The following problem was taken from Halmos's Finite Dimensional Vector Spaces:
Let $(a_0, a_1, a_2, \ldots)$ be an arbitrary sequence of complex numbers. Let $x \in P$, where $P$ is the vector space of all polynomials over $\mathbb{C}$. Write $x(t) = \sum_{i=0}^n \xi_it^i$ and $y(x) = \sum_{i=0}^n \xi_i a_i$. Prove that $y(x)$ is an element of the dual space P' consisting of all linear functionals on $P$ and that every element of P' can be obtained in this manner by a suitable choice of the $a_i$.
Now the first part of the problem is not hard to show as $y$ is the functional that takes some polynomial in the vector space, evaluates it at $1$ and to each $\xi_i$ multiplies an $a_i$, then sums all these numbers together. Then it only remains to check that for such a $y$,
$y(Ax_1 + bx_2) = Ay(x_1) + By(x_2)$, where $x_1,x_2 \in P$ and $A,B \in \mathbb{C}$.
Now for the second part, I don't know how to prove that every linear functional in the dual space must have the form of $y$ above. I can think of specific examples e.g. integrals of polynomials and see why this is true, but it is plain that this will not suffice.
There is a more promising approach I can think of. This may not be correct but if we view the \xi_i's as some "basis" (i.e. prove they are linearly independent) and somehow the linear combination represented by $y(x)$ "spans" the dual space then this may suffice.
Perhaps even related is the question If $y$ is a non-zero functional on a vector space $V$ and if $\alpha$ is an arbitrary scalar, does there necessarily exist a vector $x \in V$ such that $y(x) = \alpha$?
Please do not leave complete answers as I would like to complete this myself.
$\textbf{Note}$: The notation $y(x)$ does not mean a function evaluated at a point e.g. $y=f(x)=x^2$ but rather a functional $y$ evaluated at a vector $x$ belonging to some vector space.