Let $G$ be a quasi-split connected reductive group over a $p$-adic field $F$. Let $B$ be a Borel subgroup which is defined over $F$, with $B = TU$, $T$ defined over $F$. The choice of $T$ and $B$ gives a set of nonrestricted roots $\tilde{\Delta}$, which together with an $F$-splitting and a nontrivial unitary character $F \rightarrow S^1$ yields a unitary character $\chi$ of $U(F)$.

Let $V$ be a smooth, irreducible, admissible representation of $G(F)$. A linear functional $\lambda: V \rightarrow \mathbb{C}$ is called a **Whittaker functional** for $\chi$ if for all $u \in U(F)$ and $v \in V$, we have

$$\lambda(u \cdot v) = \chi(u) \lambda(v)$$

It is a theorem that the space of Whittaker functionals for $\chi$ is at most one dimensional. If this dimension is one, then $V$ is called **$\chi$-generic**.

Fix a nonzero Whittaker functional $\lambda$, and for $v \in V$, define a function $W_v :G(F) \rightarrow \mathbb{C}$ by $W_v(g) = \lambda(g \cdot v)$. The set $W = W_{\lambda}$ of such functions is closed under addition and scalar multiplication, and becomes a representation of $G(F)$ if we set $g \cdot W_v = W_{g\cdot v}$. Then the representation $W$ is called a **Whittaker model**, and up to $G(F)$-isomorphism it does not depend on the choice of $\lambda$, since these things are all scalar multiples of each other.

From what I can see, $V \rightarrow W_{\lambda}, v \mapsto W_v$, is an isomorphism of $G(F)$-modules. It is clearly surjective, and for injectivity, if we suppose that $0 \neq v \in V$, and $W_v(g) = 0$ for all $g \in G(F)$, then $\lambda(g \cdot v) = 0$ for all $g \in G(F)$. This is impossible, because $g \cdot v : g \in G(F)$ spans $V$, because $V$ is irreducible.

So my question is, what is the point of defining Whittaker models if the thing we defined is just isomorphic to our original representation? Why would it be important to regard $V$ as a set of functions $G(F) \rightarrow \mathbb{C}$?