I have been self studying the classical Umbral Calculus and have been reading works and papers from Rota and Roman on the material and I have a question regarding the following. The text uses $$\langle L \mid p(x) \rangle$$ to denote the action of a linear functional $L$ on a polynomial $p(x)$. $P*$ denotes the vector space of all linear functionals on $P$, the space of polynomials. Vector space operations on $P*$ are $$\langle L+M \mid p(x) \rangle=\langle L \mid p(x) \rangle+\langle M \mid p(x) \rangle$$ $$\langle cL \mid p(x) \rangle=c\langle L \mid p(x) \rangle$$ Now, $F$ is the algebra of formal power series in $t$ over $\mathbb{C}$. The formal power series $$f(t)=\sum_{k=0}^\infty \frac{a_k}{k!}t^k$$ defines a linear functional on $P$ by setting $$\langle f(t)\mid x^n\rangle=a_n$$ for all $n\ge 0$. As a consequence, $$\langle t^k\mid x^n \rangle = n!\delta_{n,k}$$ Finally we know that since if $$f_L(t)\sum_{k=0}^\infty \frac{\langle L\mid x^k\rangle}{k!}t^k$$ then it is easy to show that $L=f_L(t)$ and using the previous vector space definitions, the map $L\rightarrow f_L(t)$ is a vector space isomorphism from $P*$ to $F$.
Now, to my question, the evaluation functional is the power series $e^{yt}$, since $$\langle e^{yt} \mid x^n \rangle=y^n$$
Then the text states
$$\langle e^{yt} \mid p(x)\rangle =p(y)$$
This is the first time in the text that shows the linearity argument for the action's argument. From here, I can assume that if $p(x)=c_nx^n+c_{n-1}x^{n-1}+...+c_1x+c_0$, then
$$\langle f(t) \mid p(x) \rangle =\langle f(t) \mid c_nx^n+c_{n-1}x^{n-1}+...+c_1x+c_0\rangle$$ $$=\langle f(t) \mid c_nx^n \rangle+\langle f(t) \mid c_{n-1}x^{n-1} \rangle+...+\langle f(t) \mid c_1x \rangle+\langle f(t) \mid c_0 \rangle$$ $$=c_n\langle f(t) \mid x^n \rangle+c_{n-1}\langle f(t) \mid x^{n-1} \rangle+...+c_1\langle f(t) \mid x \rangle+c_0\langle f(t) \mid 1 \rangle$$
This linearity property was never explictly defined or shown, but I have to assume it is true because, by substituting in $f(t)=e^{yt}$, we do get $p(y)$ as our result. This would seem to imply that this action is a bilinear functional and seems to mimic the bracket notation used by Halmos in Finite Dimensional Vector Spaces. Is this an accurate assessment on my part? And how would I have determined that this linearity existed for the argument $p(x)$ prior to the revelation by the evaluation functional $e^{yt}$?