Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


share|cite|improve this question
up vote 1 down vote accepted

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 $u_A(x)\leq p(x)$.

share|cite|improve this answer
Thanks for sharing your inputs. – juniven Oct 20 '12 at 13:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.