Question: For $f \in C^{1},$ show that the Fourier coefficients, $c_n(f),~n \in \mathbb{Z}$ satisfies $$\lim_{n \rightarrow \infty} n \cdot c_n(f)=0.$$
My approach: WLOG assume that the period of $f$ to be $1.$ The Fourier series of is $$f(x)=\sum_{ n \in \mathbb{Z}}~c_n \cdot e^{2 \pi i n t}~dt$$ and the corresponding Fourier coefficients are given by $$c_n(f)=\int_{0}^{1}~e^{-2 \pi i n t} \cdot f(t)~dt.$$ Using integrating by parts with $u=f(t)$ and $dv=e^{-2 \pi i n t}~dt.$ Then $du=f'(t)~dt$ and $v=e^{-2 \pi i n t}/{-2 \pi i n}.$ Thus we have $$c_n(f)=\frac{1}{-2 \pi i n} \cdot \big\{ f(1) - f(0) \big\}+\frac{1}{2 \pi i n}\int_{0}^{1} e^{-2 \pi i n t} \cdot f'(t)~dt.$$
Now when I multiplied by the $n$ I don't get the desired result. Any help in proving this is much appreciated. Thank you.
If $f$ is periodic $f(1)=f(0)$. Then use Riemann-Lebesgue Lemma to show your second term vanishes as well since $f'\in L^1$.
If $f$ is not periodic. The conclusion does not stand. For example, $c_n[x]\sim \frac{2(-1)^n}{n\pi}$.