Talk:Lp space

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Hi, I turned five-or-so references to "Banach space"s into "wiki-hyperlinks" (they were plain text).

BTW, I noticed you refer to Sobolev spaces; you may (or may not) want to add as another reference the brand new book "Selected Works of S.L. Sobolev", ISBN 038734148X.

-- Biscay 10:33, 23 August 2007 (UTC).[reply]

Hi Biscay, it is customary to not link the same word repeatedly but only the first instance. Furthermore it is customary to add new entries to the bottom of talk pages. The + at the top of talk pages can also be used to add new sections to the bottom. --MarSch 11:57, 4 October 2007 (UTC)[reply]

Wait, I thought the problem with L^p spaces, 0<p<1, was that they weren't locally convex.

Also, who in the world decided to put the p in the subscript? When you have spaces with multiple indices like Sobolev spaces, I can understand the p landing somewhere odd, but there's only one index here, and I've never seen it downstairs... Loisel 20:49 Jan 28, 2003 (UTC)

I think you're right about the notation and we should change it; I'm not sure about the locally convex part though. Can you find it in some book? AxelBoldt 01:34 Jan 29, 2003 (UTC)

Several. Wojstaszczyk Banach Spaces for Analysts, page 9 or Rudin's Functional Analysis, page 37. By the way, I must contradict my note above now, I do know an author who puts the p downstairs. Wojstaszczyk does, but he's the only one I know. Sorry for the long delay, I never noticed your reply. Loisel 07:14, 27 Jun 2004 (UTC)

The reason you put p in the subscript can be that you write Sobolev spaces with p in the subscript, and these two p's are "things of the same kind".

Timur

Yeah, perhaps there should be a bit more emphasis here? The Sobolev space can be written, for example, W_p^{k,k'}, for some intersection of k and k' - differentiable spaces. In this case, the L^p is written L_p for consistency in the notation. The only point being, these notations are not all that rare. Cypa 06:06, 22 March 2006 (UTC)[reply]

Actually, putting the p in the subscript is a very common convention in Banach space theory; see for example several of the articles in the recent "Handbook of the Geometry of Banach Spaces" edited by Johnson and Lindenstrauss. Nonspecialists in Banach space theory (including most authors of functional analysis texts) almost always put the p as a superscript. For different people there are different reasons (besides habit and the reasons mentioned above) for preferring one convention or the other. The subscript emphasizes the relationship with the function of "raising to the pth power"; there are spaces related to more complicated functions which are sometimes analogously denoted, for example, L log L. On the other hand, when working with finite dimensional spaces, we are accustomed to seeing the dimension as a superscript, so it is convenient to put the p as a subscript to avoid confusing it with the dimension. 76.190.215.240 (talk) 16:59, 14 February 2008 (UTC)[reply]

Does anyone have any more to add about the dual of L^infinity? Lupin 15:21, 7 Feb 2004 (UTC)

I think it should be fairly easy to prove that the natural injection of L^1 into (L^infinity)** is surjective iff both of them are finite-dimensional (i.e. X is essentially a direct sum of a finite number of atoms of finite measure).

Prumpf 11:05, 10 Feb 2004 (UTC)

The injection of L^1 into (L^infinity)* (not **) is clearly surjective in the finite dimensional case. Let the measure space S be sigma-finite. Then L^infinity is not separable. If a dual space A* is separable, then A must also be separable. Hence L^1, which is separable, is not the dual of L^infinity.

However, I think there's a problem. In Banach Spaces for Analysts, by Wojtaszczyk, 1991, page 10: "The space L_infinity(Omega,mu)^* is (unless (Omega,mu) consists of a finite number of atoms) much bigger than L_1(Omega,mu) and is impossible to describe explicitly" (emphasis added.) I concur.

Loisel 05:03, 27 Jun 2004 (UTC)

What's impossible to describe explicitly supposed to mean, anyway? The Lp spaces are impossible to describe explicitly, in a way, so there doesn't seem to be any actual information added to the article by that phrase.

Also, your proof works in the sigma-finite case only, and thus it's slightly misleading to include the proof outline in the article. However, the general proof can be reduced to the sigma-finite case fairly easily.

