# Represent the generalised function by the function

Consider a function $a(x,y,\xi) \in C^{\infty}(\mathbb{R}^n \times \mathbb{R}^n \times \mathbb{R}^n)$ such that $a(x,y,\xi) \leqslant C(1 + |\xi|)^m$ and its derivatives satisfy the same inequality with $m = m - order\;of\;derivative$ for some fixed real $m > 0$. Let $a(x,y,\xi) = \mathcal{F}A^{\#}(x,y,z)$, where $\mathcal{F}$ is the Fourier transform. Let $L$ be a differential operator such that $L^{T}e^{ix\xi} = e^{ix \xi}$, where $L^{T}$ is its transposed differential operator (such operator that appears after use of the differentiation by parts, i.e. $\int L[u] v d\xi = \int u L^{T}[v] d\xi$ ). Obviously, $A(x,y) = A^{\#}(x,y,x-y)$ is the distribution on $\mathbb{R}^n \times \mathbb{R}^n$. How to show than that $$\langle A, h \rangle = (2\pi)^{-n} \int\int\int e^{i(x-y)\xi} L^{k}[a](x,y,\xi) h(x,y) \; d\xi dxdy$$ I have a problem with it because $$A(x,y) = \int e^{i(x-y)\xi} a(x,y,\xi) d\xi$$ diverges and hence the correct functional representation for $A(x,y)$ is $$A(x,y) = \int e^{i(x-y)\xi} L^{k}[a](x,y,\xi)d\xi$$ So my question is why this representation holds (we can't integrate by parts, because the function doesn't vanish at infinity).

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