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I read a statment on a book saying that the FBI transform $$\mathcal{F}_u(x,\xi)=\int_{\mathbb{R}^m} e^{i\xi.(x-y)-|\xi||x-y|^2}u(y)\,dy ,\; (x,\xi)\in \mathbb{R}^m \times \mathbb{R}^m$$ is nonlinear. I can not see how this is nonlinear.

GEdgar
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JYM
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    Non-linear in what? In $x$ and $\xi$? – GEdgar Sep 06 '18 at 13:25
  • I don't think it is in $x$ or $\xi$. My confusion is as in the Fourier transform (or any other operator) if we replace $au_1+bu_2$ in place of $u$ from the linearity of the integral it follows that it is linear. – JYM Sep 06 '18 at 13:33
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    Perhaps if you reveal the statement and the book, we can comment further. But commenting on something in a secret book seems beyond my ability. – GEdgar Sep 06 '18 at 14:54
  • S. Berhanu, D. Cardo and J. Hounie, an introduction to involutive structures (Page 226) – JYM Sep 06 '18 at 17:22

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The term ''nonlinear'' is used to indicate the non-linearity in the exponent,i.e, $|x-y|^2$

JYM
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