Fourier transform of a linear operator On a paper I am studying, they consider the following general PDE:
$ \partial_t h = \mathcal{L}\{h\} + \mathcal{N}\{h\}\, ,$
where $\mathcal{L}$ and $\mathcal{N}$ represent, respectively, a linear and a nonlinear operator. Then they say: "We take the Fourier transform, which gives the following equation for a given mode $q$ in Fourier space:"
$ \partial_t \hat{h}_q(t) = \hat{\mathcal{L}}_q\hat{h}_q(t) + \hat{\mathcal{N}}\{\hat{h}_q(t)\}\,.$
Now, what catches my attention is that the transformed linear operator $\hat{\mathcal{L}}_q$ is not acting as an operator anymore, but it just a function of $q$ multiplied for $\hat{h}_q(t)$, the Fourier transform of $h(x,t)$.
That does not surprise me completely, since, for example, if we take the linear operator to be a second derivative $\mathcal{L}\{h\} = \partial_{xx}h(x,t)$, then, in fact, the Fourier transform will be $-q^2 \hat{h}_q(t)$. The same holds for any order derivative.
My question is: how can I intuitively understand that this is true for any linear operator? I don't need a formal proof, I just want to understand it.
Thank you very much to anyone will help!
 A: This is not true for any linear operator. Consider for example the linear operator:
$$\mathcal{L}: u \mapsto a(x) \cdot u$$
where $a$ is any sufficiently regular function.
Then:
$$\widehat{\mathcal{L} u} = \widehat{a} \star \widehat{u} $$
and for a general $a$, $\widehat{\mathcal{L} u}_q$ does not depend only on $\widehat{u}_q$.
A: $\newcommand{\Lcal}{{\mathcal L}}$
Imagine that $\Lcal$ is a matrix with eigenvectors $e_q$: 
$$
\Lcal e_q = \hat\Lcal_q e_q,$$
where $\hat\Lcal_q$ is just a number. Decompose the column vector $h$ in coordinates with respect to $e_q$: 
$$h=\sum_q \hat h_q e_q.$$ 
Apply $\Lcal$ and use linearity: 
$$
\Lcal h = \sum_q \hat\Lcal_q \hat h_q e_q.$$ 
So, the projection of $\Lcal h$ onto the $q$-th vector (AKA the $q$-th mode of $\Lcal h$) is $\hat\Lcal_q \hat h_q$. 
Now replace $e_q$ with $\exp[i( x\cdot q)]$ and every sum with an integral. As noted in Delta-u's answer, if $\mathcal L$ is a translation invariant operator, then $\exp[i(x\cdot q)]$ is an eigenfunction of $\mathcal L$ and we can apply the same procedure as above.
