Talk:Discontinuous linear map

message: I wrote the article A linear functional which is not continuous only to immediately discover on its talk page a suggestion to move it to Non-continuous linear functional (darn, everybody should be drunk and sleeping this post-Christmas morning, not checking the recent changes). What do people think (if it matters at all)? Oleg Alexandrov (talk) 17:36, 26 December 2005 (UTC)
replies: text: Isn't there a mechanism for providing both names for the article? Both seem fine... Randall Holmes 17:40, 26 December 2005 (UTC)

text: By the way, I've been steadily editing through this period, and it has been awfully quiet :-)  Happy Hanukkah! Randall Holmes 17:41, 26 December 2005 (UTC)
replies: text:  Yeah, still quiet. :) Yes, there is a mechanism for providing both names, it is called a redirect, see Wikipedia:Redirect. So I guess the argument is about which is the primary meaning, for all that's worth. Oleg Alexandrov (talk) 02:31, 27 December 2005 (UTC)

text: Personally, I don't like the idea of a name for an encyclopedic article beginning with "A". Also, I think the original proposed name is too long. I'd go for Non-continuous linear functional. But maybe it's just me. --Meni Rosenfeld 12:49, 29 December 2005 (UTC)

text: I agree with Meni; initial articles are okay if they are title of a book or work of art, but I don't think they belong in a general article.  Even if it has to do with showing existence.  Isn't there some other way to word it?  Gene Nygaard 21:37, 30 December 2005 (UTC)

text: How about linear map which is not continous, dropping the leading "A" which makes people uneasy. The other option seems to be discontinous linear map, based on above. Oleg Alexandrov (talk) 00:27, 31 December 2005 (UTC)
replies: text:  Discontinuous linear map/Discontinuous linear functional and Infinitely differentiable non-analytic function seem reasonable. We anyway don't expect people to search for and bump into these articles directly, and we are probably going to use these as examples to show that being a "linear map" doesn't imply being continuous, and smoothness doesn't imply analyticity, in main articles on linear maps and analytic functions, so I guess the title is not all that crucial. Bottomline: Unless we are missing on something important, the shorter the better. deeptrivia (talk) 00:44, 31 December 2005 (UTC)

text: However, I find Infinitely differentiable non-analytic function a very clumsy name for An infinitely differentiable function that is not analytic. Oleg Alexandrov (talk) 19:57, 31 December 2005 (UTC)
thread_title:  Discussion of the name copied here from [[Wikipedia talk:WikiProject Mathematics]] 

message: So I've done a lot of structural changes to the article.  Here's a list:
# I've put the examples ahead of the proof of existence
# I've made the article primarily about linear maps instead of linear functionals.
# I didn't like Oleg's insistence on only normed vector spaces, so I chose a larger space of functions which is not normable for my derivative example.  The norm example is still there as well, because I think Fréchet spaces and Banach topologies are not so familiar to people
# I removed the norm in the proof, so that it's a map V → V instead of a map V → R.
# I added the example from locally convex topological vector space of a space for which no linear maps are continuous.
lethe talk 19:40, 27 December 2005 (UTC)
thread_title:  structural changes 

message: The introduction now states that 
:If the space is finite-dimensional, then all topologies compatible with the algebraic structure agree.
I think that this is false. If one considers on the real line the finest possible topology, when any set is open, then R becomes a linear topological space not identical to the usual topological structure on R, no?
:No. This topology makes linear operations discontinuous. You are allowed to try any topology on your linear space (this time, one-dimensional) over the field on real (or complex) numbers (the scalars), but without changing the usual topology on the field of scalars. Boris Tsirelson (talk) 20:48, 11 June 2009 (UTC)
thread_title:  Mistake in the article? 

