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I was working around with the fractional Fourier transform (FRFT) when the mathematics undergrad found out, by brute-force computations, that the derivative of the FRFT with respect to the parameter can be written as a closed form formula- explicitly, $d/da(F^a(f)(y))$$ = i(1/8d^2/dy^2-π^2y^2/2+π/4)F^a(f)(y)$.

Then I remembered that the first definition of the FRFT, using "only" a certain set (Hermite–Gaussian functions) of orthonormal eigenfunctions of the FT, is not actually "unique", especially when attempting to discretise the transform.

So I'm asking the following question instead- do unitarity, index (assumed to be real) continuity, index additivity, reduction to the identity (resp. Fourier transform) operator when the index is 0 (resp. 1), and the formula above uniquely define the FRFT, as operators on Schwartz functions?

NB: the version of the FRFT I'm using is the one used by the current revision of the Wikipedia page for the FRFT.

I was working around with the fractional Fourier transform (FRFT) when the mathematics undergrad found out, by brute-force computations, that the derivative of the FRFT with respect to the parameter can be written as a closed form formula- explicitly, $d/da(F^a(f)(y))$$ = i(1/8d^2/dy^2-π^2y^2/2$$+π/4)F^a(f)(y)$.

Then I remembered that the first definition of the FRFT, using "only" a certain set (Hermite–Gaussian functions) of orthonormal eigenfunctions of the FT, is not actually "unique", especially when attempting to discretise the transform.

So I'm asking the following question instead- do unitarity, index (assumed to be real) continuity, index additivity, reduction to the identity (resp. Fourier transform) operator when the index is 0 (resp. 1), and the formula above uniquely define the FRFT, as operators on Schwartz functions?

NB: the version of the FRFT I'm using is the one used by the current revision of the Wikipedia page for the FRFT.

I was working around with the fractional Fourier transform (FRFT) when the mathematics undergrad found out, by brute-force computations, that the derivative of the FRFT with respect to the parameter can be written as a closed form formula- explicitly, $d/da(F^a(f)(y)) = i(1/8+(π/4-π^2y^2/2)d^2/dy^2)F^a(f)(y)$.

Then I remembered that the first definition of the FRFT, using "only" a certain set (Hermite–Gaussian functions) of orthonormal eigenfunctions of the FT, is not actually "unique", especially when attempting to discretise the transform.

So I'm asking the following question instead- do unitarity, index (assumed to be real) continuity, index additivity, reduction to the identity (resp. Fourier transform) operator when the index is 0 (resp. 1), and the formula above uniquely define the FRFT, as operators on Schwartz functions?

NB: the version of the FRFT I'm using is the one used by the current revision of the Wikipedia page for the FRFT.

I was working around with the fractional Fourier transform (FRFT) when the mathematics undergrad found out, by brute-force computations, that the derivative of the FRFT with respect to the parameter can be written as a closed form formula- explicitly, $d/da(F^a(f)(y))$$ = i(1/8d^2/dy^2-π^2y^2/2+π/4)F^a(f)(y)$.

Then I remembered that the first definition of the FRFT, using "only" a certain set (Hermite–Gaussian functions) of orthonormal eigenfunctions of the FT, is not actually "unique", especially when attempting to discretise the transform.

So I'm asking the following question instead- do unitarity, index (assumed to be real) continuity, index additivity, reduction to the identity (resp. Fourier transform) operator when the index is 0 (resp. 1), and the formula above uniquely define the FRFT, as operators on Schwartz functions?

NB: the version of the FRFT I'm using is the one used by the current revision of the Wikipedia page for the FRFT.