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.

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

1 Answer 1


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 $L_2(0,1)$.


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