Talk:Bounded set (topological vector space)

Untitled

I'm moving some discussion from Talk:Bounded set to here since it clearly concerns only LCTVS. — MFH:Talk 21:31, 12 October 2006 (UTC)

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seminorms ?

I think the statement around "seminorms" is incorrect. If not, a proof or a link to a proof (at least a precise reference: Thm.X of book Y) is needed.

At least, p(S) should be defined. (a priori, for a map f:E→G, f(S)={ f(x); x∈S }; most probably, here the sup is meant.)

In the context of inductive limits, there is the notion of "regular limit" (bounded set = set contained in some space and bounded there), it seems to me that this would not make sense if the property holds.

The same (need for reference) applies to the statement "bounded lin. op. is continuous".

(How about adding a subpage ".../proofs" to all articles in mathematics?)

— MFH: Talk 16:39, 19 Apr 2005 (UTC)

Locally convex (which you inserted) and the sup was missing, but now the statement about boundedness in terms of seminorms should be correct. I will try to get a proof or reference untill tomorrow. MathMartin 17:39, 19 Apr 2005 (UTC)

I found a reference in the english translation of Bourbakis "Topological Vector Spaces". The statement is in the middle of chapter III page 2 (TVS III.2) MathMartin 20:51, 26 May 2005 (UTC)

Set is bounded in locally convex space iff bounded under each semi norm

I found no reference but the proof is quite simple (the statement is from a script I am currently reading). My main aim is to make clear that boundedness for topological vector spaces does not necessarily coincide with boundeness for metric spaces (it does when the metric is given by a norm). I will try to make this more clear in the article, but perhaps it would be best to separate the articles. Anyway here is the proof:

Statement

Given a locally convex space (X,P) with P a familiy of semi norms, then a subset A of X is bounded iff $\sup _{x\in A}p(x)<\infty$ for all p in P.

Proof

$\Rightarrow :$Let A be bounded. Using any semi norm p we can construct a neighbourhood $N_{1}^{p}$ for the zero vector with radius 1. As A is bounded there exists an $\alpha >0$ with $A\subset \alpha N_{1}^{p}$. This implies $\sup _{x\in A}p(x)\leq \alpha <\infty$.

$\Leftarrow :$Now let $\sup _{x\in A}p(x)<\infty$ for all p in P. We have to show that for any zero vector neighbourhood $N$

$\exists \alpha :A\subset \alpha N$.

It is sufficient to proof this for all neighbourhoods $B$ in a neighbourhood basis ${\mathcal {B}}$ of the zero vector. Using the family of semi norms to construct a locally convex neighbourhood basis ${\mathcal {B}}$ every neighbourhood $B$ in ${\mathcal {B}}$ can be written as

$B=\bigcap _{i=1}^{n}N_{\epsilon _{i}}^{p_{i}}\qquad n\in \mathbb {N}$ with some $\epsilon _{i}>0$ and some $p_{i}\in P$.

For a given representation of $B$ we define

$\alpha _{i}:={\frac {1}{\epsilon _{i}}}\sup _{x\in A}p_{i}(x)\quad (i=1,\ldots ,n)$

Then we can construct

$\alpha :=1+\max\{\alpha _{1},\ldots ,\alpha _{n}\}\quad (i=1,\ldots ,n)$.

and thus for our choosen neighbourhood

$A\subset \alpha B$.

MathMartin 10:21, 21 Apr 2005 (UTC)

still not completely convinced

Of course, if the topology of a locally convex space can be defined by a filtering family of seminorms, then your statement is correct (in some sense you cannot "make" the family of seminorms by taking the "gauge" of such neighborhoods). A concrete example is given (AFAIK) by Roumieu type ultradifferentiable functions, defined as inductive limit of semi-normed spaces. (Maybe also by hyperfunctions or something alike.) — MFH: Talk 17:36, 21 Apr 2005 (UTC)

P.S.: The corresponding construction in locally convex space should maybe also be checked. — MFH: Talk 17:36, 21 Apr 2005 (UTC)

I am not sure I understood your explanation. What is a filtering family of seminorms ? Anyway, I made a mistake in the proof which I fixed. If you are still not convinced, could you give some more details in your argument using more basic counterexamples (I have do not know what a Roumieu type ultradifferentiable functions is).MathMartin 20:26, 22 Apr 2005 (UTC)

Filtering family means that for p,p' you can find p" such that p" is greater or equal than both, p and p'. (aka directed set).

Roumieu type spaces are inductive limit spaces. More generally, I think that any not (semi-)metrizable space, or any space which has not a bounded open neighbourhood of zero, will provide a counter example (but maybe this is not completely true). — MFH: Talk 14:08, 27 May 2005 (UTC)

Every finite set of points is bounded?

This example is not true (at least) in the following case. In a discrete topology, the only bounded set is {0}, because the open set {0} cannot be inflated at all. — Preceding unsigned comment added by Leo-J-N-Jian (talk • contribs) 05:36, 8 March 2018 (UTC)

In the locally convex case, the statement is true again, which easily results from the fact that the origin has a local base of absorbent sets. I edited the example to elaborate on that. LMias (talk) 08:41, 6 June 2018 (UTC)

Actually, every topological vector space has a local base of absorbent sets around the origin, and this is the setting in which this article is written. The problem is that the example with the discrete topology is not a topological vector space. LMias (talk) 16:23, 11 June 2018 (UTC)