I've two definitions for an operator but I can't figure out if they are equivalent (or if one implies the other). We are on $\mathbb{R}$, let $A_g$ be the operator where $g$ is a function (I'm not asking for any regularity, if you need something you can assume it). The first definition is that $A_g$ is the operator for which \begin{align} \label{def1} \int_{-\infty}^\infty e^{-i\xi x} A_g f(x) \,dx=g(i\xi) \tilde{f}(\xi), \quad \quad (1) \end{align} where $\tilde{f}(\xi)= \int_{-\infty}^\infty e^{-i\xi x} f(x) \,dx$. So I know the Fourier transform of $A_g$.
Now I have an other definition for $A_g$, which is actually more an application: \begin{align} \label{def2} A_g e^{-kt}=g(k)e^{-kt} \quad t>0, \quad \quad (2) \end{align} where $k$ could be a parameter or also a function (not in $t$). Are these two consistent?
An example: $g(x)=x^2$, trivially \begin{align*} \partial_t^2 e^{-kt}=(-k)^2 e^{-kt}= k^2 e^{-kt}= g(k) e^{-kt} \end{align*} and \begin{align*} \int_{-\infty}^\infty e^{-i\xi x} \partial_x^2 f(x) \,dx=(i\xi)^2 \tilde{f}(\xi)=g(i\xi)\tilde{f}(\xi) \end{align*} Then in this case the two definitions are consistent, but \begin{align*} \int_0^\infty e^{-i\xi t} \partial_t^2 e^{-kt}\, dt \ne (i\xi)^2 \frac{1}{k+i\xi}. \end{align*} So, if we know only (1) how can I find (2)? Or viceversa if I know (2) how can I find (1)? If one of the two is true also the other is true?
The problem is that Fourier transform and Fourier transform on half-line (or the Laplace transform) are not equivalent, but I would like to understand how the operator defined in (1) acts on $e^{-kt}$ for $t>0$. Can someone help me? Thanks to everybody, each comment can be valuable