There dosnt seem to be any place in which $\mathcal{F}(\frac{1}{f}(x))(n)$ is being computed nor talking about its relation to $\mathcal{F}(f(x))(n)$. Prodcuts looks like they are easy to handle but is there somthing making this harder?
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Well, the transform of $1/f$ simply doesn't have any simple connection with the transform of $f$. (How are products "easy to handle"???) – David C. Ullrich Aug 06 '20 at 13:46
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@DavidC.Ullrich convolution – Aug 06 '20 at 13:47
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The main problem is that if $f \in L^1,$ (or $L^2$) then $f(x)$ must more-or-less tend to $0$ as $|x| \to \infty,$ which implies that $1/f(x)$ more-or-less tends to $\infty$ so $\frac1f \not\in L^1$ (or $L^2$).
Using distribution we can get away from this. For example, $\mathcal{F}\{x\} = 2\pi i\delta'(\xi)$ and $\mathcal{F}\{\frac1x\} = -i\pi \operatorname{sign}(\xi)$. One has $\delta'*\operatorname{sign}=\delta,$ but generally it's not easy to find the convolutional inverse of a function or distribution.
md2perpe
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That is continuous, infinitely differentiable and even analytic on all of $\mathbb{R}$. It is bounded, but its reciprocal grows too fast to have a Fourier transform. – md2perpe Aug 06 '20 at 15:27
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1Yes, a continuous function with $0 < \min |f| < \max |f| < \infty$ works. You will then have to take the Fourier transform of it as a distribution (since it's neither $L^1$ nor $L^2$). But still, finding the convolutional inverse of $\mathcal{F}{f}$ is seldom easy. – md2perpe Aug 06 '20 at 15:37