Generalization of a Real-Analysis Result to a Class of Topological Vector Spaces

TL;DR

This paper extends a fundamental real-analysis theorem to a broader class of topological vector spaces and provides a counterexample where the generalization does not hold.

Contribution

It introduces a generalization of a classical real-analysis result to topological vector spaces and identifies limitations through a specific counterexample.

Findings

The theorem is successfully generalized to certain topological vector spaces.

A specific example shows the limits of this generalization.

The paper clarifies conditions under which the result applies.

Abstract

In this paper, we generalize an elementary real-analysis result to a class of topological vector spaces. We also give an example of a topological vector space to which the result cannot be generalized.

Equations

$${\left(2x_{n+1}-x_n\right)}_{n\in \mathbb{N}}\stackrel{\tau }{⟶}2y-y=y,$$

$${\left(0_V\right)}_{n\in \mathbb{N}}={\left(\left(2x_{n+1}-x_n\right)-\left(2x_{n+1}-x_n\right)\right)}_{n\in \mathbb{N}}\stackrel{\tau }{⟶}x-y,$$

$${\left(y\right)}_{n\in \mathbb{N}}={\left(0_V+y\right)}_{n\in \mathbb{N}}\stackrel{\tau }{⟶}\left(x-y\right)+y=x.$$

$$\forall x\in V:2^p\sigma \left(x\right)$$

$$=\sigma \left(x+x\right)$$

$$\le 2\sigma \left(x\right),\left(\text{By the Triangle Inequality.}\right)$$

$$\forall x\in V,\forall \sigma \in \mathcal{S},\forall ϵ\in {\mathbb{R}}_{>0}:{\mathcal{U}}_{x,\sigma ,ϵ}\stackrel{\mathrm{df}}{=}\left\{y\in V|\sigma \left(y-x\right)<ϵ\right\}.$$

$$\left\{{\mathcal{U}}_{x,\sigma ,ϵ}\in \mathcal{P}\left(V\right)|\left(x,\sigma ,ϵ\right)\in V\times \mathcal{S}\times {\mathbb{R}}_{>0}\right\}.$$

$${\left(x_{\lambda }\right)}_{\lambda \in \Lambda }\stackrel{{\tau }_{\mathcal{S}}}{⟶}x⟺\underset{\lambda \in \Lambda }{\lim }\sigma \left(x_{\lambda }-x\right)=0\text{for all}\sigma \in \mathcal{S}.$$

$$\forall n\in \mathbb{N}:2y_{n+1}-y_n$$

$$=2x_{n+1}-2x-x_n+x$$

$$=\left(2x_{n+1}-x_n\right)-x.$$

$${\left(2y_{n+1}-y_n\right)}_{n\in \mathbb{N}}={\left(\left(2x_{n+1}-x_n\right)-x\right)}_{n\in \mathbb{N}}\stackrel{{\tau }_{\mathcal{S}}}{⟶}x-x=0_V,$$

$$\forall n\in {\mathbb{N}}_{\ge N}:\sigma \left(2x_{n+1}-x_n\right)=\sigma \left(\left(2x_{n+1}-x_n\right)-0_V\right)<\left(2^p-1\right)ϵ.$$

$$\forall k\in \mathbb{N}:\sigma \left(2^k{x}_{N+k}-2^{k-1}{x}_{N+k-1}\right)$$

$$=2^{\left(k-1\right)p}\sigma \left(2x_{N+k}-x_{N+k-1}\right)$$

$$<2^{\left(k-1\right)p}\left(2^p-1\right)ϵ.$$

$$\forall m\in \mathbb{N}:\sigma \left(2^m{x}_{N+m}-x_N\right)$$

$$\le \sum _{k=1}^m\sigma \left(2^k{x}_{N+k}-2^{k-1}{x}_{N+k-1}\right)$$

$$<\sum _{k=1}^m{2}^{\left(k-1\right)p}\left(2^p-1\right)ϵ$$

$$=\left(2^{mp}-1\right)ϵ.$$

$$\forall m\in \mathbb{N}:\sigma \left(x_{N+m}-\frac{1}{2^m}{x}_N\right)$$

$$=\frac{1}{2^{mp}}\sigma \left(2^m{x}_{N+m}-x_N\right)$$

$$<\left(1-\frac{1}{2^{mp}}\right)ϵ.$$

$$\forall m\in \mathbb{N}:\sigma \left(x_{N+m}\right)<\sigma \left(\frac{1}{2^m}{x}_N\right)+\left(1-\frac{1}{2^{mp}}\right)ϵ=\frac{1}{2^{mp}}\sigma \left(x_N\right)+\left(1-\frac{1}{2^{mp}}\right)ϵ.$$

