[proofplan] We first use compactness of $K$ and continuity of the [linear map](/page/Linear%20Map) $\ell$ to choose a point $x_0 \in K$ where $\ell$ is minimal. Since $K$ is a compact polytope, it is the convex hull of finitely many vertices, and those vertices are extreme points of $K$. We write $x_0$ as a convex combination of these vertices and use linearity of $\ell$ to express $\ell(x_0)$ as the corresponding convex combination of the vertex values. One vertex in the combination must have value no larger than $\ell(x_0)$, and minimality of $x_0$ forces equality. [/proofplan] [step:Choose a minimizer of the linear functional on the compact polytope] Since $\ell: \mathbb{R}^N \to \mathbb{R}$ is linear, it is continuous. The set $K$ is non-empty and compact, so $\ell(K) \subset \mathbb{R}$ is a non-empty compact subset of $\mathbb{R}$. Hence $\ell(K)$ has a minimum. Choose $x_0 \in K$ such that \begin{align*} \ell(x_0) = \min_{x \in K} \ell(x). \end{align*} [/step] [step:Express the minimizer as a convex combination of extreme points] Because $K$ is a compact polytope, there exist finitely many vertices $v_1,\dots,v_m \in K$, with $m \in \mathbb{N}$, such that each $v_i$ is an extreme point of $K$ and \begin{align*} K = \operatorname{conv}\{v_1,\dots,v_m\}. \end{align*} Since $x_0 \in K$, there exist coefficients $\lambda_1,\dots,\lambda_m \in [0,1]$ satisfying \begin{align*} \sum_{i=1}^m \lambda_i = 1 \end{align*} and \begin{align*} x_0 = \sum_{i=1}^m \lambda_i v_i. \end{align*} [guided] The purpose of passing to vertices is that a compact polytope has only finitely many of them, and every point of the polytope is built from them by convex averaging. More precisely, because $K$ is a compact polytope, there are vertices $v_1,\dots,v_m \in K$, where $m \in \mathbb{N}$, such that the vertices are extreme points of $K$ and \begin{align*} K = \operatorname{conv}\{v_1,\dots,v_m\}. \end{align*} This equality means that each point of $K$ can be written as a convex combination of the listed vertices. Since the minimizer $x_0$ chosen above belongs to $K$, the convex-hull representation gives coefficients $\lambda_1,\dots,\lambda_m \in [0,1]$ with total mass one: \begin{align*} \sum_{i=1}^m \lambda_i = 1. \end{align*} For these coefficients, \begin{align*} x_0 = \sum_{i=1}^m \lambda_i v_i. \end{align*} The coefficients with $\lambda_i = 0$ do not affect the sum, but keeping all $m$ vertices in the notation makes the linearity argument uniform. [/guided] [/step] [step:Use linearity to transfer the minimum value to one vertex] Applying linearity of $\ell$ to the convex representation of $x_0$ gives \begin{align*} \ell(x_0) = \ell\left(\sum_{i=1}^m \lambda_i v_i\right) = \sum_{i=1}^m \lambda_i \ell(v_i). \end{align*} We claim that there exists an index $j \in \{1,\dots,m\}$ with $\lambda_j > 0$ and $\ell(v_j) \leq \ell(x_0)$. If not, then every index $i$ with $\lambda_i > 0$ would satisfy $\ell(v_i) > \ell(x_0)$. Since the coefficients with positive weight have total sum $1$, this would imply \begin{align*} \sum_{i=1}^m \lambda_i \ell(v_i) > \sum_{i=1}^m \lambda_i \ell(x_0) = \ell(x_0), \end{align*} contradicting the displayed equality for $\ell(x_0)$. Therefore such an index $j$ exists. Since $v_j \in K$ and $x_0$ minimizes $\ell$ on $K$, we also have $\ell(x_0) \leq \ell(v_j)$. Hence \begin{align*} \ell(v_j) = \ell(x_0) = \min_{x \in K} \ell(x). \end{align*} Because $v_j$ is an extreme point of $K$, the minimum is attained at an extreme point. [/step]