Prumpf 17:31, 28 Jun 2004 (UTC)

I don't see how the general case reduces to the sigma-finite case -- are you sure of that statement? I think the proof I gave does generalize, if you replace "sigma-finite" by the appropriate notion of size for the measure space S. Say that S can be written as the disjoint union of E_a, sets of finite measure, with the index a running through an index set I. WLOG, I is of minimal size. Then, L^1 has a dense subset of cardinality #I; however, L^infinity does not (you can embed 2^I as a discrete set in L^infinity). If A* has a dense subset of a certain cardinality, then A must have a dense subset of the same cardinality. Hence we get that L^1 is not the dual of L^infinity.

I'm not quite sure what Wojtaszczyk means by "Impossible to describe explicitly." He could mean it in some formal sense (like what we mean of the Vitali set when we say it can't be constructed -- this is because it uses the axiom of choice in an essential way.) However, I suspect that's not what he means; rather, I suspect he's saying that nobody has given any good characterizations of the dual of L^infinity. I've seen it described as a space of distributions, but it's not the same space as the dual of C_c^infinity, nor is it the space of tempered distributions. Loisel 08:03, 29 Jun 2004 (UTC)

Not every measure space (not even every measure space in which there is no set $A$ with infinite measure that doesn't have a subset of nonzero finite measure; for those measure spaces it is trivial that the dual of L^infinity is larger than L^1) arises as a direct sum of finite measure spaces.

I think that if you assume that mu({x}) is finite for every x, the "trivial" proof I gave above works. If E={x|mu(x)=infinity} is not empty, I'm pretty sure you can split S into E and S\E and work from there. The proofs you give below appear kosher. Loisel 00:59, 30 Jun 2004 (UTC)

Uhm, the case in which your E isn't empty is rather simple. Still, E can be empty and I don't see how your proof would work, since that still doesn't mean S is a direct sum of finite measure spaces (trivially, consider R with infinity times the Lebesgue measure). Also, aren't there sigma-finite measure spaces for which L^1 and L^2 aren't separable (the product of infinitely many copies of the unit interval with the Lebesgue measure, for example)? Your proof seems to fail for those, I'm afraid. (And a fortiori, mine is then invalid as well, having used your proof) Prumpf 15:39, 30 Jun 2004 (UTC)

Wait, my proof is still valid, since it needs only the case where S is a countable sums of atoms, for which the separability argument holds. Prumpf 15:45, 30 Jun 2004 (UTC)

Yes, I see. Go for your proof. Loisel 19:28, 30 Jun 2004 (UTC)

Assume our measure space X is not sigma-finite, but that we can find a countable family {A_i} of disjoint measurable subsets with 0 < mu(A_i) < infinity (this covers all the cases that have been left out so far). Let Y be the measure space consisting of a countable numbers of atoms {y_i}, such that mu({y_i}) = mu(A_i). We can thus embed L^infinity(Y) in L^infinity(X) and L^1(Y) in L^1(X), and obtain a mapping (L^infinity(X))* -> (L^infinity(Y))* by contravariance. The diagram

L^1(Y) ----> (L^infinity(Y))*
  |                 ^
  v                 |
L^1(X) ----> (L^infinity(X))*

then commutes. Furthermore, the arrow in the right column is a surjection, by the Hahn-Banach theorem. Also, the arrow in the top row isn't a surjection. Assume the bottom arrow is. Choose y* in (L^infinity(Y))* that is not in the image of the top arrow, x* in (L^infinity(X))* which maps to y* under the right arrow, x in L^1(X) which maps to x*. We now modify x to be constant on the sets A_i by averaging over them, which results in a new x* but leaves y* unchanged. However, we can now choose y in L^1(Y) that maps to x, and maps to y* by commutativity, a contradiction.

