TITLE: $Hom_G(C_c^{\infty}(G),\pi)\cong Hom_{\mathbb{C}}(\pi^{\vee},\mathbb{C}) ?$
QUESTION [5 upvotes]: $G$ is an p-adic group, and $\pi$ is an irreducible representation of $G$, then do we naturally have 
$Hom_G(C_c^{\infty}(G),\pi)\cong Hom_{\mathbb{C}}(\pi^{\vee},\mathbb{C})$? I think it is true, but I do not have found the detailed proof.

REPLY [5 votes]: For two smooth representations $\pi_i$, $i=1,2$, of $G$, one has $\mathrm{Hom}_G (\pi_1 ,\pi_2 ) \simeq \mathrm{Hom}_G (\pi_2^\vee ,\pi_1^\vee )$. On the other hand the contragredient of $C_c^\infty (G)$ is $C^\infty (G)$ (the space of smooth functions with arbitrary support), the pairing being given by $\langle f, g \rangle =\int_G fg\, d\mu$, $f\in C_c^\infty (G)$, $g\in C^\infty (G)$ (for some fixed Haar measure $\mu$ on $G$).
Moreover the space $C^\infty (G)$ is the induced representation $\mathrm{Ind}_{\{ 1\}}^G {\mathbb C}$. All together, we get
$$
\mathrm{Hom}_{G}(C_c^\infty (G), \pi )\simeq \mathrm{Hom}_G(\pi^\vee ,  \mathrm{Ind}_{\{ 1\}}^G {\mathbb C}) \simeq \mathrm{Hom}_{\{ 1\}} (\pi^\vee ,{\mathbb C}),
$$
where the last isomorphism follows from Frobenius reciprocity for induction.