I am reading some lecture notes about semiparametric statistics. We are in the context of determining some basic properties about the efficient influence function, here denoted by $\tilde{\psi}_P$ where $P$ is a given probability measure. They define differentiability of a function of an infinite dimensional object $\eta \in H$ where $H$ is an Hilbert space as follows:
Definition. A map $\psi: \mathcal{P} \rightarrow \mathbb{B}$ is differentiable at $P$ relative to a given tangent set $\dot{\mathcal{P}}_P$ if there exists a continuous linear map $\dot{\psi}_P: L_2(P) \rightarrow \mathbb{B}$ such that for every $g \in \dot{\mathcal{P}}_P$ and a submodel $t \rightarrow P_t$ with score function $g$, as $t \rightarrow 0$ $$ \frac{\psi\left(P_t\right)-\psi(P)}{t} \rightarrow \dot{\psi}_P g $$
Shortly after they derive the following statement, where they take supremum over $\operatorname{lin} \dot{\mathcal{P}}_P$.
Lemma. Suppose that the functional $\psi: \mathcal{P} \rightarrow \mathbb{R}$ is differentiable at $P$ relative to the tangent set $\dot{\mathcal{P}}_P$. Then $$ \sup _{g \in \operatorname{lin} \dot{\mathcal{P}}_P} \frac{\left\langle\tilde{\psi}_P, g\right\rangle_P^2}{\langle g, g\rangle_P}=P \tilde{\psi}_P^2 $$
My question is: why taking supremum over $\operatorname{lin} \dot{\mathcal{P}}_P$ if actually the scores are in the tangent space which is taken to be a vector space?
The thing is that the tangent set is formally taken as a subset of $L^2(P)$ and it is not necessarily a linear space. By taking the closure of the linear span of this set you are sure to have a closed linear space of an Hilbert space, which is then itself Hilbert. If you don’t take the closure, by taking the linear span of the tangent set you are ensuring to have at least a linear space.