Dunkl operators realize the polynomial standard module

For $f\in \mathbb{C}[{\mathfrak{h}}^{\ast }]$ and $s\in S$, the polynomial $f-sf$ vanishes on $\ker {\alpha }_s$, since $s^{-1}$ fixes every point of that hyperplane. The ideal of this hyperplane is generated by ${\alpha }_s$, so $(1-s)f\mathrm{/}{\alpha }_s$ is a polynomial. Thus every displayed $y$-operator preserves $\mathbb{C}[{\mathfrak{h}}^{\ast }]$.

Let $R_s={\alpha }_s^{-1}(1-s)$ on the rational function field of $\mathfrak{h}$. For multiplication by $x\in {\mathfrak{h}}^{\ast }$, $[R_s,x]={\alpha }_s^{-1}(x-sx)s={\alpha }_s^{\vee }(x)s$ by [F4]. Using $[{\mathrm{\partial }}_y,x]=x(y)$ therefore gives $[{\mathrm{\partial }}_y-{\sum }_s{c}_s{\alpha }_s(y)R_s,x]=x(y)1-{\sum }_s{c}_s{\alpha }_s(y){\alpha }_s^{\vee }(x)s$, exactly [F1]. The $x$'s commute, and conjugation by $w$ sends the operator for $y$ to that for $wy$ because $c$ and the reflection forms are equivariant.

Write $T_y={\mathrm{\partial }}_y-{\sum }_s{c}_s{\alpha }_s(y)R_s$. The commutator $[T_y,T_z]$ is a differential-reflection operator on ${\mathfrak{h}}_{\mathrm{r}\mathrm{e}\mathrm{g}}$, the complement of the reflection hyperplanes. By step 1.2 and Jacobi, ${\mathrm{ad}}_{[T_y,T_z]}(x)=[T_y,\kappa (z,x)]-[T_z,\kappa (y,x)]=[T_y,\kappa (z,x)]+[T_z,\kappa (x,y)]=0$, since $\kappa (y,z)=0$ for $y,z\in \mathfrak{h}$ and [F2] applies. The commutator with a group element uses the same formula as in [F1], by the $W$-covariance just checked.

In the crossed differential-operator algebra $\mathcal{D}({\mathfrak{h}}_{\mathrm{r}\mathrm{e}\mathrm{g}})⋊W={⨁}_{w\in W}\mathcal{D}({\mathfrak{h}}_{\mathrm{r}\mathrm{e}\mathrm{g}})w$, the centralizer of all multiplication operators by $x\in {\mathfrak{h}}^{\ast }$ consists of multiplication by rational functions. Indeed, for a nonidentity $w$, choose $x$ with $wx\mathrm{\ne }x$; the top differential-order term of $[Dw,x]$ is the top term of $D$ multiplied by $wx-x$, so commutation forces $D=0$. For $w=1$, a differential operator commuting with every coordinate function has order zero, since the commutator of a positive-order operator with a coordinate has nonzero top symbol in some coordinate direction. Thus [T_y,T_z], which commutes with all $x$ by step 2.1, is multiplication by a rational function. Every $T_y$ kills $1$, so $[T_y,T_z]1=0$, and that multiplication function is zero. Hence the $T_y$ commute.

The assignments $x\mapsto m_x$, $w\mapsto w$, and $y\mapsto T_y$ now satisfy every relation in [F1]: the $x$'s commute, the $y$'s commute by step 3.1, the mixed commutator is step 1.2, and the $W$-action relation is the covariance in step 1.2. The quotient universal property therefore gives an $H_{1,c}$-module structure on $\mathbb{C}[{\mathfrak{h}}^{\ast }]$.

The constant polynomial is killed by every $T_y$, and $W$ fixes it, so induction gives an $H_{1,c}$-module map $P_c={\mathrm{\Delta }}_c(\mathbf{1})\to \mathbb{C}[{\mathfrak{h}}^{\ast }]$ sending $1\otimes 1$ to $1$. It is surjective because multiplication by ${\mathfrak{h}}^{\ast }$ generates all polynomials. By PBW [F3], both spaces have the same monomial basis and this map is the identity on that basis, hence is an isomorphism. This is the asserted polynomial representation and action.