Trying to prove that $\mathcal P(\Bbb R)'\cong\Bbb R^\infty$

Prove that $\mathcal P(\Bbb R)'$ and $\Bbb R^\infty$ are isomorphic vector spaces.

Here $\mathcal P(\Bbb R)'$ is the dual space of the vector space of polynomial functions with real coefficients $\mathcal P(\Bbb R)$. And $\Bbb R^\infty$ is the set of all real valued sequences.

This is the exercise 35 in page 132 of Linear algebra done right, 3rd edition.

The problem here is that the book at this point only teach, in almost all cases, the theory of linear algebra of finite dimensional vector spaces, but $\mathcal P(\Bbb R)$ is not finite dimensional.

My work, at the moment, is below.

Define $B_k(p):=\frac{p^{(k)}(0)}{k!}=[x^k]p(x)$ for $p\in\mathcal P(\Bbb R)$, where $p^{(k)}$ is the $k$-th derivative of $p$. Then the set defined by

$$B:=\{B_k:k\in\Bbb N_{\ge 0}\}\tag{1}$$

is linearly independent in $\mathcal P(\Bbb R)'$, however it is not a basis because functionals defined as

$$p\mapsto \sum_{k=0}^\infty c_kB_k(p),\quad c_k\in\Bbb R\tag{2}$$

belong also to $\mathcal P(\Bbb R)'$ but they arent a linear combination of $B_k$ because the above sum is not finite.

It is easy to see that any functional with the form in $(2)$ define a vector subspace $S$ of $\mathcal P(\Bbb R)'$ and the map defined by

$$h: S\to \Bbb R^\infty,\quad\sum_{k=0}^\infty c_kB_k\mapsto (c_0,c_1,\ldots,c_k,\ldots)$$

is linear and bijective. Then if we show that $S=\mathcal P(\Bbb R)'$ we are done.

Because every polynomial is a linear combination of monomials we can study all the elements of $\mathcal P(\Bbb R)'$ just studying the form of all possible linear functionals for monomials of the kind $x^k$.

For finite vector spaces $\mathcal P_m(\Bbb R)$, defined as the vector space of polynomial functions of degree at most $m$, we know that the set

$$H_m:=\{B_k:k\in\{0,\ldots,m\}\}$$

is a basis of $\mathcal P_m(\Bbb R)'$.

But Im stuck here: my main problem is that I dont know if I can prove (and how, if it would be possible) that $S=\mathcal P(\Bbb R)'$ from the finite case, that is that all the functionals of $\mathcal P_m(\Bbb R)'$ have the form

$$p\mapsto\sum_{k=0}^m c_kB_k(p),\quad c_k\in\Bbb R,p\in\mathcal P_m(\Bbb R)$$

Probably the way Im trying to prove the statement of the exercise is a dead end, but the theory in the book dont show complicated proofs or theorems, so it cannot be so complicate. Some hints will be welcome.

I think trying to work with a basis in $\mathcal P(\mathbb R)'$ is going to be a bad idea - bases of many infinite dimensional vector spaces, $\mathcal P(\mathbb R)'$ included, are impossible to right down. Thankfully, a basis of $\mathcal P(\mathbb R)$ can be written down: for each $k\in\mathbb N$ (including zero), let $e_k(x):=x^k$; then $\{e_k\}$ is a basis of $\mathcal P(\mathbb R)$. In particular, any linear functional is determined uniquely by its action on $\{e_k\}$. This turns out to be a better method.

For any sequence $a=(a_0,a_1,a_2,\ldots)$, let $\phi_a$ be the unique linear functional on $\mathcal P(\mathbb R)$ such that $\phi_a(e_k)=a_k$ for every $k\in\mathbb N$. I claim $\Phi:a\mapsto\phi_a$ is a linear isomoprhism. That $\Phi$ is linear is easy to prove. If $\phi_a=0$, then $\phi_a(e_k)=0$ for every $k$, i.e. $a_k=0$ for all $k$, so $a=0$. This shows $\Phi$ is injective. Now let $\psi\in\mathcal P(\mathbb R)'$ be any linear functional. Define $a_k:=\psi(e_k)$. By definition $\psi=\phi_a$. This shows $\Phi$ is surjective, completing the proof.

