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\|<r\right\}=:B'_{r}

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<p<1}, 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.

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One Response to Kolmogorov Normability Criterion

  1. Pingback: On the Normability of the Space of Measurable Functions | Math by Matt

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