Let $f: S^1 \rightarrow \mathbb{R}$ be Lebesgue integrable on $S^1 \cong (-\pi,\pi]$. Define the $k$th complex Fourier coefficient as
$$c_k(f):= \frac{1}{2\pi} \int_{-\pi}^{\pi} f(\theta) e^{-i k \theta}d \theta.$$
Show $c_k(f)$ is bounded and find its bound.
So do I just take its absolute value, then integrate by parts on the integrand to obtain a bound ? Here is what I have
$$\vert c_k(f) \vert := \frac{1}{2 \pi} \vert \int_{-\pi}^{\pi} f(\theta) e^{-i k \theta} d \theta \vert \leq \frac{1}{2 \pi} \int_{-\pi}^{\pi} \vert f(\theta) e^{-i k \theta} \vert d \theta$$
Then I set $u = f(\theta)$ and $dv = e^{-i k \theta}$. I then get :
$$\frac{1}{2 \pi}((-\frac{f(\theta)}{ik}e^{-i k \theta})\vert_{-\pi}^{\pi} + \int_{-\pi}^{\pi} \frac{f'(\theta)}{ik}e^{-i k \theta} d \theta)$$
Should I split my integer into positive and negative components ?