Continuity and Discontinuity of Seminorms on Infinite-Dimensional Vector Spaces

TL;DR

This paper investigates the conditions under which seminorms on infinite-dimensional vector spaces are continuous or discontinuous relative to various norm-topologies, clarifying their behavior in functional analysis.

Contribution

It provides a detailed analysis of the continuity properties of seminorms on infinite-dimensional spaces, highlighting new criteria and characterizations.

Findings

Characterization of continuous seminorms

Conditions for discontinuity of seminorms

Implications for functional analysis and topology

Abstract

Let $S$ be a seminorm on an infinite-dimensional real or complex vector space $X$. Our purpose in this note is to study the continuity and discontinuity properties of $S$ with respect to certain norm-topologies on $X$.

Equations

$$\begin{array}[]{l}N(x)>0,\quad x\neq 0,\\ N(\alpha x)=|\alpha|N(x),\\ N(x+y)\leq N(x)+N(y).\end{array}$$

$$\begin{array}[]{l}S(x)\geq 0,\\ S(\alpha x)=|\alpha|S(x),\\ S(x+y)\leq S(x)+S(y);\end{array}$$

$$N(x_n)\to 0\text{mbox}impliesS(x_n)\to 0\text{mbox}asn\to \infty .$$

$$S(x-y_n)\to 0\text{mbox}asn\to \infty .$$

$$|S(x)-S(y)|\le S(x-y),x,y\in X,$$

$$|S(x)-S(y_n)|\to 0,$$

$$N(x_n)\to 0\text{mbox}butS(x_n)\nrightarrow 0\text{mbox}asn\to \infty .$$

$$S(x_n)\ge \kappa ,n=1,2,3,\dots .$$

$$N(x_n)\le \frac{1}{n^2},n=1,2,3,\dots .$$

$$y_n=n{x}_n,n=1,2,3,\dots .$$

$$N(y_n)\to 0\text{mbox}butS(y_n)\to \infty \text{mbox}asn\to \infty .$$

$$z_n=x+y_n,n=1,2,3,\dots ,$$

$$N(x-z_n)=N(y_n)\to 0.$$

$$S(x-z_n)=S(y_n)\to \infty ,$$

$$B=\left\{\frac{h}{S(h)}:h\in H,S(h)>0\right\}\cup \{h\in H:S(h)=0\}$$

$$x=\sum_{b\in B}\alpha_{b}(x)b,\qquad\alpha_{b}(x)\in\mbox{$\mathbb{F}$},$$

$$N(x)=\sum _{b\in B}|{\alpha }_b(x)|,x\in X,$$

$$S(b)\in \{0,1\}\text{mbox}forallb\in B.$$

$$S(x)=S\left(\sum _{b\in B}{\alpha }_b(x)b\right)\le \sum _{b\in B}|{\alpha }_b(x)|S(b)\le \sum _{b\in B}|{\alpha }_b(x)|=N(x).$$

$$|S(x)-S(y)|\le N(x-y),x,y\in X,$$

$$A=\{x\in X:S(x)>0\},$$

$$\mathrm{span}A=X.$$

$$S(y+z)=S(y+z)+S(y)\ge S(y+z-y)=S(z)>0,$$

$$b_n=\frac{n{h}_n}{S(h_n)},n=1,2,3,\dots ,$$

$$S(b_n)=n,n=1,2,3,\dots .$$

$$N(b_n)=1,n=1,2,3,\dots .$$

$$x_n=\frac{b_n}{n},n=1,2,3,\dots ,$$

$$N(x_n)=\frac{1}{n}\to 0,$$

$$S(x_n)=1,n=1,2,3,\dots .$$

$$S(x)\le \tau N(x)\text{mbox}forallx\in X.$$

Full text

1Corresponding author

Throughout this note let $X$ be a vector space over a field $\mathbb{F}$, either $\mathbb{R}$ or $\mathbb{C}$. As usual, a real-valued function N\colon X\to\mbox{\mathbb{R}} is said to be a norm on $X$ if, for all $x,y\in X$ and \alpha\in\mbox{\mathbb{F}},

[TABLE]

Further, a real-valued function S\colon X\to\mbox{\mathbb{R}} is a seminorm on $X$ if, for all $x,y\in X$ and \alpha\in\mbox{\mathbb{F}},

[TABLE]

hence, a seminorm is a norm if and only if it is positive-definite.

We recall that if $X$ is finite-dimensional, then all norms on $X$ are equivalent, hence inducing on $X$ a unique norm-topology. This well-known fact leads to the following result.

Contrary to the finite-dimensional case, if $X$ is infinite-dimensional then not all norms on $X$ are equivalent, and consequently we no longer have a unique norm-topology. Indeed, our main purpose in this note is to explore the questions of continuity and discontinuity of seminorms when the assumption of the finite-dimensionality is removed.

