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

Given a vector space $V$, its dual space $V'$ 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 $({\mathcal {P}}(\mathbf {R} ))'$ 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 $(1,\mathrm {id} ,\mathrm {id} ^{2},\ldots )$ (using functional notation here for type-pedantry, where $\mathrm {id} =(x\mapsto x)$) is a basis for ${\mathcal {P}}(\mathbf {R} )$. actually, maybe we don't need bases after all.

so given $p=a_{n}\mathrm {id} ^{n}+\cdots +a_{1}\mathrm {id} +a_{0}$, we have $T(p)=a_{n}T(\mathrm {id} ^{n})+\cdots +a_{1}T(\mathrm {id} )+a_{0}T(1)$.

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.