$$\underset{n\to \infty }{\mathrm{limsup}}\sigma \left(x_n\right)=\underset{m\to \infty }{\mathrm{limsup}}\sigma \left(x_{N+m}\right)\le \underset{m\to \infty }{\mathrm{limsup}}\left[\frac{1}{2^{mp}}\sigma \left(x_N\right)+\left(1-\frac{1}{2^{mp}}\right)ϵ\right]=ϵ.$$

$$\underset{n\to \infty }{\lim }\sigma \left(x_n-0_V\right)=\underset{n\to \infty }{\lim }\sigma \left(x_n\right)=0.$$

$$\underset{\lambda \in \Lambda }{\lim }\mathsf{P}\left(\left\{\omega \in \Omega |\left|X_{\lambda }\left(\omega \right)-X\left(\omega \right)\right|>ϵ\right\}\right)=0,$$

$$\forall X,Y\in \mathsf{RV}\left(\Omega ,\Sigma \right):{\rho }_{\mathsf{P}}\left(X,Y\right)\stackrel{\mathrm{df}}{=}{\int }_{\Omega }\frac{\left|X-Y\right|}{1+\left|X-Y\right|}\mathrm{d}\mathsf{P}.$$

$${\left(X_{\lambda }\right)}_{\lambda \in \Lambda }\stackrel{\mathsf{P}}{⟶}X⟺{\left(X_{\lambda }\right)}_{\lambda \in \Lambda }\stackrel{{\tau }_{\mathsf{P}}}{⟶}X.$$

$$\forall A\subseteq \left[k\right]:{\mathsf{c}}_k\left(A\right)\stackrel{\mathrm{df}}{=}\frac{\mathsf{Card}\left(A\right)}{k},$$

$$\forall n\in \mathbb{N}:S_n\stackrel{\mathrm{df}}{=}\left\{\mathbf{v}\in \prod _{k=1}^{\infty }\left[k\right]|\mathbf{v}\left(n\right)=1\right\}.$$

$$\forall n\in \mathbb{N}:Y_n\stackrel{\mathrm{df}}{=}{2}^n{\chi }_{S_n},$$

$$\underset{n\to \infty }{\lim }\mathsf{P}\left(\left\{\omega \in \Omega |\left|Y_n\left(\omega \right)\right|>ϵ\right\}\right)=\underset{n\to \infty }{\lim }\mathsf{P}\left(S_n\right)=\underset{n\to \infty }{\lim }\frac{1}{n}=0.$$

$$\forall n\in \mathbb{N}:X_n\stackrel{\mathrm{df}}{=}\{\begin{array}{ll}{0}_{\Omega \to \mathbb{R}}& \text{if}n=1;\\ \sum _{k=1}^{n-1}\frac{1}{2^{n-k}}{Y}_k& \text{if}n\ge 2.\end{array}$$

$$\forall n\in {\mathbb{N}}_{\ge 2}:2X_{n+1}-X_n$$

Full text

1. Introduction

This paper draws its inspiration from the following result, which appears to be a popular real-analysis exam problem (see [3], for example):

Let $\left(x_{n}\right)_{n\in\mathbb{N}}$ be a sequence in $\mathbb{R}$. If $\displaystyle\lim_{n\to\infty}\left(2x_{n+1}-x_{n}\right)=x$ for some $x\in\mathbb{R}$, then $\displaystyle\lim_{n\to\infty}x_{n}=x$.

A quick proof can be given using the Stolz-Cesàro Theorem.

A natural question to ask is: Is this result still valid if $\mathbb{R}$ is replaced by another topological vector space? The answer happens to be affirmative for a wide class of topological vector spaces that includes all the locally convex ones.

We will also exhibit a topological vector space for which the result is not valid, which indicates that it is rather badly behaved.

In this paper, we adopt the following conventions:

• •

$\mathbb{N}$ denotes the set of all positive integers, and for each $n\in\mathbb{N}$, let $\left[n\right]\stackrel{{\scriptstyle\operatorname{df}}}{{=}}\mathbb{N}_{\leq n}$.

• •

All vector spaces are over the field $\mathbb{K}\in\left\{\mathbb{R},\mathbb{C}\right\}$.

2. Good topological vector spaces

Recall that a topological vector space is an ordered pair $\left(V,\tau\right)$, where:

• •

$V$ is a vector space, and

• •

$\tau$ is a topology on $V$, under which vector addition and scalar multiplication are continuous operations.

