112

Let $V$ be a vector space with infinite dimensions. A Hamel basis for $V$ is an ordered set of linearly independent vectors $\{ v_i \ | \ i \in I\}$ such that any $v \in V$ can be expressed as a finite linear combination of the $v_i$'s; so $\{ v_i \ | \ i \in I\}$ spans $V$ algebraically: this is the obvious extension of the finite-dimensional notion. Moreover, by Zorn Lemma, such a basis always exists.

If we endow $V$ with a topology, then we say that an ordered set of linearly independent vectors $\{ v_i \ | \ i \in I\}$ is a Schauder basis if its span is dense in $V$ with respect to the chosen topology. This amounts to say that any $v \in V$ can be expressed as an infinite linear combination of the $v_i$'s, i.e. as a series.

As far as I understand, if a $v$ can be expressed as finite linear combination of some set $\{ v_i \ | \ i \in I\}$, then it lies in its span; in other words, if $\{ v_i \ | \ i \in I\}$ is a Hamel basis, then it spans the whole $V$, and so it is a Schauder basis with respect to any topology on $V$.

However Per Enflo has constructed a Banach space without Schauder basis (ref. wiki). So I guess I should conclude that my reasoning is wrong, but I can't see what's the problem.

Any help appreciated, thanks in advance!

UPDATE: (coming from the huge amount of answers and comments) Forgetting for a moment the concerns about cardinality and sticking to span-properties, it has turned out that we have two different notions of linear independence: one involving finite linear combinations (Hamel-span, Hamel-independence, in the terminology introduced by rschwieb below), and one allowing infinite linear combinations (Schauder-stuff). So the point is that the vectors in a Hamel basis are Hamel independent (by def) but need not be Schauder-independent in general. As far as I understand, this is the fundamental reason why a Hamel basis is not automatically a Schauder basis.

