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I am asked to find the spectrum of the Fourier transform operator $\mathcal F:L^2(\mathbb R)\to L^2(\mathbb R)$, where $\mathcal Fu(\omega)=\int_\mathbb R \frac 1{\sqrt{2\pi}}u(x)e^{-ix\omega} dx$. I know that $\mathcal F^4=\mathrm{Id}$, and I read this thread already Two questions in spectral theory: the spectrum of the Fourier transform and the Hamiltonian of the hydrogen atom.. I'm lost in the last passage, how do I show that there is no residual spectrum?

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Because $\mathcal{F}^4=I$, it follows that $$ (\mathcal{F}-zI)^{-1}=\frac{1}{1-z^4}(\mathcal{F}^3+z\mathcal{F}^2+z^2\mathcal{F}+z^3I),\;\;\;\; z \notin \{1,i,-1,-i\}. $$ The operator on the right is bounded on $L^2$. So the spectrum $\sigma(\mathcal{F})$ satisfies $\sigma(\mathcal{F})\subseteq\{1,i,-1,-i\}$. Conversely, every point of $\{ 1,i,-1,-i\}$ is in the point spectrum; for example, $i$ is in the point spectrum because \begin{align} (\mathcal{F}-iI)(\mathcal{F}+I)(\mathcal{F}+iI)(\mathcal{F}-I)=0,\\ (\mathcal{F}+I)(\mathcal{F}+iI)(\mathcal{F}-I) \ne 0. \end{align} All spectrum of $\mathcal{F}$ is, therefore, point spectrum. There is no residual or continuous spectrum. This is generally true of any bounded operator with an annihilating polynomial.

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