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.

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

if we consider only polynomials with rational coefficients, then the set of polynomials remains countable (${\mathbf{Q}}^k$ is countable for each positive integer k (represents a finite sequence of length k of rational numbers) because a finite cartesian product of countable sets is countable, and the countable union of countable sets is countable (union over all k)). on the other hand, the linear functionals get to be all infinite sequences of rationals. it's clear this can't be countable (since even infinite sequences over a finite set is uncountable -- that's one way the real numbers can be defined). there's probably a quick way to show this for infinite sequences of Q, but idk

You might wonder, couldn't the same argument above work for the dual space of say ${\mathcal{P}}_3(\mathbf{R})$ (polynomials with degree at most 3)? After all, these are still finite-length strings, so if we have a machine with an infinite string of numbers, it would still always produce a finite result. The problem is that when the inputs are restricted in their maximum degree, some functionals "collapse" into the same thing. For example, $T(p)=5{\mathrm{i}\mathrm{d}}^3+7{\mathrm{i}\mathrm{d}}^4$ has the same behavior as $S(p)=5{\mathrm{i}\mathrm{d}}^3$ because the input will never have a 4th degree term in it. Basically, $T{|}_{{\mathcal{P}}_3(\mathbf{R})}=S{|}_{{\mathcal{P}}_3(\mathbf{R})}$, so the dual space of ${\mathcal{P}}_3(\mathbf{R})$ will be finite-dimensional (as we already know, from the fact that the dual space of a finite-dimensional vector space has the same dimension as the original space).