We begin our journey by posting the following lemma.

1. (a)

Continuity of $S$ at the origin implies ubiquitous continuity with respect to the topology induced by $N$. 2. (b)

Discontinuity of $S$ at the origin implies ubiquitous discontinuity with respect to the above mentioned topology.

Let $S$ be continuous at $x=0$ with respect to the topology induced by the given norm; hence, if $\{x_{n}\}_{n=1}^{\infty}$ is any sequence in $X$, then

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Choosing $x\in X$, and taking an arbitrary sequence $\{y_{n}\}_{n=1}^{\infty}$ for which $N(x-y_{n})\to 0$, (1) implies that,

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Thus, aided by the familiar inequality

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which follows from the subadditivity of $S$, we get,

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and part (a) of the lemma is in the bag.

As for part (b), the assumption that $S$ is discontinuous at $x=0$ implies that there exists a sequence $\{x_{n}\}_{n=1}^{\infty}$ in $X$ such that

[TABLE]

Since $S(x_{n})\not\to 0$ and $S$ is nonnegative, we can provide, by passing if necessary to a subsequence, a positive constant $\kappa$ such that

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Moreover, since $N(x_{n})\to 0$, by passing to yet another subsequence if necessary, we may assume that

[TABLE]

Now, put

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Since $N(y_{n})\leq\frac{1}{n}$ and $S(y_{n})\geq n\kappa$, we see that

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It follows that at any point $x\in X$, the sequence $\{z_{n}\}_{n=1}^{\infty}$ defined by

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satisfies

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On the other hand,

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which forces the discontinuity of $S$ at $x$. ∎

From the above lemma we may deduce the following result which hardly requires a proof.

1. (a)

$S$* is ubiquitously continuous in $X$ with respect to the topology induced by $N$ if and only if there exists some point of $X$ at which $S$ is continuous.* 2. (b)

Similarly, $S$ is ubiquitously discontinuous in $X$ with respect to the above mentioned topology if and only if there exists some point of $X$ at which $S$ is discontinuous.

We turn now to the main purpose of this note, which is to show that given a nontrivial seminorm $S$ on $X$, there exist two different norms, such that $S$ is ubiquitously continuous in $X$ with respect to one of them, and ubiquitously discontinuous with respect to the other.

1. (a)

There exists a norm with respect to which $S$ is ubiquitously continuous in $X$. 2. (b)

There exists a norm with respect to which $S$ is ubiquitously discontinuous in $X$.

[TABLE]

is a basis for $X$ as well. So every $x$ in $X$ assumes a unique representation of the form

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where $\{b\in B:\,\alpha_{b}(x)\neq 0\}$ is a finite set. With this representation at hand, we can easily confirm that the real-valued function

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is a norm on $X$. Furthermore, by the construction of $B$, we infer that

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Consequently, for every $x$ in $X$,

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By (2) and (5), therefore,

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and (a) follows.

(b) Consider the nonempty set

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and let us show that

[TABLE]

Indeed, assume on the contrary that ${\rm span}\,A\varsubsetneq X$. Then there exists an element $y\in X$ such that $y\not\in{\rm span}\,A$ and $S(y)=0$. Now, for every $z$ in $A$, we have

[TABLE]

whence, both $z$ and $y+z$ belong to $A$, so $y=(y+z)-z\in{\rm span}\,A$, a contradiction.

In view of (6), we may appeal to Corollary 4.2.2 in [K] (compare [H]), which tells us that $A$ contains a Hamel basis $H$ for $X$. So by the definition of $A$, we have $S(h)>0$ for all $h\in H$. With this in mind, we fix a sequence $\{h_{n}\}_{n=1}^{\infty}$ of distinct elements in $H$, and construct a new basis $B^{\prime}$ for $X$ by replacing $h_{n}$ by

[TABLE]

and leaving all other elements in $H$ unchanged.

Surely,

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Further, modifying the representation in (3) and the norm $N$ in (4) by replacing the underlying basis $B$ by the new basis $B^{\prime}$, we get,

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Hence, setting

[TABLE]

we obtain

[TABLE]

whereas

[TABLE]

This establishes the discontinuity of $S$ at $x=0$; so by Theorem 2(b), $S$ is discontinuous everywhere, and we are done. ∎

Falling back on Theorems 1 and 3(b), we may record now the following simple observation.

We conclude this note by recalling that a seminorm $S$ is left-equivalent to a norm $N$ on $X$ if there exists a constant $\tau>0$ such that

[TABLE]

Therefore, by (5), every seminorm $S$ on $X$ is left-equivalent to some norm on $X$; for example, to the norm in (4) (which, of course, depends on $S$). As indicated in [G], if $X$ is finite-dimensional, then $S$ is left-equivalent to every norm on X.