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.

in other words, the "infinite" sum $\sum _{i=0}^{\infty }a_{i}T(\mathrm {id} ^{i})$ turns out to be a finite sum because all the $a_{i}$ are 0 for $i>n$.