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Let $(x_n)_{n\in\Bbb N}\subseteq\mathcal L_2([a,b])$ be an orthonormal sequence. I want to prove the following:

$(x_n)_{n\in\Bbb N}$ is complete $\Leftrightarrow\sum_{n=1}^\infty \big|\int_{[a,t]}x_nd\lambda\big|^2=t-a\;\;\forall t\in[a,b]$

where $(x_n)_{n\in\Bbb N}$ being complete means that $\forall x\in\mathcal L_2([a,b])$ if $x\perp x_n\;\forall n\in\Bbb N\Rightarrow\ x=0$

So my attempt goes like this:

($\Rightarrow$)Let $(x_n)_{n\in\Bbb N}$ be complete so $(x_n)_{n\in\Bbb N}$ is a Hilbert base for $\mathcal L_2([a,b])$ and thus the Parseval's identity follows so $$ \forall x\in\mathcal L_2([a,b])\;\sum_{n=1}^\infty \big|\langle x,x_n \rangle\big|^2=\|x\|^2 $$ in particular we get that $\forall t\in[a,b]$ $$ \sum_{n=1}^\infty \big|\langle \chi_{[a,t]},x_n \rangle\big|^2=\|\chi_{[a,t]}\|^2=\langle \chi_{[a,t]},\chi_{[a,t]} \rangle=\int_{[a,b]}\chi_{[a,t]}d\lambda = \int_{[a,t]}d\lambda=t-a $$ and $$ \langle \chi_{[a,t]},x_n \rangle=\int_{[a,b]}\chi_{[a,t]}x_nd\lambda=\int_{[a,t]}x_nd\lambda $$ thereby $$ \sum_{n=1}^\infty \Big|\int_{[a,t]}x_nd\lambda\Big|^2=t-a\;\;\forall t\in[a,b] $$

($\Leftarrow$)Lets assume that $\sum_{n=1}^\infty \Big|\int_{[a,t]}x_nd\lambda\Big|^2=t-a\;\forall t\in[a,b]$ and let $x\in\mathcal L_2([a,b])$ with $x\perp x_n\;\forall n\in\Bbb N$. Lets assume $x\neq 0$ so $(x_n)_{n\in\Bbb N}\cup \{x\}$ is orthonormal, thus $$ \Big|\int_{[a,t]}xd\lambda\Big|^2+\sum_{n=1}^\infty\Big|\int_{[a,t]}x_nd\lambda\Big|^2=t-a\;\; \forall\ t\in[a,b]\\ \Rightarrow\sum_{n=1}^\infty\Big|\int_{[a,t]}x_nd\lambda\Big|^2=t-a-\Big|\int_{[a,t]}xd\lambda\Big|^2<t-a\;\; \forall\ t\in[a,b] $$ which contradicts the hypothesis, so $x=0$ and thus $(x_n)_{n\in\Bbb N}$ is complete.
What do you think? Would appreciate your comments and help if I missed something or did something wrong.

Arnulf
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    At the end you assume $\int_a^t x\ne 0$. How do you know? In fact that integral can vanish. But there must exist $t$ such that it's nonzero, because... (which is actually the whole point). – David C. Ullrich Jun 12 '16 at 01:15
  • Ok, so you say that when I assume that $x\neq 0$ I can assure then that $x\neq 0;;\ a.e.-\lambda$ which means that there $\exists;t_0\in [a,b]$ s.t. $\int_{[a,t_0]}xd\lambda\neq 0$ and since $(x_n){n\in\Bbb N}\cup{x}$ is orthonormal we get in particular that:$$\Big|\int{[a,t_0]}xd\lambda\Big|^2+\sum_{n=1}^\infty\Big|\int_{[a,t_0]}x_nd\lambda\Big|^2=t_0-a$$ so the contradiction follows and thus $x=0$. – Arnulf Jun 12 '16 at 01:32
  • What??? I didn't say any of the things you say I said. No you cannot assume that $x\ne0$ almost everywhere. You have to explain how it follows that there exists $t$ so that integral is non-zero.... – David C. Ullrich Jun 12 '16 at 02:06
  • Sorry. So I need to explain that there exists such $t_0$ where the integral is non-zero with that fact that (I assume) $x\neq 0$. But if it doesn't follow from this that $x\neq 0;;;a.e-\lambda$, any hint on how can this be possible? – Arnulf Jun 12 '16 at 02:31
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    Look at it another way. Suppose that $x\in L^2$ and $\int_a^tx=0$ for every $t$. You need to show that $x=0$. This should remind you of some Hilbert-space stuff - saying $\int_a^tx=0$ for every $t$ says something is orthogonal to something, and that implies something is zero if something is dense... – David C. Ullrich Jun 12 '16 at 14:36
  • It wouldn't be easier if just assume that $$x\perp x_n;;\forall n\in\Bbb N$$ which means that $$\int_{[a,b]}xx_nd\lambda=0;;\forall n\in\Bbb N$$ which means that $x=0;;a.e.-\lambda$ which is the zero in $\mathcal L_2$. What do you think? – Arnulf Jun 13 '16 at 00:17
  • That might or might not be a better idea. But there's still something to be proved - why does $\int xx_n=0$ for all $n$ imply $x=0$? You've really just rephrased things... – David C. Ullrich Jun 13 '16 at 00:51
  • Oh I see, I only get that $x=0;;a.e.-\lambda$ or $x_n=0;;a.e-\lambda;;\forall n\in\Bbb N$ which doesn't gives that $x=0$. – Arnulf Jun 13 '16 at 01:06
  • So getting back to the hint, if $$\int_{[a,t]}xd\lambda=0;;\forall t\in[a,b]$$ means that $$\langle \chi_{[a,t]},x\rangle =0;;\forall t\in[a,b]$$, but I don't get the dense part – Arnulf Jun 13 '16 at 01:09
  • There i something fundamental you're misunderstanding. I can't figure out exactly what it is, or I'd explain where the error was. But I never follow any of your arguments here - that's why I suspect it's something basic. You could help me try to help by explaining why you think $\int xx_n=0$ for all $n$ would imply what you say... – David C. Ullrich Jun 13 '16 at 01:09
  • So if you knew that the linear combinations of $\chi_{[a,t]}$ were dense it would finally follow that $x=0$. And they are dense, because... – David C. Ullrich Jun 13 '16 at 01:10
  • Because the set of simple functions is dense over the set of measurable functions – Arnulf Jun 13 '16 at 01:21
  • Yes, except. A simple function is a linear combination of $\chi_E$ for measurable sets $E$, and here $E$ is an interval. ("dense in the measurable functions" is meaningless, we should talk about dense in $L^2$.) – David C. Ullrich Jun 13 '16 at 01:42
  • Ok, thanks a lot for your help! I just have one doubt left, for this process I didn't see the need of the hypothesis: $\sum_{n=1}^\infty \Big| \int_{[a,t]}x_nd\lambda \Big|^2=t-a;;\forall t\in[a,b]$ – Arnulf Jun 13 '16 at 01:51

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