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).

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  • $\begingroup$ it is not clear for me how the sup beside $c_{n}$ will come. $\endgroup$ – Emptymind May 21 '17 at 6:21
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    $\begingroup$ The bound you produce is independent of $n$, so it holds for every $c_n$. Hence it holds for their supremum. $\endgroup$ – Chappers May 21 '17 at 12:21
  • $\begingroup$ but $g(x)$ that you wrote above is actually $g_{n}(x)$ and so the n is actually found in the bound also ...... so what shall I do? $\endgroup$ – Emptymind May 23 '17 at 4:33
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    $\begingroup$ But $ |e^{in\pi x/L}|=1 $ is independent of $n$. $\endgroup$ – Chappers May 23 '17 at 14:20

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