Can we correspond any linear operator $L$ to a function $\phi(x)$ such that
$$L f(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty} e^{- i \omega x}\phi(\omega) \int_{-\infty}^{+\infty}f(t)e^{i\omega t}dt \, d\omega$$
?
For instance,
$$f(x)\to f(x): \phi(\omega)=1$$
$$f(x)\to f'(x): \phi(\omega)=-i\omega$$
$$f(x)\to \int f(x) dx: \phi(\omega)=\frac1{-i\omega}$$
$$f(x)\to \Delta f(x): \phi(\omega)=e^{-i\omega}-1$$
$$f(x)\to f(x+1)-f(x-1): \phi(\omega)=-2i\sin \omega$$
etc. Can we find such $\phi(\omega)$ for any linear operator?
If well defined, your linear operator $$L f(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty} e^{- i \omega x}\phi(\omega) \int_{-\infty}^{+\infty}f(t)e^{i\omega t}dt \, d\omega$$ will satisfy
$$L(f(.+a))(x)=L(f)(x+a)$$ this is called a convolution operator. For example $L(f)(x)= f(x+\frac1{1+x^2})$ is not one.
If a convolution operator $T$ sends $L^2(\Bbb{R})\to L^2(\Bbb{R})$ and $\forall f,\|T f\|_2\le C \|f\|_2$ then (the restriction to $L^2(\Bbb{R})$ of) $T$ is given by a $\phi\in L^\infty(\Bbb{R})$. Trying to generalize this is the point of functional analysis and distribution theory.