Definition 2.1**.**

Let $\left(V,\tau\right)$ be a topological vector space, and $\left(x_{\lambda}\right)_{\lambda\in\Lambda}$ a net in $V$. Then $x\in V$ is called a $\tau$-limit for $\left(x_{\lambda}\right)_{\lambda\in\Lambda}$ — which we write as $\left(x_{\lambda}\right)_{\lambda\in\Lambda}\stackrel{{\scriptstyle\tau}}{{\longrightarrow}}x$ — if and only if for each $\tau$-neighborhood $U$ of $x$, there is a $\lambda_{0}\in\Lambda$ such that $x_{\lambda}\in U$ for all $\lambda\in\Lambda_{\geq\lambda_{0}}$.

Remark 2.2**.**

We do not assume that $\tau$ is a Hausdorff topology on $V$.

Definition 2.3**.**

A topological vector space $\left(V,\tau\right)$ is said to be good if and only if any sequence $\left(x_{n}\right)_{n\in\mathbb{N}}$ in $V$ has a $\tau$-limit whenever $\left(2x_{n+1}-x_{n}\right)_{n\in\mathbb{N}}$ has a $\tau$-limit.

A topological vector space that is not good is said to be bad.

Proposition 2.4

Let $\left(V,\tau\right)$ be a topological vector space, and $\left(x_{n}\right)_{n\in\mathbb{N}}$ a sequence in $V$ such that $\left(2x_{n+1}-x_{n}\right)_{n\in\mathbb{N}}\stackrel{{\scriptstyle\tau}}{{\longrightarrow}}x$ for some $x\in V$. Then either

• •

$\left(x_{n}\right)_{n\in\mathbb{N}}\stackrel{{\scriptstyle\tau}}{{\longrightarrow}}x$* also, or*

• •

$\left(x_{n}\right)_{n\in\mathbb{N}}$* has no $\tau$-limit.*

Proof.

If $\left(x_{n}\right)_{n\in\mathbb{N}}$ has no $\tau$-limit, then we are done.

Next, suppose that $\left(x_{n}\right)_{n\in\mathbb{N}}\stackrel{{\scriptstyle\tau}}{{\longrightarrow}}y$ for some $y\in V$. Then

[TABLE]

so $y$ is a $\tau$-limit for $\left(2x_{n+1}-x_{n}\right)_{n\in\mathbb{N}}$ in addition to $x$. It follows that

[TABLE]

which yields

[TABLE]

Therefore, any $\tau$-neighborhood of $x$ also contains $y$, giving us $\left(x_{n}\right)_{n\in\mathbb{N}}\stackrel{{\scriptstyle\tau}}{{\longrightarrow}}x$. ∎

2.4 tells us: To prove that a topological vector space $\left(V,\tau\right)$ is good, it suffices to prove that for each sequence $\left(x_{n}\right)_{n\in\mathbb{N}}$ in $V$, if $\left(2x_{n+1}-x_{n}\right)_{n\in\mathbb{N}}\stackrel{{\scriptstyle\tau}}{{\longrightarrow}}x$ for some $x\in V$, then $\left(x_{n}\right)_{n\in\mathbb{N}}\stackrel{{\scriptstyle\tau}}{{\longrightarrow}}x$ also.

Definition 2.5**.**

Let $p\in\left(0,1\right]$. A $p$-homogeneous seminorm on a vector space $V$ is then a function $\sigma:V\to\mathbb{R}_{\geq 0}$ with the following properties:

1. (1)

The Triangle Inequality: $\sigma\left(x+y\right)\leq\sigma\left(x\right)+\sigma\left(y\right)$ for all $x,y\in V$. 2. (2)

$p$-Homogeneity: $\sigma\left(kx\right)=\left|k\right|^{p}\sigma\left(x\right)$ for all $k\in\mathbb{K}$ and $x\in V$.

Remark 2.6**.**

• •

By letting $k=0$ and $x=0_{V}$ in (2), we find that $\sigma\left(0_{V}\right)=0$.

• •

A $1$-homogeneous seminorm is the same as a seminorm in the ordinary sense.

• •

No extra generality is gained by postulating that $\sigma\left(kx\right)\leq\left|k\right|^{p}\sigma\left(x\right)$ for all $k\in\mathbb{K}$ and $x\in V$. If $k\in\mathbb{K}\setminus\left\{0\right\}$, then replacing $k$ by $\dfrac{1}{k}$ gives us the reverse inequality, which leads to equality; if $k=0$, then equality automatically holds.

