Prove that the Fourier transform of an integrable function f satisfies:
(i)$sup_{n}|c_{n}(f)| \leq \frac{1}{2L}\int_{-L}^{L}|f(x)|dx$ .
Where the Fourier transform $\hat{f}$ of an integrable function $f$ is defined as the sequence $(c_{n}(f))_{n \in \mathbb{Z}}.$ Where $c_{n}(f)$ is the Fourier coefficients of a function $f \in L^{1}([-L,L])$, and is defined to be $$c_{n}(f) = \frac{1}{2L}\int^{L}_{-L} f(x) \overline{g_{n}(x)}dx = \langle f,g_{n}\rangle.$$ And for $L>0$ and for $n \in \mathbb{Z}$ we take $$g_{n}(x) = e^{in\pi x/L}. $$
Could anyone give me a hint please?
Use the trivial bound $$ \left| \int_a^b f(x) g(x) \, dx \right| \leq \sup_{x \in [a,b]} \lvert g(x) \rvert \int_a^b \lvert f(x) \rvert \, dx, $$ which is a combination of $|\int F |\leq \int |F|$ (integral triangle inequality) and $|fg|<|f|\sup{|g|}$ (obvious from definition of the supremum).