Kolmogorov's normability criterion

Remarks

If the topology comes from a norm, the open unit ball is a convex bounded neighbourhood of $0$. Conversely, a convex bounded neighbourhood of $0$ contains a convex balanced open neighbourhood $U$ of $0$, which is still bounded. Its Minkowski functional $p_U$ is a seminorm, and boundedness of $U$ shows that the sets $t U$, $t > 0$, form a neighbourhood base at $0$, so $p_U$ induces the topology. The Hausdorff property then forces $p_U(x) > 0$ for $x \ne 0$, so $p_U$ is a norm. Example: the Fréchet space $C(\R)$ with the seminorms $p_n(f) = \sup_{\abs{t} \le n} \abs{f(t)}$ is not normable, because no neighbourhood of $0$ is bounded. The space $L^p([0, 1])$ with $0 < p < 1$ is not normable either: it has bounded neighbourhoods of $0$ but no convex ones other than the whole space. The Hausdorff hypothesis is needed: $\R$ with the indiscrete topology is a topological vector space. In it, $\R$ itself is a convex bounded neighbourhood of $0$, but no norm induces this topology.

Going deeper

For a student

Local convexity means that the topology is generated by seminorms. So the criterion says that a locally convex space is normable precisely when one seminorm suffices, up to equivalence. It shows that spaces such as $C^\infty(\Omega)$, $H(\Omega)$ or the Schwartz space are not normable. In each of them a bounded neighbourhood of $0$ would be relatively compact, which happens only in finite dimension. A topological vector space with a bounded neighbourhood of $0$ is called locally bounded, so the criterion reads: normable means Hausdorff, locally convex and locally bounded. The construction of the norm is the same Minkowski functional argument that links convex sets and seminorms in the theory of locally convex spaces.