## Kolmogorov Normability Criterion

This past week I have been refreshing my knowledge of topological vector spaces (tvs) while reading some papers on generalizations of the Mazur-Ulam theorem to metrizable tvs. I intend to upload a typed set of notes on the subject which I have been putting together, but they are still mostly handwritten. For now, I want to present a theorem of A.N. Kolmogorov, the Kolmogorov normability criterion, that gives necessary and sufficient conditions for a tvs ${(X,\tau)}$ to be normable (i.e. there exists a norm ${x\mapsto\left\|x\right\|}$ such that the norm topology coincides with ${\tau}$). Two reasons motivate me to share it: 1) the proof is elegant; 2) I have spent a fair amount of time this summer studying when topological spaces, in particular function spaces, are normable.

To minimize any confusion over terminology, I include the following definitions. A notational note: we denote the line segment between two points ${x,y}$ by ${[x,y]:=\left\{tx+(1-t)y:0\leq t\leq 1\right\}}$.

Let ${X}$ be a vector space. A subset ${A}$ is…

• convex if ${[x,y]\subset A}$ for any ${x,y\in A}$.
• absorbing (or absorbent or radial) if for all ${x\in X}$, there exists a scalar ${\alpha_{0}>0}$ such that ${\left|\alpha\right|\leq\alpha_{0}\Rightarrow \alpha x\in A}$.
• circled (or balanced if ${[-x,x]\subset A}$ for any ${x\in A}$.
• bounded if for every neighborhood $U$ there exists a scalar $\alpha\in\mathbb{K}$ such that $A\subset\alpha U$.

It is a well-known that a Hausdorff tvs is metrizable if and only if it has a countable base of neighborhoods at zero (Theorem 5.12 [1]). However, there are plenty of familiar spaces which are metrizable, but normable. Consider the vector space $X$ of real sequences $a=(\alpha_{k})_{k=1}^{\infty}$ with addition and scalar multiplication defined pointwise. Define a metric $d$ by

$\displaystyle d(a,b):=\sum_{k=1}^{\infty}2^{-k}\dfrac{\left|\alpha_{k}-\beta_{k}\right|}{1+\left|\alpha_{k}-\beta_{k}\right|},\ \forall a=(\alpha_{k}), b=(\beta_{k})$

The reader may verify that that $(X,d)$ is a complete, metrizable tvs. Let $a\mapsto\left\|a\right\|$. Observe that $\left|c\alpha_{k}\right|/(1+\left|c\alpha_{k}\right|)\rightarrow 1$ as $c\rightarrow\infty$. Since a norm is positively homogeneous, $\left\|c\cdot a\right\|=\left|c\right|\left\|a\right\|\rightarrow\infty$ as $\left|c\right|\rightarrow\infty$. But then for any $r>0$, there does not exists an $\varepsilon>0$ such that

$\displaystyle B_{\varepsilon}:=\left\{a\in X:d(0,a)<\varepsilon\right\}\subset\left\{a\in X:\left\|a\right\|

Indeed, if $e_{m}$ be a standard basis vector such that $2^{-m}<\varepsilon$, then $ne_{m}\in B_{\varepsilon}$ for all $n\geq 1$, but the norm of the norm tends to $\infty$ as $n\rightarrow\infty$.

A more interesting example is the space ${L^{p}[0,1]}$ of Lebesgue measurable functions with ${\int_{0}^{1}\left|f\right|^{p}<\infty}$, where ${0, lacks a norm equivalent to its natural topology. It turns out that the non-normability is intimately connected to the lack of nontrivial convex open sets in this space, a result known as Day’s theorem. The above example might suggest to us that convexity plays a role in any criterion for normability of a tvs.

Recall that a tvs is said to be separated if for any ${x,y\in X}$, there exists a nbhd of ${x}$ which does contain ${y}$. Equivalently, ${X}$ is separated if there exists a nbhd of ${0}$ which does not contain ${y}$. The following theorem of A.N. Kolmogorov characterizes the normability of a separated (not a priori Hausdorff) tvs in terms of the existence of a bounded convex neighborhood of zero. The proof presented here is that of [2].

Theorem (A.N. Kolmogorov) A tvs ${(X,\tau)}$ is normable if and only if it is separated and contains a bounded convex nbhd of zero.

Proof: The ${\Rightarrow}$ direction is obvious. For ${\Leftarrow}$, suppose that ${X}$ is separated and ${U}$ is a bounded convex nbhd of zero. Taking a balanced nbhd of zero ${V\subset U}$ and noting that ${\text{conv}(V)\subset\text{conv}(U)=U}$, we may assume that ${U}$ is balanced, bounded, and convex.

