For any $x \in X$, define the set $\mathcal{F}(X) = \overline{\operatorname{span} \{ \delta_x : x \in X \}}$ where $\delta_x(f)=f(x)$ for all $f \in$ $\operatorname{Lip}_0(X)$.
The set $\operatorname{Lip}_0(X)$ is the set of all real-valued Lipschitz functions which vanish at $0$.
Note that $\delta_x$ is an evaluation functional on $\operatorname{Lip}_0(X)$.
Suppose that $X$ is a separable Banach space. Let $D$ to be a dense subspace of $X$ and spanned by $\{ x_n\mid n \in \mathbb{N} \}$.
In Theorem $3.1$(page $128$), in the proof, the author mentions the following:
We denote by $R_D$ the linear map from $D$ to $\mathcal{F}(X)$ which satisfies $R_D(x_n)=\phi_n$.
Question: Why such linear map exists and satisfies the equation?