José Carlos Santos
  • 427,504
Lor
  • 5,547
  • 2
  • 26
  • 54
  • For some reason, every Schauder basis is required to be countable, while a Hamel basis is allowed to be uncountable. That is why it it possible to have a Banach space with no Schauder basis. I do not know why this convention is used; presumably, though, there's a good reason. – goblin GONE Jan 07 '14 at 13:46
  • You may be interested in this question I just asked. – goblin GONE Jan 07 '14 at 13:52
  • @user18921: I don't think there's any reason for a Schauder basis to be countable (if that was the case, finding a Banach space without one wouldn't be much of an accomplishment). Certainly a Hilbert basis is a Shauder basis. What's important is that it's independent in the sense of infinite summations. Otherwise, a Schauder basis wouldn't be much of a basis. – tomasz Jan 07 '14 at 20:02
  • @tomasz Countability is part of the definition of a Schauder basis in all instances that I remember. The accomplishment is finding a separable Banach space without a Schauder basis. – Daniel Fischer Jan 07 '14 at 20:27
  • @DanielFischer Could this just be a side-effect of so many applications being rooted in separable Hilbert space? I thought I remembered that inseparable Hilbert spaces of very high dimension all had orthonormal Schauder bases... – rschwieb Jan 07 '14 at 20:37
  • @rschwieb Those are Hilbert bases, and in the non-separable cases, they are not Schauder bases per the definitions of Schauder bases I know. One difference is that for a Hilbert basis (orthonormal system, more generally) ${e_\alpha:\alpha\in A}$, the family ${\langle x,e_\alpha\rangle\cdot e_\alpha:\alpha\in A}$ is always summable, with sum $x$ (if it's merely an ONS, the sum is the projection of $x$ to the closed span). For a Schauder basis ${s_i:i\in\mathbb{N}}$ with the associated coordinate functionals $\xi_i$, the family ${\xi_i(x)\cdot s_i:i\in\mathbb{N}}$ need not be summable. – Daniel Fischer Jan 07 '14 at 20:43
  • And therefore, you need the ordering imposed by indexing with $\mathbb{N}$ to have a well-defined convergence criterion. I'm not sure you can generalise that well to larger well-ordered index sets. – Daniel Fischer Jan 07 '14 at 20:45
  • @DanielFischer That seems to indicate that a Hilbert basis has additional criterion, and that maybe Schauder bases aren't all Hilbert, but what Schauder axiom(s) are we saying a Hilbert basis doesn't satsify? – rschwieb Jan 07 '14 at 21:05
  • @rschwieb The indexing by $\mathbb{N}$. A Schauder basis is in a way more general, but you buy that generality by less generality in a different place. An unconditional Schauder basis can without problems (I think, I don't see any problems lurking, but I haven't investigated) be generalised to uncountable families, but without the unconditionality, you're bound to sum in a specific order. – Daniel Fischer Jan 07 '14 at 21:11
  • 1
    @DanielFischer Sorry, I meant something besides the indexing by $\Bbb N$ axiom (which is begging the question about how big the basis is). We're bound to countable sums, I believe that. Hamel bases are bound to finite sums, but they are not restricted to be finite themselves. Is there something wrong with taking an uncountable set and talking about their countable linear combinations? – rschwieb Jan 07 '14 at 21:14
  • 2
    @rschwieb If the family is summable, no problem. But for a Schauder basis, summability is not required. Note that for a Schauder basis you have not only countability, but a specific ordering. If ${ e_j : j \in \mathbb{N}}$ is a Schauder basis that is not unconditional, there are permutations $\pi$ of $\mathbb{N}$ such that ${ e_{\pi(j)} : j \in \mathbb{N}}$ is not a Schauder basis. – Daniel Fischer Jan 07 '14 at 21:19
  • 1
    @DanielFischer Sorry, I guess I'm still overlooking something :( The last comment still looks like you're saying "A Schauder basis does not satisfy this Hilbert axiom." I agree with that, but I'm asking the other direction. (The last sentence of that comment is super interesting though, thanks for telling me about it. I did not expect order to come into play.) – rschwieb Jan 07 '14 at 21:28
  • @DanielFischer And thanks for being patient. When I feel this close to having an answer, I'm usually keen to chase it down. I'm sure you know much more about this than I do. – rschwieb Jan 07 '14 at 21:41
  • 1
    @rschwieb A Schauder basis can lack a nice property a Hilbert basis automatically has. Therefore, we must impose a restriction on Schauder bases that we need not impose on Hilbert bases. Let us call a Friendly basis of a(n infinite dimensional Hausdorff locally convex) topological vector space $E$ a sequence ${ f_k : k\in \mathbb{N}}$ such that for every $x\in E$ there is a unique sequence ${\varphi_k(x)}$ of coefficients such that the family ${\varphi_k(x)\cdot f_k : k\in\mathbb{N}}$ is summable with sum $x$. A Friendly basis is a Schauder basis, and if $E$ is a Hilbert space and the – Daniel Fischer Jan 07 '14 at 21:45
  • 1
    family is orthonormal, also a Hilbert basis. Now, the concept of "Friendly bases" can be generalised in different directions. We can drop the countability, and we can drop the summability requirement. Dropping both leads to problems. A Schauder basis drops the summability requirement and replaces it with the convergence of the sequence of partial sums in the specified order. – Daniel Fischer Jan 07 '14 at 21:47
  • @DanielFischer : OK, I think I have the impression that there is an analytical nuance that escapes me, and that I can let it go for now. The way I had been thinking about it, Hilbert bases were just examples of Schauder bases: collections whose countable sums generated a space, and satisfied a linear independence axiom. It sounds like that is not the case. I guess I would need an example of a Hilbert base that doesn't satisfy some Schauder axiom (aside from the countability of the base.) – rschwieb Jan 07 '14 at 21:58
  • @DanielFischer: I don't understand: what keeps us from indexing Schauder bases with any ordinals, or even any (directed) posets? – tomasz Jan 07 '14 at 22:17
  • @rschwieb That's the problem, H. bases are so nice that the uncountability is the only thing that can make them fail to be a Schauder basis. H. bases are the generalisation of a subset of "Friendly bases" in one direction, Schauder bases are a generalisation in another direction. You can generalise Friendly bases in the same direction (cardinality) as Hilbert bases do for a larger class of spaces, but as far as I can see, that requires you to keep the summability. The definition of Schauder bases drops the summability requirement, and therefore - as far as I can see - must keep countability. – Daniel Fischer Jan 07 '14 at 22:17
  • @DanielFischer: and what exactly do you mean by dropping summability? – tomasz Jan 07 '14 at 22:19
  • @tomasz The family ${ \xi_j(x)\cdot e_j : j\in\mathbb{N}}$ need not be summable. That is, its sum, and whether it the sum converges at all depends on the ordering. That absolutely forbids general directed sets. It might be possible to index by ordinal numbers, but defining the sum (even with only countably many nonzero terms) is not easy (you need an uncountable number of limit operations to reach an uncountable index, if my recollection of set theory doesn't deceive me). – Daniel Fischer Jan 07 '14 at 22:26
  • @DanielFischer: For ordinals, it seems pretty simple: the sum up to [and including] $\alpha$ is the limit of all shorter partial sums [plus the $\alpha$'th term] (provided that they exist, and that the limit exists). Whether there is a countable number of nonzero terms or not should not matter much. In case of general directed posets without infinite descending sequences we can do just the same thing, as far as I can see. But I can see that this can be come rather difficult to manage and not obviously productive. Thanks for the explanation. :) – tomasz Jan 07 '14 at 22:34
  • @DanielFischer OK, thank you for putting these thoughts to me :) – rschwieb Jan 07 '14 at 22:48

