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The set of all real->real functions is still a vector space. This vector space also has a basis (even if it is not as useful): there is a (uncountably infinite

The set of all real->real functions is still a vector space. This vector space also has a basis (even if it is not as useful): there is a (uncountably infinite) subset of real->real functions such that every function can be expressed as a linear combination of a finite number of these basis functions, in exactly one way. There isn't a clean way to write down this basis, though, as you need to use Zorn's lemma or equivalent to construct it.

ttoinou 1y ago
I'd love to read more about that, he's not talking about that at all in this article though
- lmm 1y ago
  He doesn't need to talk about it (though you might like to look up the notorious Hilbert's Basis Theorem); it happens to be the case that any vector space has a basis, but even if you don't know that, a vector space is still a vector space and its elements are still vectors.
  - ttoinou 1y ago
    The choice of basis is very important for the applications, which is what the article is about
- tel 1y ago
  If you're familiar with Zorn's Lemma, the construction is just to order bases by inclusion and to consider chains created by noting that there must be an independent dimension and adding it inductively. You can upper bound each of these chains by unioning the members of the chain (which preserves linear independence). By Zorn's Lemma that means there is a maximal linearly independent system and if an element existed outside of that system's span it would contradict that maximality.
seanhunter 1y ago
If you restrict yourself to Lebesque-integrable functions, can’t you take the complex Fourier transform of the function and call the terms of the Fourier series a basis, with the coefficients being the components of the vector of the function? This is a bit above my current mathematical paygrade, so forgive me if I’m not expressing the idea accurately but I’m learning a lot both from the article and the ensuing discussion - hopefully you understand what I’m getting at. I think what I may be asking is “Does the complex Fourier transform make a Hilbert space?” but I might be wrong both about that and about that being the right question.
- eapriv 1y ago
  You can’t, because a Fourier series is not a linear combination.
  - ttoinou 1y ago
    Linear combination of what ?
- 998244353 1y ago
  No, and this is where this formal notion of basis I mentioned unfortunately diverges from what is perhaps more useful in practice. You can represent any function f: [-pi, pi] -> R as an infinite sum f(x) = sum_(k = 0 to infinity) (a_k sin(kx) + b_k cos(kx)) for some coefficients a_k and b_k as long as f is sufficiently nice (I don't remember the exact condition, sorry). This is very useful, but the functions sin(x), sin(2x), ... , cos(x), cos(2x), ... don't constitute a basis in the formal sense I mentioned above as you need an infinite sum to represent most functions. It is still often called a basis though.
  - seanhunter 1y ago
    Thanks for that. That’s the trigonometric Fourier series. A complex Fourier series (which is what I mentioned) is equivalent and works similarly except that the terms are all uniform, so f(x) = sum n=-infty to +infty C_n e^{i n x} You can derive one from the other by using the identities sin x = (e^(ix) - e^(-ix))/2i cos x = (e^(ix) + e^(-ix))/2 I specifically mentioned the complex series because I didn’t like the fact that the alternating terms use a different trig function and it seemed weird to me to have every second dimension in a space be different in that way but they are equivalent. The convergence criteria for Fourier series vary depending on how strongly you need convergence but I think basically if a function is differentiable on the interval you care about then the Fourier series provably converges on that interval and otherwise if it has jump discontinuities and that sort of thing, then depending on whether it is square-integrable or a bunch of other properties) you can prove weaker forms of convergence (absolute, pointwise etc). To address your comment I don’t see why an infinite sum prevents something being a basis. In fact I would specifically say that can’t be true because then there would never be a basis for any infinite-dimensional space- any time you want to take an inner product in such a space you need an infinite sum, and you need such an inner product to construct the basis. A sibling comment pointed me in the direction of a Hilbert basis, which seems to be what I was thinking of.

ttoinou 1y ago

I'd love to read more about that, he's not talking about that at all in this article though

lmm 1y ago
He doesn't need to talk about it (though you might like to look up the notorious Hilbert's Basis Theorem); it happens to be the case that any vector space has a basis, but even if you don't know that, a vector space is still a vector space and its elements are still vectors.
- ttoinou 1y ago
  The choice of basis is very important for the applications, which is what the article is about
tel 1y ago
If you're familiar with Zorn's Lemma, the construction is just to order bases by inclusion and to consider chains created by noting that there must be an independent dimension and adding it inductively. You can upper bound each of these chains by unioning the members of the chain (which preserves linear independence). By Zorn's Lemma that means there is a maximal linearly independent system and if an element existed outside of that system's span it would contradict that maximality.

seanhunter 1y ago

If you restrict yourself to Lebesque-integrable functions, can’t you take the complex Fourier transform of the function and call the terms of the Fourier series a basis, with the coefficients being the components of the vector of the function? This is a bit above my current mathematical paygrade, so forgive me if I’m not expressing the idea accurately but I’m learning a lot both from the article and the ensuing discussion - hopefully you understand what I’m getting at. I think what I may be asking is “Does the complex Fourier transform make a Hilbert space?” but I might be wrong both about that and about that being the right question.

eapriv 1y ago
You can’t, because a Fourier series is not a linear combination.
- ttoinou 1y ago
  Linear combination of what ?
998244353 1y ago
No, and this is where this formal notion of basis I mentioned unfortunately diverges from what is perhaps more useful in practice. You can represent any function f: [-pi, pi] -> R as an infinite sum f(x) = sum_(k = 0 to infinity) (a_k sin(kx) + b_k cos(kx)) for some coefficients a_k and b_k as long as f is sufficiently nice (I don't remember the exact condition, sorry). This is very useful, but the functions sin(x), sin(2x), ... , cos(x), cos(2x), ... don't constitute a basis in the formal sense I mentioned above as you need an infinite sum to represent most functions. It is still often called a basis though.
- seanhunter 1y ago
  Thanks for that. That’s the trigonometric Fourier series. A complex Fourier series (which is what I mentioned) is equivalent and works similarly except that the terms are all uniform, so f(x) = sum n=-infty to +infty C_n e^{i n x} You can derive one from the other by using the identities sin x = (e^(ix) - e^(-ix))/2i cos x = (e^(ix) + e^(-ix))/2 I specifically mentioned the complex series because I didn’t like the fact that the alternating terms use a different trig function and it seemed weird to me to have every second dimension in a space be different in that way but they are equivalent. The convergence criteria for Fourier series vary depending on how strongly you need convergence but I think basically if a function is differentiable on the interval you care about then the Fourier series provably converges on that interval and otherwise if it has jump discontinuities and that sort of thing, then depending on whether it is square-integrable or a bunch of other properties) you can prove weaker forms of convergence (absolute, pointwise etc). To address your comment I don’t see why an infinite sum prevents something being a basis. In fact I would specifically say that can’t be true because then there would never be a basis for any infinite-dimensional space- any time you want to take an inner product in such a space you need an infinite sum, and you need such an inner product to construct the basis. A sibling comment pointed me in the direction of a Hilbert basis, which seems to be what I was thinking of.