$L^{2}$ Approximation Error of Fourier Series of Union of Disjoint Arcs Given $N$ disjoint arcs $\{I_{\alpha}\}_{\alpha=1}^{N}\subset\mathbb{T} $,set $f=\displaystyle\sum_{\alpha=1}^{N}\chi_{I_{\alpha}}$
show that
$$\sum_{|v|>k}|\hat{f}(v)|^2\le\dfrac{CN}{k}$$
This book [C. Muscalu and W. Schlag, Classical and Multilinear Harmonic Analysis (Vol. I)] gives the following hint: "The bound $\frac{N^2}{k}$ is much easier and should be obtained first. Going from $N^2$ to $N$ then requires one to exploit orthogonality in a suitable fashion."
I can prove the hint, in other words, I can prove the bound $\dfrac{N^{2}}{k}$; but I can't prove the bound $\dfrac{cN}{k}$.
 A: Well, we saw the other day that the observation that $1/N=N^2/N$ leads to a counterexample. Nothing like making a blithering idiot of oneself for motivation.... We will take $2\pi$ and perhaps other absolute constants to have the value $1$. The letters $I$ and $J$ will always denote intervals; as usual, $|I|$ is the length of $I$.
Lightbulb: If $I$ and $J$ are intervals separated by a certain amount then $\chi_I*\phi$ and $\chi_J*\phi$ have disjoint support if $\phi$ is supported in a sufficiently small interval.
If you want to figure it out yourself you can think about that and why it helps.
Fix $k$. Let $$A=\{j\,:\,|I_j|\ge1/k\},\quad B=\{j\,:\,|I_j|<1/k\}.$$
In general write $$f_E=\sum_{j\in E}\chi_{I_j}.$$
The small intervals are trivial:
$$\sum_{n>k}|\hat f_B(n)|^2\le||f_B||_2^2=\sum_{j\in B}|I_j|\le\frac Nk.$$
Say a family of intervals is $\delta$-separated if $I\ne J$, $s\in I$, $t\in J$ imply $|s-t|\ge\delta$. We can write $$A=A_1\cup A_2\cup A_3$$in such a way that $\{I_j\,:\,j\in A_m\}$ is $1/k$-separated (if $N$ is even then two sets are enough). Let $$\phi_k=k\,\chi_{[0,1/k]}.$$First, note that there exists $\lambda\in(0,1)$, independent of $k$, such that $$|\hat\phi_k(n)|\le\lambda\quad(n\ge k>0).$$
If $I$ is any interval define $$\psi_I=\chi_I-\phi_k*\chi_I.$$Note that $|\psi_I|\le1$ and $\psi_I$ vanishes outside a set of measure no larger than $2/k$; hence $$||\psi_I||_2^2\le\frac2k.$$
And finally the point: If $I$ and $J$ are $1/k$-separated intervals then $\psi_I$ and $\psi_J$ have essentially disjoint support, hence $$\psi_I\perp\psi_J.$$
[Added later: A picture makes the last few assertions clear. A concise way to say what that picture looks like: If $I=(a,b)$ and $b-a>1/k$ then $\chi_{(a+1/k,b)}\le\phi_k*\chi_I\le\chi_{(a,b+1/k)}$.]
And now we're done. Since the intervals $I_j$ for $j\in A_m$ are $1/k$-separated we have $$\sum_{n>k}|\hat f_{A_m}(n)|^2\le\frac1{1-\lambda}\sum_{n>k}\left|\sum_{j\in A_m}\hat\psi_{I_j}(n)\right|^2
\le\frac1{1-\lambda}\left|\left|\sum_{j\in A_m}\psi_{I_j}\right|\right|_2^2
=\frac1{1-\lambda}\sum_{j\in A_m}\left|\left|\psi_{I_j}\right|\right|_2^2
\le\frac2{1-\lambda}\frac NK.$$
A: What follows isn't a solution, but some observations--perhaps trivial--that may help somebody else obtain the $O(N/k)$ bound.

Lemma. For any disjoint intervals $\left\{I_{j}=(a_{j},b_{j})\right\}_{j=1}^{N}\subset\mathbb{T}$,
$$\sum_{n>k}\left|\widehat{f}(n)\right|^{2}\lesssim N^{2}/k$$
where $f:=\sum_{j=1}^{N}\chi_{I_{j}}$.

Proof. By linearity, $\widehat{f}(n)=\sum_{j=1}^{N}\widehat{\chi}_{I_{j}}(n)$, and
$$\widehat{\chi}_{I_{j}}(n)=\dfrac{e^{-2\pi i nb_{j}}-e^{-2\pi i na_{j}}}{-2\pi i n}$$
Whence by the triangle inequality,
$$\left|\widehat{f}(n)\right|^{2}\leq\left|\sum_{j=1}^{N}\dfrac{2}{2\pi\left|n\right|}\right|^{2}\leq\dfrac{N^{2}}{\pi n^{2}}$$
Since $\sum_{\left|n\right|>k}n^{-2}=O(k^{-1})$, the conclusion follows. $\Box$
First, observe that
\begin{align*}
\sum_{\left|n\right|>k}\left|\widehat{f}(n)\right|^{2}=\sum_{j=1}^{N}\sum_{\left|n\right|>k}\left|\widehat{\chi}_{I_{j}}(n)\right|^{2}+\sum_{i\neq j}\sum_{\left|n\right|>k}\widehat{\chi}_{I_{j}}(n)\overline{\widehat{\chi}_{I_{i}}}(n)
\end{align*}
We know from the lemma that the first term is $O(N/k)$, so a suitable estimate for the second term would yield the desired result.
Second, observe that since the indicator functions $\chi_{I_{j}}$ are mutually orthogonal in $L^{2}(\mathbb{T})$. So for $i\neq j$,
$$0=\langle{\chi_{I_{j}},\chi_{I_{i}}}\rangle=\langle{\sum_{n}\widehat{\chi}_{I_{j}}(n)e_{n},\sum_{n}\widehat{\chi}_{I_{i}}(n)e_{n}}\rangle=\sum_{n}\widehat{\chi}_{I_{j}}(n)\overline{\widehat{\chi}}_{I_{i}}(n)$$
In particular,
$$\sum_{\left|n\right|>k}\widehat{\chi}_{I_{j}}(n)\overline{\widehat{\chi}}_{I_{i}}(n)=-\sum_{\left|n\right|\leq k}\widehat{\chi}_{I_{j}}(n)\overline{\widehat{\chi}}_{I_{i}}(n)$$
As I said above, I'm not sure where this get us--if anywhere at all.
