[proofplan] We prove the two structural identities directly from the definition of the support function. Positive homogeneity follows by moving a non-negative scalar through the Euclidean inner product and then through the supremum; the case $\lambda = 0$ is handled separately. Subadditivity follows because each point $x \in K$ gives an upper bound for $\langle x,u+v\rangle$ in terms of the two separate support values, and taking the supremum preserves that upper bound. [/proofplan] [step:Move a non-negative scalar through the support supremum] Fix $u \in \mathbb{R}^n$ and $\lambda \in [0,\infty)$. If $\lambda = 0$, then for every $x \in K$, \begin{align*} \langle x, \lambda u\rangle = \langle x,0\rangle = 0. \end{align*} Since $K$ is non-empty, the supremum of the constant set $\{0\}$ is $0$, hence \begin{align*} h_K(0) = 0 = 0 \cdot h_K(u). \end{align*} Now assume $\lambda > 0$. For every $x \in K$, bilinearity of the Euclidean inner product gives \begin{align*} \langle x,\lambda u\rangle = \lambda \langle x,u\rangle. \end{align*} Therefore \begin{align*} h_K(\lambda u) &= \sup_{x \in K} \langle x,\lambda u\rangle \\ &= \sup_{x \in K} \lambda \langle x,u\rangle \\ &= \lambda \sup_{x \in K} \langle x,u\rangle \\ &= \lambda h_K(u). \end{align*} The third equality uses $\lambda > 0$: multiplication by a positive scalar preserves the order on $\mathbb{R}$ and sends least upper bounds to least upper bounds. Thus $h_K(\lambda u)=\lambda h_K(u)$ for every $\lambda \in [0,\infty)$. [/step] [step:Bound the support value at a sum by the sum of support values] Fix $u,v \in \mathbb{R}^n$. For every $x \in K$, bilinearity of the Euclidean inner product gives \begin{align*} \langle x,u+v\rangle = \langle x,u\rangle + \langle x,v\rangle. \end{align*} By the definition of $h_K(u)$ and $h_K(v)$ as suprema over the same set $K$, we have \begin{align*} \langle x,u\rangle \le h_K(u), \qquad \langle x,v\rangle \le h_K(v). \end{align*} Adding these two inequalities yields \begin{align*} \langle x,u+v\rangle \le h_K(u)+h_K(v) \end{align*} for every $x \in K$. Hence $h_K(u)+h_K(v)$ is an upper bound for the set \begin{align*} \{\langle x,u+v\rangle : x \in K\}. \end{align*} Taking the least upper bound of this set gives \begin{align*} h_K(u+v) = \sup_{x \in K} \langle x,u+v\rangle \le h_K(u)+h_K(v). \end{align*} Thus $h_K$ is subadditive. [/step] [step:Conclude both asserted properties] The first step proves positive homogeneity: \begin{align*} h_K(\lambda u)=\lambda h_K(u) \end{align*} for every $u \in \mathbb{R}^n$ and every $\lambda \in [0,\infty)$. The second step proves subadditivity: \begin{align*} h_K(u+v)\le h_K(u)+h_K(v) \end{align*} for every $u,v \in \mathbb{R}^n$. These are exactly the two claimed properties of the support function $h_K$. [/step]