[proofplan] We compare the values of $f$ at two points $x,y \in K$ by restricting $f$ to the line segment from $x$ to $y$. Convexity keeps that whole segment inside $K$, while compactness and continuity of the derivative make the derivative norm uniformly bounded on $K$. Applying the one-dimensional [fundamental theorem of calculus](/theorems/632) componentwise to the curve $t \mapsto f(x+t(y-x))$ gives an integral formula, and the operator norm bound estimates the integrand uniformly. [/proofplan]

[step:Show that the derivative norm has a finite maximum on $K$] Define the function \begin{align*} M_K:K \to [0,\infty) \end{align*} by \begin{align*} M_K(z)=\|Df_z\|_{\mathrm{op}}. \end{align*} Let $\mathcal L(\mathbb R^n,\mathbb R^m)$ denote the [normed vector space](/page/Normed%20Vector%20Space) of linear maps from $\mathbb R^n$ to $\mathbb R^m$, equipped with the operator norm induced by the Euclidean norms. Since $f \in C^1(U;\mathbb R^m)$, the derivative map \begin{align*} Df:U \to \mathcal L(\mathbb R^n,\mathbb R^m) \end{align*} is continuous. Define the operator norm map \begin{align*} N:\mathcal L(\mathbb R^n,\mathbb R^m) \to [0,\infty) \end{align*} by $N(T)=\|T\|_{\mathrm{op}}$. The map $N$ is continuous, so $M_K=N\circ Df|_K$ is continuous on $K$. Because $K$ is compact and nonempty, $M_K$ attains a finite maximum. Set \begin{align*} M=\sup_{z \in K}\|Df_z\|_{\mathrm{op}}=\max_{z \in K}M_K(z)<\infty. \end{align*} [/step]

[step:Parametrize the segment from $x$ to $y$ inside $K$]Fix $x,y \in K$. Define the affine path \begin{align*} \gamma:[0,1]\to K \end{align*} by \begin{align*} \gamma(t)=x+t(y-x). \end{align*} This is well-defined because $K$ is convex: for each $t \in [0,1]$, \begin{align*} \gamma(t)=(1-t)x+ty \in K. \end{align*} Since $K \subset U$, the composition \begin{align*} g:[0,1]\to \mathbb R^m \end{align*} defined by \begin{align*} g(t)=f(\gamma(t))=f(x+t(y-x)) \end{align*} is a $C^1$ map. By the chain rule, for every $t \in (0,1)$, \begin{align*} g'(t)=Df_{\gamma(t)}(y-x). \end{align*} The map $t \mapsto Df_{\gamma(t)}(y-x)$ extends continuously to $[0,1]$.[/step]

[guided]Fix two arbitrary points $x,y \in K$. The goal is to estimate $|f(y)-f(x)|$ using derivative information only on $K$, so the natural path between the two points is the straight line segment. Define \begin{align*} \gamma:[0,1]\to K \end{align*} by \begin{align*} \gamma(t)=x+t(y-x). \end{align*} The codomain is really $K$, not merely $U$, because convexity gives \begin{align*} \gamma(t)=(1-t)x+ty \in K \end{align*} for every $t \in [0,1]$. This is the point where convexity is used: without it, the segment might leave $K$, and the supremum of $\|Df_z\|_{\mathrm{op}}$ over $K$ would not control the derivative along the path. Now define \begin{align*} g:[0,1]\to \mathbb R^m \end{align*} by \begin{align*} g(t)=f(\gamma(t))=f(x+t(y-x)). \end{align*} Since $f \in C^1(U;\mathbb R^m)$, since $\gamma([0,1]) \subset K \subset U$, and since $\gamma$ is a $C^1$ map, the composition $g=f\circ\gamma$ is $C^1$. The derivative of $\gamma$ is the constant vector $y-x \in \mathbb R^n$. Therefore the chain rule gives, for every $t \in (0,1)$, \begin{align*} g'(t)=Df_{\gamma(t)}(y-x). \end{align*} The function $t \mapsto Df_{\gamma(t)}(y-x)$ is continuous on $[0,1]$, so this derivative has the continuous extension to the endpoints needed for the integral formula. This converts the multivariable estimate into a one-dimensional estimate along a curve.[/guided]

[step:Apply the integral formula along the segment and estimate the integrand] Let $\mathcal L^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $[0,1]$. For each component index $i \in \{1,\dots,m\}$, let \begin{align*} g_i:[0,1]\to \mathbb R \end{align*} denote the $i$-th component of $g$. Since $g_i$ is $C^1$ and $g_i'$ is the continuous extension of the derivative from $(0,1)$ to $[0,1]$, the one-dimensional fundamental theorem of calculus gives \begin{align*} g_i(1)-g_i(0)=\int_0^1 g_i'(t)\,d\mathcal L^1(t). \end{align*} Applying this componentwise yields the vector identity \begin{align*} f(y)-f(x)=g(1)-g(0)=\int_0^1 Df_{\gamma(t)}(y-x)\,d\mathcal L^1(t). \end{align*} Taking Euclidean norms and using the triangle inequality for vector-valued integrals, \begin{align*} |f(y)-f(x)| \leq \int_0^1 |Df_{\gamma(t)}(y-x)|\,d\mathcal L^1(t). \end{align*} For every $t \in [0,1]$, the point $\gamma(t)$ lies in $K$, and the definition of the operator norm gives \begin{align*} |Df_{\gamma(t)}(y-x)| \leq \|Df_{\gamma(t)}\|_{\mathrm{op}}|y-x| \leq M|y-x|. \end{align*} Therefore \begin{align*} |f(y)-f(x)| \leq \int_0^1 M|y-x|\,d\mathcal L^1(t)=M|y-x|. \end{align*} Since $M=\sup_{z \in K}\|Df_z\|_{\mathrm{op}}$ and $|y-x|=|x-y|$, this is exactly \begin{align*} |f(x)-f(y)|\le \left(\sup_{z \in K}\|Df_z\|_{\mathrm{op}}\right)|x-y|. \end{align*} The points $x,y \in K$ were arbitrary, so the estimate holds for all $x,y \in K$. [/step]