Note: the notation $\mathbb R^\infty$ is not standard, and in fact some authors use it to refer to the space of sequences for which only finitely many entries are non-zero. (This space is actually isomorphic to $\mathcal P(\mathbb R)$ - can you prove it?) Writing $\mathbb R^\mathbb N$ is more standard, and is a special case of the general notation $X^Y:=\{\text{functions }f:Y\to X\}$.

• Yes, I generally use the notation $\Bbb R^{\Bbb N}$ but Im following the notation of the book. I will need some time to digest the answer completely. May 17 '17 at 17:21
• I think I get it. Indeed we can define each linear functional $\phi_a$ by a series of the form of $(2)$ in my question, right? May 17 '17 at 17:42
• Well... as written, $B_k$ maps $\mathcal P(\mathbb R)$ to itself. If you instead take $B_k(p)=\frac{p^{(k)}(0)}{k!}$, then $B_k(p)$ is simply the coefficient of $x^k$ in $p$, so something like that should work. May 17 '17 at 17:56
• I had a typo in the original text, yes $B_k(p):=[x^k]p(x)$. I had written my own answer completing the path of my question. May 17 '17 at 18:01

$\mathcal{P}(\mathbb{R})$ has basis $\{1,x,x^2,\dots,x^n,\dots\}$. Since any linear functional in $\mathcal{P}(\mathbb{R})'$ is determined completely by its action on the basis, we may construct an injective linear map from $\mathcal{P}(\mathbb{R})'$ to $\mathbb{R}^\infty$ via the assignment $$\ell \mapsto (\ell(x^n))_{n=0}^\infty$$ Surjectivity is also quick to see.

• what means $\ell(x^n)$? The concept of action is not shown in this book (at this moment). May 17 '17 at 17:02
• The phrase "its action on" may be replaced with "what it does to". By $\ell(x^n)$ I mean the real number one gets when applying a linear functional $\ell$ to $x^n$. May 17 '17 at 17:06
• I see but, how you knows that the map is surjective? It is hard to imagine that for each real valued sequence exists a linear functional $\ell$ that with the map $(\ell(x^n))_{n\in\Bbb N}$ define this sequence. May 17 '17 at 17:11
• Let $(a_n)_{n\in \mathbb{N}}$ be any real valued sequence. Now define $\ell$ on $\mathcal{P}(\mathbb{N})$ by $\ell(x^n) = a_n$ for all $n$. By defining $\ell$ on the basis we have actually defined it everywhere: any polynomial in our space may be written $p = \sum_{j=1}^n \lambda_j x^j$ for some real scalars $\lambda_j$. We may force linearity of $\ell$ by setting $\ell(p) = \sum_{j=1}^n \lambda_j a_j$. May 17 '17 at 17:14
• @Masacroso The map he means is a bijection is the map that associates to each linear functional $\ell$ the sequence $(\ell(x^{n})_{n=0}^{\infty}$. May 17 '17 at 17:14

Just for the record I will add my own solution to complete the path of my question (based in the previous answers).

Any possible functional $f_{c,k}$ over $x^k$ can be written as

$$f_{c,k}(x^k)=c\tag{3}$$

for any $c\in\Bbb R$. But observe that if we define $$f_{c,k}(x^k):=c B_k(x^k)$$

then the $f_{c,k}$ are linear, and because $\{x^k:k\in\Bbb N_{\ge 0}\}$ is a basis of $\mathcal P(\Bbb R)$ then any functional of $\mathcal P(\Bbb R)'$ have the form in $(2)$, hence $S=\mathcal P(\Bbb R)'$.$\Box$