Question: Let $G$ be a locally compact group and $H$ be a closed subgroup of $G$. given $0 \neq g \in C_c(G/H)$, the support of $g$, $C:= \text{spt}(g) \subset G/H$ is compact. Then, there exists compact $K \subset G$ so that $C \subset \pi(K)$. By Urysohn's lemma, there exists $\phi \in C_c^+(G)$ so that $\phi \geq0, \phi \equiv 1$ on $K$. Define $$f := \frac{g \phi}{\phi^H}$$ in the support of $g$ and $0$ otherwise. Here $$\phi^H(x) :=\int_H \phi(xh) \, dh $$ for any $x \in G$. Is there a way to show that $\phi^H$ (in fact $\phi^H >0$) is non-vanishing in the support of $g$?
Attempt: (Hints/Solns provided in the comments) Let $W := (L_{x^{-1}}\phi)^{-1}(\frac{1}{2}, \frac{3}{2})$ where $L_{x^{-1}}\phi(y) := \phi(xy)$, then $W$ is an open set in $G$. Then, $W \cap H$ is a nonempty open set in $H$ since given any $x \in \text{spt}(g)$, it is clear that $xH \subset \text{spt}(g)$ hence $xH \subset KH$ in $G$ and there exists $h_0 \in H$ so that $xh_0 \in K$. Thus, $L_{x^{-1}}\phi(h_0) = \phi(xh_0) = 1$ so $h_0 \in H \cap W$. With that, $$\phi^H(x) := \int_{H} \phi(xh) \, dh = \int_H L_{x^{-1}}\phi(h) \, dh \geq \int_{W \cap H} L_{x^{-1}}\phi(h) \, dh >0$$ since $\phi\geq0$ and $\mu_H(W \cap H)>0$ where $\mu_H$ is the Haar measure on $H$. Here we use the result that if $U$ is a nonempty open set in $H$ then the measure $\mu_H(U) >0$.
Source: Page 18 of Deitmar et al's Principles of Harmonic Analysis.