orthonormal sequence in $L^2[0,1]$ - how to prove these following equivalent terms? I've been asked this following very interesting question and would like some help figuring out why it is true :)
Let $u_n$ be an orthonormal sequence in $L^2[0,1]$
Prove that the following are equivalent:


*

*$u_n$ is complete (orthonormal basis)

*for every $x\in [0,1]$ : $x=\sum_{n=1}^{\infty} |\int_{0}^{x} u_n(t)dt|^2 $

*$0.5=\sum_{n=1}^{\infty} \int_{0}^{1} |\int_{0}^{x} u_n(t)dt|^2 dx $
My thoughts so far:
a. if $u_n$ is complete, I could use Parseval's identity with functions $f=1_{[0,x]}$ and get condition 2, and integrate it to get condition 3 (p.s: why could I change the order of the inifite sum with that $\int_{0}^{1}$?)
b. The other way aronud is harder. I figured out that using the definition of orthonormal basis would be a little difficult, So probably I'm supposed to show the Parsevals' identity, or something like that. no luck there.
I would like some guide if you could.
Thanks!
Edit: I tried this question for a couple more hours and couldn't solve. A more direct hint would be appreciated.
 A: The span $S$ of the functions $\{ \chi_{[0,x]} : 0 \le x \le 1\}$ is a dense linear subspace of the Hilbert space $L^{2}[0,1]$. To see that the span is dense, it is enough to show that $(f,\chi_{[0,x]})=0$ for all $x$ and some $f \in L^{2}[0,1]$ implies $f=0$. By the Lebesgue differentiation theorem, the following holds a.e.:
$$
                  \frac{d}{dx}(f,\chi_{[0,x]})=\frac{d}{dx}\int_{0}^{x}f(t)\,dt = f(x).
$$
So it follows that $S$ is dense in $L^{2}[0,1]$.
Let $\{ u_{n} \}$ be an orthonormal subset of $L^{2}[0,1]$. Parseval's equality holds for a given $f$ with respect to this set iff $f$ is in the closed linear subspace $\mathcal{U}$ generated by this orthonormal subset, which means that $\{ u_{n} \}$ is a complete orthonormal subset iff $\mathcal{U}$ includes $S$. That is, $\{ u_{n} \}$ is complete iff
$$
\begin{align}
          x & = \int_{0}^{x}|\chi_{[0,x]}(t)|^{2}\,dt \\
            & = \sum_{n}^{\infty}|(\chi_{[0,x]},u_{n})|^{2} \\
            & =\sum_{n} \left|\int_{0}^{x}u_{n}(t)\,dt\right|^{2},\;\;\; 0 \le x \le 1.
\end{align}
$$
So that takes care of (1) $\iff$ (2).
For (2) $\iff$ (3), note that Bessel's inequality always holds:
$$
        \sum_{n}\left|\int_{0}^{x}u_{n}(t)\,dt\right|^{2} \le \int_{0}^{1}|\chi_{[0,x]}(t)|^{2}\,dx = x.
$$
Let $P$ be the orthogonal projection of $L^{2}[0,1]$ onto the closed linear span $\mathcal{U}$ of the $\{ u_{n} \}$. Then $x\mapsto P\chi_{[0,x]}$ is continuous because $x\mapsto \chi_{[0,x]}$ is continuous from $[0,1]$ into $L^{2}[0,1]$ (dominated convergence,) and because $P$ is continuous on $L^{2}[0,1]$. The left side above is $\|P\chi_{[0,x]}\|^{2}$, which must be continuous in $x$. The left side must equal $x$ for all $x$ in order for $\mathcal{U}$ to be complete. It follows that equality holds for all $x$ iff
$$
          \int_{0}^{1} \sum_{n}\left|\int_{0}^{x}u_{n}(t)\,dt\right|^{2}\,dx = \frac{1}{2}.
$$
By monotone convergence, you are allowed to interchange summation and integration in order to conclude that the above holds iff
$$
                \sum_{n}\int_{0}^{1}\left|\int_{0}^{x}u_{n}(t)\,dt\right|^{2}\,dx = \frac{1}{2}.
$$