Not as concise as I had hoped, but this really should work for nasty measure spaces (for which L^infinity presumably isn't the dual of L^1, but whatever) Prumpf 11:08, 29 Jun 2004 (UTC)

Slightly simpler: define a map L^1(X)->L^1(Y) by averaging over A_i, then prove the diagram

L^1(Y) ----> (L^infinity(Y))*
  ^                 ^
  |                 |
L^1(X) ----> (L^infinity(X))*

commutes. Assuming the bottom arrow is surjective, the diagram in fact looks like this

L^1(Y) ----> (L^infinity(Y))*
  ^                 ^
  ^                 ^
  |                 |
L^1(X) --->> (L^infinity(X))*

(this can all be done in Set), from which it follows that the top arrow was surjective as well, contradicting the result about sigma-finite measure spaces. Prumpf 11:40, 29 Jun 2004 (UTC)

I'm not sure I see the problem with identifying the dual of L^infinity with the ba space. Certainly, every linear functional phi: L^infinity(X) -> R gives rise to a finitely additive (signed) measure mu on X by defining mu(A) = phi(1_A). Since |mu(A)| = |phi(1_A)| <= |phi|, mu is bounded, and |phi| is just the variation of mu, the maximal integral of a measurable function f with |f(x)| = 1 for all x in X.

OTOH, if mu is a f.a. bounded signed measure, we see that there is exactly one linear functional corresponding to it (if

${\displaystyle f=\sum _{n\in \mathbb {Z} }2^{n}\times 1_{A_{n}}}$, then ${\displaystyle \phi (f)=\sum _{n\in \mathbb {Z} }2^{n}\times \mu (A_{n})}$. It is bounded since mu is.

Looks good. Loisel 19:42, 30 Jun 2004 (UTC)

Thus, (L^infinity)* is isometrically isomorphic to the ba space, assuming it is defined to consist only of those measures which keep all of the original measure's null sets. That seems a fairly explicit description to me, if possibly not a very useful one.

Note that by the Radon-Nikodym theorem, L^1 can be identified with the subspace of the ba space consistion of all sigma-additive measures. Thus, the natural endomorphism L^1->(L^infinity)* is an isomorphism if and only if all finitely additive measures on X which keep all null sets are automatically sigma-finite, and it is trivial to see that it is exactly the finite-dimensional case where that is true. This is a much simpler proof than the one I've given above. Prumpf 16:02, 30 Jun 2004 (UTC)

In reference to the above, I vote we remove all lies from our math articles, now and forever, in the name of the great turtle A'tuin. Amen. Loisel 07:25, 27 Jun 2004 (UTC)

The section Further properties states:

The map sending f to ||f||_p is a quasi-norm, and L ^p is a quasi-Banach space, that is, a complete quasi-normed vector space.

This statement is true/confusing/contradictory with the rest of the article. Here's why: The Lp norm will assign a norm of zero for functions that are not identically zero; thus by definition, it is a semi-norm not a norm. Thus, "raw Lp space" is a Kolmogorov space topologically speaking (and that article has a good discussion of this). Only after the Kolmogorov quotient is taken, where we place in an equivalence class all functions having the same norm, do we get the "true" Banach space. So, somehow, the above sentence needs to be re-written and incorporated more directly into the discussion. linas 20:30, 10 December 2005 (UTC)[reply]

Is a quasi-norm the same as a semi-norm, i.e., something which satisfies all the axioms except that ||x|| = 0 does not imply x = 0? In that case, what's wrong with the discussion in the section L^p spaces? -- Jitse Niesen (talk) 20:57, 10 December 2005 (UTC)[reply]

Okay, I found out that a quasi-norm is something different. Now I don't understand Linas' comment at all :( As I understand it, "raw Lp space" is not a Kolmogorov space, because the functions which differ in one point are topologically indistinguishable. After taking the quotient, you get a Kolmogorov space. I have no idea what quasi-norms have to do with this, but I've never worked with p < 1. -- Jitse Niesen (talk) 21:20, 10 December 2005 (UTC)[reply]

Oh, well, I thought that quasi-norm is the same thing as semi-norm, since they both redirect to the same article (and that article doesn't otherwise define quasi-norm). My complaint wasn't about other discussions, it was about the quoted sentence in particular. Its not clear what it has to do with p < 1, if anything. I retract my complaint if we can clear up what a quasi-norm is.

