Let $(V,\pi)$ be an irreducible, admissible, supercuspidal representation of $G = \operatorname{GL}_n(F)$ for $F$ a $p$-adic field. Let $B = TU$ be the usual Borel subgroup, maximal torus, and unipotent radical of $G$. Let $f_{v^{\ast},v}(g) = \langle v^{\ast}, \pi(g)v \rangle$ be a matrix coefficient for $v \in V$ and $v^{\ast} \in V^{\ast \infty}$. Then $f = f_{v^{\ast},v}$ is locally constant and compactly supported modulo the center $Z$ of $G$. If $\chi$ is a generic character of $U$, the integral

$$\int\limits_U f(ug)\chi(u^{-1}) du $$

can be shown to converge absolutely for every $g \in G$. Also, a change of variables shows that for fixed $v^{\ast}$,

$$v \mapsto \int\limits_U f_{v^{\ast},v}(u)\chi(u^{-1})du \tag{1} $$

defines a linear functional $\lambda: V \rightarrow \mathbb{C}$ which satisfies $\lambda(\pi(u_1)v) = \chi(u_1)\lambda(v)$ for all $u_1 \in U$. However, this linear functional might be the zero functional.

In general, $\pi$ is called *generic* if there exists a nonzero linear functional satisfying the property of the previous paragraph for $\chi$. If $\pi$ is generic for $\chi$, it is also generic for every other generic character.

1 . Is every irreducible, admissible supercuspidal representation of $G$ generic?

2 . If $\pi$ is generic, does there exist a smooth linear functional $v^{\ast}$ such that the map (1) is not the zero map? In other words, if there is a nonzero Whittaker functional, can it always be defined by an integral?