User:IssaRice/Linear algebra/Dual space of vector space of polynomials

Given a vector space $V$, its dual space $V^{\prime }$ is the set of all linear functionals $V\to \mathbf{F}$, i.e. all linear functions that "evaluate" each vector into a constant.

For the space $\mathcal{P}(\mathbf{R})$ of single-variable polynomials on $\mathbf{R}$, an element $T$ of $(\mathcal{P}(\mathbf{R}){)}^{\prime }$ takes a polynomial $p$ and returns a real number $T(p)$. What does this space look like?

If $p$ is defined by $p(x)=a_n{x}^n+\cdots +a_{1}x+a_0$, we want to say something like that T "acts linearly" on p. except... i don't think axler defines what a basis is for an infinite-dimensional space. want to say that $(1,\mathrm{i}\mathrm{d},{\mathrm{i}\mathrm{d}}^2,\dots )$ (using functional notation here for type-pedantry, where $\mathrm{i}\mathrm{d}=(x\mapsto x)$) is a basis for $\mathcal{P}(\mathbf{R})$. actually, maybe we don't need bases after all.

so given $p=a_n{\mathrm{i}\mathrm{d}}^n+\cdots +a_1\mathrm{i}\mathrm{d}+a_0$, we have $T(p)=a_{n}T({\mathrm{i}\mathrm{d}}^n)+\cdots +a_{1}T(\mathrm{i}\mathrm{d})+a_{0}T(1)$.

now the interesting thing is that even though each polynomial must have finite length (because not all infinite sums converge), each machine eating up a polynomial can have infinite length -- it can get away with this because each input is finite. mentally, when i think about this situation, it's sort of like the inputs have chickened out of being infinite, so the linear functionals can "afford to" stay infinite.