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The boundedness condition for linear operators on normed spaces can be restated. An operator is bounded if it takes every bounded set to a bounded set, and here is meant the more general condition of boundedness for sets in a topological vector space (TVS): a set is bounded if and only if it is absorbed by every neighborhood of 0. Note that the two notions of boundedness coincide for locally convex spaces.

I was wondering what "the two notions of boundedness" are referred to? In other words, their definitions?

Are they for operators between TVSes, or for subsets of TVS?

Thanks and regards!

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The first is probably: "a bounded linear operator is a linear transformation L between normed vector spaces X and Y for which the ratio of the norm of L(v) to that of v is bounded by the same number, over all non-zero vectors v in X" where you somehow use seminorms (see en.wikipedia.org/wiki/Locally_convex_spaces) instead of the norm. And the second is the one you cite. –  savick01 Feb 25 '12 at 21:29
    
It shows 'how': en.wikipedia.org/wiki/… –  savick01 Feb 25 '12 at 21:35
    
@savick01: Thanks! That makes sense. –  Tim Feb 25 '12 at 21:44
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2 Answers

up vote 4 down vote accepted

As far as I can tell, the citation you give from Wikipedia is ambiguous at best, and mistaken under one natural interpretation.

Here is a brief explanation of the relationship between continuity and boundedness for linear maps between topological vector spaces:

For a linear map $L$ between normed vector spaces spaces $V$ and $W$, there are (at least) two natural conditions you could impose: that $L$ is continuous, in the usual topological sense, or that $L$ is bounded, i.e. takes bounded subsets in $V$ to bounded subsets in $W$. It turns out that these two conditions coincide (this is an easy exercise, using the definition of the metric in terms of the norm, and the linearity of $L$).

On an arbitrary topological vector space, one can also define the notion of bounded subsets, even though the topology is not defined in terms of a norm, or even a metric, in general. The definition is the one given in your Wikipedia citation: a subset $B$ is bounded if given any neighbourhood $U$ of the origin, there is a scalar $\lambda$ such that $B \subset \lambda U$. (This is often phrased as in the paragraph you cite: $B$ is absorbed by $U$.)

In a normed space, it is easy to check that this notion coincides with the notion of bounded subsets as defined in terms of the norm.

Given a linear map between topological vector spaces, we call it bounded if it takes bounded sets in the domain to boudned sets in the codomain. One easily checks that continuous linear maps are necessarily bounded. But it's not true in general that a bounded linear map between topological vector spaces is continuous.

A locally convex topological vector space $V$ is called bornological if any bounded linear map from $V$ to another locally convex topological vector space $W$ is necessarily continuous. The above discussion shows that normed spaces are bornological. (But as was already indicated, not all locally convex topological vector spaces are bornological, although most common ones are; e.g. Frechet spaces are, and locally convex inductive limits of bornological spaces are again bornological.)

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+1 Thanks! (1) Does the definition of bounded operator in terms of taking bounded subsets to bounded subsets apply to a mapping that is not necessarily linear? I try to compare this definition to the other one in terms of the range of the mapping being a bounded subset. (2) Is a bornological space required to be a locally convex TVS or just a TVS as in Vobo's reply? –  Tim Feb 26 '12 at 10:32
    
@Tim: Dear Tim, I don't know what terminology people use for nonlinear maps, since I've never had cause to think about the non-linear situation. Also, I've only ever had to use the concept of bornological spaces in the locally convex setting; while the definition makes sense in the non-locally convex setting, I don't know whether or not people use it there. Regards, –  Matt E Feb 26 '12 at 12:32
    
Thanks! (1) I was wondering if "bornological spaces for locally convex TVSes" and "general bornological sets" as in Wiki page you linked are related concepts? (2) Can bornological spaces be defined alternatively as those locally convex TVSes such that the class of bounded subsets defined in TVS sense forms a bornology? (If you feel comments may not be proper to address my questions, shall I turn them into a new post?) –  Tim Feb 26 '12 at 15:46
    
(3) Is the class of all bounded subsets of a metric space a bornology, isn't it? Is the class of all bounded subsets of a TVS a bornology, isn't it? –  Tim Feb 26 '12 at 16:58
    
@Tim: Dear Tim, I think that the concept of bornology arose from the same functional analysis considerations that gave rise to the concept of bornological TVSes, but unfortunately I've never become fully versant enough with the general concept of bornologies to answer your questions. I haven't looked at exactly what wikipedia says on this point, but if it doesn't answer your questions precisely, then I suggest asking them as a separate question. Regards, –  Matt E Feb 27 '12 at 2:45
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They are for subsets of normed linear spaces.

  • In a normed linear space a set $X$ is bounded, iff there is some $C>0$ with $||x|| < C$ for all $x \in X$.
  • Equivalently, a set is bounded iff it is absorbed by every neighbourhood of 0.

So a linear operator on normed spaces is bounded, iff it maps bounded sets to bounded sets.

This motivates the definition for more general topological vector spaces: An operator is called bounded, iff it maps bounded sets to bounded sets.

Such an operator does not need to be continuous, in contrast to normed spaces. Topological vector spaces where all bounded operators are continuous are called bornological spaces.

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Wiki says "the two notions of boundedness coincide for locally convex spaces". However you say that it is for normed spaces. It does not suit each other. Anyway your equivalence works for metric linear spaces. –  savick01 Feb 25 '12 at 22:05
    
As Matt said, wiki is ambiguous in this point. This specific sentence "the two notions of boundedness coincide for locally convex spaces" is completely superfluous in that section. –  Vobo Feb 25 '12 at 22:14
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