[proofplan] We prove the contraction estimate directly along the line segment joining two arbitrary points of $U$. Convexity ensures that this segment stays inside $U$, so the derivative bound applies at every point of the segment. Applying the one-dimensional [fundamental theorem of calculus](/theorems/632) componentwise to the curve $t \mapsto f((1-t)x+ty)$ expresses $f(y)-f(x)$ as an integral of derivatives, and the operator norm bound estimates that integral by $c|x-y|$. [/proofplan] [step:Parametrize the line segment inside the convex domain] Fix arbitrary points $x,y \in U$. Define the line-segment map \begin{align*} \gamma:[0,1]\to U \end{align*} by $\gamma(t)=(1-t)x+ty$. Since $U$ is convex and $x,y \in U$, we have $\gamma(t) \in U$ for every $t \in [0,1]$. Define also the composed curve \begin{align*} \varphi:[0,1]\to \mathbb{R}^n \end{align*} by $\varphi(t)=f(\gamma(t))$. Since $\gamma([0,1])\subset U$ and $U$ is open, for each $t_0\in[0,1]$ the map $f\circ\gamma$ is defined and differentiable on a neighbourhood of $t_0$ relative to $[0,1]$, with one-sided derivatives at the endpoints. Since $f\in C^1(U;\mathbb{R}^n)$ and $\gamma$ is affine, these derivatives depend continuously on $t$, so $\varphi$ is a $C^1$ map on $[0,1]$. For every $t \in [0,1]$, the chain rule gives \begin{align*} \varphi'(t)=\mathrm{d}f_{\gamma(t)}(y-x). \end{align*} [/step] [step:Represent the difference $f(y)-f(x)$ by integrating the derivative along the segment] Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. By the one-dimensional Fundamental Theorem of Calculus for $C^1$ functions, applied componentwise to the $C^1$ curve $\varphi:[0,1]\to\mathbb{R}^n$, we obtain \begin{align*} f(y)-f(x)=\varphi(1)-\varphi(0)=\int_0^1 \varphi'(t)\,d\mathcal{L}^1(t). \end{align*} Using the formula for $\varphi'(t)$ from the previous step, this becomes \begin{align*} f(y)-f(x)=\int_0^1 \mathrm{d}f_{\gamma(t)}(y-x)\,d\mathcal{L}^1(t). \end{align*} [guided] The goal is to turn the pointwise derivative bound into a global estimate between $f(x)$ and $f(y)$. The natural path from $x$ to $y$ is the straight line $\gamma(t)=(1-t)x+ty$, because convexity guarantees that this entire path remains in $U$. Thus the derivative bound on $Df_z$ is available at every point $z=\gamma(t)$. We define the curve \begin{align*} \varphi:[0,1]\to\mathbb{R}^n \end{align*} by $\varphi(t)=f(\gamma(t))$. Since $\gamma([0,1])\subset U$ and $U$ is open, the composition $f\circ\gamma$ is differentiable on a neighbourhood of each parameter value relative to $[0,1]$, with one-sided derivatives at $0$ and $1$. Since $f$ is $C^1$ on $U$ and $\gamma$ is affine, these derivatives vary continuously with $t$, so $\varphi$ is $C^1$. The chain rule gives, for each $t \in [0,1]$, \begin{align*} \varphi'(t)=\mathrm{d}f_{\gamma(t)}(y-x). \end{align*} Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. Now apply the one-dimensional Fundamental Theorem of Calculus componentwise to the $C^1$ map $\varphi:[0,1]\to\mathbb{R}^n$. This gives \begin{align*} \varphi(1)-\varphi(0)=\int_0^1 \varphi'(t)\,d\mathcal{L}^1(t). \end{align*} Since $\varphi(1)=f(y)$ and $\varphi(0)=f(x)$, substituting the chain-rule expression for $\varphi'(t)$ yields \begin{align*} f(y)-f(x)=\int_0^1 \mathrm{d}f_{\gamma(t)}(y-x)\,d\mathcal{L}^1(t). \end{align*} This identity is the bridge between the infinitesimal hypothesis $\|Df_z\|_{\mathrm{op}}\le c$ and the desired global Lipschitz estimate. [/guided] [/step] [step:Estimate the integral using the operator norm bound] Taking Euclidean norms and applying the triangle inequality for the vector-valued integral gives \begin{align*} |f(y)-f(x)|\le \int_0^1 |Df_{\gamma(t)}(y-x)|\,d\mathcal{L}^1(t). \end{align*} For each $t \in [0,1]$, the definition of the operator norm gives \begin{align*} |\mathrm{d}f_{\gamma(t)}(y-x)|\le \|\mathrm{d}f_{\gamma(t)}\|_{\mathrm{op}}\,|y-x|. \end{align*} The hypothesis applies because $\gamma(t)\in U$, so \begin{align*} |\mathrm{d}f_{\gamma(t)}(y-x)|\le c|y-x|. \end{align*} Therefore \begin{align*} |f(y)-f(x)|\le \int_0^1 c|y-x|\,d\mathcal{L}^1(t)=c|y-x|. \end{align*} [/step] [step:Conclude that the self-map is a contraction] The estimate \begin{align*} |f(x)-f(y)|\le c|x-y| \end{align*} holds for the arbitrary points $x,y \in U$. Since $f(U)\subset U$, the map $f$ is a self-map \begin{align*} f:U\to U. \end{align*} Since $c\in[0,1)$, this is precisely the contraction estimate for the Euclidean metric on $U$. Hence $f:U\to U$ is a contraction. [/step]