[proofplan] We first regard the linear functional $u$ as a continuous real-valued function on the compact set $C$. Compactness then gives a point $x_0 \in C$ at which $u$ attains its maximum. The defining inequality for the maximum places all of $C$ in the closed half-space $H^-(u,u(x_0))$, while the maximizing point itself lies on the hyperplane $H(u,u(x_0))$, giving the required contact. [/proofplan] [step:Show that the linear functional is continuous on $C$] Let $(e_1,\dots,e_n)$ denote the standard basis of $\mathbb{R}^n$. Define the constant \begin{align*} A := \left(\sum_{i=1}^n |u(e_i)|^2\right)^{1/2}. \end{align*} For any $x=(x_1,\dots,x_n) \in \mathbb{R}^n$, linearity of $u$ gives \begin{align*} u(x)=\sum_{i=1}^n x_i u(e_i). \end{align*} Hence, for any $x,y \in \mathbb{R}^n$, the [Cauchy-Schwarz inequality](/theorems/432) in $\mathbb{R}^n$ gives \begin{align*} |u(x)-u(y)| &= |u(x-y)| \\ &= \left|\sum_{i=1}^n (x_i-y_i)u(e_i)\right| \\ &\leq \left(\sum_{i=1}^n |x_i-y_i|^2\right)^{1/2} \left(\sum_{i=1}^n |u(e_i)|^2\right)^{1/2} \\ &= A|x-y|. \end{align*} Therefore $u:\mathbb{R}^n \to \mathbb{R}$ is Lipschitz continuous, and its restriction \begin{align*} u|_C:C \to \mathbb{R} \end{align*} is continuous. [guided] We need a maximum of $u$ on $C$, so the first point is to verify that $u$ is continuous. Let $(e_1,\dots,e_n)$ be the standard basis of $\mathbb{R}^n$, and define \begin{align*} A := \left(\sum_{i=1}^n |u(e_i)|^2\right)^{1/2}. \end{align*} For a point $x=(x_1,\dots,x_n) \in \mathbb{R}^n$, linearity of $u$ gives the coordinate formula \begin{align*} u(x)=u\left(\sum_{i=1}^n x_i e_i\right) =\sum_{i=1}^n x_i u(e_i). \end{align*} Thus, for $x,y \in \mathbb{R}^n$, \begin{align*} |u(x)-u(y)| &= |u(x-y)| \\ &= \left|\sum_{i=1}^n (x_i-y_i)u(e_i)\right|. \end{align*} Applying the Cauchy-Schwarz inequality to the two vectors $((x_i-y_i))_{i=1}^n$ and $(u(e_i))_{i=1}^n$ in $\mathbb{R}^n$, we obtain \begin{align*} |u(x)-u(y)| &\leq \left(\sum_{i=1}^n |x_i-y_i|^2\right)^{1/2} \left(\sum_{i=1}^n |u(e_i)|^2\right)^{1/2} \\ &= A|x-y|. \end{align*} This is a Lipschitz estimate for $u$, so $u:\mathbb{R}^n \to \mathbb{R}$ is continuous. Consequently the restricted map \begin{align*} u|_C:C \to \mathbb{R} \end{align*} is continuous as well. [/guided] [/step] [step:Choose a point where $u$ attains its maximum on $C$] Since $C$ is compact and $u|_C:C \to \mathbb{R}$ is continuous, the image set \begin{align*} u(C):=\{u(x):x\in C\} \end{align*} is compact in $\mathbb{R}$. Since $C$ is non-empty, $u(C)$ is non-empty. Therefore $u(C)$ has a largest element. Let \begin{align*} m := \max u(C). \end{align*} By the definition of image set, there exists $x_0 \in C$ such that $u(x_0)=m$. Equivalently, \begin{align*} u(x_0)=\max_{x\in C}u(x). \end{align*} [guided] Now we use compactness. The function under consideration is the restricted map \begin{align*} u|_C:C \to \mathbb{R}. \end{align*} From the previous step, this map is continuous. Since $C$ is compact, its continuous image \begin{align*} u(C):=\{u(x):x\in C\} \end{align*} is a compact subset of $\mathbb{R}$. The hypothesis that $C$ is non-empty ensures that $u(C)$ is also non-empty. A non-empty compact subset of $\mathbb{R}$ contains its supremum, so it has a largest element. Define that largest element by \begin{align*} m := \max u(C). \end{align*} Because $m$ belongs to the image set $u(C)$, there exists a point $x_0 \in C$ such that $u(x_0)=m$. Rewriting the definition of $m$ gives \begin{align*} u(x_0)=\max_{x\in C}u(x). \end{align*} This is the point of contact for the supporting hyperplane. [/guided] [/step] [step:Verify that the maximizing hyperplane supports $C$] Let \begin{align*} \alpha := u(x_0). \end{align*} For every $x \in C$, the maximality of $x_0$ gives \begin{align*} u(x)\leq u(x_0)=\alpha. \end{align*} Therefore $x \in H^-(u,\alpha)$ for every $x \in C$, and hence \begin{align*} C \subset H^-(u,\alpha). \end{align*} Also $u(x_0)=\alpha$, so $x_0 \in H(u,\alpha)$. Since $x_0 \in C$, we have \begin{align*} x_0 \in C \cap H(u,\alpha), \end{align*} and therefore \begin{align*} C \cap H(u,\alpha) \neq \varnothing. \end{align*} Substituting $\alpha=u(x_0)$, this proves that $H(u,u(x_0))$ supports $C$. [/step]