Spherical transform

Remarks

For a convolution $f_1 * f_2$ of two such functions, $\widehat{f_1 * f_2} = \hat f_1 \hat f_2$, and each map $f \mapsto \hat f(\varphi)$ is a character of the algebra $L^1(K \backslash G / K)$. Every nonzero character of this algebra arises in this way from exactly one $\varphi \in \Phi$, so the spherical transform is the Gelfand transform of the algebra.

For a connected noncompact semisimple Lie group $G$ with finite centre and maximal compact subgroup $K$, the spherical functions are the $\varphi_\lambda$ of Harish-Chandra's formula, and $\varphi_\lambda(g^{-1}) = \varphi_{-\lambda}(g)$. In this case one writes $\hat f(\lambda) = \int_G f(g) \varphi_{-\lambda}(g) \, dg$ for $\lambda \in \mathfrak a^*_\C$, a $W$-invariant function of $\lambda$.

Example: for $G = \R$ and $K = \set{0}$, the bounded spherical functions are $x \mapsto e^{i \xi x}$ with $\xi \in \R$, and $\hat f(\xi) = \int_\R f(x) e^{-i \xi x} \, dx$ is the Fourier transform.

Non-example: for the same $(G, K)$, the function $x \mapsto e^{x}$ is a spherical function, but it is unbounded, so it is not in $\Phi$. Indeed the integral $\int_\R f(x) e^{-x} \, dx$ need not converge for $f \in L^1(\R)$.