$G$ 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.



$Hom_G(C_c^{\infty}(G),\pi)\cong Hom_{\mathbb{C}}(\pi^{\vee},\mathbb{C}) ?$
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. 