5 Answers5

66

People keep mentioning the restriction on the size of a Schauder basis, but I think it's more important to emphasize that these bases are bases with respect to different spans.

For an ordinary vector space, only finite linear combinations are defined, and you can't hope for anything more. (Let's call these Hamel combinations.) In this context, you can talk about minimal sets whose Hamel combinations generate a vector space.

When your vector space has a good enough topology, you can define countable linear combinations (which we'll call Schauder combinations) and talk about sets whose Schauder combinations generate the vector space.

If you take a Schauder basis, you can still use it as a Hamel basis and look at its collection of Hamel combinations, and you should see its Schauder-span will normally be strictly larger than its Hamel-span.

This also raises the question of linear independence: when there are two types of span, you now have two types of linear independence conditions. In principle, Schauder-independence is stronger because it implies Hamel-independence of a set of basis elements.

Finally, let me swing back around to the question of the cardinality of the basis. I don't actually think (/know) that it's absolutely necessary to have infinitely many elements in a Schauder basis. In the case where you allow finite Schauder bases, you don't actually need infinite linear combinations, and the Schauder and Hamel bases coincide. But definitely there is a difference in the infinite dimensional cases. In that sense, using the modifier "Schauder" actually becomes useful, so maybe that is why some people are convinced Schauder bases might be infinite.

And now about the limit on Schauder bases only being countable. Certainly given any space where countable sums converge, you can take a set of whatever cardinality and still consider its Schauder span (just like you could also consider its Hamel span). I know that the case of a separable space is especially useful and popular, and necessitates a countable basis, so that is probably why people tend to think of Schauder bases as countable. But I had thought uncountable Schauder bases were also used for inseparable Hilbert spaces.

rschwieb
  • 153,510
  • 2
    Your answer is probably the best here, but I think it would be even better if you cleared up the whole misconception about the Schauder basis being countable. – tomasz Jan 07 '14 at 20:03
  • This is the kind of things I'm looking for. Please correct me if I'm wrong: if I take a Schauder basis then its Schauder-span will be the whole V (by def). Of course its Hamel-span will normally be smaller, that's ok. Let's consider on the other hand a Hamel basis for V. Then again by def its Hamel-span is the whole V. I think we can say that its Schauder-span (with respect to any nice enough topology on V) is again the whole V...this is what I meant in the question with "so it is a Schauder basis with respect to any topology on V". What am I missing? – Lor Jan 07 '14 at 20:09
  • 1
    @Lor Let's consider on the other hand a Hamel basis for V. Then again by def its Hamel-span is the whole V. I think we can say that its Schauder-span (with respect to any nice enough topology on V) is again the whole V I think the last thing is that you would have to be certain the Schauder linear independence condition was satisfied, but otherwise yes, I agree with that. The Schauder independence condition is, in principle, stronger, although I don't have any informative examples :S – rschwieb Jan 07 '14 at 20:16
  • Ok, so for example "nice enough topology" means that V shoud at least be complete with respect to that topology. – Lor Jan 07 '14 at 20:22
  • @tomasz I think your idea is a good one, but I don't know if what I added did justice to it. I hope to hear again from you: thanks! – rschwieb Jan 07 '14 at 20:24
  • @Lor Yeah, so that infinite sums can be defined :) – rschwieb Jan 07 '14 at 20:33
  • 2
    Informative examples have already come up: In $\ell^2,$ compare your favorite not-eventually-zero sequence to the standard basis vectors, say, the harmonic sequence $(1/n)$. This is Hamel-independent of the set ${e_i}$ because a Hamel combination of the basis vectors only has finitely many nonzero coordinates. But it's not Schauder independent, because of course $(1/n)=\sum 1/n e_n$. – Kevin Carlson Jan 07 '14 at 21:08
  • 1
    @rschwieb: I think I'm rather confused myself. I've only had a brief experience with Schauder bases, and I've been told to think that they're like Hilbert bases only for non-Hilbert spaces. Perhaps I was mistaken, so I'll leave it to others for now. – tomasz Jan 07 '14 at 22:23
  • @tomasz I'm in the same boat! Thanks for priming that discussion. I'm glad it happened. – rschwieb Jan 07 '14 at 22:48
30

The problem is that an element of a Hamel basis might be an infinite linear combination of the other basis elements. Essentially, linear dependence changes definition.

