Define $u(\\sigma) = f(a + \\sigma e_i + t e_j) - f(a + \\sigma e_i)$ for $\\abs{\\sigma} \u003c r$. Then $\\Delta(s, t) = u(s) - u(0)$.\n8. The partial derivative $\\partial_i f$ exists on $U$, so $u$ is differentiable on $(-r, r)$, with $u'(\\sigma) = \\partial_i f(a + \\sigma e_i + t e_j) - \\partial_i f(a + \\sigma e_i)$. In particular $u$ is continuous there.\n9. By the [[mean value theorem|0BARQY52N5]] on the interval between $0$ and $s$, there is $\\sigma_1$ strictly between $0$ and $s$ with $\\Delta(s, t) = s \\, u'(\\sigma_1)$.\n10. Define $v(\\tau) = \\partial_i f(a + \\sigma_1 e_i + \\tau e_j)$ for $\\abs{\\tau} \u003c r$. Then $u'(\\sigma_1) = v(t) - v(0)$.\n11. The function $\\partial_j \\partial_i f$ exists on $U$, so $v$ is differentiable with $v'(\\tau) = \\partial_j \\partial_i f(a + \\sigma_1 e_i + \\tau e_j)$.\n12. By the mean value theorem again, there is $\\tau_1$ strictly between $0$ and $t$ with $v(t) - v(0) = t \\, \\partial_j \\partial_i f(p)$, where $p = a + \\sigma_1 e_i + \\tau_1 e_j$.\n13. Steps 9 and 12 give $\\Delta(s, t) = s t \\, \\partial_j \\partial_i f(p)$.\n14. The expression $\\Delta(s, t)$ does not change if we exchange the roles of $(i, s)$ and $(j, t)$. So the same argument, with these roles exchanged, gives a point $q = a + \\sigma_2 e_i + \\tau_2 e_j$ with $\\abs{\\sigma_2} \u003c \\abs{s}$, $\\abs{\\tau_2} \u003c \\abs{t}$ and $\\Delta(s, t) = s t \\, \\partial_i \\partial_j f(q)$.\n15. Since $s t \\ne 0$, steps 13 and 14 give $\\partial_j \\partial_i f(p) = \\partial_i \\partial_j f(q)$.\n16. The points $p$ and $q$ depend on $s$ and $t$, and both satisfy $\\norm{p - a} \\le \\abs{s} + \\abs{t}$ and $\\norm{q - a} \\le \\abs{s} + \\abs{t}$.\n17. Take $s = t = 1/k$ for integers $k \u003e 1/r$. Then the points $p_k$ and $q_k$ tend to $a$ as $k \\to \\infty$.\n18. The functions $\\partial_j \\partial_i f$ and $\\partial_i \\partial_j f$ are [[continuous|0S9EK5MQPE]] at $a$, because $f$ is of class $C^2$. So $\\partial_j \\partial_i f(p_k) \\to \\partial_j \\partial_i f(a)$ and $\\partial_i \\partial_j f(q_k) \\to \\partial_i \\partial_j f(a)$.\n19. By step 15 the two sequences are equal, so their limits are equal: $\\frac{\\partial^2 f}{\\partial x_j \\, \\partial x_i}(a) = \\frac{\\partial^2 f}{\\partial x_i \\, \\partial x_j}(a)$.\n\n## Idea\n\nThe double difference $\\Delta(s, t)$ is symmetric in the two directions. Two uses of the mean value theorem write it as $s t$ times a mixed partial at a nearby point, once in each order. Dividing by $s t$ and letting $s, t \\to 0$, continuity of the mixed partials makes the two values meet at $a$.\n",
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