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.


After some thought, I decided to see what happens in the Fourier case. Since $j_{\alpha}$ is an eigenfunction of $D_{\alpha}$, it stood to reason that we should look to the following PDE in the Fourier case:

$$ \frac{\partial}{\partial x}u(x,y) = \frac{\partial}{\partial y}u(x,y) \tag{1}$$

since the exponential is an eigenfunction of the derivative operator and subject to the "initial" condition (if we view this as an evolution equation) $u(x,0) = f(x)$. The method of characteristics gives us that $u(x,y) = g(x+y)$ for some $g$. Based on our initial condition, it must be the case that $u(x,y) = f(x+y)$ and so the usual displacement operator naturally falls out of these considerations. We could also Fourier transform $(1)$ and again the same result pops out. This approach seems more promising for the case of Bessel translation operator since the method of characteristics isn't applicable.

This is a pretty interesting way to view displacement operators I feel, but very surprising. I would have naturally expected the Bessel translation operator should be defined by $j_{\alpha}(D_{\alpha})$ (or something quite similar) as a parallel to the Fourier displacement operator which can be viewed as $\exp(D)$. However after some thought, I can understand why the PDE approach is more desirable than considering $j_{\alpha}(D_{\alpha})$. The analysis of this operator is hardly trivial and very technical, whereas the PDE definition is very natural and easily analyzed (by comparison).

  1. The generalized translation operator helps to move singularity for PDE from zero to any value.
  2. 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.
  3. 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.
  4. Google gives 9 480 000 references for "generalized translation operator " -enough to satisfy any curiosity and questions, not so?
  • $\begingroup$ I'm well aware with why we come up with generalized translation. It allows for a generalization of a convolution theorem, positive-definite functions, reproducing kernel Hilbert space related ideas, etc. That was not my question. My question was why we chose that particular definition but in light of my answer above, it is quite natural and reasonable. Initially one might have been tempted to do something like $j_{\alpha}(D_{\alpha})$ which I think is what Delsarte does but that operator is quite difficult to analyze; whereas the translation defined by Levitan is easy to analyze. $\endgroup$ – Cameron Williams Apr 2 '15 at 23:10
  • $\begingroup$ It is also easy to generalize in this particular PDE form, just change a pair of differential operators $(B_\alpha,D^2)$ to another pair. $\endgroup$ – Sergei Apr 3 '15 at 4:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.