Talk:Uniform boundedness principle
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What the hell?! --JensMueller
[edit] The orginal non-Latex text of the proof
- For n = 1,2,3, ... let Xn = { x : ||T(x)|| ≤ n (∀ T ∈ F) } . By hypothesis, the union of all the Xn is X.
- Since X is a Baire space, one of the Xn has an interior point, giving some δ > 0 such that ||x|| < δ ⇒ x ∈ Xn.
- Hence for all T ∈ F, ||T|| < n/δ, so that n/δ is a uniform bound for the set F.
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- I am reverting to the non-latex version, because the latex version adds nothing. (It has actually subtracted something: a rather important "implies" sign.) See Wikipedia:How_to_write_a_Wikipedia_article_on_Mathematics. Andrew Kepert 22:46, 14 Mar 2005 (UTC)
[edit] consistency
The second theorem, about the pointwise limit, could be stated a bit better. The operators should be 'linear', no? I know this is probably implicity assumed, but this is not the place for implicit assumptions. And the article interchanges 'continuous' and 'bounded' when referring (presumably) to linear operators. While they might be the same in Banach Spaces (they are, no?), they are still separate ideas and are not always the same in all topological vector spaces. So let's pick one, and stick with it. Also, I like the non-latex version of the proof on the talk page better than the one on the main page. Lavaka 00:28, 21 September 2006 (UTC)
- As the second theorem is a consequence of the first, I'd guess that the assumptions are the same: The T_n are linear operators from a Banach space to a normed vector space. -- Jitse Niesen (talk) 06:39, 6 October 2006 (UTC)
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- Looking back at the proof, the wording is very bizarre: "Besides for all z in X such ..." What does that mean? Lavaka 23:28, 13 March 2007 (UTC)
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- "Besides" means "furthermore" here. Anyway, I rewrote the proof in an attempt to clarify it; does that help? -- Jitse Niesen (talk) 04:20, 14 March 2007 (UTC)
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