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seen Jan 10 '13 at 9:25

Oct
16
awarded  Popular Question
Aug
28
asked To show Taylor series of a Fourier transform $\hat{f }$ converges to $\hat{f}$
Aug
28
awarded  Editor
Aug
28
comment find a certain $L^1(R)$ function
I forgot to mention it's compactly supported
Aug
28
revised find a certain $L^1(R)$ function
added 45 characters in body
Aug
28
asked find a certain $L^1(R)$ function
Aug
26
comment convolution of a function with itself equals itself
sorry again ...that's just riemannn-lebesgue lemma ..
Aug
26
comment convolution of a function with itself equals itself
sorry , what I want to ask is : how do you get $\hat{f}$ goes to zero at infinity ? and why fourier transform maps a L1 funciton to a contiuous one ? is that some propositon ? Is it possible for you to show the solution ?
Aug
26
comment convolution of a function with itself equals itself
Hi , I've added some working above , still need more help though..
Aug
26
comment convolution of a function with itself equals itself
I cant see the relation between $\hat{f}$ is continuous and f is zero function ..
Aug
26
comment convolution of a function with itself equals itself
Now If I can put the limit under the integral and show $lim_{B\to \infty}\frac{sin(B2\pi(x-t))}{\pi(x-t)}$ is in both L1(R) and L2(R), then by replacing $f(x)$ by $lim_{B\to \infty}\frac{sin(B2\pi(x-t))}{\pi(x-t)}$, can I get the conclusion ? The reason that I need to show the function is in L1(R) is because to get the first equality , I assumed $f(x)$ is absolutely integrable.
Aug
26
comment convolution of a function with itself equals itself
For part (2) , now my thinking is : $f(x) =\int\hat{f(w)}e^{2\pi iwx} = lim_{B\to \infty}\int_{-B}^B\hat{f(w)}e^{2\pi iwx} $ . After a few lines of change order of integral , this is equal to $lim_{B\to \infty}\int_Rf(t)\frac{sin(B2\pi(x-t))}{\pi(x-t)}dt $ .
Aug
26
awarded  Student
Aug
26
asked convolution of a function with itself equals itself