# Uniform density of polynomials in $C([0, 1])$ implies $L^2$ density of derivatives

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?

• 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) . – Pietro Majer Oct 13 at 18:17
• @PietroMajer why by $2m$ functions? If $m=1$, $b_1=1$, $a(x)=x$, then both spans are dense. – Fedor Petrov Oct 13 at 18:47
• 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. – Nik Weaver Oct 13 at 19:17
• "I've encountered". Can you provide a reference? – Piotr Hajlasz Oct 13 at 19:43
• @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'? – potionowner Oct 13 at 22:11