Duality on $R_3[X]$ Again and again, I post here because I don't succeed to do something ! Even if I should be able to do that, I don't know why but I don't succeed to do that.
So, that's the notation :
$x_1, x_2, x_3 \in \mathbb{R}$, all distincts, and we are on $R_3[X]$. We consider : $\phi_i : P \rightarrow P(x_i)$, $\psi : P \rightarrow P'(x_i)$ and finally : $\nu : P \rightarrow \int_{x_0}^{x_2} P(t) \mathrm{d}t$.
First, we want to show that $(\phi_0, \phi_1,\phi_2, \psi_{1})$ and $(\phi_0, \phi_1,\psi_0, \psi_{2})$ are some basis of $R_3[X]^*$. I did it by showing the vectors are linearly independent. But the next question is : Show that $\nu = \frac{x_2-x_0}{2}(\phi_0 + \phi_2) + \frac{(x_2-x_0)^2}{2}(\psi_0+ \psi_2)$. 
As I didn't succeed to do that directly, I've just shown this assertion for the polynoms $1, X, X^2, X^3$, and conclude using linearity. But I would like to prove it directly, I thought about using duality, but the calculation to find the basis such that $(\phi_0, \phi_1,\psi_0, \psi_{2})$ are really long. So, I've try to determine the coordinate by evaluating $\nu$ on diferent polynoms, but what I found is just : $x_2-x_1 = \lambda_2 + \lambda_0$ and $\frac{(x_0-x_2)^2}{2} = \lambda_2(x_2-x_0) + \lambda'_0 + \lambda'_2$, where $\nu = \lambda_0 \phi_0 + \lambda_2 \phi_2 + \lambda'_0 \psi_0 + \lambda'_2 \psi_2$. Now, I would like to have $\lambda_2=\lambda_0$ and $\lambda'_0 = \frac{(x_0-x_2)^2}{2}$, but I don't succeed to do that.
Someone could help me ?
 A: If $P(x)=a_0+a_1x+a_2x^2+a_3x^3$, then\begin{align}\nu(P)&=\int_{x_0}^{x_2}P(t)\,\mathrm dt\\&=a_0(x_2-x_0)+\frac{a_1}2\left({x_2}^2-{x_0}^2\right)+\frac{a_2}3\left({x_2}^3-{x_0}^3\right)+\frac{a_3}4\left({x_2}^4-{x_0}^4\right)\\&=(x_2-x_0)\left(a_0+\frac{a_1}2(x_2+x_0)+\frac{a_2}3({x_2}^2+x_2x_0+{x_0}^2)+\frac{a_3}4({x_2}^3+{x_2}^2x_0+x_2{x_0}^2+{x_0}^3\right).\end{align}On the other hand,\begin{align}\frac{x_2-x_0}2(\phi_2+\phi_0)&=\frac{x_2-x_0}2(a_0+a_1x_0+a_2{x_0}^2+a_3{x_0}^3+a_0+a_1x_2+a_2{x_2}^2+a_3{x_2}^3)\\&=\frac{x_2-x_0}2\bigl(2a_0+a_1(x_0+x_2)+a_2({x_0}^2+{x_2}^2)+a_3({x_0}^3+{x_2}^3)\bigr)\end{align}and\begin{align}\frac{{x_2}^2-{x_0}^2}2(\psi_0+\psi_2)&=\frac{{x_2}^2-{x_0}^2}2(a_1+2a_2x_0+3a_3{x_0}^2+a_1+2a_2x_2+3a_3{x_2}^2)\\&=\frac{{x_2}^2-{x_0}^2}2\bigl(2a_1+2a_2(x_0+x_2)+3a_3({x_0}^2+{x_2}^2)\bigr)\\&=\frac{x_2-x_0}2\bigl(2a_1+2a_2(x_0+x_2)^2+3a_3({x_0}^2+{x_2}^2)(x_0+x_2)\bigr).\end{align}Now, sum these two last expressions and see that what you get is the first one.
