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Can someone give an example of an inverse Fourier transform that's not unique and show it? Thank you for any help.

Thomas
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  • Its definition is unique as far as I know. Certainly for the discrete version. – Pieter21 Dec 04 '16 at 12:28
  • A Fourier transform, Ff(t), is invertible if f(t) and f'(t) are smooth and f(t) is Fourier transformable. 1/(2Pi) (lim(R->+infinity) int(F(w)exp(iwt) dw , -R..R) )= 1/2 [f(t+) + f(t-)]. (Sorry for my notations) – Thomas Dec 04 '16 at 12:59
  • I understand your argument. Thank you. –  Dec 04 '16 at 12:59
  • @Thomas what do you mean exactly with $f,f'$ are smooth ? The Schwartz space I mentioned ? And as I wrote you can prove the Fourier inversion theorem when $f,\hat{f} \in L^1$, and for that a sufficient condition is that $f,f' \in L^1$ because it implies $f \in L^1, f' \in L^2$ so that $2i \pi \xi \hat{f}(\xi) \in L^2$ – reuns Dec 04 '16 at 13:07
  • @user1952009, Yes I mean the Schwartz space. – Thomas Dec 04 '16 at 13:08
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    @Thomas in that case consider $f(x) = 0$ and $g(x) = 0$ except $g(0) = 1$ then clearly $\hat{f}(\xi) = \hat{g}(\xi) =0$ so in some sense $f,g$ both are the inverse Fourier transform of the zero function (now in $L^1$ : $f,g$ are in the same "class of equivalence" of functions) – reuns Dec 04 '16 at 13:11
  • @user1952009 I will give an example: F(w) = 2a sin(aw)/(aw), the Fourier inverse is Y(a-abs(t)), if t not equal to +/- a, and 1/2 if t = +/- a. Why is this not an unique Fourier inverse? – Thomas Dec 04 '16 at 13:13
  • @Thomas It is unique in $L^2$ where we don't care of the value at a single point, because $L^p$ spaces are spaces of class of equivalence of functions (with the equivalence being $f \equiv g$ if $|f-g|_p = 0$) – reuns Dec 04 '16 at 13:15
  • @user1952009 Oh I understand. Thank you for your help – Thomas Dec 04 '16 at 13:16
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    And $h(x) = \frac{\sin(\pi x)}{\pi x}$ is not in the Schwartz space : it is smooth but its decay as $x \to \infty$ isn't fast enough. It is not even in $L^1$ but it is in $L^2$. – reuns Dec 04 '16 at 13:19

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The big question is : how do you define the Fourier transform and the inverse Fourier transform ? I ask this because you can define several set of functions (or class of equivalence of functions) where the Fourier and inverse Fourier transform is well-defined. Outside of those sets, the Fourier transform is not so well and uniquely defined.

In general you start from $\hat{f}(\xi) = \int_{-\infty}^\infty f(x)e^{-2i \pi \xi x}dx$ well-defined whenever $f \in L^1$. You prove the convolution theorems and the Fourier inversion theorem $f(x) = \int_{-\infty}^\infty \hat{f}(\xi)e^{-2i \pi \xi x}d\xi$ for $f,\hat{f} \in S(\mathbb{R})$ (the Schwartz space in particular for $f,\hat{f}$ Gaussian) and extend it to $f,\hat{f} \in L^1$.

Then, using that $L^1\cap L^2$ is dense in $L^2$, you can extend all this to $f \in L^2$ (a Hilbert space where the Fourier transform is a unitary operator, see the Parseval's theorem).

Finally from $S(\mathbb{R})$ and the Fourier transform on $S(\mathbb{R})$ you can define the tempered distributions $S'(\mathbb{R})$ (containing the Dirac delta $\delta$ and its derivatives, useful for solving ordinary differential equations) and the Fourier transform and the Fourier inverison theorem on $S'(\mathbb{R})$.

reuns
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