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Every function $f \in L^2([0, 2\pi])$ can be written as an infinite linear combination of the linearly independent set $S = \{\cos(nx), \sin(nx): n \in \mathbb{Z}\}$. But there are functions in $L^2([0, 2\pi])$ which cannot be written as a finite combination of these basis functions and therefore $S$ is not a Hamel basis, it is a Schauder basis.

However, I have read that it is possible to have a Hamel basis for this space and that it is 'much larger' than the set $S$.

  1. So what does such a Hamel basis for this space look like..how would we represent $f \in L^2([0, 2\pi])$ using a Hamel basis.
  2. Why exactly is the Hamel basis larger than $S$ which seems pretty large already in that it contains an infinite number of functions?
ManUtdBloke
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  • No one knows, it requires the axiom of choice. 2. Your set is countable, and the dimension of $L^2$ as a vector space is uncountable (equal to $2^{\aleph_0}$). Related: http://math.stackexchange.com/questions/194189/a-hamel-basis-for-l-p?rq=1
    • – Jonas Meyer Mar 16 '17 at 07:21
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    On the cardinality of the dimension of such spaces, consider the simpler space of continuous functions here, especially Robert Israel's answer: http://math.stackexchange.com/questions/664084/the-dimension-of-the-real-continuous-functions-as-a-vector-space-over-mathbbr – symplectomorphic Mar 16 '17 at 07:26
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    Well, the much easier question "How does a basis of $\Bbb R^7$ which extends [write your favourite set of $4$ vectors] look like?" is not particularly informative either, in my opinion. –  Mar 16 '17 at 07:27
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    You could be interested in this question: http://math.stackexchange.com/questions/630142/what-is-the-difference-between-a-hamel-basis-and-a-schauder-basis – Luigi M Mar 16 '17 at 07:31
  • @JonasMeyer For i) can we not just take a Hamel basis $B$, which we know exists, and project $f$ onto it so that $f = \sum_{i=0}^N (f, x_i) x_i$ for $x \in B$? What is the problem with doing that? Can we have explicit representations for vectors $x \in B$ explicit (like the cosines and sines for the Schauder basis)? – ManUtdBloke Mar 16 '17 at 07:40
  • @G.Sassatelli I don't understand your point which seems to be talking about extending $\mathbb{R}^4$ to $\mathbb{R}^7$. Obviously $\mathbb{R}^4$ cannot form a basis for $\mathbb{R}^7$, but we can form a Hamel basis for $L^2([0, 2\pi])$. – ManUtdBloke Mar 16 '17 at 07:43
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    I was not talking about "extending $\Bbb R^4$ to $\Bbb R^7$" (whatever it means). I was talking about "finding a basis of $\Bbb R^7$ which contains $4$ given vectors". It's something that, well, yes you can do it, but... Not in a particularly interesting way, without invoking additional structure on $\Bbb R^7$. –  Mar 16 '17 at 08:02
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    @eurocoder: No it is not explicit, and requires axiom of choice to even exist. Once it exists, no the coefficients are not given by the inner products in that way, in part because it is not an orthonormal basis. All you know comes form the theory, which is that coefficients exist such that each element is a finite linear combination of the basis elements. – Jonas Meyer Mar 16 '17 at 15:02