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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?

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  • $\begingroup$ 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. $\endgroup$
    – Vincent
    Nov 30, 2020 at 20:03
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    $\begingroup$ But for linear operators from sufficiently nice spaces to sufficiently nice spaces this is a really interesting question! $\endgroup$
    – Vincent
    Nov 30, 2020 at 20:04

1 Answer 1

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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.

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