Assume that $f$ is a nice function from $\mathbb R^{n}$ to $\mathbb R$ (say smooth), with compact support inside $B_{R}(0)$, for some $R>0$, say large. One may take its Fourier transform $\hat{f}(z) = \int_{x\in\mathbb R^{n}} f(x)e^{2\pi i \langle x,z\rangle}dx$, where the integral converges as $f\in L^{1}(\mathbb{R}^n)$. Under appropriate assumptions, one may recover $f(x)$ by means an inverse Fourier transform by $f(x) = \int_{z\in\mathbb{R}^n} \hat{f}(z) e^{-2\pi i \langle x,z\rangle}dz$.
In particular, one might be interested in calculating say the integral of $f(x)$ over the ball say (or any other type of convolutional operator).
Writing the convolution operator as $Af(x) = \int f(x+y)dK(y)$ for some kernel $K$, we have $Af(x) = \int_{z\in \mathbb{R}^{n}}\hat{f}(z)\int e^{-2\pi i \langle x+y,z\rangle}dy dz = \int_{z\in\mathbb R^n}\hat{f}(z)e^{-2\pi i \langle x,z\rangle}\hat{K}(z) dz$ where $\hat{K}(z) = \int_{y} e^{-2\pi i \langle y,z\rangle}dK(y)$.
For the case of the integral over the ball, one recovers some Bessel function as $\hat{K}(z) = \omega(1)\cdot \frac{J_{n/2}(1\cdot \lVert z\rVert)}{(1\cdot \lVert z\rVert)^{n/2}}$, where $\omega(1)$ is the volume of the unit ball in $\mathbb{R}^n$.
My question is - how one shows that the right hand side (namely the Fourier representation) indeed equal to the physical $Af(x)$.
My main issue is that I don't manage to prove the convergence of the later term (the Fourier side expression of $Af(x)$), as the kernel $\hat{K}$ is NOT in $L^{1}$. One may use some rapid decay properties of $\hat{f}$ in order to mitigate that, but even with using that, I cannot seem to recover the value of $Af(x)$.