The Question
Show that if we view $f_N = 2^N\chi_{[0,2^{-N}]}$ as a periodic function on $[0,1),$ with $M^{th}$ partial Fourier sum $S_M(f_N)(x)=\sum_{|m|\leq M}\hat{f_N}(m)e^{2\pi imx},$ then the following rate of decay holds: $||f_N-S_M(f_N)||_{L^2([0,1))}\sim 1/\sqrt{M}.$
My Understanding
What I tried is first calculate the Fourier transform of $\hat{f}_N(m)= \frac{2^{N-1}i}{m\pi} (e^{-2^{-N+1}im\pi}-1)$ which I am sure that is correct. There's no trick in calculating the transform. Then I plug this into the given norm form, which has nothing to do with $1/\sqrt{M}$.
\begin{equation*} \begin{aligned} ||f_N-S_M(f_N)||_{L^2([0,1))} &= ||2^N\chi_{[0,2^{-N}]} - \sum_{|m|\leq M,m\not=0} \frac{2^{N-1}i}{m\pi} (e^{-2^{-N+1}im\pi}-1) e^{2\pi imx} -1||_{L^2([0,1))}\\ &= ||2^N\chi_{[0,2^{-N}]} -1||_{L^2([0,1))} \end{aligned} \end{equation*} I also tried another way that is to expand the norm into different components and calculate separately. However, I ended up with messy form and did not find $1/\sqrt{M}$ as well. I think I might on the wrong track.