# Orthonormal sequence in L$^2(0,1)$

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.

• What was it that did not work with your approach for the second direction? Nov 13, 2021 at 19:45
• @Thomas I wasn't able to show that if $\langle f,\phi_n\rangle=0$ for every $n$ then $f=0$
– GBA
Nov 13, 2021 at 19:46
• 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 $\phi=0$. Nov 13, 2021 at 19:51
• ... 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? Nov 13, 2021 at 19:55
• 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 $a,b$. Nov 13, 2021 at 19:58

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 $$||\phi||_{L^2} = 1$$.
• 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).
• @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. Nov 13, 2021 at 20:38