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  "body": "\n## Proof\n\nWe apply the one-variable mean value theorem to the real function $\\varphi(t) = w \\cdot f(x + t (y - x))$, where $w = f(y) - f(x)$.\n\n1. Let $U$, $f$, $x$, $y$ and $M$ be as in the statement. Put $w = f(y) - f(x)$. If $w = 0$, the inequality holds, since its right side is $\\ge 0$. From now on, suppose that $w \\ne 0$.\n2. Let $z \\in U$ and $L = Df(z)$. By [[linear maps on finite-dimensional spaces are continuous|0C13E623WD]] and [[bounded linear maps are continuous|0F3K4C7XYW]], there is $C \\ge 0$ with $\\|L h\\| \\le C \\|h\\|$ for each $h$.\n3. Since $f$ is [[differentiable|0135XZZQ8M]] at $z$, there is $\\delta \u003e 0$ with $\\|f(z + h) - f(z) - L h\\| \\le \\|h\\|$ for $\\|h\\| \u003c \\delta$ and $z + h \\in U$. By the [[triangle inequality|0FJZMBX0ZZ]], $\\|f(z + h) - f(z)\\| \\le (C + 1) \\|h\\|$ for these $h$. So $f$ is continuous at $z$.\n4. Define $\\gamma(t) = x + t (y - x)$ for $t \\in [0, 1]$. By hypothesis $\\gamma(t) \\in U$. Also $\\|\\gamma(t) - \\gamma(t')\\| = |t - t'| \\, \\|y - x\\|$, so $\\gamma$ is continuous.\n5. Put $\\varphi(t) = w \\cdot f(\\gamma(t))$ for $t \\in [0, 1]$, with the dot product of [[Euclidean space|0XPR1E85W0]]. By the [[Cauchy–Schwarz inequality|0RFWW56Y22]], $|w \\cdot v - w \\cdot v'| \\le \\|w\\| \\, \\|v - v'\\|$, so $v \\mapsto w \\cdot v$ is continuous.\n6. By steps 3, 4 and 5, $\\varphi$ is a [[composition of continuous maps|05NK26G0DR]]. So $\\varphi$ is continuous on $[0, 1]$.\n7. For $t \\in (0, 1)$ and small $s \\ne 0$, $(\\gamma(t + s) - \\gamma(t)) / s = y - x$. So $\\gamma$ is differentiable on $(0, 1)$ with $\\gamma'(t) = y - x$.\n8. By the [[chain rule for a composition with a curve|0A8TJ0VHC7]] on the open interval $(0, 1)$, $f \\circ \\gamma$ is differentiable at each $t \\in (0, 1)$. Its derivative is $(f \\circ \\gamma)'(t) = Df(\\gamma(t))(y - x)$.\n9. For $t \\in (0, 1)$ and small $s \\ne 0$, $(\\varphi(t + s) - \\varphi(t)) / s = w \\cdot q(s)$, where $q(s) = ((f \\circ \\gamma)(t + s) - (f \\circ \\gamma)(t)) / s$. By steps 5 and 8, this tends to $w \\cdot Df(\\gamma(t))(y - x)$.\n10. So $\\varphi$ is differentiable on $(0, 1)$ with $\\varphi'(t) = w \\cdot Df(\\gamma(t))(y - x)$.\n11. By the [[mean value theorem|0BARQY52N5]], there is $c \\in (0, 1)$ with $\\varphi(1) - \\varphi(0) = \\varphi'(c)$.\n12. The left side is $w \\cdot f(y) - w \\cdot f(x) = w \\cdot w = \\|w\\|^2$. By step 10 and the Cauchy–Schwarz inequality, $\\|w\\|^2 \\le \\|w\\| \\, \\|Df(\\gamma(c))(y - x)\\|$.\n13. Let $T = Df(\\gamma(c))$ and $v = y - x$. If $v \\ne 0$, then $v / \\|v\\|$ has norm $1$. By the definition of the [[operator norm|0ZTEH055YF]], $\\|T v\\| = \\|v\\| \\, \\|T(v / \\|v\\|)\\| \\le \\|T\\| \\, \\|v\\|$. If $v = 0$, both sides are $0$.\n14. The point $\\gamma(c)$ lies on the segment $[x, y]$, so $\\|T\\| \\le M$ by hypothesis. Steps 12 and 13 give $\\|w\\|^2 \\le \\|w\\| \\, M \\, \\|y - x\\|$.\n15. Dividing by $\\|w\\| \u003e 0$ gives $\\|f(y) - f(x)\\| \\le M \\, \\|y - x\\|$.\n\n## Idea\n\nThe mean value theorem has no equality form for vector-valued maps, but it does apply to one real component. The trick is to take the component along the direction $w = f(y) - f(x)$ of the total change. The change of that component is $\\|w\\|^2$, and Cauchy–Schwarz bounds its derivative by $\\|w\\|$ times $M \\|y - x\\|$.\n",
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