$|e^{2\pi i (x+h)\xi} - e^{2\pi i (x)\xi}| \leq 2sin(\pi |h\xi|)$
Original inequation is
$\int \hat{f(\xi)}(e^{2\pi i (x+h)\xi} - e^{2\pi i (x)\xi})d\xi \leq \int |\hat{f(\xi)}|2sin(\pi |h\xi|)d\xi$
Thank you very much!
2026-03-27 14:21:36.1774621296
How to prove this inequation in Fourier Analysis?
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Well $$ e^{2\pi i (x+h)\xi} - e^{2\pi i x \xi} = e^{2\pi i x \xi}(e^{2\pi i h \xi} - 1) $$ So the absolute value of the left-hand side is $$ |e^{2\pi i h \xi} - 1|... $$ can you go from here (I was going to just write full answer but better respect the fact that Scientifica was giving hints)