• •

We do not consider $p\in\left(2,\infty\right)$ because

[TABLE]

so if $\sigma$ is non-trivial, then $2^{p}\leq 2$, which implies that $p\in\left(0,1\right]$ if $p\in\mathbb{R}_{>0}$.

Let $V$ be a vector space, and $\mathcal{S}$ a collection of $p$-homogeneous seminorms on $V$ where $p\in\left(0,1\right]$ may not be fixed. Define a function $\mathcal{U}:V\times\mathcal{S}\times\mathbb{R}_{>0}\to\mathcal{P}\left(V\right)$ by

[TABLE]

Then let $\tau_{\mathcal{S}}$ denote the topology on $V$ that is generated by the sub-base

[TABLE]

Proposition 2.7

The following statements about $\tau_{\mathcal{S}}$ hold:

1. (1)

$\tau_{\mathcal{S}}$* is a vector-space topology on $V$.* 2. (2)

Let $\left(x_{\lambda}\right)_{\lambda\in\Lambda}$ be a net in $V$. Then for each $x\in V$, we have

[TABLE]

Proof.

One only has to imitate the proof in the case of locally convex topological vector spaces that the initial topology generated by a collection of seminorms is a vector-space topology. We refer the reader to Chapter 1 of [2] for details. ∎

Proposition 2.8

$\left(V,\tau_{\mathcal{S}}\right)$* is a good topological vector space.*

Proof.

Let $\left(x_{n}\right)_{n\in\mathbb{N}}$ be a sequence in $V$. Suppose that $\left(2x_{n+1}-x_{n}\right)_{n\in\mathbb{N}}\stackrel{{\scriptstyle\tau_{\mathcal{S}}}}{{\longrightarrow}}x$ for some $x\in V$. Then without loss of generality, we may assume that $x=0_{V}$. To see why, define a new sequence $\left(y_{n}\right)_{n\in\mathbb{N}}$ in $V$ by $y_{n}\stackrel{{\scriptstyle\operatorname{df}}}{{=}}x_{n}-x$ for all $n\in\mathbb{N}$, so that

[TABLE]

Hence,

[TABLE]

so if we can prove that $\left(y_{n}\right)_{n\in\mathbb{N}}\stackrel{{\scriptstyle\tau_{\mathcal{S}}}}{{\longrightarrow}}0_{V}$, then $\left(x_{n}\right)_{n\in\mathbb{N}}\stackrel{{\scriptstyle\tau_{\mathcal{S}}}}{{\longrightarrow}}x$ as desired.

Let $\sigma\in\mathcal{S}$ and $\epsilon>0$, and suppose that $\sigma$ is $p$-homogeneous for some $p\in\left(0,1\right]$. Then by (2) of 2.7, there is an $N\in\mathbb{N}$ such that

[TABLE]

By $p$-homogeneity, we thus have

[TABLE]

Next, a telescoping sum in conjunction with the Triangle Inequality yields

[TABLE]

Then by $p$-homogeneity again,

[TABLE]

Applying the Triangle Inequality and $p$-homogeneity once more, we get

[TABLE]

Consequently,

[TABLE]

As $\epsilon>0$ is arbitrary, we obtain

[TABLE]

Finally, as $\sigma\in\mathcal{S}$ is arbitrary, (2) of 2.7 says that $\left(x_{n}\right)_{n\in\mathbb{N}}\stackrel{{\scriptstyle\tau_{\mathcal{S}}}}{{\longrightarrow}}0_{V}$. ∎

By 2.8, the class of good topological vector spaces includes:

• •

All locally convex topological vector spaces.

• •

All $L^{p}$-spaces for $p\in\left(0,1\right)$, which are generally not locally convex.

In the next section, we will give an example of a bad topological vector space.

3. A bad topological vector space from probability theory

Before we present the example, let us first fix some probabilistic terminology.

Definition 3.1**.**

Let $\left(\Omega,\Sigma,\mathsf{P}\right)$ be a probability space.

• •

A measurable function from $\left(\Omega,\Sigma\right)$ to $\left(\mathbb{R},\mathscr{B}\left(\mathbb{R}\right)\right)$ is called a random variable.111$\mathscr{B}\left(\mathbb{R}\right)$ denotes the Borel $\sigma$-algebra generated by the standard topology on $\mathbb{R}$.

• •

The $\mathbb{R}$-vector space of random variables on $\left(\Omega,\Sigma\right)$ is denoted by $\mathsf{RV}\left(\Omega,\Sigma\right)$.