message: I did some rewriting in this article. I would like to say that I removed the part about discontinuous linear maps over the field of rationals. While they may be interesting, they are not relevant to functional analysis, where any serious work is done over the real or complex numbers for good enough reasons.
Now, about the following text:
: The derivative operator is a linear map on a space of differentiable functions which in most cases will not be continuous.  For example, it is not continuous in the Fréchet space D∞(R) of smooth functions on the real line with its inductive limit of Banach topologies.  
I could not prove that the derivative will be discontinuous. Actually, if by those fancy Banach topologies is meant the seminorms given by 
: p_{K, n}(f)=\sup_{0\le m\le n, x\in K}|f^{(m)}(x)|
where K are compact sets and n is integer (that is, uniform convergence over compact sets for sequences of functions and their derivatives), then it looks to me that the derivative operator is actually continuous. But I don't have much experience in this spaces, I may be wrong. Comments? Oleg Alexandrov (talk) 19:01, 1 January 2006 (UTC)
thread_title:  Issues 

message: I have a question about the text:
:The space of measurable functions on the unit interval with quasinorm given by 
::||f|| = \int_I \frac{|f(x)|}{1+|f(x)|}dx
:is not a locally convex space, and the only continuous functional on this space is 0.
I wonder, which is the range of those functionals? I mean, you can always find a nonzero functional from the space to itself, that being the identity. So, do you mean scalar-valued functionals? Oleg Alexandrov (talk) 16:07, 3 January 2006 (UTC)
replies: text: I suppose you meant to say codomain, not range, since the range of the zero function or functional is always the singleton {0}.  Anyway, we're talking about functionals here, not functions, so the codomain should be the field of scalars, which I believe in this case is meant to be R.  -lethe talk 18:57, 3 January 2006 (UTC)
replies: text: Or maybe you meant to ask what is the codomain of f, the elements of the space?  Those are measurable functions on the unit interval, so it's their domain and codomain. -lethe talk 19:00, 3 January 2006 (UTC)
replies: text:  OK, so now you see why I named this article originally a linear 'functional'... Because I did mean scalar-valued transformations. But I realized I don't like the word functional anymore, so I am going to reword the article. (And yes, I did mean codomain of the functional, thanks!) Oleg Alexandrov (talk) 22:18, 3 January 2006 (UTC)

text: See the new sentence, which is my attempt to address this issue. -lethe talk 23:50, 3 January 2006 (UTC)
thread_title:  One more issue 

message: Oleg, you say "it looks to me that the derivative operator is actually continuous" in the space of smooth functions of compact support.  I want to investigate this matter.  I poked around in my Reed and Simon last night, and it seems that we have a boundedness criterion for continuity in TVSes in terms of their seminorms very like the familiar one we have for Banach spaces.
Firstly, there's an easier space than the space of smooth functions of compact support with its inductive limit; namely the functions of rapid decrease, smooth functions all of whose derivatives fall off faster than any polynomial.  This is the guy whose dual space is the tempered distributions.  I think it might also be called the Schwarz space, but I'm not sure; R&S don't mention it and Wikipedia doesn't seem to know either.  Anyway, this is a Fréchet space with a family of seminorms given by 
:\| f \|_{a,b}=\sup | x^a D^bf|.
So the derivative map will be continuous in this space If for all (c,d) there is some finite collection (ai, bi) such that 
: \|Df\|_{c,d} \le M(\|f\|_{a_1,b_1}+\dotsb + \|f\|_{a_n,b_n})
which will be the case if we can find an M such that
:x^cD^{d+1}f \le x^aD^bf.
But simply take a=c, and b to be d+1, and the inequality is saturated with M=1.  Basically, since the derivative is already in the definition of the seminorm, all it does is increment the index to the next seminorm in the family.  So I conclude that yes, Oleg, you're right, the derivative is continuous in this top.  Unless the quantifiers go the other way round, it wasn't crystal clear in the text. -lethe talk 09:21, 5 January 2006 (UTC)
replies: text: Right. So, the derivative is not such a bad operator. It is not continuous, but close. :) And both on Banach spaces and Frechet spaces. Oleg Alexandrov (talk) 21:48, 5 January 2006 (UTC)
thread_title:  derivative in Frechet space 

