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  "body": "\n## Proof\n\nWe compute the limit of the second difference $\\Delta(t) / t^2$ in two ways, with the mean value inequality.\n\n1. Let $U$, $f$ and $a$ be as in the statement. Let $E$ be the space of linear maps $\\mathbb{R}^n \\to \\mathbb{R}^m$, with the [[operator norm|0ZTEH055YF]]. So $\\|T w\\| \\le \\|T\\| \\, \\|w\\|$ for $T \\in E$ and $w \\in \\mathbb{R}^n$.\n2. All norms on the finite-dimensional space $E$ are [[equivalent|03MC70ZRNR]]. A change to an equivalent norm changes the quotient in the definition of the [[Fréchet derivative|08EVNYZF42]] by bounded factors only. So the hypothesis on $Df$ holds for the operator norm.\n3. Put $B = D(Df)(a)$, a linear map $\\mathbb{R}^n \\to E$. By the definition of the [[iterated derivative|0EY05Z9RPN]], $D^2 f(a)(u, v) = (B u)(v)$. This is linear in $u$ since $B$ is linear, and linear in $v$ since $B u \\in E$. So $D^2 f(a)$ is [[bilinear|0SDMPMCN43]].\n4. For $z$ with $a + z \\in U$, put $R(z) = Df(a + z) - Df(a) - B z \\in E$. By the [[little-o form of the derivative|0MN882VJ4P]], for each $\\varepsilon \u003e 0$ there is $\\delta \u003e 0$ with $\\|R(z)\\| \\le \\varepsilon \\|z\\|$ for $\\|z\\| \u003c \\delta$ and $a + z \\in U$.\n5. Since $U$ is [[open|01ZQQDCERP]], there is $\\rho \u003e 0$ such that the [[open ball|0DC3MPR7C9]] $B(a, \\rho)$ lies in $U$.\n6. Fix $u, v \\in \\mathbb{R}^n$ and put $K = \\|u\\| + \\|v\\| + 1$. For $0 \u003c t \u003c \\rho / K$, put $\\Delta(t) = f(a + t u + t v) - f(a + t u) - f(a + t v) + f(a)$. The four points lie in $B(a, \\rho)$, since $\\|t u + t v\\| \\le t K \u003c \\rho$.\n7. Fix such a $t$, and put $r = \\rho - t \\|u\\| \u003e 0$. Let $V$ be the open ball of radius $r$ about $0$. For $y \\in V$, the [[triangle inequality|0FJZMBX0ZZ]] gives $a + y \\in B(a, \\rho)$ and $a + t u + y \\in B(a, \\rho)$.\n8. Define $G \\colon V \\to \\mathbb{R}^m$ by $G(y) = f(a + t u + y) - f(a + y) - (B(t u))(y)$.\n9. Let $p \\in \\mathbb{R}^n$ with $p + y \\in U$. The map $y' \\mapsto f(p + y')$ has the same error quotient at $y$ as $f$ at $p + y$. So it is differentiable at $y$ with derivative $Df(p + y)$.\n10. The map $y \\mapsto (B(t u))(y)$ is linear, so its [[derivative|0CYYAF8AD2]] is $B(t u)$ at each point. By step 9 and the [[linearity of the derivative|0CXZ217QDP]], $G$ is differentiable on $V$ with $DG(y) = Df(a + t u + y) - Df(a + y) - B(t u)$.\n11. By step 4, $Df(a + t u + y) = Df(a) + B(t u + y) + R(t u + y)$ and $Df(a + y) = Df(a) + B y + R(y)$. As $B$ is linear, step 10 gives $DG(y) = R(t u + y) - R(y)$.\n12. We have $G(0) = f(a + t u) - f(a)$ and $G(t v) = f(a + t u + t v) - f(a + t v) - t^2 (B u)(v)$. So $G(t v) - G(0) = \\Delta(t) - t^2 (B u)(v)$.\n13. For $0 \\le s \\le 1$, $\\|s t v\\| \\le t \\|v\\| \u003c r$, because $t K \u003c \\rho$. So the segment from $0$ to $t v$ lies in $V$.