• •

Let $\left(X_{\lambda}\right)_{\lambda\in\Lambda}$ be a net in $\mathsf{RV}\left(\Omega,\Sigma\right)$, and let $X\in\mathsf{RV}\left(\Omega,\Sigma\right)$. Then $\left(X_{\lambda}\right)_{\lambda\in\Lambda}$ is said to converge in probability to $X$ (for $\mathsf{P}$) if and only if for each $\epsilon>0$, we have

[TABLE]

in which case, we write $\left(X_{\lambda}\right)_{\lambda\in\Lambda}\stackrel{{\scriptstyle\mathsf{P}}}{{\longrightarrow}}X$.

The following theorem says that convergence in probability is convergence with respect to a vector-space topology on the vector space of random variables.

Theorem 3.2

Let $\left(\Omega,\Sigma,\mathsf{P}\right)$ be a probability space, and define a pseudo-metric $\rho_{\mathsf{P}}$ on $\mathsf{RV}\left(\Omega,\Sigma\right)$ by

[TABLE]

Then the topology $\tau_{\mathsf{P}}$ on $\mathsf{RV}\left(\Omega,\Sigma\right)$ generated by $\rho_{\mathsf{P}}$ has the following properties:

• •

$\tau_{\mathsf{P}}$* is a vector-space topology.*

• •

Let $\left(X_{\lambda}\right)_{\lambda\in\Lambda}$ be a net in $\mathsf{RV}\left(\Omega,\Sigma\right)$. Then for each $X\in\mathsf{RV}\left(\Omega,\Sigma\right)$, we have

[TABLE]

Proof.

Please refer to Problems 6, 10 and 14 in Section 5.2 of [1]. ∎

Now, for each $k\in\mathbb{N}$, define a probability measure $\mathsf{c}_{k}$ on $\left(\left[k\right],\mathcal{P}\left(\left[k\right]\right)\right)$ by

[TABLE]

and let $\left(\Omega,\Sigma,\mathsf{P}\right)$ denote the product probability space $\displaystyle\prod_{k=1}^{\infty}\left(\left[k\right],\mathcal{P}\left(\left[k\right]\right),\mathsf{c}_{k}\right)$. Define a sequence $\left(S_{n}\right)_{n\in\mathbb{N}}$ in $\Sigma$ by

[TABLE]

Then $\mathsf{P}\left(S_{n}\right)=\dfrac{1}{n}$ for all $n\in\mathbb{N}$, and the $S_{n}$’s form mutually-independent events.

Next, define a sequence $\left(Y_{n}\right)_{n\in\mathbb{N}}$ in $\mathsf{RV}\left(\Omega,\Sigma\right)$ by

[TABLE]

where $\chi_{S_{n}}$ denotes the indicator function of $S_{n}$. Then we get for each $\epsilon>0$ that

[TABLE]

The first equality is obtained because, for each $\epsilon>0$, we have $2^{n}>\epsilon$ for all $n\in\mathbb{N}$ large enough. Consequently, $\left(Y_{n}\right)_{n\in\mathbb{N}}\stackrel{{\scriptstyle\mathsf{P}}}{{\longrightarrow}}0_{\Omega\to\mathbb{R}}$.

Define a new sequence $\left(X_{n}\right)_{n\in\mathbb{N}}$ in $\mathsf{RV}\left(\Omega,\Sigma\right)$ by

[TABLE]

Then $2X_{2}-X_{1}=2X_{2}=Y_{1}$, and

[TABLE]

It follows that $\left(2X_{n+1}-X_{n}\right)_{n\in\mathbb{N}}=\left(Y_{n}\right)_{n\in\mathbb{N}}\stackrel{{\scriptstyle\mathsf{P}}}{{\longrightarrow}}0_{\Omega\to\mathbb{R}}$.

Gathering what we have thus far, observe that

[TABLE]

As the $S_{k}$’s are mutually independent, their complements are as well, so

[TABLE]

Hence, $\left(X_{n}\right)_{n\in\mathbb{N}}$ does not converge to $0_{\Omega\to\mathbb{R}}$ in probability. By 3.2:

Proposition 3.3

$\left(\mathsf{RV}\left(\Omega,\Sigma\right),\tau_{\mathsf{P}}\right)$* is therefore a bad topological vector space.*

By 2.4, $\left(X_{n}\right)_{n\in\mathbb{N}}$ does not, in fact, converge in probability at all.

4. Acknowledgments

The author would like to express his deepest thanks to Dr. Jochen Wengenroth for communicating his example above of a bad topological vector space.