In traditional Fourier theory, the Dirac delta plays the role of an "identity" for the $L^1$ algebra with respect to the usual convolution. The convolution is traditionally built out of group structures by translating by group elements. The Fourier-Bessel transform of $f\in L^1(\mathbb{R}^+)$ is given by
$$\mathcal{F}_{\nu}f(y) = \int_{\mathbb{R}^+} j_{\nu}(2\pi xy)f(x)\,d\mu_{\nu}(x),$$
where $-\frac{1}{2} < \nu$, $d\mu_{\nu}(x) = \frac{2\pi^{\nu+1}}{\Gamma(\nu+1)} x^{2\nu+1}\,dx$, and $j_{\nu}(x) = 2^{\nu} \Gamma(\nu+1)x^{-\nu}J_{\nu}(x)$. (Here $J_{\nu}$ is the Bessel function of the first kind.)
The Fourier-Bessel transform has a notion of translation associated to it which is not generated by a group; instead, the translation is given by
$$T_y^{\nu} f(x) = \frac{\Gamma(\nu+1)}{\sqrt{\pi}\Gamma\left(\nu+\frac{1}{2}\right)} \int_{-1}^1 f\left(\sqrt{x^2-2\alpha xy + y^2}\right)(1-\alpha^2)^{\nu-\frac{1}{2}} \,d\nu.$$
With this generalized translation, a generalized convolution can be defined:
$$ (f\ast_{\nu} g)(y) = \int_{\mathbb{R}^+} T_y^{\nu}f(x)g(x)\,d\mu_{\nu}(x).$$
The generalized translation and convolution share many properties with the usual (Fourier) translation and convolution operators:
$$ T_y^{\nu} (j_{\nu}(\lambda x)) = j_{\nu}(\lambda x) j_{\nu}(\lambda y) \tag{1} $$
$$ \mathcal{F}_{\nu}(T_y^{\nu}f)(x) = j_{\nu}(2\pi xy) \mathcal{F}_{\nu}f(x) \tag{2} $$
$$ \mathcal{F}_{\nu}(f\ast_{\nu} g) = \mathcal{F}_{\nu}f \cdot \mathcal{F}_{\nu} g \tag{3} $$
$$ f\ast_{\nu} g = g\ast_{\nu} f \tag{4}$$
The first is analogous to the property that $e^{-i\omega(t-t')} = e^{-i\omega t}e^{-i\omega t'}$. The second is analogous to the property that $\mathcal{F}(f(\cdot - t)) = e^{-i\omega t}\mathcal{F}f$. The third is analogous to the usual convolution theorem. The last is a statement of the commutativity of convolution.
With the numerous analogues, it is natural to consider what plays the role of the Dirac delta for the generalized convolution for the Fourier-Bessel transform.
The usual Dirac delta is not an identity for the generalized convolution. To see this, let's see what the generalized translation of the Dirac delta is.
$$T_y^{\nu}\delta(x) = \frac{\Gamma(\nu+1)}{\sqrt{\pi} \Gamma \left(\nu+ \frac{1}{2}\right)}\int_{-1}^1 \delta\left(\sqrt{x^2-2\alpha xy+y^2} \right)(1-\alpha^2)^{\nu-\frac{1}{2}}\,d\alpha.$$
(This is of course meant as a distributional integral.) The Dirac delta has the following property:
$$\delta(g(z)) = \sum_{z_0}\frac{\delta(z-z_0)}{g'(z_0)},$$
where the $z_0$ are the zeroes of $g$. Here $g(\alpha) = \sqrt{x^2-2\alpha xy+y^2}$. $g(\alpha) = 0$ when $\alpha = \frac{x^2+y^2}{2xy}$. Moreover, $g'(\alpha) = -\frac{xy}{g(\alpha)}$ and so
$$\delta\left(\sqrt{x^2-2\alpha xy+y^2}\right) = -\frac{g\left(\frac{x^2 +y^2}{2xy}\right)}{xy}\delta\left(\alpha - \frac{x^2+y^2}{2xy}\right) = 0.$$
The generalized translation of the Dirac delta actually vanishes! This is more or less because $\frac{1}{g'} \propto g$.
With standard translation operators, by definition we have
$$\delta \ast f = \langle \delta_y,f\rangle = f(y),$$
Replacing this with the generalized translation, this would read
$$\delta\ast_{\nu} f = \langle T_y^{\nu}\delta, f\rangle \stackrel{?}{=} f(y).$$
However the above analysis shows that $\langle T_y^{\nu}\delta, f\rangle = 0$ since $T_y^{\nu}\delta$ is zero and so the Dirac delta is not the identity with regards to the generalized convolution. Since the usual Dirac delta is not an identity with regards to convolution, my question is: what is? I've thought about this for a while but nothing jumps out at me. I would think it's some sort of variation on the Dirac delta but it's not clear at all due to the nonlinear nature (in terms of coordinates) of the generalized translation.
Much of this is based on this paper by Ghobber and Jaming.