The Dunkl intertwining operator $V_k$ on $C(\mathbb{R}^d)$ - MathOverflow most recent 30 from http://mathoverflow.net2013-05-18T18:07:41Zhttp://mathoverflow.net/feeds/question/95942http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/95942/the-dunkl-intertwining-operator-v-k-on-c-mathbbrdThe Dunkl intertwining operator $V_k$ on $C(\mathbb{R}^d)$Abdelmajid Khadari2012-05-04T01:15:45Z2012-05-09T00:27:55Z
<p>The Dunkl intertwinig operator $V_k$ on $C(\mathbb{R}^d)$ is defined by:</p>
<p>$$V_k f(x)=\int_{\mathbb{R}^d}f(y)d\mu_x(y),$$</p>
<p>where $d\mu_x$ is a probability measure on $\mathbb{R}^d$ with support in the closed ball $B(0,||x||)$ of center $0$ and radius $||x||$. For all $x\in\mathbb{R}^d$, $z\in\mathbb{C}^d$, we have:</p>
<p>$$K(x,z)=V_k(e^{<.,z>})(x).$$</p>
<p>Let $^tV_k$ the operator on $D(\mathbb{R}^d)$ satisfying for all $f\in D(\mathbb{R}^d)$ and $g\in C(\mathbb{R}^d)$,</p>
<p>$$\int_{\mathbb{R}^d} {^tV_k(f)(y)g(y)dy}=\int_{\mathbb{R}^d}V_k(g)f(x)w_k(x)dx.$$</p>
<p>Then there exists a positive measure $\nu_y$ on $\mathbb{R}^d$ with support in the set $\{ x\in \mathbb{R}^d, ||x||\geq ||y||\} $ for which</p>
<p>$$^tV_k(f)(y)=\int_{\mathbb{R}^d}f(x)d\nu_y(x).$$</p>
<p>This operator $^tV_k$ is called the dual Dunkl intertwining operator.</p>
<p>The operators $V_k$ and $^tV_k$ satisfy the flolowing properties:</p>
<p>For all $f\in \xi(\mathbb{R}^d)$ $T_iV_k(f)=V_k(\frac{\partial}{\partial y_j}f(x))$,
and for all $f\in D(\mathbb{R}^d)$, $^tV_k(T_jf)(y)=\frac{\partial}{\partial y_j} {^tV_k}(f)(y).$</p>
<p>Where $T_j$ the Dunkl operators, $\xi(\mathbb{R}^d)$ the space of $C^{\infty}$-functions on $\mathbb{R}^d$ and $D(\mathbb{R}^d)$ the space of $C^{\infty}$-functions on $\mathbb{R}^d$ with compact support, from this two properties cames the importance of the Dunk intertwining operator.</p>
<p>My question is, there is an explicit formula for Dunk intertwining operator, in general ?
if the answer in no, then there an explicit formula for some special cases ?</p>