In the book Automorphic Forms representations and L-functions(Corvallis), Borel and Jacquet wrote a document with the title "Automorphic Forms and automorphic Representations". Proposition 4.5 reads as follows.
${\bf Proposition\, 4.5}$ Let $f$ be a smooth function on $G(A)$ satisfying 4.2 (a),(b),(d). Then the following conditions are equivalent:
1) $f$ is an automorphic form.
2) For each infinite place $v\in V,$ the space $f \ast \mathcal{H}_v$ is an admissible $\mathcal{H}_v$-module.
3) For each place $v\in V,$ the space $f \ast \mathcal{H}_v$ is an admissible $\mathcal{H}_v$-module.
4) The space $f\ast \mathcal{H}$ is an admissible $\mathcal{H}$-module.
The condition 4.2 (a) means that $f$ is invariant on the left under $G(Q)$. Condition 4.2 (b) means that $f*\xi=f$, for some $\xi \in \mathcal{H}_{\infty}$ and d) is the growth condition. So in order to establish that this is an automorphic form we need to verify 4.2 c), namely there exists an ideal $J$ of finite codimension on $\mathcal{Z}(\mathfrak{g})$ such that $f\ast J=0$.
My question: Why admissibility implies the existence of an ideal $J$ as above? Jacquet Borel say it is trivial but it is not trivial for me.
Admissibility (more precisely, admissibility at primes above infinity, i.e. condition 2) in your list) means that $f$ generates an admissible representation under the real points of the group $G$, or, more-or-less equivalently, under the action of the Lie algebra $\mathfrak g$, generates an admissible $(\mathfrak g,K)$-module.
But a finitely generated admissible $(\mathfrak g, K)$-module $M$ is always annihilated by a finite codimension ideal $J \subset \mathfrak Z(\mathfrak g)$, i.e. always has a generalized central character.
This follows from the fact that $M$ is generated (by assumption) by finitely many of its $K$-isotypic pieces, each of which is finite-dimensional. Thus $M$ is generated by a finite-dimensional $\mathfrak Z( \mathfrak g)$-invariant subspace. But the annihilator of such a subspace is actually the sought-after ideal $J$.