Show that the dual space of the vector space of all polynomials is isomorphic to the infinite-dimensional Euclidean vector space over the reals May I please ask how to show the dual space of the vector space of all polynomials is isomorphic to the infinite-dimensional Euclidean vector space over the reals?
(i.e. Show that $(\Bbb{R}[X])^*$ and $\Bbb{R}^{∞}$ are isomorphic where $\Bbb{R}$ means the set of all reals)
 A: A linear functional $f$ on the vector space of real polynomials is determined by the real numbers $f(1), f(x), f(x^2),\dots$ and for any sequence of real numbers $\{a_n\}$ we can define a linear functional $f$ so that $f(x^n)=a_n$. Therefore an isomorphism between the two vector spaces is given by the map
$$f\mapsto (f(1),f(x),f(x^2),\dots)$$
A: We know that: $\mathbb{R}[X]\cong \bigoplus_{n=0}^{\infty}\mathbb{R}X^{n}$, so that: $(\mathbb{R}[X])'=Hom_{\mathbb{R}}(\bigoplus_{n=0}^{\infty}\mathbb{R}X^{n},\mathbb{R})\cong \prod_{n=0}^{\infty}Hom_{\mathbb{R}}(\mathbb{R}X^{n},\mathbb{R})\cong \prod_{n=0}^{\infty}\mathbb{R}=\mathbb{R}^{\infty}$.
A: Let $\mathcal{P}(\mathbf{R})$ be the linear space of polynomials and $\big(\mathcal{P}(\mathbf{R})\big)'$ its dual space.
Define $\Lambda:\big(\mathcal{P}(\mathbf{R})\big)'\to\mathbf{R}^\infty$ by
\begin{equation}
    \Lambda(\varphi) = \big(\varphi(1),\varphi(x),\varphi(x^2),\dotsc\big).
\end{equation}
Let $\varphi_1,\varphi_2\in\big(\mathcal{P}(\mathbf{R})\big)'$ and $\lambda\in\mathbf{R}$. Then
\begin{multline}
    \Lambda(\varphi_1+\lambda\varphi_2)=\big(\varphi_1(1)+\lambda\varphi_2(1), \varphi_1(x)+\lambda\varphi_2(x), \dotsc\big) = \\
    \big(\varphi_1(1), \varphi_1(x), \dotsc\big) + \lambda\big(\varphi_2(1), \varphi_2(x), \dotsc\big) = \Lambda(\varphi_1)+\lambda\Lambda(\varphi_2).
\end{multline}
Therefore, $\Lambda$ is a linear map.
Suppose $\Lambda\varphi = 0$. Then
\begin{equation}
    \varphi(1)=0\quad\varphi(x)=0\quad\varphi(x^2)=0\quad\dotsb.
\end{equation}
Multiplying each equation by arbitrary constants and adding up a finite number of them to make a polynomial
\begin{align}
    a_0\varphi(1)+a_1\varphi(x)+a_2\varphi(x^2)+\dotsb+a_m\varphi(x^m) &= 0\\
    \varphi(a_0+a_1x+a_2x^2+\dotsb+a_mx^m) &= 0.
\end{align}
Because $\varphi(p)=0$ for any $p\in\mathcal{P}(\mathbf{R})$, $p(x)=a_0+a_1x+a_2x^2+\dotsb+a_mx^m$, then $\varphi=0$ and $\Lambda$ is injective.
Now let $(x_0,x_1,\dotsc)$ be an arbitrary member of $\mathbf{R}^\infty$. Then, choose $\varphi\in\big(\mathcal{P}(\mathbf{R})\big)'$ such that
\begin{equation}
    \varphi(1)=x_0\quad \varphi(x)=x_1\quad \varphi(x^2)=x_2\quad\dotsb.
\end{equation}
Such $\varphi$ exists because it is simply a linear map from $\mathcal{P}(\mathbf{R})$ to $\mathbf{R}$. Then
\begin{equation}
    \Lambda\varphi = (x_0,x_1,\dotsc),
\end{equation}
which implies $\Lambda$ is surjective.
Finally, because $\Lambda$ is linear, injective and surjective, it is an isomorphism from $\big(\mathcal{P}(\mathbf{R})\big)'$ onto $\mathbf{R}^\infty$. $\square$
