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I've encountered the following statement:

Let $a, b_1, \ldots, b_m\in P([0, 1])$ be polynomials over $[0, 1]$. If $\mathrm{span}\{a^k b_j, j=1, \ldots, m, k = 0, 1, \ldots\}$ is uniformly dense in $C([0, 1])$, then $\mathrm{span}\{(a^k b_j)', j = 1, \ldots, m, k = 0, 1, \ldots\}$ is dense in $L^2([0, 1])$ with respect to the $L^2$-norm.

I guess the statement is false. But I can't figure out a counter-example. Can anyone give me some idea of proving or disproving the statement?

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  • $\begingroup$ If I understand the question, the sets you are considering are spanned by 2m functions. But then they can't be dense (so the statement would be vacuously true) . $\endgroup$ – Pietro Majer Oct 13 at 18:17
  • $\begingroup$ @PietroMajer why by $2m$ functions? If $m=1$, $b_1=1$, $a(x)=x$, then both spans are dense. $\endgroup$ – Fedor Petrov Oct 13 at 18:47
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    $\begingroup$ If the second span fails to be dense in $L^2[0,1]$ then every $(a^kb_j)'$ must be orthogonal to some nonzero $\phi \in L^2[0,1]$. I want to apply integration by parts and use the fact that the first span is dense in $C[0,1]$ to infer that $\phi = 0$ ... not sure how to formalize this. $\endgroup$ – Nik Weaver Oct 13 at 19:17
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    $\begingroup$ "I've encountered". Can you provide a reference? $\endgroup$ – Piotr Hajlasz Oct 13 at 19:43
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    $\begingroup$ @NikWeaver Probably issue of the pointwise evaluation is more important. Namely, there exists $g\in L^2([0, 1])$ such that we have $\int_0^1 g df = fg\big\vert_0^1 - \int_0^1 f dg$ for all $f\in\mathrm{span}\{a^k b_j\}$. But $fg\big\vert_0^1$ is not well-defined here since we cannot evaluate an arbitrary $L^2$ function pointwisely. Or is there any theorem claiming that we can find out a $g$ 'good enough'? $\endgroup$ – potionowner Oct 13 at 22:11

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