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Let $(S,d)$ be the space of all sequences in $\mathbb{R}$ with the metric

$$d(\mathbf{x},\mathbf{y})=\sum_{i=1}^{\infty}\dfrac{1}{2^i}\dfrac{|\xi_i-\eta_i|}{1+|\xi_i-\eta_i|}$$ where $\mathbf{x}=(\xi_i)$ and $\mathbf{y}=(\eta_i)$.

This is a complete metric space, but the metric does not come from a norm. Therefore the topology of $S$ cannot be defined by a norm.

My question is: does there exist any complete norm on the underlying vector space $S$ of all sequences in $\mathbb{R}$?

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    $\begingroup$ @Surb Exactly, I couldn't think of any norm for this space, but I thought that maybe there would be some way because it is complete in this metric. $\endgroup$
    – Mrcrg
    May 25, 2020 at 19:38
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    $\begingroup$ The topology defined by this metric is the product topology on $S = \mathbb{R}^{\mathbb{N}}$. Every closed and bounded (in the sense used in topological vector space theory) subset of $S$ is compact. In an infinite-dimensional normed space, the closed unit ball contains an infinite sequence with no Cauchy subsequence, and so is not compact. Therefore the topology of $S$ cannot be defined by a norm. $\endgroup$ May 26, 2020 at 2:06
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    $\begingroup$ Or are you asking if there exists any complete norm on the underlying vector space $S$, defining some other topology? This is true because $S$ is just a real vector space of (Hamel) dimension $2^{\aleph_0}$, so is (discontinuously) isomorphic to e.g. $\ell^2$, which is complete in its norm. $\endgroup$ May 26, 2020 at 2:09
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    $\begingroup$ @RobertFurber Would be great if you could transform your comments into an answer. $\endgroup$
    – Surb
    May 26, 2020 at 8:48
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    $\begingroup$ @Surb OK, I'll turn them into an answer with references to textbooks. I wanted to see which interpretation Mrcrg wanted, but it seems people are interested in both, so I'll put both. $\endgroup$ May 26, 2020 at 17:42

3 Answers 3

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There are two interpretations of the question possible:

  1. Is there a norm $\|\cdot\|$ on $S$ such that the topology defined by $\|\cdot\|$ agrees with the topology defined by $d$, and $(E, \|\cdot\|)$ is a Banach space?
  2. Is there a norm $\| \cdot \|$ on the vector space $S$ such that $(S, \|\cdot\|)$ is a Banach space?

I will show, with references, that the answer to 1 is no, even if we do not require that $(E, \|\cdot\|)$ be complete, and the answer to 2 is yes, using the axiom of choice in the form of "every vector space has a basis".


Answer to number 1:

The first thing to observe is that the topology defined by $d$ is the same as the product topology on $S$ (equivalently, the topology of pointwise convergence). Continuity of the identity map from $(S,d)$ to $\mathbb{R}^{\mathbb{N}}$ is most easily proved by showing that the projection mappings $\pi_n : S \rightarrow \mathbb{R}$ defined by $\pi_n((x_i)) = x_n$ are continuous. Proving that the identity mapping is continuous in the other direction is most easily done using the observation that $0 \leq \frac{|x|}{1 + |x|} < 1$ for all real $x$, which means $d(0,(x_i))$ can be bounded by bounding only finitely many coordinates.

Then we need the fact that every closed bounded subset of $S$ is compact. Let's recall the definitions. Recall that in a topological vector space $E$, a 0-neighbourhood $N \subseteq E$ is a set such that there exists an open set $U$ such that $0 \in U \subseteq N$. A set $B \subseteq E$ is bounded if for all 0-neighbourhoods $N$ there exists a real number $\alpha > 0$ such that $B \subseteq \alpha N$. If $f : E \rightarrow F$ is a continuous linear map between topological vector spaces, and $B \subseteq E$ is bounded, then $f(B)$ is bounded (see e.g. Schaefer's Topological Vector Spaces I.5.4, or prove it directly using the definition of continuity of linear maps in terms of 0-neighbourhoods). So if $B \subseteq S$ is bounded, for all $n \in \mathbb{N}$, $\pi_n(S) \subseteq \mathbb{R}$ is bounded. Let $B$ be a closed, bounded set. Since $[-1,1]$ is a 0-neighbourhood in $\mathbb{R}$, for each $n \in \mathbb{N}$, the set $\pi_n(B) \subseteq [-\alpha_n,\alpha_n]$ for some real $\alpha_n > 0$. It follows that $B \subseteq \prod_{i=1}^\infty [-\alpha_n,\alpha_n]$, which is compact in $S$ by Tychonoff's theorem (because $(S,d)$ has the product topology). As $B$ is closed, it is also compact.

Now, in a normed space $(E, \|\cdot\|)$, the closed unit ball $U$ is bounded (every 0-neighbourhood $N$ contains an open ball of radius $\epsilon > 0$, so $U \subseteq (\epsilon^{-1} + 1)N$). By Riesz's lemma, if $E$ is infinite-dimensional, then $U$ contains a sequence for which the distance between the elements is bounded below by some number, which therefore has no Cauchy subsequence, and therefore no convergent subsequence (even in an incomplete metric space, every convergent sequence is Cauchy). In a compact metric space, every sequence has a convergent subsequence, so this proves $U$ is not compact. Therefore there is no norm defining the topology of $(S,d)$.


