Bounded Nbhd Criterion for a Hausdorff LCTVS to be Normable
Bounded Neighbourhood Criterion for a Hausdorff LCTVS to be Normable
The following theorem gives us a nice criterion for when a Hausdorff locally convex topological vector space is normable, based on bounded neighbourhoods of the origin.
| Theorem 1: Let $E$ be a Hausdorff locally convex topological vector space. If $E$ has a bounded neighbourhood of the origin, then $E$ is normable. |
- Proof: Since $E$ is a locally convex topological vector space, it has a base of (closed) absolutely convex (and absorbent) neighbourhoods of the origin. Let $U$ be an absolutely convex neighbourhood of the origin that is contained in the given bounded neighbourhood of the origin. Then $U$ itself is a bounded neighbourhood of the origin, because subsets of bounded sets are bounded as proven on the Subsets, Scalar Multiples, Finite Unions, and Arbitrary Intersections of Bounded Sets in a LCTVS page.
- Thus, for every absolutely convex neighbourhood $V$ of the origin, there exists a $\lambda_V > 0$ such that:
\begin{align} \quad U \subseteq \lambda_V V \end{align}
- Or equivalently, $\lambda_V^{-1} U \subseteq V$. Since this is true for all absolutely convex neighbourhoods $V$ of the origin, we see that $\{ \epsilon U : \epsilon > 0 \}$ is a base of absolutely convex neighbourhoods of the origin.
- So the locally convex topology on $E$ is generated by the set which contains only the seminorm $p_U$.
- But since $E$ is Hausdorff, we have by theCriterion for the Coarsest Topology Determined by a Set of Seminorms to be Hausdorff, that for each $x \neq o$, we must have that $p_U(x) > 0$. Hence $p_U(x) = 0$ if and only if $x = o$, and so $p_U$ is actually a norm on $E$.
- Thus the Hausdorff and locally convex topology on $E$ is determined by the norm $p_U$, and so $E$ with this topology is normable. $\blacksquare$
| Corollary 2: Let $E$ be a Hausdorff locally convex topological vector space. If $E$ is not normable, then every neighbourhood of the origin is unbounded. |
| Corollary 3: Let $(E, d)$ be a metric space. If $(E, d)$ is not normable, then for all $\epsilon > 0$, $B(o, \epsilon)$ is unbounded. |
- Proof: The metric space $(E, d)$ with the topology determined by the collection $\{ B(a, \epsilon) : a \in E, \epsilon > 0 \}$ is a Hausdorff locally convex topological vector space, and $\{ B(o, \epsilon) : \epsilon > 0 \}$ is a base of absolutely convex neighbourhoods of the origin.
- Thus if $(E, d)$ is not normable, by Corollary 2, for all $\epsilon > 0$, $B(o, \epsilon)$ is unbounded. $\blacksquare$