Re: continious operator in first-countable spaces

On Tue, 11 Apr 2006 16:01:22 EDT, eugene <jane1806@xxxxxxx> wrote:

On 11-04-2006 17:12, eugene wrote:

Prove that in a vector topological space with
first-countability axiom
every bounded linear operator is continious.

What's the meaning of "bounded linear operator" in
this context? I'm
aware of the meaning in the context of linear
operators between normed
vector spaces; if T is such an operator, we then say
that it is bounded
if there's some K > 0 such that, for every vector _v_
on the domain of
T, ||T(v)|| <= K.||v||. Then, yes, bounded =>
continuous.
Best regards,

Jose Carlos Santos

Yes, if you are dealing with normed vector spaces, it is to prove this fact and i know how to do it. But here is more general consept: vector topological space, here we have a topological structure and vector space structure on a set, such that (x,y)->x+y and (k,x)->kx are continious(is a sense of topology)-here is in vector topological space(you can search for surely better definition).

So, in vector topological spaces the notion of boundedness is slighly different: the set E is bounded is for every neighborhood of 0 there is a number s such that E \in tV for all t>s.

And boundedness here in this sense.

You haven't said what a bounded linear operator is!

As for example normed space is a space with first-ciuntability axiom we can recongize here a well known fact that the notions of boundedness and continuity of linear operator in the normed space are equivalent.

Thanks


************************

David C. Ullrich
.

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