Proof. Clearly $\norm{0}{K}=0$ and $\norm{x}{K}\geq 0$ for any $x\in K$. Let $\e>0$ such that $B(0,\e)\suf K$ so $\fall x\neq 0,\exists t^{*}>0$ such that $\norm{\frac{x}{t^{*}}}{2}\leq \e$ then $\frac{x}{t^{*}}\in B(0,\e)\suf K$. Thus $\norm{x}{K}<\infty$. By boundedness of $K$, $\norm{x}{K}=0\implies x=\vec{0}$.

For $\lam\neq 0$ set $\lam=\e\abs{\lam}$ then

hence

Let $x,y\in\R^{,}\e>0$. By definition there are $s\leq \norm{x}{K}+\e,t\leq \norm{y}{K}+\e$ with $\frac{x}{s},\frac{y}{t}\in K$. Since $K$ is convex

so $\norm{x+y}{K}\leq s+t\leq\norm{x}{K}+\norm{y}{K}+2\e$.

If $x\in K$ then $x/1\in K$ so $\norm{x}{K}\leq 1$. Conversely if $\norm{x}{K}\leq 1$ then pick a non-increasing sequence $t_{j}\to \norm{x}{K}$ with $x/t_{j}\in K$. If $\norm{x}{K}<1$ we may assume $t_{j}\leq 1$ so $x=t_{j}(x/t_{j})+(1-t_{j})0\in K$ while $\norm{x}{K}=1\implies x/t_{j}\to x\in K$ by closedness.$\square$