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Let $u:[0,2\pi)\to \mathbb{R}$ be the step function $$u(x) = \begin{cases} 1 & \text{if } x \in [0,\pi), \\ 0 & \text{if } x \in [\pi,2\pi) \end{cases}$$ By a direct computation, one discovers that \begin{equation}\tag{$*$} \|u-S_N u\|_{L^2} \leq C N^{-1/2}, \end{equation} where $S_N u $ is the truncated Fourier series of $u$: $$S_N u(x) = \sum_{k = -N}^N c_k e^{i k x}, \quad c_k = \frac{1}{2\pi}\int_0^{2\pi} u(x) e^{-ikx} dx.$$

I am quite sure that a bound of the form $(*)$ holds for generic piecewise continuous functions on $[0,2\pi)$, but I didn't manage to find anything in the literature... The books that I consulted usually explain the Gibbs phenomenon for discontinuous functions, but $L^2$ error estimates are provided only when $u$ is regular.

Please give me a reference or maybe a counterexample, if I am wrong. Thank you!

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    $\begingroup$ The Fourier series of an $L^2$ function on the circle converges in $L^2$. (See e.g. Katznelson's An introduction to harmonic analysis.) Do you need an estimate for the rate of convergence when the function happens to be piecewise continuous? $\endgroup$
    – Joonas Ilmavirta
    Commented Oct 6, 2014 at 9:47
  • $\begingroup$ @JoonasIlmavirta I edited the question, maybe now it's more clear that what I need is an estimate. $\endgroup$
    – Paglia
    Commented Oct 6, 2014 at 9:56
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    $\begingroup$ Since continuous functions are dense in L^2, I don't think their rates of L^2 convergence can be any better than for generic L^2 functions, unless you are more precise about what the constant C depends on $\endgroup$
    – Yemon Choi
    Commented Oct 6, 2014 at 10:19
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    $\begingroup$ Well, as my previous remark observed, you can get continuous functions (i.e. NO jumps) whose Fourier series converge slower than any prescribed rate of convergence. Note that your example is not just piecewise continuous, it is piecewise smooth, and so you should probably look for C depending on the number of jumps and the smoothness of the pieces $\endgroup$
    – Yemon Choi
    Commented Oct 6, 2014 at 11:55
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    $\begingroup$ Kahane's book on random series of functions has a proof of the theorem that given any sequence $a_n\in\ell^2$, $a_n\ge 0$, there is a continuous function with $|\widehat{f}_n|\ge a_n$. $\endgroup$
    – Christian Remling
    Commented Oct 6, 2014 at 17:43

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Such an assertion is close enough to being an "exercise" that there may not be a really clear "reference" for it... but such a result can be explained easily and shortly enough, I think:

Finite sums of (dilates of translates of) derivatives of $\cos x/2$ can be subtracted from a given (finitely-) piecewise smooth periodic function to give a $C^2$ periodic function (meaning $C^2$ on the circle, so imposing a matching condition at $0$ and $2\pi$). The derivatives of $\cos x/2$ are essentially itself and $\sin x/2$, so direct computation shows that the Fourier coefficients of any derivative of $\cos x/2$ decay like $1/n$, so that the $L^2$ norm of tails are like $1/\sqrt{N}$, as desired.

(The same conclusion is a little less obvious if the discontinuities are approximated by piecewise-polynomial functions, because then the boundary terms in integration by parts play a role...)

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