Consider the following set of polynomials: $S := \{x^{2m}: m \geq 0\}$. Is there a non-zero element $f \in L^2([0,1])$ such that $\int_0^1 fx^{2m} = 0$ for each $m \geq 0$? Note that the answer is negative if $[0,1]$ is replaced by $[-1,1]$.
More generally, given a set of polynomials, is there any standard procedure to check if its linear span is dense in $L^2(X)$?
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2$\begingroup$ Stone Weierstrass theorem is the way to go. The set is is clearly a unital subalgebra so all that is left to show is that it separates points. This is exactly what fails for $[-1,1]$. On$ [0,1]$ though it is simple to show that S seperates points. $\endgroup$– AliCommented Nov 5, 2016 at 13:50
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2$\begingroup$ If you are interested in whether a linear hull of a family of monomials is dense, here is an answer: en.wikipedia.org/wiki/M%C3%BCntz%E2%80%93Sz%C3%A1sz_theorem (the $L^p$ criterion is just the same). $\endgroup$– Ilya BogdanovCommented Nov 5, 2016 at 14:06
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$\begingroup$ Your use of "basis" seems to be vague. For people working on Banach and Hilbert spaces, the words "basis" does not merely mean: a lin indep set whose span is dense. I suggest amending your question to take account of this $\endgroup$– Yemon ChoiCommented Nov 5, 2016 at 16:37
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1$\begingroup$ To my knowledge, Pietro Majer's the best one can get for the first part of your question. It is basically a variation of Muntz theorem, see alsohere By they way, by translation and rescaling you can easily get monomials in $(ax+b)$ spanning $L^2[\alpha,\beta]$. I am not aware of any neat result discussing the second part. $\endgroup$– BigMCommented Nov 6, 2016 at 5:33
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1$\begingroup$ @auniket no, the definition of Schauder basis is much stronger, as would hopefully be explained in some of the sources you found. Among those working with Banach spaces, "basis" usually means "Schauder basis". If you want a basis in the purely algebraic sense (i.e. a lin indep spanning set) this would be called a "Hamel basis" $\endgroup$– Yemon ChoiCommented Nov 6, 2016 at 22:06
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2 Answers
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You want a version of the classical Müntz–Szász theorem for the space $L^2([0,1])$ (which is, incidentally, the case considered initially by Szász). Here is a nice paper on the situation for $L^p([0,1])$ spaces, always for the span of monomials (with real exponents allowed) http://www.math.tamu.edu/~terdelyi/papers-online/Fullmuntz.pdf
answered Nov 5, 2016 at 14:26
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2$\begingroup$ This is only for monomials (powers of $x$), not for arbitrary polynomials. $\endgroup$ Commented Nov 5, 2016 at 14:52
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$\begingroup$ The statement of the Müntz–Szász theorem is beautiful - thanks! $\endgroup$– pinakiCommented Nov 6, 2016 at 21:14
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Basis in a Banach space is a different thing usually.
The answer to your first question is affirmative. Indeed, by Weierstrass theorem the polynomials in $x^2$ are dense in the space $C[0,1]$ of continuous functions with max-norm, hence in $L_2[0,1]$. Thus only zero element of $L_2$ is orthogonal to them all.
answered Nov 5, 2016 at 13:51