[proofplan] The strategy is to reduce the problem to Theorem 1097 by showing that the uniform gradient bound makes the family uniformly Lipschitz. For each $f \in \mathcal{F}$ and each pair $x, y \in U$, convexity of $U$ guarantees that the line segment $[x,y]$ lies in $U$, so the [Mean Value Inequality](/theorems/328) applies and yields $|f(x) - f(y)| \le M|x - y|$. This is a uniform Lipschitz condition with constant $L = M$, so [Equicontinuity of Lipschitz Families](/theorems/1097) gives uniform equicontinuity with modulus $\delta = \varepsilon / M$. [/proofplan] [step:Apply the Mean Value Inequality along line segments in the convex domain] Fix $f \in \mathcal{F}$ and two points $x, y \in U$. Since $U$ is convex, the line segment \begin{align*} [x, y] := \{x + t(y - x) : t \in [0,1]\} \end{align*} lies entirely in $U$. Define the auxiliary function \begin{align*} g_{x,y}: [0,1] &\to \mathbb{R} \\ t &\mapsto f(x + t(y - x)). \end{align*} Since $f \in C^1(U; \mathbb{R})$ and $[x, y] \subset U$, the function $g_{x,y}$ is continuously differentiable on $[0,1]$ by the chain rule, with \begin{align*} g_{x,y}'(t) = \nabla f(x + t(y - x)) \cdot (y - x). \end{align*} By the [Fundamental Theorem of Calculus](/theorems/632) applied to $g_{x,y}$ on $[0,1]$, \begin{align*} f(y) - f(x) = g_{x,y}(1) - g_{x,y}(0) = \int_0^1 g_{x,y}'(t) \, d\mathcal{L}^1(t) = \int_0^1 \nabla f(x + t(y - x)) \cdot (y - x) \, d\mathcal{L}^1(t). \end{align*} Taking absolute values and applying the Cauchy--Schwarz inequality on $\mathbb{R}^n$ to the dot product $\nabla f(z) \cdot (y - x)$ with $z = x + t(y - x)$, \begin{align*} |f(y) - f(x)| &\le \int_0^1 |\nabla f(x + t(y - x))| \cdot |y - x| \, d\mathcal{L}^1(t) \\ &\le \int_0^1 M \cdot |y - x| \, d\mathcal{L}^1(t) \\ &= M |y - x|, \end{align*} where the second inequality uses the hypothesis $|\nabla f(z)| \le M$ for all $z \in U$. Since $f \in \mathcal{F}$ was arbitrary, every member of $\mathcal{F}$ is Lipschitz with the same constant $M$. [guided] The goal of this step is to show that the pointwise gradient bound $|\nabla f| \le M$ on the convex domain $U$ produces a **uniform** Lipschitz estimate $|f(y) - f(x)| \le M|y - x|$ for all $f \in \mathcal{F}$. Why is convexity needed? Because we integrate $\nabla f$ along the line segment $[x,y]$, and the integral representation is valid only if the entire segment lies in $U$. Without convexity, two points $x, y \in U$ might be connected only by a curve that exits $U$, where $\nabla f$ is not controlled (or not defined). Fix $f \in \mathcal{F}$ and $x, y \in U$. Since $U$ is convex, $[x, y] \subset U$. Define the one-variable function \begin{align*} g_{x,y}: [0,1] &\to \mathbb{R} \\ t &\mapsto f(x + t(y - x)). \end{align*} The chain rule gives $g_{x,y}'(t) = \nabla f(x + t(y - x)) \cdot (y - x)$, and $g_{x,y}$ is $C^1$ on $[0,1]$ because $f \in C^1(U; \mathbb{R})$ and the path $t \mapsto x + t(y-x)$ maps $[0,1]$ into the open set $U$. By the [Fundamental Theorem of Calculus](/theorems/632), \begin{align*} f(y) - f(x) = \int_0^1 \nabla f(x + t(y - x)) \cdot (y - x) \, d\mathcal{L}^1(t). \end{align*} Taking absolute values and applying the Cauchy--Schwarz inequality $|a \cdot b| \le |a| |b|$ on $\mathbb{R}^n$ to the integrand, \begin{align*} |f(y) - f(x)| &\le \int_0^1 |\nabla f(x + t(y - x))| \cdot |y - x| \, d\mathcal{L}^1(t). \end{align*} The factor $|y - x|$ is constant in $t$, so it exits the integral. The hypothesis $|\nabla f(z)| \le M$ for all $z \in U$ gives $|\nabla f(x + t(y-x))| \le M$ for each $t \in [0,1]$ (since $x + t(y-x) \in U$). Therefore \begin{align*} |f(y) - f(x)| \le M |y - x| \int_0^1 1 \, d\mathcal{L}^1(t) = M|y - x|. \end{align*} This holds for every $f \in \mathcal{F}$ and every $x, y \in U$: the family $\mathcal{F}$ is uniformly $M$-Lipschitz. [/guided] [/step] [step:Conclude uniform equicontinuity via the Lipschitz family theorem] The previous step established that every $f \in \mathcal{F}$ satisfies $|f(x) - f(y)| \le M |x - y|$ for all $x, y \in U$, where $M$ is independent of $f$. Viewing $U$ with the Euclidean metric $d(x,y) = |x - y|$ and $\mathbb{R}$ with the standard metric, this is exactly the hypothesis of [Equicontinuity of Lipschitz Families](/theorems/1097) with Lipschitz constant $L = M$. That theorem yields uniform equicontinuity of $\mathcal{F}$ on $U$ with modulus $\delta = \varepsilon / M$. [/step]