# Can all linear operators be represented as functions?

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?

• You should impose some conditions on $f$ (or equivalently: on the domain of your linear operator) and moreover on the co-domain otherwise there are bound to be some stupid pathological cases that prove that the answer is no while not actually answering your question. Nov 30, 2020 at 20:03
• But for linear operators from sufficiently nice spaces to sufficiently nice spaces this is a really interesting question! Nov 30, 2020 at 20:04

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.