\n14. Let $\\varepsilon \u003e 0$, take $\\delta$ from step 4, and suppose also $t \u003c \\delta / K$. Let $y$ be on the segment from $0$ to $t v$. Then $\\|y\\| \\le t \\|v\\| \u003c \\delta$ and $\\|t u + y\\| \\le t (\\|u\\| + \\|v\\|) \u003c \\delta$.\n15. By steps 4, 11 and 14, $\\|DG(y)\\| \\le \\varepsilon t (\\|u\\| + \\|v\\|) + \\varepsilon t \\|v\\| \\le 2 \\varepsilon t K$.\n16. By the [[mean value inequality|02CYB184RC]] for $G$ on $V$, with steps 13 and 15, $\\|G(t v) - G(0)\\| \\le 2 \\varepsilon t K \\cdot t \\|v\\| \\le 2 \\varepsilon K^2 t^2$.\n17. By steps 12 and 16, $\\big\\| \\Delta(t) / t^2 - (B u)(v) \\big\\| \\le 2 \\varepsilon K^2$ for each $t$ with $0 \u003c t \u003c \\min(\\rho, \\delta) / K$.\n18. The formula for $\\Delta(t)$ and the number $K$ do not change when $u$ and $v$ are swapped. So steps 6 to 17 with $u$ and $v$ swapped give $\\big\\| \\Delta(t) / t^2 - (B v)(u) \\big\\| \\le 2 \\varepsilon K^2$ for the same $t$.\n19. Fix one such $t$. By the triangle inequality and steps 17 and 18, $\\|(B u)(v) - (B v)(u)\\| \\le 4 \\varepsilon K^2$.\n20. This holds for each $\\varepsilon \u003e 0$, and $K$ does not depend on $\\varepsilon$. So $(B u)(v) = (B v)(u)$, that is, $D^2 f(a)(u, v) = D^2 f(a)(v, u)$.\n21. Now let $i, j \\in \\{1, \\dots, n\\}$, with standard basis vectors $e_i$ and $e_j$. For $y \\in U$, [[the derivative gives the directional derivatives|0A6PMR20N4]], so $\\partial f / \\partial x_j (y) = Df(y)(e_j)$.\n22. For real $t \\ne 0$ with $|t| \u003c \\rho$, step 21 and the definition of $R$ give $\\dfrac{\\partial_j f(a + t e_i) - \\partial_j f(a)}{t} - (B e_i)(e_j) = \\dfrac{R(t e_i)(e_j)}{t}$.\n23. By steps 1 and 4, the norm of the right side is at most $\\|R(t e_i)\\| / |t| \\le \\varepsilon$ when $|t| \u003c \\min(\\rho, \\delta)$. So $\\partial_i (\\partial_j f)(a)$ exists and equals $(B e_i)(e_j)$.\n24. By the definition of the [[second partial derivative|02RSXVM1ZN]], applied to each component, $\\dfrac{\\partial^2 f}{\\partial x_i \\, \\partial x_j}(a) = D^2 f(a)(e_i, e_j)$. In the same way, $\\dfrac{\\partial^2 f}{\\partial x_j \\, \\partial x_i}(a) = D^2 f(a)(e_j, e_i)$.\n25. By step 20 with $u = e_i$ and $v = e_j$, the two mixed partial derivatives at $a$ are equal.\n\n## Idea\n\nThe second difference $\\Delta(t)$ is symmetric in $u$ and $v$ by its formula. We show that $\\Delta(t) / t^2$ tends to $D^2 f(a)(u, v)$, by viewing $\\Delta(t)$ as the change of an auxiliary map $G$ along the segment from $0$ to $t v$. The derivative of $G$ is only a difference of two error terms of $Df$, so the mean value inequality makes the remaining error small. Swapping $u$ and $v$ gives the same limit, which forces the symmetry.\n",
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