Let $\mathcal C$ be the space of real continuous $2\pi$-periodic functions bounded in absolute value by 1.
Can we compute
$$c_s := \sup_{f\in\mathcal C} \left|\int_0^{2\pi} e^{-isx}f(x)dx\right|$$
for any $s\in\mathbb Z$? In other words, what is the biggest $s$-Fourier coefficient in the space?
Obviously all $c_s$ are upper bounded by $2\pi$ and $c_0 =2\pi$. Moreover we can restrict $\mathcal C$ to the functions whose maximum is exactly 1 (that probably is not compact, pity).
Moreover, by choosing the probably most obvious candidate $f(x) = \cos(sx)$, one finds that $c_s\ge \pi$ for every $s$.
Notice that for complex-valued $f(x)$ we get $c_s=2\pi$ for any $s$, but can we prove that for real functions the best is half of that?
Let's give an elementary proof that $c_n \le 4$ and one can get $c_n=4-\epsilon$ for every $\epsilon >0$, so for fixed $N \ge 1$ find $f$ continuous, periodic, real and bounded by $1$ for which $c_N=4-\epsilon$, while it will be clear that $4$ cannot be attained with continuous functions.
Let's fix $N \ge 1$ and assume $c_N \ne 0$ and let $t_0$ its argument so $e^{-it_0}\int_0^{2\pi} e^{-iNx}f(x)dx =e^{-it_0}c_N=|c_N|>0$
But this means that $\int_0^{2\pi} e^{-iN(x+t_0/N)}f(x)dx >0$ or with $g(x)=f(x+t_0/N)$ we have $\int_0^{2\pi} e^{-iNx}g(x)dx >0$ using the change of variable $x \to x+t_0/N$ and the periodicity of $f$.
(Added per comments) But the integral above being real (and positive) means that its real part is positive and equal to it and its imaginary part is zero. Using that $g$ is real (here is crucial) we get that:
$\int_0^{2\pi} e^{-iNx}g(x)dx =\int_0^{2\pi} g(x) \cos Nx dx >0$ (real part) and $\int_0^{2\pi} g(x) \sin Nx dx =0$ (imaginary part)
But $|g(x)| \le 1$ means that $|g(x) \cos Nx| \le |\cos Nx|$ so $|\int_0^{2\pi} e^{-iNx}g(x)dx| \le \int_0^{2\pi}|\cos Nx|dx=4$ hence $|c_N| \le 4$ and we clearly get equality for $f(x)=\text{sign} \cos Nx$.
Now $f$ is not continuous but clearly can be approximated as well as we want by continuous functions for which $c_N$ will be as close to $4$ as we wish.