Proposition 10.1.2. Let $E$ be a separated TVS over $K \in \RC$, then the following are equivalent:
There exists $U \in \cn^{o}(0)$ bounded and convex.
There exists a norm $\norm{\cdot}_{E}: E \to [0, \infty)$ that induces the topology on $E$.
Proof. (1) $\Rightarrow$ (2): Using Proposition 8.1.11, assume without loss of generality that $U$ is also circled. Let $V \in \cn^{o}(0)$, then there exists $\lambda \in K$ such that $\lambda V \supset U$ and $\abs{\lambda}^{-1}U \subset V$.
Let $\norm{\cdot}_{E}: E \to [0, \infty)$ be the gauge of $U$. For each $r > 0$, let $B_{E}(0, r) = \bracs{x \in E|\norm{x}_E < r}$, then
\[\bracs{\lambda U|\lambda > 0}= \bracs{B_E(0, r)|r > 0}\]
is a fundamental system of neighbourhoods at $0$ for $E$. Therefore $\norm{\cdot}_{E}$ induces the topology on $E$.$\square$