[proofplan] The forward implication follows directly from monotonicity in the definition of the [support function](/page/Support%20Function) under enlargement of the set over which the supremum is taken. For the converse, we prove the contrapositive: if a point of $K$ lies outside $L$, [compactness](/page/Compact%20Set) and [convexity](/page/Convex%20Set) of $L$ produce a supporting direction $u_0 \in S^{n-1}$ that strictly separates that point from $L$. Evaluating the two support functions in this direction gives $h_K(u_0) > h_L(u_0)$, contradicting the assumed support-function inequality. [/proofplan] [step:Compare the suprema when $K$ is contained in $L$] Assume $K \subset L$. Fix $u \in S^{n-1}$. Since $K \subset L$, the set \begin{align*} \{\langle x,u\rangle : x \in K\} \end{align*} is a subset of \begin{align*} \{\langle x,u\rangle : x \in L\}. \end{align*} Both sets are non-empty subsets of $\mathbb{R}$, and compactness of $K$ and $L$ ensures that the displayed suprema are finite. Therefore \begin{align*} h_K(u) = \sup_{x \in K}\langle x,u\rangle \le \sup_{x \in L}\langle x,u\rangle = h_L(u). \end{align*} Since $u \in S^{n-1}$ was arbitrary, $h_K(u) \le h_L(u)$ for every $u \in S^{n-1}$. [/step] [step:Construct a separating direction from a point outside $L$] Assume now that $h_K(u) \le h_L(u)$ for every $u \in S^{n-1}$. We prove $K \subset L$ by contradiction. Suppose there exists $x_0 \in K \setminus L$. Define the continuous function \begin{align*} \Phi: L &\to \mathbb{R} \\ y &\mapsto |x_0-y|^2 . \end{align*} Since $L$ is [compact](/page/Compact%20Set) and non-empty and $\Phi$ is continuous, the extreme value theorem applies to $\Phi$ on $L$ and gives a point $y_0 \in L$ at which $\Phi$ attains its minimum. Define \begin{align*} v_0 := x_0-y_0 \in \mathbb{R}^n . \end{align*} Because $x_0 \notin L$ and $y_0 \in L$, we have $v_0 \ne 0$. Define the unit vector \begin{align*} u_0 := \frac{v_0}{|v_0|} \in S^{n-1}. \end{align*} We claim that \begin{align*} \langle z-y_0,v_0\rangle \le 0 \end{align*} for every $z \in L$. Fix $z \in L$. For $t \in [0,1]$, [convexity](/page/Convex%20Set) of $L$ gives \begin{align*} y_t := (1-t)y_0+tz \in L. \end{align*} Since $y_0$ minimizes $\Phi$ on $L$, \begin{align*} |x_0-y_0|^2 \le |x_0-y_t|^2 . \end{align*} Using $x_0-y_t=v_0-t(z-y_0)$, this becomes \begin{align*} |v_0|^2 \le |v_0-t(z-y_0)|^2 = |v_0|^2 - 2t\langle v_0,z-y_0\rangle + t^2|z-y_0|^2 . \end{align*} For every $t \in (0,1]$, \begin{align*} 0 \le -2\langle v_0,z-y_0\rangle + t|z-y_0|^2 . \end{align*} Letting $t \to 0^+$ gives \begin{align*} \langle v_0,z-y_0\rangle \le 0. \end{align*} Thus, for every $z \in L$, \begin{align*} \langle z,u_0\rangle \le \langle y_0,u_0\rangle . \end{align*} [guided] The point $x_0$ lies outside the compact convex set $L$, so we choose the point of $L$ closest to $x_0$. Formally, define \begin{align*} \Phi: L &\to \mathbb{R} \\ y &\mapsto |x_0-y|^2 . \end{align*} The function $\Phi$ is continuous, and $L$ is non-empty and [compact](/page/Compact%20Set). Therefore the extreme value theorem applies to $\Phi$ on $L$, so there exists $y_0 \in L$ such that $\Phi(y_0) \le \Phi(y)$ for every $y \in L$. Define \begin{align*} v_0 := x_0-y_0 . \end{align*} Because $x_0 \notin L$ while $y_0 \in L$, the vector $v_0$ is non-zero. Hence \begin{align*} u_0 := \frac{v_0}{|v_0|} \end{align*} is a well-defined element of $S^{n-1}$. We now show that $u_0$ is a supporting direction for $L$ at $y_0$. Fix $z \in L$. Since $L$ is [convex](/page/Convex%20Set), every point on the line segment from $y_0$ to $z$ remains in $L$; that is, for each $t \in [0,1]$, \begin{align*} y_t := (1-t)y_0+tz \in L. \end{align*} The minimality of $y_0$ gives \begin{align*} |x_0-y_0|^2 \le |x_0-y_t|^2 . \end{align*} Substituting $v_0=x_0-y_0$ and $y_t=y_0+t(z-y_0)$ yields \begin{align*} x_0-y_t = x_0-y_0-t(z-y_0) = v_0-t(z-y_0). \end{align*} Therefore \begin{align*} |v_0|^2 \le |v_0-t(z-y_0)|^2 = |v_0|^2 - 2t\langle v_0,z-y_0\rangle + t^2|z-y_0|^2 . \end{align*} Subtracting $|v_0|^2$ and dividing by $t>0$ gives \begin{align*} 0 \le -2\langle v_0,z-y_0\rangle + t|z-y_0|^2 . \end{align*} Now let $t \to 0^+$. The term $t|z-y_0|^2$ tends to $0$, so \begin{align*} \langle v_0,z-y_0\rangle \le 0. \end{align*} Dividing by the positive number $|v_0|$ gives \begin{align*} \langle z-y_0,u_0\rangle \le 0, \end{align*} or equivalently \begin{align*} \langle z,u_0\rangle \le \langle y_0,u_0\rangle . \end{align*} Thus every point of $L$ lies on one side of the hyperplane through $y_0$ with normal vector $u_0$. [/guided] [/step] [step:Use the separating direction to contradict the support-function inequality] From the previous step, for every $z \in L$, \begin{align*} \langle z,u_0\rangle \le \langle y_0,u_0\rangle . \end{align*} Taking the supremum over $z \in L$ gives \begin{align*} h_L(u_0) \le \langle y_0,u_0\rangle . \end{align*} On the other hand, since $x_0 \in K$, \begin{align*} h_K(u_0) \ge \langle x_0,u_0\rangle . \end{align*} Using $x_0=y_0+v_0$ and $u_0=v_0/|v_0|$, we compute \begin{align*} \langle x_0,u_0\rangle = \langle y_0,u_0\rangle + \langle v_0,u_0\rangle = \langle y_0,u_0\rangle + |v_0|. \end{align*} Since $|v_0|>0$, \begin{align*} h_K(u_0) \ge \langle x_0,u_0\rangle = \langle y_0,u_0\rangle+|v_0| > \langle y_0,u_0\rangle \ge h_L(u_0). \end{align*} This contradicts the assumption that $h_K(u) \le h_L(u)$ for every $u \in S^{n-1}$. Hence no point $x_0 \in K \setminus L$ exists, and therefore $K \subset L$. [/step]