# orthonormal subset of $L_2(0,1)$ is complete

Let $$\{v_k\}_{k=1}^{\infty} \subset L_2[0,1]$$ be orthonormal.

Assuming for each $$x\in [0,1]$$ , $$x = \sum_{k=1}^{\infty} |\int_0^xv_k(t) dt|^2$$. I want to show that $$\{v_k\}_{k=1}^{\infty}$$ is complete.

I showed the reversed statement, and I noticed that if we define $$f_x(t) = 1$$ for $$t \le x$$ and $$f_x(t) = 0$$ otherwise we get the $$x = ||f_x||^2 = \sum_{k=1}^{\infty} |(f_x,v_k) |^2$$ , so Parseval's equality hold for $$f_x \in L_2[0,1]$$.

So now I want to show Parseval's equality holds for each $$f\in L_2[0,1]$$, but I cant find a way to do that(maybe $$\{f_x\}_{x\in [0,1]}$$ are dense in $$L_2[0,1]$$ ? ).

Thanks for helping.

• Why? $||f_x||^2= \int_0^1 |f_x|^2 = x$ @nicomezi
– user335501
Nov 16, 2018 at 14:24
• Hmm, you are right, I computed $(\int_0^1 f_x dx )^2$ . Nov 16, 2018 at 14:26

Indeed $$\{f_x\}_{x\in [0, 1]}$$ spans a space that is dense in $$L_2(0,1)$$ :
For $$a\lt b$$, we have $$f_b - f_a = 1_{]a,b]}$$ (where $$1_I$$ is the function that values $$1$$ in $$I$$ and $$0$$ elsewhere).
Therefore the closure of the span contains the set of functions that are continuous by part. This last set is dense in $$L_2(0,1)$$.