Brian Rushton
  • 13,255
  • 11
  • 59
  • 93
  • But, it can still be written as $1$ times itself, without having to use the other basis elements. – detnvvp Jan 07 '14 at 13:31
  • 11
    Exactly; this gives two different representations of the element, so representations aren't unique. – Brian Rushton Jan 07 '14 at 13:33
  • Ah, ok, I see the point. – detnvvp Jan 07 '14 at 13:36
  • 1
    Dear: @BrianRushton : I don't think I get the comment about nonunique representations. Doesn't the analogous definition of "linearly independent" guarantee representations are unique? (I'm not sure what detnvvp is getting at in the first comment, either.) – rschwieb Jan 07 '14 at 19:04
  • 2
    @rschwieb: the point is that a Hamel basis can't always serve as a Schauder basis. Certainly every element of a vector space can be written as an infinite linear combination of elements of a Hamel basis, since it can be written even as a finite such combination. But there's no reason for that infinite representation to be unique. – Kevin Carlson Jan 07 '14 at 21:01
  • @KevinCarlson Oh sorry, I wasn't clear. I understand the solution given above completely, just not the representation remark. Is the $\sum \alpha_i b_i=0\implies \alpha_i=0$ axiom not assumed? (This might be stuff I learned about Hilbert spaces seeping in... maybe I do not have the requisite experience with more general Schauder bases.) – rschwieb Jan 07 '14 at 21:08
  • 1
    I take it you $b_i$ are elements of a Hamel basis? Then the point is simply that your independence axiom does not have to hold for infinite sums just because it holds for finite ones. – Kevin Carlson Jan 07 '14 at 21:10
  • @KevinCarlson No, not a Hamel basis. You apparently are expecting me to go down a path of thought that I am not going down. I'm asking if the infinite version of independence is assumed or not. I was expecting it to be assumed. – rschwieb Jan 07 '14 at 21:11
  • I'm sorry that I failed to guess your intent correctly. The infinite version of independence is one of the conditions for the $b_i$ to form a Schauder basis-is that what you mean? If I'm still wrong, it'd be helpful if you'd specify the context in which you're expecting the assumption to be made. – Kevin Carlson Jan 07 '14 at 21:13
  • 1
    @KevinCarlson In the meantime, I've just gone and looked up the wiki. It contains the phrase A Schauder basis is a sequence ${b_n}$ of elements of V such that for every element v ∈ V there exists a unique sequence ${α_n}$ of scalars in F so that... – rschwieb Jan 07 '14 at 21:18
  • I am just saying that IF a basis element is a combination of other basis elements, we get two representstions for that element, which is a contradiction. – Brian Rushton Jan 07 '14 at 21:18
  • @KevinCarlson The question is: "Given a Schauder basis ${b_i\mid i\in I}$ and a defined sum $\sum \alpha_ib_i=0$ (the sum is not restricted to be finite) isn't it required that all the $\alpha_i=0$?" – rschwieb Jan 07 '14 at 21:19
  • @BrianRushton Oh, if that's the case, then I might just be worried about nothing :) Thanks for letting me know. – rschwieb Jan 07 '14 at 21:40
25

Maybe a good point to start is this useful corollary of Baire Cathegory Theorem

the cardinality of an Hamel base of a Banach Space can be finite or uncountable. It can't be countable

The proof is a delightful application of Baire theorem.

