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Is the inverse Fourier transform on $L^2$ computable? That is, \begin{align}F^{-1}: &L^2 \rightarrow L^2\\ & f \mapsto \check{f} \end{align} $(\delta_{L^2}, \delta_{L^2})$ computable? Here, we know that $\check{f}(x) = \hat{f}(-x)$ almost everywhere in $L^2(\mathbb{R})$ where $\hat{f}$ is the Fourier Transform of $f$, that is, $\hat{f}(\xi)= \int f(x)e^{-2\pi\iota\xi x}dx$ . We already know that the Fourier Transform $F: L^2\rightarrow L^2$ given by $f \mapsto \hat{f}$ is $(\delta_{L^2}, \delta_{L^2})$ computable by the paper "Type-2 computability on spaces of Integrable functions" by Daren Kunkle.

I think yes because given $\delta_{L^2}$ name of $f$, we can computably find $\delta_{L^2}$ name of $\hat{f}$. Now, if $\lVert \hat{f}-\sum c_j\chi_{(a_j,b_j)}\rVert < 2^{-n}$ then $\lVert \check{f}-\sum c_j\chi_{(-b_j,-a_j)}\rVert < 2^{-n}$ and hence we can get $\delta_{L^2}$ name of $\check{f}$.

Poonam
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  • $F : L^2 \to L^2$ is the same. What means computable exactly ? How did you define $f$ in a first place ? – reuns Jul 20 '17 at 16:56
  • Also what is $\delta_{L^2}$ and $\delta_{L^2}$ name of $f \in L^2$ ? – reuns Jul 20 '17 at 17:32
  • Sir, the question is from the topic computable analysis. Here f is an element of L^2 and δL2 is the Cauchy representation of the space L^2. – Poonam Jul 22 '17 at 13:40
  • What means Cauchy representation exactly ? – reuns Jul 22 '17 at 20:36
  • Basically, Cauchy representation of an element in an effective Hilbert space is finding a name of that element that encodes a dense sequence in the space converging rapidly to the element. The exact definition can be seen in the paper mentioned in the question. – Poonam Jul 25 '17 at 16:09

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