What is the rate of convergence of $f^N(x) = \sum_{n=-N}^N f_n e^{in x}$ to $f(x) = \sum_{n=-\infty}^\infty f_n e^{in x}$ in $L^2([0,2\pi])$?

Question

Let $f \in L^2([0,2\pi])$ if $2$$\pi$ periodic. Then $f$ can be decomposed using a Fourier series as $$ f(x) = \sum_{n=-\infty}^\infty f_n e^{in x}. $$

Now suppose we approximate $f$ by $f^N$ where $f^N$ is given by $$ f^N(x) = \sum_{n=-N}^N f_n e^{in x}. $$ I am wondering what the rate of convergence is of this approximation in $L^2([0,2\pi])$? I assume it should be exponential (spectral) as it is based on a Fourier series. I.e.

$$ \begin{align} ||f - f^N||_{L^2([0,2\pi])}^2 & = \int_0^{2\pi} |f - f^N|^2 dx \\ & = \int_0^{2\pi} |\sum_{n=-\infty}^\infty f_n e^{in x} - \sum_{n=-N}^N f_n e^{in x}|^2 dx \\ & = \int_0^{2\pi} |\sum_{n=-\infty}^\infty f_n e^{in x} - \sum_{n=-N}^N f_n e^{in x}|^2 dx \\ & = \int_0^{2\pi} |\sum_{n=-\infty}^{-N-1} f_n e^{in x} + \sum_{n=N+1}^{\infty} f_n e^{in x}|^2 dx \\ & \le \int_0^{2\pi} \left(|\sum_{n=-\infty}^{-N-1} f_n e^{in x}| + |\sum_{n=N+1}^{\infty} f_n e^{in x}|\right)^2 dx \\ & \le \int_0^{2\pi} \left(\sum_{n=-\infty}^{-N-1} |f_n| + \sum_{n=N+1}^{\infty} |f_n|\right)^2 dx \\ & \le C \left(\sum_{n=-\infty}^{-N-1} |f_n| + \sum_{n=N+1}^{\infty} |f_n|\right)^2 \\ \end{align} $$ But this doesn't seem to be leading anywhere..I want to end up with something like $$||f - f^N||_{L^2([0,2\pi])}^2 = Ce^{-N}. $$ I don't have much experience with rate of convergence so what is the actual rate of convergence for this approximation and what is the technique used to obtain it?

Answer 1

There's nothing that can be said about the rate of convergence without further assumptions on $f$; in fact $||f-f^N||$ can tend to $0$ arbitrarily slowly.

Roughly speaking, a faster rate of convergence implies some sort of smoothness for $f$. For example, If $||f-f^N||\to0$ exponentially then $f_n\to0$ exponentially, which implies that $f$ is infinitely differentiable.

Edit: Someone supplied a reference for the "arbitrarily slowly" statement. This is much more trivial than that:

Prop If $s_0,s_1,s_2,\dots$ is a sequence decreasing to zero then there exists $f\in L^2$ with $||f-f^N||^2=s_N$.

Proof. Let $a_n^2=s_{n-1}-s_n$, and define $f(t)=\sum_{n=1}^\infty a_ne^{int}$. Then $$||f-f^N||_2^2=\sum_{n=N+1}^\infty a_n^2=s_N.$$