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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 ocw.mit.edu.

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!

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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 $u(B(0,\eps)) \subset u(U)=(-1,1)$.

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