This is a followup to my previous question here.
Let $G = \operatorname{GL}_n(F)$ for $F$ a $p$-adic field. Let $Z$ be the center of $G$, and let $f: G \rightarrow \mathbb{C}$ be a function which is locally constant and compact modulo $Z$.
This means that there exists a compact set $\Omega \subseteq G$ such that $\{ g \in G : f(g) \neq 0 \}$ is contained in the product set $Z.\Omega$. Equivalently, the image of the closure of $\{g \in G : f(g) \neq 0 \}$ in $G/Z$ is compact.
Assume also that there exists a continuous homomorphism $\omega: Z \rightarrow \mathbb{C}^{\ast}$ such that $f(zg) = \omega(z)f(g)$ for all $g \in G, z \in Z$.
Let $U$ be the group of upper triangular unipotent matrices in $G$, and define
$$F(g) = \int\limits_U f(ug)du$$
I want to understand why the integral $F$ converges absolutely. Already I have a proof, found in my previous question, based on the fact that a certain homomorphism $C_c^{\infty}(G) \rightarrow \textrm{c-Ind}_Z^G(\omega)$ is surjective. But I don't want to use this fact.
The paper I'm reading claims that the integral converges "since the orbit $UZg$ is closed in $G$." I would like to understand why we know the integral converges on account of this explanation. I would appreciate any hint on how to approach this.
I use your notation. Take $v\in U$ in the support of $u\mapsto f(ug)$, then $vg\in Z\Omega$. So there exists $z\in Z$ such that $zv\in \Omega g^{-1}\cap ZU$, a compact subset of $ZU$. Since $Z\times U\longrightarrow ZU$ is a homeomorphism, the projection ${\rm pr}_U$: $ZU\longrightarrow U$ is continuous. Hence $v\in {\rm pr}_U (\Omega g^{-1})$, a compact subset of $U$.