It is known that if $p$ is a seminorm on a real vector space $X$, then the set
$A= \{x\in X: p(x)<1\}$ is convex, balanced, and absorbing. I tried to prove that the Minkowski functional $u_A$ of $A$ coincides with the seminorm $p$.
Im interested on proving that $u_A$ less or equal to $p$ on $X$. My idea is as follows. We let $x \in X$. Then we can choose $s>p(x)$. Then $p(s^{-1}x) = s^{-1}p(x)<1$. This means that $s^{-1}x$ belongs to $A$. Hence $u_A(x)$ is less or equal to $s$. From here, how can we conclude that $u_A(x)$ is less than or equal to $p(x)$?
Thanks in advance
juniven
Best Answer
What you proved is that $u_A(x)\leq s$ for every $s\in(p(x),\infty)$. In other words, $$ u_A(x)\leq p(x)+\varepsilon $$ for every $\varepsilon>0$. This implies that $u_A(x)\leq p(x)$.