Talk:Bounded operator

message: Does the Laplacian map onto L^2? ie is L^2 this the range or is it simply the codomain? Lupin 14:19, 2 Mar 2005 (UTC)
replies: text: Good point. I think it could be proved that under certain conditions the Laplacian is actually surjective, but this could be hard and is not the point of the example. So I reworded the example to not imply the Laplacian is surjective. Oleg Alexandrov 16:04, 2 Mar 2005 (UTC)

text: Of course it is wrong in the L2 norm. But the article says "\Delta:H^2({\mathbf R}^n)\to L^2({\mathbf R}^n)" and emphasizes "its domain is a Sobolev space"; if you do not know the norm of the Sobolev space, follow the link. Boris Tsirelson (talk) 07:31, 14 December 2011 (UTC)

message: I think you are making a mistake here. An operator which is defined on its domain is not necessarily bounded on that domain. It is essential that the  M in the estimate must be independent of the  v.
: Do you refer to the first paragraph? In the inequality there, I think it was clear that M is independent of v. I also made a small change for clarification. I wonder what you think. Thanks. Oleg Alexandrov 15:49, 15 Mar 2005 (UTC)
thread_title:  Defined versus bounded 

message: ::There exists M such that BLAH BLAH BLAH for all x.
::There exists M such that (BLAH BLAH BLAH for all x).
::(There exists M such that BLAH BLAH BLAH) for all x.
::There exists M such that for all x BLAH BLAH BLAH.
::For all x there exists M such that BLAH BLAH BLAH.
The first version above is ambiguous!  It can mean either the second version or the third.  The last two are unambiguous.  That is why my recent edit is important.  That may explain the misunderstanding above. Michael Hardy 01:49, 16 Mar 2005 (UTC)
replies: text:  Thanks, looks better the new way. Oleg Alexandrov 01:59, 16 Mar 2005 (UTC)
thread_title:  Continued

message: A reader emailed the following comments to Wikipedia.  Please review and correct the article, as appropriate:
I think there is a mistake in the article at
Bounded operator.
It is claimed there, in the fourth example, that
"if a linear operator L has as domain and range
Banach spaces, then it will be bounded".
This is false, as shows the following (quasi-explicit)
example: consider l^2, the vector space of square summable
sequences (u_n)_{n\in\mathcal{N}} (of real numbers); it has
a natural scalar product which makes it a Hilbert space, and
a natural Hilbert basis (the canonical basis (e^k)_{k\in\mathcal{N}},
e^k_n = 0 if n is different from k, 1 if n=k).
The axiom of choice allow to consider an algebraic basis of l^2
containing (e^k)_{k\in\mathcal{N}}.
Now, the linear map A from l^2 to \mathcal{R} which sends e^k to k+1
and sends other elements of the basis to zero is well-defined, linear,
one-to-one from l^2 to \mathcal{R}, but not continuous.
I suppose that the mistake of the author comes from Banach's theorem,
which asserts "any linear one-to-one continuous map from a Banach space to
another has continuous inverse".
Let me add an erratum to my previous mail.
In the end of the counter-example I gave, don't read
"the linear map A from l^2 to \mathcal{R} which sends e^k to k+1
and sends other elements of the basis to zero is well-defined, linear,
one-to-one from l^2 to \mathcal{R}, but not continuous",
but replace "one-to-one" with "onto" --I think this is now a correct
counter-example.
(Please excuse any errors in the markup; I don't understand a word of this.) Kelly Martin 14:31, 5 October 2005 (UTC)
replies: text: It was many times when I thought of that example 4. It looked suspicious to me and I never managed to prove it. But for the record, it was not me who wrote it, I just moved it from operator norm. See this version of operator norm, before I ever contributed anything to that article. Oleg Alexandrov 16:34, 5 October 2005 (UTC)
replies: text: Actually, I doubt that the linear map in the mail quoted by Kelly is well defined. Consider the sequence (u_n) with u_n = 1/(n+1)^2. This sequence is in l^2 (and even in l^1), but Au = \sum_{n=0}^\infty (n+1)u_n = \sum_{n=0}^\infty \frac{1}{n+1} is not defined. -- Jitse Niesen (talk) 17:14, 5 October 2005 (UTC)

text: Hmm, you have a point. I ignored the expression "algebraic basis", which I did not understand, but I see now that this is what I'd call a Hamel basis. -- Jitse Niesen (talk) 18:16, 5 October 2005 (UTC)
thread_title:  Error in article? 