Sorry for the confusing term "raw Lp space"; I was trying to refer to that thing that has the finer topology, "the space of functions that are p-integrable but has a topology that distinguishes functions that differ at a point." I don't know if that space has a name, or what the name would be. linas 21:44, 10 December 2005 (UTC)[reply]

Yes, the redirect from quasi-norm also misled me, so I removed it. A web search suggested that a quasi-norm satisfies ||x + y|| ≤ C(||x|| + ||y||) instead of the triangle equality. This makes more sense in the context. I also tried to clarify the article. -- Jitse Niesen (talk) 23:03, 10 December 2005 (UTC)[reply]

Isn't it necessary in defining a vector space to define addition in that space?Loodog 04:27, 7 March 2006 (UTC)[reply]

Well, in Lp spaces, which are function spaces, the addition is just sum of functions. I guess nobody bothered to write that in the article, and it is rather clear from the context. Oleg Alexandrov (talk) 04:57, 7 March 2006 (UTC)[reply]

I would agree, but the function space article also fails to define addition (I think) on account of the definition of function space being too general. Also, citing vectors in Euclidean space as an Lp space in the article, it seems that the domain can be things other than functions. Maybe there should be a brief note saying, "Lp space is a function space. Addition in Lp space is the usual addition under any function space" or something like that.Loodog 20:41, 7 March 2006 (UTC)[reply]

Vectors are lp with a little "l", functions are Lp with a big "L". linas 00:53, 8 March 2006 (UTC)[reply]

The article isn't complete without a full definition of the vector space structure, so I've added the information you've requested. -lethe ^{talk +} 02:32, 8 March 2006 (UTC)[reply]

This page is a mess -- "capital L"^p referes to Lebesgue spaces, not general normed vector spaces. And, the part on L^p in terms of Lebesgue starts with a measure space with some measure mu, then talks about Lebesgue integrability. What's going on? Please be more careful! —The preceding unsigned comment was added by 131.215.242.202 (talk • contribs) 04:22, 10 April 2006 (UTC2)

I guess the term "Lebesgue integral" can mean two things: 1. the supremum of simple functions less than the integrand over sets measurable with respect to any measure, or 2. the same supremum using the Lebesgue measure. If you understand the term in the first way, the more general way, there is no problem understanding Rⁿ with the l^p norm as a Lebesgue space (use the counting measure on the set n). There is also no problem with the phrase you are complaining about. I claim that we have to understand the term in its more general sense, at least for the scope of this article, since such spaces as L^p(M,μ), integrable over an aribtrary space with respect to an arbitrary measure, are found in the literature thusly denoted. This is the place to treat such spaces. -lethe ^{talk +} 21:27, 10 April 2006 (UTC)[reply]

Is it really necessary to assume that the measure is sigma-finite to conclude that "the dual of L¹(S) is isomorphic to L^∞(S)"? Theorem 2.34 of Adams, Sobolev Spaces, states that

${\displaystyle [L^{1}(\Omega )]'\cong L^{\infty }(\Omega )}$

if Ω is an open, but not necessarily bounded, subset on R^n (the underlying measure is the Lebesgue measure). So the statement is true for more measures; what I'm wondering is whether it's true for all measures. I guess not, given the number of people that have contributed to this page. -- Jitse Niesen (talk) 09:38, 19 July 2006 (UTC)[reply]

Not true in general, unless you define L^\infty weirdly when \mu is not sigma-finite. I believe the correct definition would be {f|f restricted to E is bounded, for every sigma-finite E}. Note that the Lebesgue measure on R^n is sigma-finite, Adams's statement is a special case, not a generalization. Loisel 10:17, 19 July 2006 (UTC)[reply]

Oh dear, I forgot the difference between finite and sigma-finite measures. How embarrassing … Anyway, thanks for your quick answer. -- Jitse Niesen (talk) 10:39, 19 July 2006 (UTC)[reply]

Sullivan, I understand what you're saying but it's just a question of notation, so I suggest it stays the way I put it now. If one understands that f d\mu is the measure such that f d \mu(E)=\int_E d d\mu then the notation is consistent.

