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I was reading the following question, A function is $L^2$-differentiable if and only if $\xi\widehat{f}(\xi) \in L^2$., and in the accepted answer there is a bound $$\left\lvert\frac{e^{2\pi ih\xi}-1}{h} \right\rvert \leqslant 2\pi \lvert \xi\rvert$$ and although it may not be complicated i got stuck on showing this bound, i do see that

$$ \lim_{h \to 0} \left\lvert\frac{e^{2\pi ih\xi}-1}{h} \right\rvert = 2\pi \lvert \xi\rvert$$

So i thought of showing this function is an increasing function but all i got was a mess on calculations, is there an easier way to see that bound?

Edit: i dont know what my idea was but definetely increasing function is not a way to show the bound

Daniel Moraes
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  • don't forget that $|e^{ir}| = 1$ for real $r$, so your LHS is bounded above by $2/h$. And as $h \to 0$ we have that $e^{2\pi i h \xi} \to 1$ so that might help. – Prime Mover Apr 09 '21 at 22:23
  • @PrimeMover i just need the upper bound in order to apply the dominated convergence theorem, but it must be in the whole domain, my goal with the limit is to show where i think the bound appears – Daniel Moraes Apr 09 '21 at 22:30
  • Maybe mean value theorem with a=h and b=0 – Daniel Moraes Apr 09 '21 at 23:14

1 Answers1

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$|e^{2\pi i\xi}-1|=2\pi |\xi||\int_0^{h} e^{2\pi i t\xi} dt|\leq 2\pi |\xi||\int_0^{h} 1 dt| \leq 2\pi |\xi|$ if $|h| \leq 1$.

Kavi Rama Murthy
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