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Suppose $X$ is an infinite-dimensional Banach space.

Is there a natural topology on $$C(X)=\{f:X\to\mathbb{R}: \text{ $f$ is continuous}\}?$$

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@NateEldredge Oops, I did not read the first line... –  1015 Feb 2 '13 at 18:33
    
@NateEldredge, exactly. It is not clear what would be a family of seminorms that gives $C(X)$ a Frechet space topology. –  Cantor Feb 2 '13 at 18:33
    
@Cantor Sorry, my comment was absolutely pointless, I had not read the question carefully. –  1015 Feb 2 '13 at 18:36

1 Answer 1

up vote 2 down vote accepted

There are a few, one example would be the weak* topology:

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

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So, what is one natural weak topology on $C(X)$?! –  Cantor Feb 2 '13 at 18:35
    
@Cantor: Sorry, meant to say the weak* topology. It's the weak topology with respect to the functions $\{\phi\mapsto \phi(x) \ | \ x \in X\}$. –  Jim Feb 2 '13 at 18:41
    
I think weak topology was correct. –  Cantor Feb 2 '13 at 18:51
    
Well, the weak* topology is a weak topology. It's just that if you say weak topology that's a topology you put on a space using a set of functions on that space. We have a space and a set of functions but we don't want a topology on $X$, we want a topology on $C(X)$, so it's a little confusing to just say weak topology. If we want the topology to be on $C(X)$ we have to specify the functions on $C(X)$ that we are taking the weak topology with respect to. By saying weak* topology I'm just being specific about what those functions are. –  Jim Feb 2 '13 at 18:56
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I don't want to make this into a long discussion, but I still think the term weak topology is correct. We are putting a weak topology on C(X). Weak* topology is the weak topology that is put on the dual of a space, i.e. the space of continuous, 'linear' functionals on that space. –  Cantor Feb 2 '13 at 19:01

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