Loisel, I understand "${\displaystyle f\,\mathrm {d} \mu }$" notation. My comment that "notation is not definition" refers to the fact that a space is not defined by the notation used to describe it. The weighted ${\displaystyle L^{p}}$ space is defined to be a space of functions with certain properties. It is not defined to be ${\displaystyle L^{p}(S,w\,\mathrm {d} \mu )}$ or ${\displaystyle L_{w}^{p}(S,\mu )}$. These are just notations, and I deliberately included both in my revision of the article. Also, I must note that your revision itself used inconsistent notation for the space (different notation in body text and displayed equation)! Sullivan.t.j 16:12, 1 October 2006 (UTC)[reply]

I'm not sure if I'm understanding you right, but you shouldn't redo the L^p theory when you add a weight function; that's wasted effort. If \nu is the measure such that d\nu=f d\mu, then the weighted L^p space should be defined in terms of the already-defined space L^p(\nu). Otherwise, you have two possibly inconsistent definitions, and the reader may not be sure which theorems of L^p spaces apply to weighted L^p spaces.

If it's clear that a weighted L^p space is nothing more than an L^p space with a different measure, the reader understands that the theorems for L^p spaces and weighted L^p spaces are the same. Loisel 18:57, 1 October 2006 (UTC)[reply]

On further reflection, I agree that using ${\displaystyle w\,\mathrm {d} \mu }$ re-uses notation in a fairly nice way and saves space. Nonetheless, I think that it should be more explicitly stated: my first thought was that you meant something like ${\displaystyle L^{p}(S,w,\mu )}$ but had made a typo. So, my latest revision uses more notation but (hopefully) makes the reasoning more obvious. I have also fixed the subscript on the norm in the displayed equation (it used ${\displaystyle L_{w}^{p}(S)}$-notation). Sullivan.t.j 08:29, 2 October 2006 (UTC)[reply]

OK. Loisel 21:09, 2 October 2006 (UTC)[reply]

Under motivation it says "It turns out that this definition indeed satisfies the properties of a length function (or norm), which are that only the length of the zero vector is zero (snip)."

If I understand how length is defined in Minkowski space than 4-vectors of length zero exist that are not the zero vector. Maybe a better description of length function is needed? Unmasked 14:09, 8 October 2006 (UTC)[reply]

I have modified the wording of the offending paragraph to make it a little clearer. However, you are right to note that in Minkowski space there are non-zero vectors with zero "length"; the "length" function on Minkowski space is not a norm. Sullivan.t.j 16:26, 8 October 2006 (UTC)[reply]

I seem to see things such as ${\displaystyle L_{2}^{1}(X)=\{f:\int _{X}|f(x)|(1+x^{2})\,\mathrm {d} x<\infty \}}$ a lot. Is that some variant of Lp spaces? (Is there some such thing as "L^p_q space", defined as something along the lines of ${\displaystyle L_{q}^{p}(X)=\{f:\int _{X}|f(x)|^{p}(1+x^{q})\,\mathrm {d} x<\infty \}}$?) -- 129.78.64.102 03:19, 31 May 2007 (UTC)[reply]

These are weighted Lp spaces, from what I know, with the weight being that function you multiply |f(x)| by. Oleg Alexandrov (talk) 03:40, 31 May 2007 (UTC)[reply]

Why are these L spaces? Why that letter? 75.67.167.110 17:31, 26 July 2007 (UTC)[reply]

L is for Henri Lebesgue and his integral, which is the sense in which the integrals defining the L^p norms are defined. Sullivan.t.j 21:53, 26 July 2007 (UTC)[reply]

I'd really love to have a good proof of completeness in the article instead of the "follows from convergence theorems" there is now. proof that L_p spaces are complete (at PlanetMath) might also help with accomplishing that. --MarSch 12:00, 4 October 2007 (UTC)[reply]

In the properties of L^p spaces part, the functional G(f) is said to be in the dual of L^p, but it is in fact an antilinear map, so we should talk about the antidual of L^p and not its dual, if one want to use the map G.

Thus, we could say that the antidual of L^p is L^p' where p^-1 + p^-1 = 1, the antidual of L^1 is L^\infty and we recover the famous fact that L^2 is its own antidual, as any Hilbert space.

