Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My goal for this question is to understand the implications of continuity its relationship to normed spaces. This question is based off of the notes for 18.155 available at

Let $u$ be a linear functional $u:V\rightarrow R$ on a normed space $V$. Let's assume we know that $u$ is continuous at $0$. This implies that $u^{-1}(-1, 1)$ is a neighborhood of $0 \in V$.

The text continues to say that this implies that $\exists \epsilon>0$ such that $u(\{f \in V: ||f|| < \epsilon\}) \subset (-1,1)$. Simply put, I don't understand why this is implied. So two questions: Why is this implied? Why does the value of the norm have anything to do with a neighborhood based on the distance function?

Any help at all would be appreciated. Thanks much!

share|cite|improve this question

1 Answer 1

$\newcommand{\eps}{\epsilon}$A set $U$ in a metric space $(X,d)$ is open if for every $x \in U$ there exists an $\epsilon>0$ such that $B(x,\epsilon) \subset U$. A normed vector space is a metric space with metric given by $d(v,w)=||v-w||$. In your question we have that $U=u^{-1}((-1,1))$ is an open set containing $0$ thereby there is an $\epsilon>0$ such that $B(0,\eps) \subset U$ but

$$B(0,\eps)=\{f \in V : ||f-0|| < \eps\},$$

in particular $||f-0||=||f||$. Finally we see that since $B(0,\eps) \subset U$ so when we push forward under $u$ we must have that $u(B(0,\eps)) \subset u(U)=(-1,1)$.

share|cite|improve this answer

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.