Answer to number 2:

Since we are disregarding the original topology of $S$ in this section, we will be using the non-topological notions of basis and dimension. If $E$ is a vector space over $\mathbb{R}$, we say that a family $(x_i)_{i \in I}$ is a (Hamel) basis if:

  1. $(x_i)_{i \in I}$ spans $E$, i.e. for each $x \in E$ there is a finite set $K \subseteq I$ and a family $(\alpha_i)_{i \in K}$ with each $\alpha_i \in \mathbb{R}$ such that $x = \sum_{i \in K} \alpha_i x_i$.
  2. $(x_i)_{i \in I}$ is linearly independent, i.e. for all finite sets $K \subseteq I$ and families $(\alpha_i)_{i \in K}$ with $\alpha_i \in \mathbb{R}$ such that $\sum_{i \in K}\alpha_i x_i = 0$, we have $\alpha_i = 0$ for all $i \in K$.

The key facts are that every linearly independent set can be extended to a basis, all bases of a vector space $E$ have the same cardinality (called the dimension of $E$), and vector spaces are linearly isomorphic iff they have the same dimension. Unfortunately, standard references on linear algebra usually only prove these facts for the finite-dimensional case, but the general case is treated in Chapter IX of volume II of Jacobson's Lectures in Abstract Algebra.

It is clear that the dimension of $E$ is $\leq |E|$, because a linearly independent set in $E$ is a subset of $E$. We have $$ |S| = |\mathbb{R}|^{|\mathbb{N}|} = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0 \times \aleph_0} = 2^{\aleph_0}, $$ so the dimension of $S$ is $\leq 2^{\aleph_0}$. For any separable Banach space $E$, such as $\ell^2$, every element is the limit of a sequence from a countable dense subset, so $\mathbb{N}^{\mathbb{N}}$ maps surjectively onto $E$, so $E$ also has cardinality and therefore dimension $\leq 2^{\aleph_0}$.

In his paper On Infinite-dimensional Linear Spaces, Mackey proves in Theorem I.1 that every infinite-dimensional Banach space is of dimension $\geq 2^{\aleph_0}$. He does this in two steps. The first is to show that for every infinite-dimensional Banach space $E$, there exists an injective linear map $f : \ell^\infty \rightarrow E$. The second is the observation that the $(0,1)$-indexed family $((\alpha^i)_{i \in \mathbb{N}})_{\alpha \in (0,1)}$ in $\ell^\infty$ is linearly independent, making the dimension of $E$ greater than or equal to $|(0,1)| = 2^{\aleph_0}$. Therefore any separable infinite-dimensional Banach space has dimension exactly $2^{\aleph_0}$. Since $\ell^\infty$ is a linear subspace of $S$, this also proves that $S$ has dimension exactly $2^{\aleph_0}$.

So by mapping a basis of $S$ onto a basis of $\ell^2$ and extending by linearity, we can define a linear isomorphism $f : S \rightarrow \ell^2$. Let's use $\| \cdot \|_2$ for the norm on $\ell^2$ and define, for $x \in S$, $\|x\|_1 = \|f(x)\|_2$. The linearity of $f$ ensures that $\|\cdot\|_1$ is a norm. If $(x_i)_{i \in \mathbb{N}}$ is $\|\cdot\|_1$-Cauchy, define $y_i = f(x_i)$, and expanding the definitions shows that $(y_i)_{i \in \mathbb{N}}$ is $\|\cdot\|_2$-Cauchy, so converges to some $y \in \ell^2$. Since $f$ is an isomorphism, there exists $x \in S$ such that $f(x) = y$, and expanding the definitions and using linearity of $f$ shows that $(x_i)_{i \in \mathbb{N}}$ converges to $x$ in $\|\cdot\|_1$. Therefore $(S,\|\cdot\|_1)$ is a Banach space.

Some examples in mathematics are mainly used to show why things are defined the way they are, rather than for practical use. This is one of those examples.

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This is not an answer, but it is too long for a comment.

I don't know if a Banach norm $\lVert\cdot\rVert$ exists on $S$. However, if it exists, it must be a "crazy" norm, because

$\lVert\cdot\rVert$ cannot be compatible with pointwise convergence.

By this, I mean that the following cannot be true; $$\tag{*} \lVert x(n)-x\rVert \to 0 \Rightarrow \lvert x_k(n)-x_k\rvert\to 0,\ \forall k\in\mathbb N.$$ (Here I denote a sequence by $x$, its $k$-th entry by $x_k$, and a sequence of sequences by $x(n)$).

Indeed, if (*) is true, then the identity map $$I\colon (S, \lVert\cdot\rVert)\to (S, d)$$ would be a continuous linear operator. Since it is also obviously bijective, by the open mapping theorem it would be an isomorphism. But this cannot be true, because $(S, d)$ is not normable (see this comment of Robert Furber).

