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See this old question: Does C[a,b] has a dense Hamel basis?

The answer provided by David C. Ullrich proves that every Banach space $X$ with $\dim X=\infty$ has a dense Hamel basis.

The question now is: can I construct a concrete example of such dense basis? The basis used in Fourier expansion is clearly not dense in $C[a,b]$.

Actually, I am not limiting myself to $C[a,b]$. Any Banach space can be fine.

Ma Joad
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    I suspect this is not possible to explicitly show since Hamel basis existence is given by the axiom of choice. To make matters worse, it would be uncountable which further complicates the concrete nature. – Cameron Williams Aug 09 '19 at 11:40
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    See this thread. I suspect similar arguments can be made for $C[a,b]$. – Robert Israel Aug 09 '19 at 12:38
  • The "basis" used in Fourier expansion isn't a basis. If you have a basis, then any element may be written as a finite linear combination of basis elements. Fourier analysis usually requires an infinite number of terms. – Arthur Aug 09 '19 at 12:54
  • @RobertIsrael: Indeed, Asaf's answer there works for any separable Banach space. – Nate Eldredge Aug 09 '19 at 18:53

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Of course there's no definitive answer to the question without a definition of "concrete". But, assuming it's like pornography, "can't give a definition but I know it when I see it": Surely there is no such basis. Obtaining any Hamel basis for an infinite-dimensional Banach space uses the axiom of choice, extremely non-concrete.

You say "The basis used in Fourier expansion is clearly not dense in $C[a,b]$." It seems likely this indicates a major misunderstanding, perhaps explaining why you don't appreciate what's so difficult here:

Taking $[a,b]=[0,2\pi]:=I$ it appears you're talking about $B=(e_n)_{n\in\Bbb Z}$, where $e_n(t)=e^{int}$. This is not just not dense in $C(I)$, it's not a Hamel basis for $C(I)$ in the first place.

It sounds like you've forgotten the definition. Because it's utterly and completely totally obvious that $B$ is not a Hamel basis for $C(I)$, just because there exists a continuous function that is not a trigonometric polynomial.

In fact, although this is less trivial, $B$ is not even a Schauder basis for $C(I)$; if it were that would say the Fourier series for any continuous function converges uniformly, which is not so. $B$ is a Schauder basis for $L^2(I)$, although again not a Hamel basis.

It seems clear to me that no infinite-dimensional Banach space has a "concrete" Hamel basis. As mentioned, we can't prove that until we define "concrete", but the following is well known and easy:

Fact If $E$ is an infinite-dimensional Banach space then any Hamel basis is uncountable.

Proof: In fact any spanning set must be uncountable. Say $b_1,b_2,\dots\in E$. Let $V_n$ be the span of $b_1,\dots, b_n$. Then $V_n$ is closed and has empty interior, so the Baire Category Theorem implies $$\bigcup_n V_n\ne E.$$

David C. Ullrich
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