message: A minor typographical proposal. Can we use the letter A to denote a linear operator in the article as par operator norm. I also think A is more common than L. (I noticed people use L to denote a set of linear operators or for the matter, bounded linear operators but not a linear operator in general). -- Taku 22:51, 15 January 2006 (UTC)
replies: text: I don't care either way. As long as you change all, to keep consistent. Oleg Alexandrov (talk) 02:11, 16 January 2006 (UTC)
thread_title:  A to denote linear operator instead of L 

message: I did this revert, as to me it appears that introducing one more notion, of Lipschitz continuity is not helpful. The introduction I think should be a simple as possible, and the point of the two original short sentences was to 
* Make it clear that bounded operator ≠ bounded function
* linear bounded operator is the same as linear continuous operator
I don't see how the new concept of Lipschitz continuity helps with that; the reader may not even know what that is, and I think it distracts from the point the intro is trying to make. Comments? Oleg Alexandrov (talk) 03:02, 10 March 2006 (UTC)
replies: text: The reason I made the change is this: the article currently states that the continuity is a consequence of the linearity of the bounded locally bounded function.  This is not correct, or at least it's misleading.  When one recognizes that the boundedness condition is simply a strong continuity condition, then continuity is automatic.  Now, if you want to keep the intro as simple as possible, why is there this stuff about locally bounded functions?  I consider locally boundedness more obscure than Lipschitz which seems somewhat common to me.  Maybe you want to mention local boundedness so that the name bounded doesn't confuse people.  I want to mention Lipschitz so that continuity seems natural.  But I could be persuaded to move it down the article.  You're not arguing for complete noninclusion, are you? -lethe talk [ +] 04:06, 10 March 2006 (UTC)

text: A consequence of the linearity property is that a linear operator is bounded if and only if it is continuous.
It says just that, a linear operator is bounded if and only if it is continuous.  Now I removed the words "A consequence of the linearity property", so I think it is more clear.
Yeah, including something about Lipschitz further down may be good, although I don't see why you would do that. Maybe except if you wish to provide a complete proof of linearity+local boundedness= linearity + continuity. Oleg Alexandrov (talk) 04:24, 10 March 2006 (UTC)

text: I think it's a nice thing to include, but I agree with Oleg that it is not as important as local boundedness. Local boundedness needs to be mentioned early on because it answers the question "why is it called bounded if it is not bounded as a function?" I added a sentence on Lipschitz continuity further down. I'm not quite happy with the location though. -- Jitse Niesen (talk) 05:46, 10 March 2006 (UTC)
thread_title:  Lipschitz 

message: I was reading the discussion above about the apparently erroneous statement that all linear operators between Banach spaces are bounded.  After all my recent work with discontinuous linear map, I feel I understand these sorts of questions better; clearly the statement is wrong if we assume AC, and it appears to do so, since the Baire category theorem uses AC.  So this isn't a constructivism issue.  The material was originally added to operator norm by Charles Matthews.  I'd be pleased if we could fix his work and find a home for it.  In particular, this notion of "condensing singularities" sounds fun to me.  Now, as for what he was going for with this, the closest I could find is the uniform boundedness theorem, which relies on the Baire category theorem, and says that a family of maps is pointwise bounded if and only if the family is uniformly bounded.  For a single map, I guess this says that a map is bounded if and only if it's bounded, so this isn't what he was going for. -lethe talk [ +] 11:34, 10 March 2006 (UTC)
replies: text: At that point I was probably trying to reconstruct something that appears in one of the Halmos books, for which you need to find another Halmos book to get the proof. This point has been discussed before though, and I thought was corrected. I wonder where that was ... Charles Matthews 11:42, 10 March 2006 (UTC)
replies: text: Would that probably be his Hilbert space book then? -lethe talk [ +] 11:46, 10 March 2006 (UTC)
thread_title:  Old error 

message: I'm tempted to move it, but I'm not a mathematician. Qwfp (talk) 15:54, 24 January 2011 (UTC)
replies: text: Mathematicians usually say "operator" meaning "linear operator", and "nonlinear operator" meaning "operator". It is rather illogical, but it happens, because linear operators are much more investigated and important. --Boris Tsirelson (talk) 19:17, 24 January 2011 (UTC)

text: Also, "bounded linear operator", being longer than just "bounded operator", is equivalent to it, since we have no useful notion of "bounded" for nonlinear operators. --Boris Tsirelson (talk) 19:20, 24 January 2011 (UTC)
thread_title:  Why is the title of this article not 'Bounded linear operator'? 

message: Where you have "||L(v)|| = 2*Π*n → ∞ as n → ∞" I think it should be
"||L(v)|| = n -> ∞ as n -> ∞" Rgilbarco (talk) 00:45, 6 December 2011 (UTC)
replies: text: Agree. Fixed. Boris Tsirelson (talk) 16:02, 6 December 2011 (UTC)
thread_title:  Error in "Linearity and boundedness" 

message: The current revisions of the article says: "A bounded linear operator is generally not a bounded function; the latter would require that the norm of L(v) be bounded for all v, which is not possible unless Y is the zero vector space." I think it should be zero function (i.e., L(v)=0 for each v) instead of zero space. Since the function L is not required to be surjective, L[X] rather than Y influences whether it is a bounded function or not. --Kompik (talk) 04:30, 26 August 2017 (UTC)
replies: text: Yes, I agree. Now fixed. Boris Tsirelson (talk) 05:15, 26 August 2017 (UTC)
thread_title: Zero space or zero function