For $N\geq 1$ the discrete Fourier Coefficient of $f$ are defined by \begin{align} a_N(n)=\frac{1}{N}\sum_{k=1}^{N}f(e^{2\pi i k/N})e^{-2\pi i k n/N}, \quad \forall n\in \mathbb{Z} \end{align} Suppose $f\in C^1$ function on circle, how to prove there exist $c>0$ such that $|a_N(n)|\leq \frac{c}{|n|}$ whenever $0<|n|\leq \frac{N}{2}$. Hint: Write \begin{align} a_N(n)[1-e^{2\pi i\ell n/N}]=\frac{1}{N}\sum_{k=1}^{N}[f(e^{2\pi i k/N})-f(e^{2\pi i(k+\ell)/N})]e^{-2\pi i k n/N} \end{align} and choose $\ell$ so that $\ell n/N$ nearly 1/2.
Source: Fourier Analysis An Introduction, Elias M.Stein & Rami Sacharchi. Chapter 7 problem 2.
I am not understand Hint of the book. How we prove LHS=RHS in the hint.
My approach for proof the hint.
If $\ell=0$, it's so clear.
Suppose $1\leq \ell< N$ and $\ell\in \mathbb{Z}$, \begin{align*} f(e^{2\pi i(k+\ell)/N})e^{-2\pi i k n/N}= \begin{cases} f(e^{2\pi i(k+\ell)/N})e^{-2\pi i k n/N}, \quad 1\leq k\leq N-\ell\\ f(e^{2\pi i(k-(N-\ell))/N})e^{-2\pi i k n/N}\quad, N-\ell< k\leq N \end{cases} \end{align*} and we get \begin{align*} \frac{1}{N}\sum_{k=1}^{N} f(e^{2\pi i(k+\ell)/N})e^{-2\pi i k n/N}&=\frac{1}{N}\sum_{k=1}^{N-\ell} f(e^{2\pi i(k+\ell)/N})e^{-2\pi i k n/N}\\&+\frac{1}{N}\sum_{k=N-\ell+1}^{N}f(e^{2\pi i(k-(N-\ell))/N})e^{-2\pi i k n/N}\\ &=\frac{1}{N}\sum_{k=\ell+1}^{N}f(e^{2\pi i k/N})e^{-2\pi i(k-\ell)n/N}+\frac{1}{N}\sum_{k=1}^{\ell}f(e^{2\pi ik/N})e^{-2\pi i (k-l)n/N}\\ &=\frac{1}{N}\sum_{k=1}^{N}f(e^{2\pi i k/N})e^{-2\pi i(k-\ell)n/N}\\ &=a_n(N)e^{2\pi i \ell n/N} \end{align*}
If $-1\leq \ell <-N$ \begin{align*} f(e^{2\pi i(k+\ell)/N})e^{-2\pi i k n/N}= \begin{cases} f(e^{2\pi i(k-(-\ell-N))/N})e^{-2\pi i k n/N}, \quad 1\leq k\leq-\ell\\ f(e^{2\pi i(k+\ell)/N})e^{-2\pi i k n/N}\quad, -\ell< k\leq N \end{cases} \end{align*} \begin{align*} \frac{1}{N}\sum_{k=1}^{N} f(e^{2\pi i(k+\ell)/N})e^{-2\pi i k n/N}&=\frac{1}{N}\sum_{k=1}^{-\ell} f(e^{2\pi i(k-(-\ell-N))/N})e^{-2\pi i k n/N}\\&+\frac{1}{N}\sum_{k=-\ell+1}^{N}f(e^{2\pi i(k+\ell)/N})e^{-2\pi i k n/N}\\ &=\frac{1}{N}\sum_{k=N+\ell+1}^{N}f(e^{2\pi i k/N})e^{-2\pi i(k+\ell)n/N}+\frac{1}{N}\sum_{k=1}^{N+\ell}f(e^{2\pi ik/N})e^{-2\pi i (k+\ell)n/N}\\ &=\frac{1}{N}\sum_{k=1}^{N}f(e^{2\pi i k/N})e^{-2\pi i(k+\ell)n/N}\\ &=a_n(N)e^{2\pi i \ell n/N} \end{align*}