message: I changed a bit the general example to be less non-constructive. :DD —The preceding unsigned comment was added by Scineram (talk • contribs) .
: And I undid your change. I believe that it is an unnecessary complication to insist on constructivism. The axiom of choice is a well-established tool in functional analysis, and I see no need to avoid making use of it. As such, any infinite dimensional space has an infinite sequence of linearly independent elements, no need to restrict oneself only to vector spaces for which such a sequence can be explicitely constructed. Oleg Alexandrov (talk) 19:18, 27 February 2006 (UTC)
replies: text: Except as I guess you already know, I disagree with you, Oleg, and agree with Scineram that this page may have interest not only for analysts (and also think that even analysis articles should note the use of AC).  It's on my to-do list. -lethe talk [ +] 19:53, 27 February 2006 (UTC)
replies: text:  Enjoy working on that. :) Oleg Alexandrov (talk) 19:59, 27 February 2006 (UTC)
thread_title:  more ZF friendly 

message: I've done some changes.  I restored the reals over rationals example, as I think it's important.  I've added a section near the bottom about choice considerations, and I tried to refrain from mentioning the axiom by name as much as possible throughout the rest of the article.  Was it enough? -lethe talk [ +] 17:46, 8 March 2006 (UTC)
replies: text: I wanted to incorporate scineram's recent comment into the section on AC that I've added to the article, but I can't convince myself that the comments are right.  He claims that given an infinite sequence of independent vectors, define the operator to be 0 outside the sequence.  But does this work?  It contradicts my statement that there are no constructible examples over a complete domain.  On the other hand, I don't see why it shouldn't work.  scineram, you're not still around, are you? -lethe talk [ +] 18:15, 8 March 2006 (UTC)
replies: text: So you need choice to extend a linear map on a subspace to a linear map on the whole space.  You need choice even if you want it to be 0 outside the subspace.  So I think scineram's comments for making the construction ZF friendly are not correct. -lethe talk [ +] 18:33, 8 March 2006 (UTC)
thread_title:  well 

message: I really don't like inline tex, but I don't know how to make an overbar without it.  So I used it in my new addition.  I guess maybe there are alternative notoations for closures.  -lethe talk [ +] 23:17, 14 March 2006 (UTC)
replies: text: Inline TeX is fine, as long as it does not become inline PNG. The math style manual about avoiding inline PNGs is just a recommendation, so if PNGs can't be avoided, then so be it. :) Oleg Alexandrov (talk) 00:38, 15 March 2006 (UTC)
replies: text: Well I perversely have my settings set to "always render PNG", which perhaps explains why I dislike inline TeX so much.  Also, did you see that in that WP:MSM, they say that the phrase "note that..." or "it should be noted that..." are to be avoided?  I use those all the time.  Yep, I'm definitely a bad person. -lethe talk [ +] 02:08, 15 March 2006 (UTC)
replies: text:  I don't argue that you are not a bad person, you are indeed. :) But the reasons are not the ones you present. You don't like the axiom of choice, that's what makes you baddest person. :) Oleg Alexandrov (talk) 02:30, 15 March 2006 (UTC)
thread_title:  inline tex 

message: The paragraph:
:In functional analysis, one often considers operators defined only on a proper subset of the domain.  Every discontinuous linear operator on an incomplete space can also be viewed as a densely defined operator on the completion of the space.  It will still be discontinuous, but not everywhere defined.  Discontinuous operators on complete spaces that are defined everywhere on the domain require a little more work.
from the text is in my view somewhat confusing. I don't see the need to view a disctontinuous operator on an incomplete space as an operator on a bigger complete space on which it is still discontinous. I would suggest all this paragraph be replaced by one sentence saying that now we will construct discontinuous operators on complete spaces; without going in the complications of not everywhere defined operators and completions. Oleg Alexandrov (talk) 23:12, 19 March 2006 (UTC)
replies: text: Yes, I see your point, and I agree.  I guess I introduced that text because of a point you made to me in our discussion: the derivative example isn't quite as interesting, since it's not defined on a complete space.  I then learned about the theorem, which implies that the derivative will never be defined on a complete space, which surprised me a little.  I guess the business about viewing it as an operator on its completion was my attempt to convey the dichotomy between the complete and incomplete cases.  But the way the article is now, that idea is discussed later on.  I don't remember if that was the case when I introduced that part, but anyway, you're right, it doesn't really belong there.  Take a look at my attempt to redress your concern.  By the way, how do you feel about the section on closed operators? -lethe talk [ +] 09:50, 20 March 2006 (UTC)
thread_title:  A few questions 

