Fourier series of function f: $f(x)=\sum_{s=-\infty}^{\infty}f_{s}\exp(2\pi isx)$
Suppose $f_{s}\sim\frac{1}{s^{p}}$. What can we say about $f(x)$? Can we find some bounds for $f(x)$ like $f(x)
Fourier series of function f: $f(x)=\sum_{s=-\infty}^{\infty}f_{s}\exp(2\pi isx)$
Suppose $f_{s}\sim\frac{1}{s^{p}}$. What can we say about $f(x)$? Can we find some bounds for $f(x)$ like $f(x)
The size of the Fourier coefficients of a function is related to the smoothness of the function, not to its size. The smoother the function (i.e. the more derivatives it has), the faster the convergence to $0$ of its Fourier coefficients.
If $f$ is a periodic function of period $2\pi$, integrable on an interval of length $2\pi$, then, by the Riemann-Lebesgue theorem, $\hat f_s\to0$, where $\hat f_s$ is the $s$-th Fourier coefficient of $f$. If moreover $f$ is $k$ times differentiable and $f^{(k)}$ is integrable, then $|\hat f_s|\le C|s|^{-k}$ (even more, $|\hat f_s||s|^{k}\to0$.)
Conversely, if $|\hat f_s|\le C|s|^{-k}$ for some $k\ge2$, then $f$ is continuous with $k-2$ continuous derivatives. But there is nothing you can say about the size of $f$. For instance, if $f=M$ is constant, then $\hat f_s=0$ for all $s\ne0$.