Now to give an explicit example, we can consider the space $\ell^2 $ which has the standard base $ M:=$ $\lbrace e_n \rbrace $ which is not an Hamel base, but an Hilbert base (or Schauder, in this case the two coincide). To see the differences consider the linear span of $ M $. It's trivial to see that it is $ c_{00} $ but (using orthonormality property of $ M $) each vector $ v \in \ell^2 $ can be expressed as $ v=\sum_{k=1}^{\infty}(v, e_k) e_k $

In fact the restriction to FINITE linear combinations is a strong restriction. Let me show you another similar example. Consider $ c_0 $ the Banach space of the sequences convergent to 0. $ M$ is a Schauder base of it (verify it) but given for example $ u= \lbrace \frac {1}{n}_n \rbrace $ you can't express u as a finite linear combination of elements of M . So changing the meaning of the base in fact change "how big is its span"

Martin Sleziak
  • 53,687
Riccardo
  • 7,401
  • I'll add link to another question about the cardinality of Hamel basis: http://math.stackexchange.com/questions/217516/let-x-be-an-infinite-dimensional-banach-space-prove-that-every-basis-of-x-is (More links can be found there.) – Martin Sleziak Jan 07 '14 at 15:23
  • Heyy, really like your answer, if you could help with an exercise, would really appreciate it :) https://math.stackexchange.com/questions/3096022/normed-vector-space-schauder-basis-exercise – Homaniac Feb 02 '19 at 05:51
4

In the case of an infinite dimensional complete space, if you have a Banach space, then any Hamel basis is not countable. On the other hand, any Schauder basis has to be countable.

detnvvp
  • 8,237
4

There are two extra requirements on a Schauder basis:

(a) it must be countable;

(b) it needs not only to be linearly independent, but to satisfy the infinitary analogue of this property: given any infinite linear dependency $\sum_{i \in \mathcal{I}} a_i \mathbf{v}_i = 0$, we must have $a_i = 0$ for each $i$.

Both of these will fail for any Hamel basis of an infinite-dimensional Banach space.

Peter LeFanu Lumsdaine
  • 9,434
  • There is a reason for a Schauder basis to be countable, @tomasz. Let $\xi_j$ be the coordinate functionals for the Schauder basis ${e_j}$. The family ${ \xi_j(x)\cdot e_j }$ is not required to be summable. – Daniel Fischer Jan 07 '14 at 20:34
  • Is there a difference between your (b) and the "extension" they do in the definition on wikipedia? https://en.wikipedia.org/wiki/Basis_(linear_algebra) – user123124 Sep 20 '15 at 15:43
  • @Peter can you give an example of (b)? – noname1014 Aug 25 '16 at 16:45
  • 1
    @noname1014: consider the Banach space $\ell^\infty$ — i.e. the space of all absolutely convergent infinite sequences of reals. This has a Schauder basis given by the sequences $e_0 = (1,0,0,0…), e_1 = (0,1,0,0,…), e_2 = (0,0,1,0,…)$ and so on. Then this basis, together with the extra sequence (1,1/2,1/4,1/8,…), is linearly independent in the sense of finite linear combinations, but is not linearly independent in the Schauder sense, where infinite linear combinations are allowed, since (1,1/2,1/4,…) is expressible as $\sum_i 2^{-i} e_i$. – Peter LeFanu Lumsdaine Aug 25 '16 at 16:56
  • @PeterLeFanuLumsdaine I think in(a) countable should be "at most countable" because in finite space we can also define Schaular basis. – noname1014 Aug 25 '16 at 18:41
  • @noname1014: I follow the convention that “countable” means “$\leq \aleph_0$”, so includes finite, and I explicitly say “countably infinite” when I mean “$= \aleph_0$”. (Also, as described in rschwieb’s answer, some authors do explicitly require Schauder bases to be infinite, and this doesn’t lose too much, since in the finite-dimensional case they’re the same as ordinary bases. But I agree with you that the finite case should be included — I generally believe trivial cases shouldn’t be excluded form definitions.) – Peter LeFanu Lumsdaine Aug 25 '16 at 19:20
  • 1
    @Peter in the above comments you mentioned that $(e_{n}) $ is basis (schauder and Hamel) for $l^ \infty $. Is that true really? As much as I know the space $l^ \infty $ is not having any schauder basis. I am confused please answer – Prince Khan Dec 12 '16 at 15:42
  • 1
    @PrinceKhan: yes, you’re right, I meant $\ell^1$ not $\ell^\infty$. In $\ell^1$, the family $(e_n)$ form a Schauder basis (but not a Hamel basis). Whereas $\ell^\infty$ is not-separable and so has no Schauder basis, as you say. Thanks for catching that! – Peter LeFanu Lumsdaine Dec 12 '16 at 15:49
  • 2
    Oh thats fine as $l^{p} $ space has schauder basis and thatis what you have presented above. Sir! As every vector space has Hamel basis so the space $l^ \infty $ too. Can we have a presentation of elements of Hamel basis for $l^ \infty $ ? – Prince Khan Dec 12 '16 at 16:15