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Let $\{\phi_n\}_{n=1}^{\infty}$ be an orthonormal sequence in L$^2(0,1)$. Prove that $\{\phi_n\}_{n=1}^{\infty}$ is an orthonormal basis iff $\forall a\in [0,1]$, $a=\sum_{n=1}^{\infty}|\int_0^a\phi_ndx|^2$.

For the first direction, this is just using Parseval's identity with $\chi_{(0,a)}$, because: $$a=\|\chi_{(0,a)}\|^2=\sum_{n=1}^{\infty}|\int_0^1\phi_n\chi_{(0,a)}dx|^2=\sum_{n=1}^{\infty}|\int_0^a\phi_ndx|^2$$.

For the second direction, I tried using the fact that $\{\phi_n\}$ is orthonormal basis iff $\{\{\phi_n\}_{n=1}^{\infty}\})^{\bot}=\{\vec 0\}$, but this didn't work. Also tried to show that Parsavel's equality must hold but also got stuck there.

Any hint would be appreciated.

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  • $\begingroup$ What was it that did not work with your approach for the second direction? $\endgroup$
    – Thomas
    Nov 13, 2021 at 19:45
  • $\begingroup$ @Thomas I wasn't able to show that if $\langle f,\phi_n\rangle=0$ for every $n$ then $f=0$ $\endgroup$
    – GBA
    Nov 13, 2021 at 19:46
  • $\begingroup$ It's been a while, but I think it is true that, if $\phi \in L^2(0,1)$ and $\int_a^b \phi (x) dx = 0$ for every $0\le a < b \le 1$, then $\phi=0$. $\endgroup$
    – Thomas
    Nov 13, 2021 at 19:51
  • $\begingroup$ ... because, for almost every $x\in (0,1)$ we have $\phi(x) = \lim_{\varepsilon \rightarrow 0} \frac{1}{2\varepsilon}\int_{x-\varepsilon}^{x+\varepsilon}\phi(y)\, dy$. Do you happen to know that? $\endgroup$
    – Thomas
    Nov 13, 2021 at 19:55
  • $\begingroup$ Finally, if $(\phi_n)_{n\in \mathbb{N}}$ is not complete, I believe you should be able to show that there is $\phi\in L^2(0,1)$ such that $||\phi||= 1 $ and $\int_a^b \phi=0$ for every $a,b$. $\endgroup$
    – Thomas
    Nov 13, 2021 at 19:58

1 Answer 1

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As I told you in the comments, it's been a while, so I suggest you try to find the error in my ways instead of just believing me.

If $B:= (\phi_n)_{n\in \mathbb{n}}$ is not complete, you will find $\phi \in L^2(0,1)$ such that $B$ together with $\phi$ is an orthonormal system. By Bessels inequality, $$|\int \phi fdx| + |\sum_n \int \phi_n fdx| \le ||f||^2$$ for every $f\in L^2$. By assumption you have equality for $f=\chi_a$ for every $a\in (0,1)$, even if you omit the first term in the sum $|\int \phi \chi_a|\,dx$, so $$\int_0^a \phi(x) \,dx = 0$$ for every $a\in (0,1)$ So $$\int_a^b \phi(x) \,dx = \int_0^b \phi(x) \,dx - \int_0^a \phi(x) \,dx = 0$$ But then, $$\phi(x_0) = \lim_{\delta\rightarrow 0}\frac{1}{2\delta}\int_{x_0-\delta}^{x_0+\delta}\phi(x) \,dx =0 $$ for almost every $x_0$, contradicting the fact that $||\phi||_{L^2} = 1$.

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  • $\begingroup$ I'm not familiar with the last result you wrote. What I did is replaced it with the following proposition: $\int_{0}^a\phi(x)dx=0$ for every $0<a<1$, so $\phi(x)=0$ almost everywhere (standard result in measure theory). $\endgroup$
    – GBA
    Nov 13, 2021 at 20:29
  • $\begingroup$ @GBA well, if you know that already, then that's also fine. But what is the problem, then? The result I used is easy to show, also - you start with continuous $\phi$ and use the mean value theorem. Then you approximate $L^1 $ functions by continuous functions. Since for bounded intervals, $L^2 \subset L^1$ you get the result in the case it is needed here. $\endgroup$
    – Thomas
    Nov 13, 2021 at 20:38
  • $\begingroup$ I see. Thank you very much $\endgroup$
    – GBA
    Nov 13, 2021 at 20:41
  • $\begingroup$ @GBA the answer is proving this fact you quote from measure theory, in fact. $\endgroup$ Nov 13, 2021 at 20:45

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