Problem using Bessel's inequality to prove that an orthonormal system is complete in Vitali-Dalzell theorem.

Question

I have a problem understanding the proof of the Vitali-Dalzell theorem. Here I write the statement and the proof I am studying:

Let $\{u_n\}$ be a orthonormal system of $L^2[a,b]$. Then:

$ (i) \quad$ if $\Lambda$ is dense in $[a,b]$ and $\sum_n \Big | \hat{\mathbb{1}_{[a,\lambda]}}(n)\Big | ^2 = \lambda - a $ for all $\lambda \in \Lambda$ then $\{u_n\}$ is complete.

$(ii) \quad$ if $\Lambda \subseteq [a,b]$ is such that $\mu([a,b] \setminus \Lambda) = 0$ and $\sum_n \int_a^b \Big | \hat{\mathbb{1}_{[a,\lambda]}}(n) \Big | ^2 \ d\lambda = \frac{(b-a)^2}{2}$ then $\{u_n\}$ is complete.

I have no doubt about the proof of point $(i)$, and the proof we gave of $(ii)$ is to show that under hypothesis $(ii)$ we can reduce to the case $(i)$.

Here how the proof of $(ii)$ goes:

Since $L^2$ is an Hilbert space we can use Bessel's inequality: $\sum_n |\hat f (n)|^2 \le \|f\|^2_2$.

Thus for all $\lambda \in \Lambda$, $$ \sum_n \Big| \hat{\mathbb{1}_{[a,\lambda]}}(n)\Big | ^2 \le \| \mathbb{1}_{[a,\lambda]}\|^2_2 = \lambda -a .$$

The assumptions imply that $$ \int_a^b \Big[ \lambda - a - \sum_n \Big| \hat{\mathbb{1}_{[a,\lambda]}}(n)\Big|^2\Big] \ d\lambda = 0.$$

This, in turn, implies that $$\lambda - a = \sum_n \Big| \hat{\mathbb{1}_{[a,\lambda]}}(n)\Big|^2$$ for almost every $\lambda \in [a,b]$. We can now apply $(i)$ to conclude.

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Questions:

• I do not see how Bessel's inequality is useful for this proof

• I can't get the passage "The assumptions imply that ..."

Can anyone add some details so that I can understand better this proof please?

Answer 1

Second bullet point: just check that $\frac{(b-a)^2}2 = \int_a^b (\lambda - a)d\lambda $. So after exchanging $\int$ with $\sum$, the integral assumption immediately rearranges to $$\int_a^b(\lambda - a) - \sum_n |\widehat {\unicode{x1D7D9}_{[a,\lambda]}}(n)|^2 d\lambda = 0. \tag{*} $$

First bullet point: Bessel's inequality shows that the function $$ F(\lambda):= (\lambda - a) - \sum_n |\widehat {\unicode{x1D7D9}_{[a,\lambda]}}(n)|^2$$ is non-negative, i.e. $$ F(\lambda)\ge 0.\tag{**}$$ Note that (*) says precisely that $$ \int F(\lambda) d\lambda = 0.\tag{*'}$$ From both (*') and (**), it follows that $F(\lambda) = 0 $ a.e., which in turn gives the required reduction.