Talk:Continuous linear extension

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Please update this rating as the article progresses, or if the rating is inaccurate. Click to show/hide comments. Please add to or update the comments to suggest improvements to the article. Improving the typography and math formatting would help. Geometry guy 22:44, 16 June 2007 (UTC)

I think this is not a great article title. It's

• non-standard -- googling for BLT "bounded linear transformation" returns few hits and I've never heard of it :P
• contains a cryptic acronym, so is hard to refer to without explaining what the acronym means.

I'd suggest changing it to one of continuous linear extension, continuous linear extension theorem and redirecting from BLT theorem and BLT-theorem to that. You use the word bounded in place of continuous, too.

Nice article, though!

-- Lupin 10:43, 9 Feb 2004 (UTC)

I agree that it should be moved. Why do you prefer continuous? The texts which I checked (admittedly only three), all talk about bounded linear operators; is continuous more common? (I know that both terms are equivalent for linear operators.) It did make me think though; the construction seems to work just as for continuous, nonlinear operators (except that the extension is of course no longer linear). Any idea why Reed and Simon single out the linear case? -- Jitse Niesen 14:38, 9 Feb 2004 (UTC)

I think I've seen it referred to in texts as "let T be the continuous linear extension of the densely defined operator S". The extension process is also referred to as "extend T by continuity to the whole of <vector space>". I think that gives more of a flavour of the proof that such an operator does actually exist. This accounts for my (slight) preference for continuous over bounded, although I actually meant to type "You could use the word bounded in place of continuous" above.

If you take the function defined on the reals minus the origin by f(x)=1/x, then you'll see that bounded and linear are pretty important -- although f is continuous, there's no continuous extension of f to the whole real line although the domain of f is dense in the real line. In that case, the proof doesn't work because of boundedness: the sequence {L(x_n)} is not Cauchy for a general Cauchy sequence converging to 0.

There may well be good generalisations to nonlinear functions, but you need to rule out things like this.

-- Lupin 16:09, 9 Feb 2004 (UTC)

Indeed, "bounded linear extension" somehow does not sound quite right. Thanks for the (embarassingly simple) counterexample. I thought continuous operators transform Cauchy sequences to Cauchy sequences, but that is not true, they only transform convergent sequences to Cauchy sequences. -- Jitse Niesen 18:36, 9 Feb 2004 (UTC)

[edit] Matrix?

There was mentioned a matrix norm in this article, although there are no matrices to be found, the norm should probably be an operator norm.—The preceding unsigned comment was added by 131.159.71.77 (talk • contribs).

Yes, that was a mistake. Fixed, thanks! Oleg Alexandrov (talk) 03:33, 6 February 2007 (UTC)