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Consider a function $f_p : \mathbb{R} \to \mathbb{R}$ which is continuously differentiable in $(0,2\pi)$ except at two points $x = x_c$ and $x = x_o$. At $x = x_c$, $f_p(x)$ has a jump discontinuity. At the point $x = x_o$, $f_p'(x_o^+)$ and $f_p'(x_o^-)$ exist and are not equal.

Now define a function $f$ equal to $f_p$ on $(0,2\pi)$ and let it be a periodic function with period $2\pi$. Let $\hat{f}_k$ be the Fourier series coefficients of $f$. What I would like to know is whether the Fourier series defined by the coefficients $ik\hat{f}_k$ converge to $\frac{f'(x_o^+)+f'(x_o^-)}{2}$ at $x = x_o$ ?

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    @user1551 : you are probably right. I remember reading that it diverges at such points. I don't have the reference book in hand right now (Trig. Series by Zygmund) and hence this question...someone pleas confirm in form of an answer, even without any proof.2011-09-02

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It will diverge, but if I am not mistaken, can be Cesàro summed.

Your function can be written as a sum of a Lipschitz function that is continuously differentiable except at (at most) two points, and a sawtooth wave. The first component has Fourier series that converges, and has the property that the Fourier series of the derivative is $ik\hat{f}_k$. The second component, however, gives the problem. A sawtooth wave has Fourier series expansion (up to a normalising constant) $\hat{f}_k = \frac{(-1)^k}{k}$. So the main issue is handling the sum of the form

$ \sum (-1)^k\sin(k x) $

This of course is not absolutely convergent. But away from the singularity, you can take the Cesàro sum: the partial sum of the above expression is more or less the Dirichlet kernel (shifted by a constant). And so the partial Cesàro sum is (roughly) just the Fejér kernel, which is known to converge point-wise away from the singularity.

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    It is Cesaro summable $(C,r)$ away from the singularity, only for r > 1. (given in the book Trignometric Series by Walker Page no. 55 section 3.5)...but by imposing a condition that the singularity to be that of a jump type, would we be able to make it $(C,1)$ summable ? could there be any such possibility ?2011-09-10