message: A few days ago, someone came to the math reference desk (eventually it'll be in the March 2006 archive) looking for help on finding a discontinuous linear map.  I directed him or her to this article, but he was particular about having one for the space of smooth functions of compact support (the space whose dual is the space of distributions).  This space is an LF-space, so would be an example of a discontinuous linear functional on a space more general than a Banach space.  After a few days of discussion, he came up with this example.  I don't feel particularly compelled to include it in the article, (and I don't expect you will either Oleg).  It takes a bit longer to construct than the corresponding Banach space, and in its essence, is no different from the one we already have: make a convergent sequence of independent vectors, then extend to a basis.  Nevertheless, I think it's a nice example, and once you unpack the construction, it's pretty simple.  So I thought I'd include it on the talk page here, for the sake of posterity.  Who knows, maybe at some point in the future, it will make sense to put it in the article. -lethe talk [ +] 16:33, 20 March 2006 (UTC)
replies: text: I agree that it is not worth including, at least for now. I think it just provides a concrete way of constructing a sequence of linearly independent vectors in a space a bit more general than Banach, everything else seems to be the same; that if I read it right. Oleg Alexandrov (talk) 02:53, 21 March 2006 (UTC)
thread_title:  twma's example 

message: The article now has the sentence:
:All of the comments for the existence of discontinuous linear maps on normed spaces apply to any topological vector space; given an appropriate basis, one can define an unbounded operator, and unbounded operators are discontinuous for any TVSes.
I am not sure that's true. Take an infinite-dimensional real vector space, and consider the discrete topology on it, such that every set is open. Then, any linear map from that space (actually, any map at all) is continuous. No? Oleg Alexandrov (talk) 15:43, 23 April 2006 (UTC)
replies: text: Yeah, I guess all maps are continuous on a discrete space.  I guess the problem is that a discrete space is not locally convex. -lethe talk [ +] 23:43, 23 April 2006 (UTC)
replies: text:  It is locally convex, isn't it? The singletons form a basis for the topology and are convex. A Geek Tragedy 19:13, 5 March 2007 (UTC)

text: Careful!  There is no such thing as a discrete topological vector space over the real or complex numbers, other than the zero space.  The reason is that if the space is discrete, then scalar multiplication won't be continuous: take any nonzero vector v, and consider the sequence (1/n)v for positive integers n.  If scalar multiplication were continuous, then this sequence would converge to 0.  But then, since the set {0} is open in your vector space, this implies that (1/n)v=0 for all sufficiently large n, and that's impossible.  I am currently trying to prove that every infinite-dimensional TVS admits a discontinuous linear operator for a course I'm taking.  I assume that it's true since it's been assigned as a problem, but maybe there's been an error. 24.150.43.97 (talk) 01:33, 10 January 2009 (UTC)
thread_title:  Beyond normed spaces 