For ${x\neq 0}$, define a set of scalars ${A(x):=\left\{\lambda:x\notin\lambda U\right\}}$; define ${A(0):=\left\{0\right\}}$. I claim that ${A(x)}$ contains nonzero scalars when ${x\neq 0}$. Indeed, for ${x\neq 0}$, there exists an open nbhd ${V}$ of zero such that ${x\notin V}$. Since ${U}$ is bounded, there exists ${\alpha\in\mathbb{K}}$ such that ${U\subset\alpha V\Leftrightarrow \alpha^{-1}U\subset V}$. As ${x\notin V}$, ${x\notin\alpha^{-1}U\Leftrightarrow\alpha^{-1}\in A(x)}$. Although we know that ${A(x)}$ contains nontrivial scalars, it is not obvious that ${A(x)}$ contains an interval or disk when ${x\neq 0}$. We will establish this result below in order to define a norm structure on ${X}$.

Define a nonnegative (possibly) extended-valued function ${x\mapsto\left\|x\right\|:=\sup\left\{\left|\lambda\right|:\lambda\in A(x)\right\}}$. We will show that ${\left\|\cdot\right\|}$ is a norm. It is clear that ${\left\|0\right\|=0}$ and the preceding claim shows that ${x\neq 0\Rightarrow\left\|x\right\|>0}$. I claim that

$\displaystyle {\left\{\lambda:\left|\lambda\right|<\left\|x\right\|\right\}\subset A(x)} \ \forall x\in X.$     (1)

If ${x=0}$, then the LHS is empty. If $x\neq 0$, then there exists a sequence of scalars $\lambda_{n}\in A(x)$ increasing to $\left\|x\right\|$. For $\left|\lambda\right|<\left|\lambda_{n}\right|$, we have $\lambda/\lambda_{n}\cdot U\subset U$, as $U$ is balanced. Hence if $\lambda_{n}^{-1}x\notin U$, then $\lambda_{n}^{-1}x\notin(\lambda/\lambda_{n})\cdot U\Leftrightarrow\lambda x\notin U$. As a consequence, we obtain ${\left\|x\right\|<\infty}$ for all ${x}$. Indeed, note that ${U}$ is a nbhd of zero, therefore absorbing. So given ${x}$, there exists ${\alpha_{0}>0}$ such that ${0\leq\left|\alpha\right|\leq\alpha_{0}\Rightarrow \alpha x\in U}$. It follows that ${\left\|x\right\|\leq\alpha_{0}^{-1}}$.

For homogeneity, let $\alpha$ be a nonzero scalar. Observe that by (1), $x\notin\lambda U\Leftrightarrow\alpha x\notin\alpha\cdot\lambda U$ if and only if $\left|\lambda\right|<\left\|x\right\|$. It follows that $\left|\alpha\right|\left\|x\right\|\leq\left\|\alpha x\right\|$. If $\left|\lambda\right|>\left\|x\right\|$, then (1) implies $\left|\alpha\lambda\right|>\left\|\alpha x\right\|$. Letting $\left|\lambda\right|\downarrow\left\|x\right\|$, we obtain the reverse inequality.

So far we have not used the hypothesis that ${U}$ is convex, but we will need it to show that ${x\mapsto\left\|x\right\|}$ satisfies the triangle inequality. Fix ${x,y\in X}$. If either ${x=0}$ or ${y=0}$, then the triangle inequality is trivial, so assume otherwise. Take ${\varepsilon>0}$ sufficiently small so that ${\max\left\{\left\|x\right\|,\left\|y\right\|\right\}+\varepsilon<\left\|x\right\|+\left\|y\right\|}$. Then ${x\in(\left\|x\right\|+\varepsilon)\cdot U}$ and ${y\in(\left\|y\right\|+\varepsilon)\cdot U}$. Since ${U}$ is convex,

$\displaystyle\dfrac{\left\|x\right\|+\varepsilon}{\left\|x\right\|+\left\|y\right\|}\cdot\dfrac{1}{\left\|x\right\|+\varepsilon}x+\dfrac{\left\|y\right\|+\varepsilon}{\left\|x\right\|+\left\|y\right\|}\cdot\dfrac{1}{\left\|y\right\|+\varepsilon}y\in U,$

whence ${x+y\in (\left\|x\right\|+\left\|y\right\|)U}$. As ${\left\{\left|\lambda\right|<\left\|x+y\right\|\right\}\subset A(x+y)}$, it follows that ${\left\|x+y\right\|\leq\left\|x\right\|+\left\|y\right\|}$.

Lastly, we need to show that norm topology coincides with ${\tau}$. Let ${B_{\varepsilon}:=\left\{x\in X:\left\|x\right\|<\varepsilon\right\}}$ be a ball of radius ${\varepsilon>0}$ centered at zero. If ${x\in (\varepsilon/2)\cdot U}$, then ${\left\|x\right\|\leq\varepsilon/2}$, whence ${B_{\varepsilon}}$ is ${\tau}$-open. Conversely, suppose ${V}$ is an open nbhd of zero. Since ${U}$ is bounded, there is ${\alpha\in\mathbb{K}}$ such that ${U\subset\alpha V\Leftrightarrow \alpha^{-1}U\subset V}$. If ${\left\|x\right\|<\alpha^{-1}}$, then ${x\in\alpha^{-1}U}$, whence ${B_{\alpha^{-1}}\subset V}$. $\Box$

1. S.K. Berberian, Lectures in Functional Analysis and Operator Theory, Spring Verlag 1974.