Prove that the Fourier transform of an integrable function f satisfies:
(i) $\hat{f}$ is bounded.
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?
Since your $f$ is in $L^1[-L,L]$ by assumption, the result should be obvious, once you notice that $$ \left| \int_{-L}^L fg_n \right| \leq \int_{-L}^L |fg_n | = \int_{-L}^L |f |.$$
(Though usually Fourier series are defined for functions in $L^2[-L,L]$, where the $\langle . , . \rangle$ you wrote down is the natural inner product. In this case, you could use Bessel's inequality or Parseval's theorem.)