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    $\begingroup$ The solution is actually much simpler hat what has been stated. The space in the problem at hand can be described easily in terms of convergence in probability on the space $(\mathbb{N},2^{\mathbb{N}},\mu)$ where $\mu(\{n\})=2^{-n}$. It turns out that $(S,d)$ is the same as the space $L_0$, the space of real measurable functions equipped with the topology of convergence in probability. It is not difficult to show that in the present setting, the ball $B(0;1/2)$ in the given metric is not bounded. $\endgroup$
    – Mittens
    May 28, 2020 at 4:49
  • $\begingroup$ @OliverDiaz: your interesting reasoning proves that $(S, d)$ is not normable, if I understand correctly. I have one question, though. What do you mean by "bounded"? In my head, a ball in a metric space is always bounded, by definition. I guess you mean to say that $B(0, 1/2)$ contains a line? (EDIT: now I see that you modified your answer, I understand what you mean). $\endgroup$ May 28, 2020 at 13:35
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    $\begingroup$ bounded in the sense of topological linear spaces, that is $E$ is bounded if for any open neighborhood $W$ of $0$, there is $s_0>0$ such that $E\subset sW$ for all $s.s_0$. (It is enough with $E\subset s_0$ but that irrelevant at this point) The punchline is that in for the linear $F$ space (topological linear space with translation invariant metric) $X$, a set $E\subset X$ is bounded iff $\lim_{\lambda\rightarrow0}\sup_{x\in E}d(\lambda x,0)=0$. $\endgroup$
    – Mittens
    May 28, 2020 at 13:50
  • $\begingroup$ There is another interesting example along the lines of this problem. Consider $L_0$ in the Lebesgue space $((0,1),\mathscr{M},\lambda)$. $d(X,Y)=\mathbb{E}[|X-Y|\wedge1]$ is also a complete translation invariant metric (once one identify functions that are $\lambda$--a.s. equal). Convergence in $d$ is equivalent to convergence in probability. The pathology here is that space $L_0(\lambda)$ is not locally convex. In the problem that was asked, $\ell_0$ (to make a distinction) is locally convex. This is another instance of $L_0$ not being normable. $\endgroup$
    – Mittens
    May 28, 2020 at 14:56
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The answer to your question is no.


Some background:

There are linear topological spaces that are metrizable (with complete metric) but not normable. A typical example is the space of real measurable functions $L_0$ in a probability space $(\Omega,\mathscr{F},\mathbb{P})$ with the topology of convergence in probability. This is generated by the metric $d(X,Y)=\mathbb{E}[\rho(X,Y)]$ where $\phi$ is any complete bounded metric in $\mathbb{R}$, which happens to be complete. In infinite dimensional spaces, $L_0$ fails to be either locally convex or locally bounded. The key point is that any normable space is locally convex and locally bounded.


In the example of interest, consider the probability space $(\mathbb{N},2^{\mathbb{N}},\mathbb{P})$ where $\Pr[\{n\}]=2^{-n}$ for $n\in\mathbb{N}$. Notice that $S=L_0(\mathbb{N},2^{\mathbb{N}},\mathbb{P})$. Take for instance, $\rho(x,y)=\frac{|x-y|}{1+|x-y|}$ defines a complete metric on $\mathbb{R}$. For any random variable (same as real-valued sequences in this case) $\mathbf{x},\mathbf{y}$ $$ d(\mathbf{x},\mathbf{y})=\mathbb{E}_\mu[\rho(\mathbf{x},\mathbf{y})]=\sum^\infty_{n=1}2^{-n}\rho(\mathbf{x}(n),\mathbf{y}(n)) $$

As a side note, convergence in probability is the same as point wise convergence in this case.

Since the topology of $L_0$ is generated by the balls $B(0;r)$, $r>0$, to see that $L_0$ is not normable, it suffices to check that no ball is bounded (in the sense of topological linear spaces). Consider the family of sequences $E_n=\{f_{c,n}=c\mathbb{1}_{\{(n,\infty)}:c>0\}$ where $n\in\mathbb{N}$. Since $\frac{x}{1+x}\leq x\wedge1\leq 2\frac{x}{1+x}$, the metric $d_1(\mathbf{x},\mathbf{y})=\mathbb{E}[|\mathbf{x}-\mathbf{y}|\wedge1]$ is equivalent to $d$. Notice that

$$d_1(f_{c,n},0)=\mathbb{E}[f_{n,c}\wedge1]=\left\{\begin{array}{lc} 2^{-n} & \text{if} & c\geq1\\ c2^{-n} &\text{if} & c\leq1 \end{array} \right. $$ and so, $E_n$ is contained in the unit ball $B_{d_1}(0;2^{-n})$. However,

$$ \lim_{\lambda\rightarrow0}\sup_{c>0}d_1(\lambda f_{c,n},0)=2^{-n}\neq0 $$ Meaning that no $E_n$, and thus no ball $B(0;2^{-n})$, is bounded (in the topological linear sense).

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    $\begingroup$ How does that answer the question? He asks about a specific case, not a counterexample. $\endgroup$ May 25, 2020 at 21:48

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