$\phi_{i}(P)=P\left(x_{i}\right)$, $\psi_{i}(P)=P^{\prime}\left(x_{i}\right)$ for distinct $x_{1}, \ldots, x_{n}$ is basis of$({R}_{2 n-1}[x])^{*}$ I know that 2n distinct evaluation functionals $\varphi_{i}(P)=P\left(x_{i}\right)$ for 2n $x_{i}$ is a basis for $(\operatorname{R}_{2 n-1}[x])^{*}$, but I have no idea how to prove when there are n evaluation functionals and n derivation at a point functional.
 A: The following is a key observation.
Claim: If $\phi_i(P) = \psi_i(P) = 0$ for all $1 \leq i \leq n$, then $P = 0$.
Proof: Note that if $P$ is such that $\phi_i(P) = \psi_i(P) = 0$, then $(x-x_i)^2$ divides $P$. Thus, since $\phi_i(P) = \psi_i(P) = 0$ for all $i$, the degree $2n$ polynomial $(x-x_1)^2 \cdots (x- x_n)^2$ divides $P$. Because $P$ has degree at most $2n-1$, this can only happen if $P=0$.
With that, the remainder of the proof is entirely linear algebra. Multiple approaches exists; here is one:
Define $\Phi:\Bbb R_{2n-1}[x] \to \Bbb R^{2n}$ by
$$
\Phi(P) = (\phi_1(P),\dots,\phi_n(P),\psi_1(P),\dots,\psi_n(P)).
$$
By the claim above, we see that $\ker \Phi = 0$. Because $\dim(\Bbb R_{2n-1}[x]) = \dim(\Bbb R^{2n})$, we can conclude that $\Phi$ is an isomorphism of vector spaces.  It follows that for any vector $c \in \Bbb R^{2n}$, the map $\Phi^*(c) = \Phi_c:\Bbb R_{2n-1}[x] \to \Bbb R$ defined by $\Phi_c(P) = c^T\Phi(P)$ will only be the zero map if $c = 0$.  In other words, if $c_1,\dots,c_{2n}$ are such that
$$
c_1 \phi_1 + \cdots + c_n \phi_n + c_{n+1}\psi_{1} + \cdots + c_{2n}\psi_n = 0,
$$
then $c_1 = \cdots = c_{2n} = 0$. Since the functionals are linearly independent and there are $2n$ of them, we conclude that they indeed form a basis. Because we have shown that $\Phi$ is injective,

Alternatively, we could have shown that the $\phi_i,\psi_i$ span the dual space.  Equivalently, we want to show that for an arbitrary $\alpha \in (\Bbb R_{2n}[x])^*$, there is a $c \in \Bbb R^n$ for which $\Phi^*(c) = \alpha$. This holds because for maps over finite dimensional spaces, $\Phi^*$ is surjective whenever $\Phi$ is injective.
