Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $x = x_o$ ?

share|improve this question
1  
$g=f_p'$ is piecewise continuous and the Fourier series of $g$, when evaluated at $x_0$, should converge to the average of its left and right limits at $x_0$. So you are essentially asking if $is\hat{f}(s)=\hat{g}$. But if this is true, we should also have $is\widehat{(f+H)}(s)=\hat{g}$, where $H$ is the Heaviside function, because $f'=(f+H)'$. This implies that $\hat{H}(s)=0$, which is weird. So I guess the answer to your question is negative. –  user1551 Sep 2 '11 at 6:40
    
@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. –  Rajesh D Sep 2 '11 at 8:25
add comment

1 Answer

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.

share|improve this answer
    
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 ? –  Rajesh D Sep 10 '11 at 6:45
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.