Шар в Банаховом пространстве

[Solunac]

Does ball in a Banach space always have an infinite number of elements of the space?
Is there an exception?

Спасибо.

[Профессор Снэйп]

Solunac в сообщении #257665 писал(а):

Is there an exception?

Yes, there is. $\mathbb{R} = \mathbb{R}^1$ is a Banach space with two-element ball

-- Пн ноя 02, 2009 22:18:40 --

P. S. Oh, sorry It is an example for sphere, not for ball. The only exception is the ball of radius zero

P. P. S. ball = шар $\neq$ бал = dancing party

[Circiter]

2Solunac

Цитата:

Is there an exception?

Let $\Sigma$ be a discrete space such that any finite subset of $\Sigma$ is countable (e.g. $\Sigma$ can be discrete subset of an Euclidean space). Since the product of a countable number of copies of $\Sigma$ is complete (up to homeomorphism, at least), the space $\Sigma^N$ for any $N\in\mathbb{N}$ can be viewed as a Banach space (with respect to an apropriate norm).

So, $\Sigma^N$ is Banach and any ball in this space have only a finite number of points (due to its discreteness). But I can be mistaken...

[id]

I guess the answer depends heavily on the field, over which our linear space is defined.
If the field $\mathbb{F}$ is an infinite, complete normed field (like $\mathbb{R} , \mathbb{C}$), it is hard to imagine a counterexample.

However, there are finite discrete fields as well.

[Solunac]

Thanks, I see... So it is possible to construct space $L^1(X,M,\mu)$ that for some measurement it has a ball with only finite number of points?

[id]

If the $\sigma$-algebra is very poor ( finite, for instance ) and the field $\mathbb{F}$ is finite as well ( functions must return values from this field, too), it looks possible.

Though I am just uncertain about this as I am about the usefullness of this counterexample.

[Circiter]

2Solunac

Цитата:

So it is possible to construct space $L^1(X,M,\mu)$ that for some measurement it has a ball with only finite number of points?

What if we define $L^1(X,M,\mu)$ over measure space of positive integers with the counting measure?

Perhaps, it's also possible to construct the desired space (i.e., Banach space with the finite countable balls) as the collection of all sequences in some apropriate set with the distance (metric) defined as $1/n$, where $n$ is the index of first distinct items in the two sequences. Such space is obviously complete, but it's not clear whether this space can be linear or not.

[Профессор Снэйп]

id в сообщении #257745 писал(а):

I guess the answer depends heavily on the field, over which our linear space is defined.

Banach spaces are always over $\mathbb{R}$ or $\mathbb{C}$.

[AGu]

Solunac в сообщении #257665 писал(а):

Does ball in a Banach space always have an infinite number of elements of the space?
Is there an exception?

As soon as the field $F$ of scalars is infinite (and $F$ is traditionally $\mathbb R$ or $\mathbb C$), the only exception is provided by the zero space $\{0\}$. This is evident, since every ball is convex. Therefore, if a ball contains two distinct elements $x,y$, then it also includes the segment $[x,y]=\{(1-\alpha)x+\alpha y:\alpha\in[0,1]\}$ which is infinite (since $[0,1]$ in $F$ is infinite and $(1-\alpha)x+\alpha y$ are distinct for distinct $\alpha$).

[Профессор Снэйп]

AGu в сообщении #257846 писал(а):

As soon as the field $F$ of scalars is infinite (and $F$ is traditionally $\mathbb R$ or $\mathbb C$), the only exception is provided by the zero space $\{0\}$. This is evident, since every ball is convex. Therefore, if a ball contains two distinct elements $x,y$, then it also includes the segment $[x,y]=\{(1-\alpha)x+\alpha y:\alpha\in[0,1]\}$ which is infinite (since $[0,1]$ in $F$ is infinite and $(1-\alpha)x+\alpha y$ are distinct for distinct $\alpha$).

Hmm... What about the next problem?

Let $X$ be a normed vector space over $\mathbb{R}$ of dimension $\geqslant 2$ and $S = \{ x \in X : \| x \| = 1 \}$. Could the set $S$ be of less then continuum cardinality?

[id]

Цитата:

and $F$ is traditionally $\mathbb R$ or $\mathbb C$

By the way, why?
I mean, if a normed space is defined as a vector space over some field $\mathbb{F}$ and Banach space is defined as a normed space which topology is complete, why can't we consider spaces over finite fields as well? I can't see any obvious contradictions.

Well, this might become a problem somewhere where Archimedes property is needed; and, sure, such spaces aren't of much use.

[Профессор Снэйп]

id в сообщении #258158 писал(а):

By the way, why?

By the defininion

http://en.wikipedia.org/wiki/Banach_space

Цитата:

...a Banach space is a vector space V over the real or complex numbers...

If Wiki does not have enough autority for you open any textbook on functional analisys.

[id]

Профессор Снэйп
Yep, that's what I did at first.
For instance, in Helemsky's textbook the definition is given the way I posted it here, with a comment afterwards: "Hereafter we assume that our field is $\mathbb{R}$ or $\mathbb{C}$".

So what's that: a definition - or just an assumption, helping to avoid considering "weird field" cases?

[AGu]

Профессор Снэйп в сообщении #258133 писал(а):

Let $X$ be a normed vector space over $\mathbb{R}$ of dimension $\geqslant 2$ and $S = \{ x \in X : \| x \| = 1 \}$. Could the set $S$ be of less then continuum cardinality?

The answer is "no."

Let $(X,\|{\cdot}\|)$ be a normed space and let $x,y\in X$ be such that $x\ne y$ and $\{x,y\}$ is linearly independent. For $\alpha\in\mathbb R$ put $z_\alpha:=(1-\alpha)x+\alpha y$. Since $x$ and $y$ are linearly independent, it is clear that $z_\alpha\ne0$ and therefore $\|z_\alpha\|\ne0$. Put $u_\alpha:=\frac1{\|z_\alpha\|}z_\alpha$. Then $u_\alpha\in S:=\{u\in X:\|u\|=1\}$ for all $\alpha\in\mathbb R$. It remains to observe that $u_\alpha$ are distinct for distinct $\alpha\in\mathbb R$. Indeed, let $\alpha,\beta\in\mathbb R$ and suppose that $u_\alpha=u_\beta$. Then $z_\alpha=\lambda z_\beta$, where $\lambda=\frac{\|z_\alpha\|}{\|z_\beta\|}$. We thus have $(1-\alpha)x+\alpha y=\lambda(1-\beta)x+\lambda\beta y$. Since $x$ and $y$ are linearly independent, we conclude that $1-\alpha=\lambda(1-\beta)$ and $\alpha=\lambda\beta$, which easily implies $\alpha=\beta$.

[Профессор Снэйп]

id в сообщении #258158 писал(а):

I mean, if a normed space is defined as a vector space over some field $\mathbb{F}$

What are axioms of norm?

One of the axioms is $\| \lambda x \| = |\lambda| \| x \|$. What is $|\lambda|$ if $\lambda$ is not from $\mathbb{C}$?