Kirillov model: integral vanishes for some $n$, or for all large $n$? Let $G = \operatorname{GL}_2(F)$ for a $p$-adic field $F$, and $(\pi,V)$ an irreducible, admissible representation of $G$.
Let $\tau$ be a continuous character of $F$.  For $n \geq 1$ and $\xi \in V$, consider the vector valued integral
$$\int\limits_{\mathfrak p^{-n}} \overline{\tau(x)} \pi \begin{pmatrix} 1 & x \\ & 1 \end{pmatrix} \xi \space dx $$
The integrand is a locally constant function on the compact group $\mathfrak p^{-n}$, taking values in $V$.   It takes only finitely many values, so we may interpret the integral as a finite sum.  Now, let 
$$V_0 = \{ \xi \in V :  \int\limits_{\mathfrak p^{-n}} \overline{\tau(x)} \pi \begin{pmatrix} 1 & x \\ & 1 \end{pmatrix} \xi \space dx = 0  \textrm{ for } n \textrm{ sufficiently large }  \}$$
This is a space defined in Roger Godement's notes on Jacquet-Langlands theory.  I am confused on a couple of points:
1 .  In general, how are we to interpret the integral $\int\limits_{\mathfrak p^{-n}} \overline{\tau(x)} \pi \begin{pmatrix} 1 & x \\ & 1 \end{pmatrix} \xi \space dx \in V$ as $n$ becomes larger and larger?  Is this a principal value integral, i.e. one whose value in $V$ becomes constant once $n$ reaches a sufficiently large value?  (that large value depending of course on $\pi, \tau,$ and $\xi$)
2 .   If $\int\limits_{\mathfrak p^{-n}} \overline{\tau(x)} \pi \begin{pmatrix} 1 & x \\ & 1 \end{pmatrix} \xi \space dx = 0 $ for some $n$, does it follow that $\int\limits_{\mathfrak p^{-n}} \overline{\tau(x)} \pi \begin{pmatrix} 1 & x \\ & 1 \end{pmatrix} \xi \space dx = 0$ for all $n' \geq n$?
 A: For (2): the answer is yes and this is Jacquet-Langlands Prop 2.8.: It follows from the factorization of the Haar integral: We have $\textit{Avg}: C_{c}^{\infty}(G) \to C_{c}^{\infty}(G/H)$ surjective with $f \mapsto \textit{Avg} (f)=\overline{f} : x \mapsto \int\limits_{H} f(xh)dh $, and when the measures play nice (certainly in the abelian case) $\int\limits_{G} f dg = \int\limits_{G/H} \overline{f}d\overline{g}$.  
So with $G=\mathfrak{p}^{-m}$, $H=\mathfrak{p}^{-n} $, the integral over $\mathfrak{p}^{-m}$ is a finite sum of $\overline{\tau(y)} \pi (\begin{pmatrix} 1 & y \\ & 1 \end{pmatrix})$-translates of the integral over $\mathfrak p^{-n}$ where the $y$ are coset representatives of $\mathfrak p^{-m}/\mathfrak p^{-n}$:
$$ 
\int\limits_{\mathfrak p^{-m}/{\mathfrak{p}^{-n}}} \int\limits_{\mathfrak{p}^{-n}}  \overline{\tau(y+x)} \pi (\begin{pmatrix} 1 & y+x \\ & 1 \end{pmatrix}) \xi \space dxdy \space= \\ 
\int\limits_{\mathfrak p^{-m}/{\mathfrak{p}^{-n}}} \overline{\tau}(y) \pi (\begin{pmatrix} 1 & y \\ & 1 \end{pmatrix})   \int\limits_{\mathfrak{p}^{-n}}  \overline{\tau(x)} \pi (\begin{pmatrix} 1 & x \\ & 1 \end{pmatrix}) \xi \space dx dy = \\ \sum\limits_{\textrm{finite}} \overline{\tau}(y) \pi (\begin{pmatrix} 1 & y \\ & 1 \end{pmatrix}) \int\limits_{\mathfrak{p}^{-n}}  \overline{\tau(x)} \pi (\begin{pmatrix} 1 & x \\ & 1 \end{pmatrix}) \xi \space dx $$
which is zero if the original integral is. 
I'm not sure about (1).
