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  "body": "\n## Proof\n\nWe show that $(A - B) v = 0$ for each $v \\in V$.\n\n1. Let $V$, $W$, $U$, $f$, $x$, $A$ and $B$ be as in the statement. For $h \\in V$ with $x + h \\in U$, put $r_A(h) = f(x + h) - f(x) - A h$ and $r_B(h) = f(x + h) - f(x) - B h$.\n2. Subtracting the two definitions gives $(A - B) h = r_B(h) - r_A(h)$.\n3. Let $\\varepsilon \u003e 0$. Both $A$ and $B$ are [[Fréchet derivatives|08EVNYZF42]] of $f$ at $x$. By the [[little-o form of the derivative|0MN882VJ4P]], there is $\\delta \u003e 0$ with $\\|r_A(h)\\| \\le \\varepsilon \\|h\\|$ and $\\|r_B(h)\\| \\le \\varepsilon \\|h\\|$. This holds for $\\|h\\| \u003c \\delta$ with $x + h \\in U$. We take $\\delta$ as the smaller of the two numbers given by $A$ and by $B$.\n4. Since $U$ is [[open|01ZQQDCERP]], there is $\\rho \u003e 0$ such that the [[open ball|0DC3MPR7C9]] $B(x, \\rho)$ lies in $U$.\n5. Let $v \\in V$. If $v = 0$, then $(A - B) v = 0$. So suppose that $v \\ne 0$. Fix $\\varepsilon \u003e 0$, and take $\\delta$ as in step 3.\n6. Choose a real $t \u003e 0$ with $t \\|v\\| \u003c \\min(\\delta, \\rho)$, and put $h = t v$. Then $\\|h\\| \u003c \\rho$, so $x + h \\in U$. Also $\\|h\\| \u003c \\delta$.\n7. By steps 2 and 3 and the [[triangle inequality|0FJZMBX0ZZ]], $t \\|(A - B) v\\| = \\|(A - B) h\\| \\le \\|r_B(h)\\| + \\|r_A(h)\\| \\le 2 \\varepsilon t \\|v\\|$.\n8. Dividing by $t \u003e 0$ gives $\\|(A - B) v\\| \\le 2 \\varepsilon \\|v\\|$.\n9. This holds for each $\\varepsilon \u003e 0$, so $\\|(A - B) v\\| = 0$ and $A v = B v$.\n10. As $v$ was arbitrary, $A = B$.\n\n## Idea\n\nThe difference of two good linear approximations is a linear map whose values are tiny compared with $\\|h\\|$. Scaling a fixed vector $v$ down to a short step $t v$ and back up shows that this linear map vanishes on $v$. Openness of $U$ is what lets us step from $x$ in every direction.\n",
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