$\int_{\mathbb R}(I-\triangle_{\xi})[e^{ix\xi}] \phi (\xi) d\xi = \int_{\mathbb R}e^{ix\xi} (I-\triangle_{\xi})[\phi (\xi)] d\xi$? 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 ?  

 A: 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.
