Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

$\operatorname{sinc} : \mathbb{R} \to \mathbb{R} \;\;$ is defined by $\;\; \operatorname{sinc}(x) \; = \; \begin{cases} 1 & \text{if }\:\;x=0 \\ \\ \frac{\operatorname{sin}(x)}x & \text{else} \end{cases} \;\; $ .


For what square-summable sequences $\langle x_0,x_1,x_2,x_3,...\rangle$ of complex numbers is it the case that

$\displaystyle\lim_{h\to 0^+} \left(\displaystyle\sum_{n=0}^{\infty} \: (\operatorname{sinc}(n\cdot h)\cdot x_n)\right) \; $ exists and is finite?


(In other words, when is it the case that the Lebesgue mean at zero of a member of $\; \operatorname{L}_2[-\pi,\pi] \;$
exists and is finite, in terms of the coefficients of that member's exponential fourier series?)

share|improve this question
    
I know that the answer is at least the absolutely summable sequences. –  Ricky Demer Oct 27 '11 at 23:48
    
When you ask for the existence of the limit, is the infinite sum being understood in the usual sense of $\lim_{N\to \infty} \sum_{n=0}^N$? –  Yemon Choi Oct 28 '11 at 0:40
1  
Why don't we go from the other end: what condition would you be happy with? If you formulate it, we can prove or disprove its sufficiency. As stated, the question can be given many answers all of which can easily miss the point you have in mind entirely. –  fedja Oct 28 '11 at 1:02
1  
@Yemon: The series converges unconditionally for all square-summable sequences and positive $h$, so I don't think it matters. –  Ricky Demer Oct 28 '11 at 1:32
1  
Note that any continuous at $0$ square summable function would do and those can have terrible Fourier partial sums at $0$, so the class is huge and includes many series that diverge in the classical sense. –  fedja Oct 28 '11 at 1:56
show 2 more comments

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.