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.
Best Answer
The following steps lead to a solution:
(1) Prove that the tuple $(1,x,x^2,\dots)$ is a basis for the complex vector space $V$ of all polynomials (in the variable $x$) with coefficients in $\mathbb{C}$.
(2) If $p(x)=\sum_{i=0}^n a_ix^i$ and if $\Lambda$ is a linear functional on $V$, prove that $\Lambda(p(x))=\sum_{i=0}^n a_i\Lambda(x^i)$. (Hint: the linearity of the functional $\Lambda$ is, of course, relevant.)
(3) Therefore, the linear functional $\Lambda$ is completely determined by the values $\Lambda(x^i)$ for $i\geq 0$ an integer.
The following exercises are relevant:
Exercise 1: Prove the Riesz representation theorem for finite dimensional Hilbert spaces:
(Hint: note that $H$ has an orthonormal basis.)
Exercise 2: Let $V$ be a finite dimensional vector space over $\mathbb{C}$ and let $(v_1,\dots,v_n)$ be a basis of $V$. Prove that if $(a_1,\dots,a_n)$ is a tuple of complex numbers, then the map $\Lambda:V\to\mathbb{C}$ given by $\Lambda(\sum_{i=1} c_iv_i)=\sum_{i=1}^n c_ia_i$ (where $c_i\in\mathbb{C}$ for all $1\leq i\leq n$) is a linear functional on $V$. Conversely, prove that all linear functionals on $V$ have this form.
Exercise 3: Let $V$ be a finite dimensional vector space over $\mathbb{C}$ and let $(v_1,\dots,v_n)$ be a basis of $V$. If $1\leq i\leq n$, let $\Lambda_i:V\to \mathbb{C}$ be defined by the rule $\Lambda_i(v_i)=1$ and $\Lambda_i(v_j)=0$ if $j\neq i$. Prove that $\Lambda_i$ is a linear functional on $V$ for all $1\leq i\leq n$. If $V^{*}$ is the dual space of $V$ (= vector space of all linear functionals on $V$), prove that $(\Lambda_1,\dots,\Lambda_n)$ is a basis of $V^{*}$.
Exercise 4: Let $V$ be the vector space of all polynomials (in the variable $x$) with coefficients in $\mathbb{C}$. If $i\geq 0$, let $\Lambda_i:V\to \mathbb{C}$ be defined by the rule $\Lambda_i(x^i)=1$ and $\Lambda_i(x^j)=0$ if $j\neq i$. Prove that $\Lambda_i$ is a linear functional on $V$ for all $i\geq 0$. Is the tuple $(\Lambda_0,\Lambda_1,\Lambda_2,\dots)$ a basis of the dual space of $V$?
I hope this helps!