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Let $\phi$ be be a compactly supported smooth function on $\mathbb R.$ We put $I=$ the identity operator and $\triangle_{\xi} =\frac{\partial^{2}}{\partial \xi^{2}}$ (Laplace operator). In Fourier analysis , people keep using the following step: $$\int_{\mathbb R}(I-\triangle_{\xi})[e^{ix\xi}] \phi (\xi) d\xi = \int_{\mathbb R}e^{ix\xi} (I-\triangle_{\xi})[\phi (\xi)] d\xi. $$

My question: How to justify the above step ? Is function in the question compactly supported necessary ?

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    $\begingroup$ Use \Delta. $ $ $\endgroup$ – Did Jan 6 '15 at 8:28
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Lets split the terms: \begin{align} \int_{\mathbb R}(I-\triangle_{\xi})[e^{ix\xi}] \phi (\xi) d\xi &= \int_{\mathbb R}e^{ix\xi} \phi (\xi)- \Delta_\xi [e^{ix\xi}] \phi (\xi) d\xi \\ &= \int_{\mathbb R}e^{ix\xi} \phi (\xi)d\xi - \int_{\mathbb R}\Delta_\xi [e^{ix\xi}] \phi (\xi) d\xi \\ \end{align}

Now since $\phi$ has compact support, integrating twice by parts w.r.t. $\xi$ yields \begin{align} \int_{\mathbb R}\Delta_\xi [e^{ix\xi}] \phi (\xi) d\xi = \int_{\mathbb R}e^{ix\xi}\Delta_\xi [ \phi (\xi)] d\xi \end{align} and thus \begin{align} \int_{\mathbb R}e^{ix\xi} \phi (\xi)d\xi - \int_{\mathbb R}\Delta_\xi [e^{ix\xi}] \phi (\xi) d\xi &= \int_{\mathbb R}e^{ix\xi} \phi (\xi)d\xi - \int_{\mathbb R}e^{ix\xi}\Delta_\xi [ \phi (\xi)] d\xi \\ &= \int_{\mathbb R}e^{ix\xi} \phi (\xi)-e^{ix\xi}\Delta_\xi [ \phi (\xi)] d\xi \\ &= \int_{\mathbb R}e^{ix\xi} (I-\Delta_\xi) [ \phi (\xi)] d\xi \end{align} We need the compact support and smoothness of $\phi$ when we integrate by parts, otherwise boundary terms during the integration would remain.

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