message: The equivalence of norms in the proof that all maps on finite-dimensional spaces are continuous is used for convenience only.  it is not essential to the proof, which could be modified to work with any norm with more work.  I just thought the proof was simpler, easier to understand, using the l1 norm.  The real reason the proof fails for infinite dimensions is that the norm can't be bounded, it has no supremum.  Given any norm, any linear map for which the supremum does exist will be bounded and therefore continuous.  Only maps for which this supremum does not exist will be discontinuous.  This is seen in the construction of discontinuous linear maps as well, where you have to construct a sequence for which sup ||f(ei|| does not exist.   In short, finite sets have suprema, infinite sets need not. -lethe talk [ +] 05:37, 13 May 2006 (UTC)
replies: text: I wonder how to modify the proof.  It seems to me that the l1 norm is the largest norm, so I don't know how to make the inequality hold for smaller norms.  Maybe this is another way to beat the proof?  I stand by my comment that the failure of the supremum is the "important" reason, whatever that might mean. -lethe talk [ +] 05:53, 13 May 2006 (UTC)
replies: text:  Good points. I tired to tweak the proof a bit. Wonder what you think. Oleg Alexandrov (talk) 06:03, 13 May 2006 (UTC)
replies: text: I like the new edit.  It does seem a little unnatural to assume a particular norm at the outset, and this makes it clearer exactly what that norm does for us.  But I'm still curious, can we beat the proof this way?  Is there a linear map which is unbounded in the l1 norm, but bounded in the l∞ norm, say?  I'm gonna go to the bar and think about it. -lethe talk [ +] 06:12, 13 May 2006 (UTC)
thread_title:  equivalence of norms 

message: Per this edit, any linear map defined on some infinite-dimensional normed space with real values is continuous. But that's not true, I thought that this article was providing a counterexample. :) Oleg Alexandrov (talk) 23:17, 14 May 2006 (UTC)
replies: text: Geez, I don't know what happened there.  Gone for a six. -lethe talk [ +] 10:52, 15 May 2006 (UTC)
thread_title:  Finite rank 

message: It says that BP is a weak negation of AC. But not-LM and not-BP are weak consequences of AC, thus BP and LM are STRONG negations of AC. They are at the bottom of the tree.
http://www.math.vanderbilt.edu/~schectex/ccc/excerpts/acchart.gif
:(I've copied this here from User talk:Lethe; this is probably the more appropriate forum.)
::There are a couple of really odd things about the way that section is worded. For one thing, the Baire property is not any form of negation of the axiom of choice; it's a property of sets of reals (or more generally, subsets of a topological space). The "strong" negation of AC would be that every set of reals has the Baire property; the negation is not the Baire property itself.
::Perhaps worse is the line about how
:::Solovay's result shows that it is not necessary to assume that all infinite-dimensional vector spaces admit discontinuous linear maps,
::What's that supposed to mean, exactly? I'd say it shows almost the opposite—namely that you do have to assume something, beyond ZF, if you want to prove there are always such maps. (It certainly doesn't imply, by itself, that there don't have to be such maps, for if you believe AC is true—as Solovay, of course, does—then you'll believe there always are such maps.)
::Finally, there's a problematic remarks about how such maps always have to be "nonconstructible", without specifying what "constructible" means. --Trovatore 04:16, 7 August 2006 (UTC)
replies: text:  I strongly have no clue what you guys are talking about (admittedly, that's a weak point). Oleg Alexandrov (talk) 15:45, 7 August 2006 (UTC)
thread_title:  strong vs weak 

message: "It turns out that the derivative of a function at a certain point is really a CONTINUOUS linear map."  I'm quoting from the Functional Analysis article. (I added the caps to continuous.)  Is the quoted sentence true?Rich 07:30, 19 September 2006 (UTC)
replies: text: that short paragraph is somewhat outta place, IMHO. the Frechet derivative of a map between Banach spaces at a given point, if it exists, is by definition a bounded operator. on the other hand, the differential operator on, say, C[0,1] with the sup norm, is definitely not bounded. Mct mht 07:59, 19 September 2006 (UTC)
thread_title: Functional Analysis Article

message: In section #Axiom of choice:
 ...was led to adopt ZF + DC + BP (dependent choice is a weakened form
 and the Baire property is a strong negation of AC) as his axioms.
I'm not a specialist but that phrase seems inconsistent: shouldn't it read
 ...and the Baire property is a negation of strong AC
? (This seems to me the minimal change necessary so that it could possibly make sense.) — MFH:Talk 22:02, 4 September 2007 (UTC)
replies: text: No, "strong negation of AC" was what was intended -- if every set of reals has the Baire property then AC is false, and you have more information in addition to AC being false. So your change is not correct, but the original wording is not that good either (the Baire property is not an axiom or proposition at all, but rather a property of sets in a topological space). --Trovatore 22:23, 4 September 2007 (UTC)
thread_title:  AC - strong negation?? 

