$\DeclareMathOperator\Ind{Ind}$Let $G$ be a reductive group over a $p$-adic local field $F$, and $P=MN$ a parabolic subgroup.

Let $\sigma$ be an irreducible representation of $M(F)$ and consider its unnormalized induced representation $\Ind_P^G(\sigma)$. Let $\pi$ be a subrepresentation of $\Ind_P^G(\sigma)$. For an arbitrary element $v$ in $\sigma$, we can choose $f \in \pi$ such that $f(e)=v$.

Let $U^-$ be the unipotent radical of the opposite parabolic subgroup $P^-$ of $P$. Then I am wondering whether we can choose $U^-$-invariant $f \in \pi$ such that $f(e)=v$? If $\pi$ is the full induced representation, it is possible because $P \cap U^- = \{e\}$. But I don't know whether it holds for proper sub-representation of $\Ind_P^G(\sigma)$.

Furthermore, I am also wondering whether we can choose $f\in \pi$ so that $f(e)=v$ and $f$ has small support near $e$. This also holds when $\pi$ is the full induced representation. But I don't know whether it is possible for proper sub-representation $\pi$.

Thank you very much!

(PS: I am especially think of the case when $\pi$ is the image of the local intertwining operator $M(s) : \Ind_P^G(\sigma\cdot |\det|^s) \to \Ind_P^G( \sigma\cdot |\det|^{-s})$ defined by $M(s)f(s)(g)= \int_{N}f(wng)dn$, where $w$ is the longest Weyl element. If it is difficult to consider general $\pi$, how about the case when $\pi=M(s)(\Ind_P^G(\sigma\cdot |\det|^s) )$?)