TITLE: Motivating the Bessel translation operator
QUESTION [1 upvotes]: In a paper I am reading on the Hankel transform (this paper to be exact), I've come across a somewhat peculiar definition for a generalized translation operator. The operator is designed with a convolution theorem for the Hankel transform in mind. The Hankel transform, as given in the paper, is defined to be (for suitable $f$)
$$\mathcal{F}_{\alpha}f(y) = \int_0^{\infty} j_{\alpha}(2\pi xy)f(x)\,d\mu_{\alpha}(x),$$
where $d\mu_{\alpha}(x) = \frac{2\pi^{\alpha+1}}{\Gamma(\alpha+1)}x^{2\alpha+1}$ and
$$j_{\alpha}(x) = \Gamma(\alpha+1)\sum_{l=0}^{\infty} \frac{(-1)^l}{l!\Gamma(\alpha+l+1)}\left(\frac{x}{2}\right)^{2l}.$$
I understand the philosophy behind defining such a generalized translation however the intuition for how it it should be defined in this case is lost on me. Here's the definition that is given in the paper:

Definition Let $f\in C^2(\mathbb{R}^+)$, define the generalized Bessel translation operator
  $$T^{\alpha}_y f(x) = u(x,y), \quad x,y\in\mathbb{R}^+$$
  where $u(x,y)$ is a solution of the following Cauchy problem
  $$\left(\frac{\partial^2}{\partial x^2} + \frac{2\alpha+1}{x}\frac{\partial}{\partial x}\right)u(x,y) = \left(\frac{\partial^2}{\partial y^2} + \frac{2\alpha+1}{y}\frac{\partial}{\partial y}\right)u(x,y)$$
  with initial conditions $u(x,0) = f(x)$ and $\frac{\partial}{\partial x}u(x,0) = 0$.

With this definition, one can define a generalized convolution:
$$(f\ast_{\alpha} g)(y) = \int_0^{\infty} T^{\alpha}_y f(x)g(x)\,dx$$
for suitable $f$ and $g$. Then we have that $\mathcal{F}_{\alpha}(f\ast_{\alpha}g) = \mathcal{F}_{\alpha}f\cdot\mathcal{F}_{\alpha}g$ as desired.
The function $j_{\alpha}$ is an eigenfunction of the operator $D_{\alpha} = \frac{\partial^2}{\partial x^2}+\frac{2\alpha+1}{x}\frac{\partial}{\partial x}$ (with eigenvalue $-1$) so its inclusion in the definition for the generalized translation is reasonable to me. Especially considering that translation in the Fourier case is nothing more than the exponential of the derivative operator (for which the Fourier kernel is an eigenfunction). However beyond this triviality, it isn't clear to me why we should define generalized translation in such a way.

REPLY [0 votes]: The generalized translation operator helps to move singularity for PDE from zero to any value.
It helps to define generalized positively defined functions and so on, analogous to use of standard shift in classical harmonic analysis. Now it is important for Dunkl operators.
Via this PDE form it was originally defined by J. Delsarte and in books of B.M.Levitan. Levitan construct general theory of generalized translation operators on groups. 
Google gives  9 480 000 references for "generalized translation operator " -enough to satisfy any curiosity and questions, not so?
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