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I read in Kolmogorov-Fomin's (p. 430 here) the statement, sadly left without a proof, that if function $f:(a,b)\to\mathbb{C}$, measurable almost everywhere on $(a,b)$, where $-\infty\leq a<b\leq\infty$, is not equal to 0 [i.e. $\forall x\in(a,b)\quad f(x)\ne 0$, as Daniel, whom I deeply thank, explains in his answer] and satisfy the condition $|f(x)|\leq Ce^{-\delta|x|}$ with $\delta>0$, then the system of functions $\{x^nf(x)\}_{n\in\mathbb{N}_{\geq 0}}$ is complete [as a system of vectors] in $L_2(a,b)$.

The text says that it is a consequence of the precedingly stated lemmas (pp. 428-430), which are ($F[f](\lambda):=\int_{\mathbb{R}}f(x)e^{-i\lambda x} d\mu_x$ is the Fourier transform):

  1. If $f,f',...,f^{(k-1)}\in L_1(\mathbb{R})$ and $f^{(k-1)}$ is absolutely continuous on any finite interval then $F[f'](\lambda)=(i\lambda)^k F[f](\lambda)$.
  2. If $f,f',f''\in L_1(\mathbb{R})$ and $f$ is absolutely continuous on any finite interval then $F[f]\in L_1(\mathbb{R})$.
  3. If $f,...,x^nf(x)\in L_1(\mathbb{R})$ [where $xf(x)$ is a somewhat unrigourous notation for the function $x\mapsto xf(x)$ as Kolmogorov-Fomin's uses] then $F[f]$ has the $n$-th order derivative everywhere and $\frac{d^n}{d\lambda^n}F[f](\lambda)=F[(-ix)^n f(x)](\lambda)$.
  4. If $f, e^{\delta|x|}f\in L_1(\mathbb{R})$ for a certain $\delta>0$ then $z\mapsto\int_{\mathbb{R}}f(x) e^{-i x z}d\mu_x$ (where $\mu_x$ is the Lebesgue linear measure) is analytic in $\{z\in\mathbb{C}:|\text{Im z}|<\delta\}$.

Can anybody prove the the theorem (in italics) or give a link to any resources proving it? I thank you so much!!!

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Self-teaching worker
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    Maybe this can work: extend functions to $\mathbb{R}$ using zeros values outside $(a,b)$, use Parseval's theorem, you measure the $L^2$ distance of $\sum_{k=0}^n a_k x^k f(x)$ and any $g\in L^2(a,b)$ by the $L^2$ distance of their Fourier tansforms. Since you know both Fourier tansforms are analytical, you can evaluate their derivatives and find $a_k$'s such that their first $n$ derivatives are identical at $\lambda=0$ – Petite Etincelle Nov 21 '14 at 21:01
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    Sorry, @Davide, I was a bit rash. I thought the exponential decay of the bound on $f$ would make dominated convergence straight, but it's not so easy (unless $(a,b)$ is bounded or $\delta > 1/2$). I would have changed the answer to deduce that the Fourier transform of $\overline{g}\cdot f$ is zero by the vanishing of all derivatives at $0$, but T.A.E. already has done that, so I just deleted my answer. – Daniel Fischer Nov 21 '14 at 23:52
  • @DanielFischer No problem at all! On the contrary, I thank you for the interest in the question and the interesting points you've raised! – Self-teaching worker Nov 22 '14 at 12:54

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Suppose $f$ is as stated with exponential bound $Ce^{-\delta|x|}$, and suppose that $\overline{g} \in L^{2}(\mathbb{R})$ is orthogonal to the functions $\{ x^{n}f(x) \}_{n=0}^{\infty}$ in the $L^{2}$ inner-product. Define $$ H(\lambda) = \int_{-\infty}^{\infty}g(x)f(x)e^{-i\lambda x}\,dx. $$ Suppose that $|\Im\lambda| \le \delta' < \delta$. Then $$ |fge^{-i\lambda x}| \le |g(x)|e^{-(\delta-\delta')|x|}\in L^{2}\cdot L^{2} \subseteq L^{1}. $$ So $H$ is defined and continuous in the strip $|\Im\lambda| < \delta$ by the Lebesgue dominated convergence theorem. By Morera's theorem, $H$ is holomorphic in this strip because, for any triangle $\Delta$ contained in this strip, Fubini's Theorem allows you interchange order of integration in order to conclude that $\oint_{\Delta}H(\lambda)\,d\lambda = 0$. If $C$ is a positively-oriented simple closed piecewise smooth curve in the strip, and if $z$ is inside $C$, then the Cauchy integral formula gives $$ H^{(n)}(\lambda) = \frac{n!}{2\pi i}\oint_{C}\frac{1}{(\lambda'-\lambda)^{n+1}}H(\lambda')\,d\lambda'. $$ Using Fubini's Theorem, it is possible to interchange the order of integration with the defining integral for $H$ in order to the put derivatives onto the integrand: $$ H^{(n)}(\lambda) = \int_{-\infty}^{\infty}g(x)f(x)e^{-i\lambda x}(-ix)^{n}\,dx $$ By the assumption that $g$ is orthogonal to $x^{n}f(x)$ for all $n=0,1,2,3,\cdots$, it follows that $H$ and all of its derivatives vanish at $\lambda=0$, which means that $H$ must be identically $0$ in the strip $|\Im\lambda| < \delta$ because this is a connected domain. Therefore, $$ \int_{-\infty}^{\infty} g(x)f(x)e^{-i\lambda x}\,dx $$ is identically $0$ for $|\Im\lambda| < \delta$. It is enough to know this fact for $\lambda\in\mathbb{R}$ in order to conclude that $fg =0$ a.e.. because $fg \in L^{2}$ and the Fourier transform is unitary on $L^{2}$. Assuming $f$ is non-zero a.e., then $g=0$ a.e., which proves that $\{ x^{n}f(x) \}_{n=0}^{\infty}$ spans a dense subspace of $L^{2}$.

Disintegrating By Parts
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  • I'm not sure I understand a thing: if $H^{(n)}(0)=0$, why $H^{(n)}(z)=0$ for all $z$ in the connected set? Because the Taylor series is 0 in the sphere centred in 0, and then also in the ball centred in any $z_1\in B(0,\delta)$ and contained in the connected domain, and so on covering all the connected open domain? I heartily thank you again! – Self-teaching worker Nov 22 '14 at 16:11
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    If $H^{(n)}(0)=0$ for all $n=0,1,2,3,\cdots$ (not just for one n,) then the power series at $0$ is identically $0$ in a neighborhood of $0$. By the identity theorem, $H$ must be identically $0$ on the entire connected open region where $H$ is holomorphic. – Disintegrating By Parts Nov 22 '14 at 16:18
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    @DavideZena : You're welcome. Thanks for the problem. When I looked at the Kolmogorov-Fomin hints, and found it was so elegant, I decided it was a nice idea to post a solution following their plan as an alternative to any other solutions. I learned some elegant tricks from their hints. – Disintegrating By Parts Nov 22 '14 at 16:40