Let us define $R_n$ by:$$\min_{a,b \in \mathbb{C}}\bigg[\frac{1}{2\pi}\int_{-\pi}^\pi\bigg|\sqrt{|x|^3}-a\sin(nx)-b\sin([n+1]x)\bigg|^2dx\bigg]$$ Calculate $\lim_{n\to\infty}R_n$
My attempt:
Let us define: $g(x)=\sqrt{|x|^3}-\frac{\hat f(n)-\hat f(-n)}{2i}\sin(nx)-\frac{\hat f(n+1)-\hat f(-(n+1))}{2i}\sin([n+1]x)$
Where $\hat f(n)$ is the $n-$th Fourier coefficient.
$g(x) \perp \text{span}\{\sin(nx)\},$ so by Pythagoras:
$\begin{align} \bigg\|\sqrt{|x|^3}-a\sin(nx)-b\sin([n+1]x)\bigg\|^2 &=\bigg\|g(x)\bigg\|^2+\bigg\|\bigg(\frac{\hat f(n)-\hat f(-n)}{2i}-a\bigg)\sin(nx)\bigg\|^2\\ &+\bigg\|\bigg(\frac{\hat f(n+1)-\hat f(-(n+1))}{2i}-b\bigg)\sin([n+1]x)\bigg\|^2\\ &=\bigg\|g(x)\bigg\|^2+\bigg\|\bigg(\frac{\hat f(n)-\hat f(-n)}{2i}-a\bigg)\bigg\|^2\\ &+\bigg\|\bigg(\frac{\hat f(n+1)-\hat f(-(n+1))}{2i}-b\bigg)\bigg\|^2\\ \end{align}$
Therefore the minimum is obtained when $a=\frac{\hat f(n)-\hat f(-n)}{2i}, b=\frac{\hat f(n+1)-\hat f(-(n+1))}{2i}$
By Riemann–Lebesgue lemma: $\hat f(n) \xrightarrow{n\to \infty}0$
Hence, $$\lim_{n\to \infty} R_n = \frac{1}{2\pi}\int_{-\pi}^\pi\bigg|\sqrt{|x|^3}\bigg|=\frac{\pi^3}{4}$$
Is that correct?
Any help appreciated.