Given a $p$-adic reductive group $G$ with Grothendieck group $R(G)$ and $f$ an element of the Hecke Algebra $H(G)$ we can consider the function $x: R(G) \to \mathbb{C}$ given by $\pi \mapsto trace \pi(f)$ (for $f \in H(G)$). It is stated 


(condition (i) in 1.2 here, for instance: <https://publications.ias.edu/sites/default/files/Number55.pdf>)


that $x$ restricted to the Bernstein variety of $G$ is a regular function. This should be easy, but I would appreciate an explanation of why this is true.



Why are Trace characters regular functions on the Bernstein Variety?
As in [my comment](https://mathoverflow.net/questions/262136/why-are-trace-characters-regular-functions-on-the-bernstein-variety#comment646829_262136), Theorem 2 of [van Dijk - Computation of certain induced characters of $p$-adic groups](http://link.springer.com/article/10.1007%2FBF01429876) ([MR](http://www.ams.org/mathscinet-getitem?mr=338277)) says, with the notation of that result, that
\begin{align\*}
\Theta\_{\mathrm{ind}\_P^G(\rho\chi)}(f)
& = \int\_M \chi(m)\int\_K \int\_N f(k m n k^{-1})\delta\_P(m)^{1/2}\theta\_\rho(m)\mathrm dn\,\mathrm dk\,\mathrm dm \\
& = \sum\_{m \in M/K \cap M} \chi(m)\operatorname{meas}(K \cap M)\int\_K \int\_N f(k m n k^{-1})\delta\_P(m)^{1/2}\theta\_\rho(m)\mathrm dn\,\mathrm dk.
\end{align\*}
(Actually the stated integral is in a different order, but each integral is compactly supported, so the interchange is OK.) The sum is finite because $f$ is compactly supported.


