what is a basis for $\ell^p$?

Question

For the familiar vector space $\mathbb{R}^n$ we have the following standard orthonormal basis: $\{e_1, e_2, \ldots, e_n\}$. Where $e_1 = (1, 0, \ldots, 0)$ and so on...

For $\ell^p$ would we have $\{x_1 = (1, 0, 0, \ldots), x_2 = (0, 1, 0, \ldots), \ldots, x_n = (0,\ldots,0,1,0,\ldots), \ldots\}$?

Would this be orthonormal as well?

See this question here https://math.stackexchange.com/questions/194189/a-hamel-basis-for-l-p. — Arbiter, May 31 '17 at 21:41
Would the ideas from $\mathbb{R}^n$ not work here? I was hoping in the same sense. — Yabbadule, May 31 '17 at 21:41
Dimensions act more or less like cardinality. Since when the things you know about cardinality of finite sets make sense about cardinality of infinite sets? — Asaf Karagila, May 31 '17 at 21:43

Duncan Ramage · Answer 1 · 2017-05-31T21:49:31.197

3

No, this is not a basis.

Consider $p = 2$, and the sequence $s_n = \frac{1}{n}$. Then, this is in $l^p$, but this is not expressible as a linear combination of your basis, as any finite sum of your $x_i$ will eventually have every coordinate be $0$, but every coordinate of $s$ is nonzero.

edited May 31 '17 at 21:49

answered May 31 '17 at 21:44

Duncan Ramage

6,928
1
20
38

Generally speaking, how would you construct a basis for one of these infinite spaces? – Yabbadule May 31 '17 at 21:45
Huh??? How would that be a basis for $\ell^\infty$? – Asaf Karagila May 31 '17 at 21:48
I wouldn't. You cannot write an explicit basis for this space; the question Dimitris proved explains why. – Duncan Ramage May 31 '17 at 21:49
@AsafKaragila, it's not, I mixed up $l^\infty$ with the vector space of compact sequences. – Duncan Ramage May 31 '17 at 21:52

score 0 · Answer 2 · answered May 31 '17 at 21:54

This set does not "span" the space according to one of the standard definitions of "span", i.e. not every point in the space is a linear combination of finitely many members of this set. Nor is it a "basis" when that term is defined by relying on that notion of "span".

But it is what is called an "orthonormal basis" in the case in which $p=2$. As far as being orthonormal is concerned, just compute the inner products. That's all there is to that.

The definition of "orthonormal basis" requires that every member of $\ell^2$ be representable as a sum of an infinite linear combination of members of the set. Whether a sum of infinitely many things exists depends on which kind of convergence you're talking about. The fact that various different kinds exist is the reason why people are somewhat careful and cautious about these things. In $\ell^p,$ the kind of convergence you're talking about is defined by the $\ell^p$ norm of the difference between the limit and the thing approaching the limit going to $0.$

In order to speak of orthonormality in a normed space, it must be an inner product space. Norms that come from inner products satisfy the parallelogram law. The $\ell^p$ norm does not satisfy the parallelogram law except when $p=2.$

score 0 · Answer 3 · answered Jun 01 '17 at 08:47

0

It is no basis, because every basis of an infinite-dimensional Banach space is uncountable as a consequence of the Baire category Theorem, See https://en.wikipedia.org/wiki/Baire_category_theorem

answered Jun 01 '17 at 08:47

Michael Hoppe

18,103
3
32
49

what is a basis for $\ell^p$?

3 Answers3