Its seems to me that if we use G( \bar f) instead of G(f) the previous properties still hold by replacing antiduality by duality, but one should keep in mind that if we consider Bochner spaces of functions with values in a complex Hilbert space, we only have G(f) at our disposal, since no \bar f is canonically defined. So the natural thing in the complex setting is antiduality, not duality. Please tell me if you agree or disagree.

HB. —Preceding unsigned comment added by 139.124.7.126 (talk) 06:15, 9 December 2007 (UTC)[reply]

You are correct that G (not G(f); that is a number) is antilinear in the complex case. I think you are also correct in saying that L^p is isomorphic to its antidual and anti-isomorphic to its dual in the complex case. We should probably just treat the complex case, or maybe use *-algebras and mention that the real numbers with the trivial involution trivially form a *-algebra and the complex numbers with conjugation also form a *-algebra.

I don't know much about Bochner spaces. Why can the involution of a complex Hilbert space not be pulled back to each Bochner space of functions to that Hilbert space? --MarSch (talk) 18:07, 9 December 2007 (UTC)[reply]

OK but it seems to me that there is no natural involution in an abstract complex Hilbert space, even if there is one in \C^n. So we have a lot possible involutions and none are canonical, so it is probably better to keep the antidual formulation. As you say, we can always recover the real case by noting that in this case antidual=dual. Another nice feature of antidual formulation is that the antiduality bracket is a natural extension of the scalar product of L^2, contrarily to the duality bracket. I don't know if these remarks are somewhere on Wikipedia, I guess it would be useful to mention them in this article.

139.124.7.126 (talk) 02:52, 10 December 2007 (UTC) HB[reply]

I think that this sentence is not correct: how can a set grow larger? TomyDuby (talk) 02:08, 24 July 2008 (UTC)[reply]

By getting in additional elements, I presume. Oleg Alexandrov (talk) 02:16, 24 July 2008 (UTC)[reply]

If ${\displaystyle p<q}$, then ${\displaystyle \ell ^{p}\subsetneqq \ell ^{q}}$. Loisel (talk) 02:22, 24 July 2008 (UTC)[reply]

Um, except that, for the special cases of sequences, one does have ${\displaystyle \ell ^{1}\subset \ell ^{p}\subset \ell ^{q}\subset \ell ^{\infty }}$ when ${\displaystyle 1<p<q<\infty }$, and similarly for the set of measurable functions defined on the closed unit interval. This doesn't hold for general measurable functions on general measurable spaces, nor does it hold for measurable functions on open intervals of the real line. linas (talk) 18:29, 7 August 2008 (UTC)[reply]

Linas, I think this is incorrect. The sequence ${\displaystyle a_{k}=1/k}$ is in ${\displaystyle \ell ^{p}}$ for every p>1, but not in ${\displaystyle \ell ^{1}}$.

On the other hand, for function, e.g., of the interval, you have that ${\displaystyle 1/{\sqrt {x}}}$ is in ${\displaystyle L^{p}(0,1)}$ for every ${\displaystyle p<2}$, but not in ${\displaystyle L^{p}}$ for every ${\displaystyle p\geq 2}$.

I will now go through your edits carefully to see which of your criticisms are founded. Loisel (talk) 21:37, 7 August 2008 (UTC)[reply]

And so, in the section on "embeddings", I read

L^p(S) is not contained in L^q(S) iff S contains sets of arbitrarily small measure, and

which sure as heck does not seem to make sense to me, as, the closed unit interval has sets of arbitrarily small measure, and yet one has ${\displaystyle L^{p}\subset L^{q}}$ when ${\displaystyle p<q}$ for the closed unit interval! I mean, its been a while, and my memory is failing with age, but jeez, surely there's a bug here!? I removed the above sentence; someone please clarify the intended meaning! linas (talk) 18:45, 7 August 2008 (UTC)[reply]

Am I crazy? The few books/papers I have use subscripts, not superscripts: so L_p and not L^p, so it was jarring to see superscripts used in this article. Is superscript really the predominant practice today? Perhaps subscripts date back to an earlier era?linas (talk) 18:29, 7 August 2008 (UTC)[reply]

Many of the recent books use superscript (e.g. Rudin), but there are many others that use subscript. Loisel (talk) 21:37, 7 August 2008 (UTC)[reply]