message: The paragraph 'General existence theorem' says:
axiom of choice, which was not needed for the concrete example in the previous section.
but the construction of this hamel basis needs the AC - or am I wrong? 134.109.40.117 (talk) 14:00, 13 August 2009 (UTC)
thread_title:  Hamel Basis vs. AC 

message: In the article there is an error. It says that every on every infinite dimensional locally convex topological vector space, there are discontinuous functionals. This can not be true. Let E be an arbitrary infinite dimensional vector space (without topology). Then take the topology generated by its algebraic dual. Then E has the finest topology with the property that all linear functionals are continuous. Topologies of this form are always locally convex since they are generated by the seminorms |phi| for all linear functionals phi. --79.196.201.6 (talk) 14:03, 17 January 2010 (UTC)
replies: text: Yes, you are right! Thanks. Boris Tsirelson (talk) 15:49, 17 January 2010 (UTC)

text: Really, I do not know what to do with the "Beyond normed spaces" section. If we remove the wrong claim, it will contain nothing about discontinuous linear functionals. Probably this section should be merged to something else. Boris Tsirelson (talk) 15:55, 17 January 2010 (UTC)
thread_title:  Generalization to locally convex spaces 

message: In the section "Beyond normed spaces" it is claimed that  : 
: The argument for the existence of discontinuous linear maps on normed spaces can be generalized to all metrisable topological vector spaces,... 
If i understand correctly, this means that the whole argument can be generalised, by replacing "norm" by "quasinorm" (the one corresponding to the metric). But the key argument in the normed case is that "continuous implies bounded", or (which is what is used) that "non-bounded implies non-continuous". 
I have some doubt about that for the non-normed case. Indeed, if one considers linear operators (instead of linear forms), then for example on some Frechet spaces, it is easy to construct continuous operators which are NOT bounded (with respect to the metrix). 
I suspect this is also the case for forms, though i don't have a example in mind...  —Preceding unsigned comment added by Roy.nico (talk • contribs) 09:33, 5 March 2010 (UTC)
replies: text: I guess that a discontinuous linear functional on a Frechet space exists. We take a sequence of linearly independent vectors and construct the functional on their linear span so that its value on the n-th vector is too large for the 1/n-neighborhood of the origin. Then we extend it to the whole space (using the choice axiom). But this is only a guess. It would be nice to find it in a book. Otherwise, "citation needed" should be put to the article. Boris Tsirelson (talk) 12:56, 5 March 2010 (UTC)
thread_title:  Generalization to metrisable spaces 

message: This article would be far more useful if individual references were given for the claims made.  Citing two large books at the end of the article is not sufficient.  128.84.234.217 (talk) 23:03, 21 July 2010 (UTC)
thread_title:  References 

message: Someone replaced "instead" with "insted". What could it mean? Boris Tsirelson (talk) 06:34, 21 March 2012 (UTC)
thread_title:  "insted" instead of "instead" 

message: Is this proof correct? The definition of a bounded operator requires an a constant M independent of x such that ||f(x)|| <= M ||x|| for all x, while the article's proof has ||f(x)|| <= C*M*||x|| where C depends on x.  — Preceding unsigned comment added by 198.84.149.241 (talk) 00:43, 1 December 2014 (UTC)
replies: text: No, C does not depend on x; C is the constant of equivalence of two norms. Boris Tsirelson (talk) 06:52, 1 December 2014 (UTC)
thread_title:  A linear map from a finite-dimensional space is always continuous 

message: I've added a clarification in the article where one was requested:
: Examples of discontinuous linear maps are easy to construct in spaces that are not complete; on any Cauchy sequence e_i of linearly independent vectors which does not have a limit, there is a linear operator T such that the quantities \|T(e_i)\|/\|e_i\| grow without bound. In a sense, the linear operators are not continuous because the space has "holes".
This could be improved in various ways, but I think it's what the original comment was getting at.
John Baez (talk) 01:51, 28 June 2021 (UTC)
thread_title: Clarification