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 Aug 17 awarded Notable Question 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