Kolmogorov's normability criterion

In mathematics, Kolmogorov's normability criterion is a theorem that provides a necessary and sufficient condition for a topological vector space to be normable; that is, for the existence of a norm on the space that generates the given topology. The normability criterion can be seen as a result in same vein as the Nagata–Smirnov metrization theorem and Bing metrization theorem, which gives a necessary and sufficient condition for a topological space to be metrizable. The result was proved by the Russian mathematician Andrey Nikolayevich Kolmogorov in 1934.

Statement of the theorem

Kolmogorov's normability criterion — A topological vector space is normable if and only if it is a T₁ space and admits a bounded convex neighbourhood of the origin.

Because translation (that is, vector addition) by a constant preserves the convexity, boundedness, and openness of sets, the words "of the origin" can be replaced with "of some point" or even with "of every point".

Definitions

It may be helpful to first recall the following terms:

• A topological vector space (TVS) is a vector space $X$ equipped with a topology $\tau$ such that the vector space operations of scalar multiplication and vector addition are continuous.
• A topological vector space $(X,\tau )$ is called normable if there is a norm $\|\cdot \|:X\to \mathbb {R}$ on $X$ such that the open balls of the norm $\|\cdot \|$ generate the given topology $\tau .$ (Note well that a given normable topological vector space might admit multiple such norms.)
• A topological space $X$ is called a T₁ space if, for every two distinct points $x,y\in X,$ there is an open neighbourhood $U_{x}$ of $x$ that does not contain $y.$ In a topological vector space, this is equivalent to requiring that, for every $x\neq 0,$ there is an open neighbourhood of the origin not containing $x.$ Note that being T₁ is weaker than being a Hausdorff space, in which every two distinct points $x,y\in X$ admit open neighbourhoods $U_{x}$ of $x$ and $U_{y}$ of $y$ with $U_{x}\cap U_{y}=\varnothing$; since normed and normable spaces are always Hausdorff, it is a "surprise" that the theorem only requires T₁.
• A subset $A$ of a vector space $X$ is a convex set if, for any two points $x,y\in A,$ the line segment joining them lies wholly within $A,$ that is, for all $0\leq t\leq 1,$ $(1-t)x+ty\in A.$
• A subset $A$ of a topological vector space $(X,\tau )$ is a bounded set if, for every open neighbourhood $U$ of the origin, there exists a scalar $\lambda$ so that $A\subseteq \lambda U.$ (One can think of $U$ as being "small" and $\lambda$ as being "big enough" to inflate $U$ to cover $A.$)

See also

• Locally convex topological vector space – A vector space with a topology defined by convex open sets
• Normed vector space – Vector space on which a distance is defined
• Topological vector space – Vector space with a notion of nearness