# Proving that given functionals span the dual space of all polynomials of 4-th degree (and finding basis for which given functionals are dual basis)

Let's call $\Bbb{R[x]}_4$ a linear space of all polynomials of 4-th degree with real coefficients. We are given functionals $\phi_j \in (\Bbb{R[x]}_4)^*$:

• $\phi_0(p)=p(-1)$
• $\phi_1(p)=p(0)+p(1)$
• $\phi_2(p)=p(0)-p(1)$
• $\phi_3(p)=p(2)$
• $\phi_4(p)=p(-2)$

The problems asks us to prove that given functionals are basis for $(\Bbb{R[x]}_4)^*$. Then we are to find coefficients $a_0,a_1,a_2,a_3,a_4 \in \Bbb{R}$ such that if given functionals are dual basis to polynomials $p_0,p_1,p_2,p_3,p_4$ and $p(t)=t$ then $p=\sum_{i=0}^4a_ip_i$.

When it comes to the 1st part of this problem I thought of picking functionals $\beta_0=p(0)$, $\beta_1=p(1)$, $\beta_2=p(-1)$, $\beta_3=p(2)$, $\beta_4=p(-2)$, and these functionals must be dual basis for Lagrange basis constructed from them. Then you can show that every $\beta$ function can be written as a linear combination of $\phi$ functions, so their span is equal, and as we know that $\beta$ functions must be basis of $(\Bbb{R[x]}_4)^*$ so the same must be true for $\phi$ functions.

When it comes to the 2nd part of the problem I thought of computing the Lagrange basis for $\beta$ functions and then do some manipulations to it to get the Lagrange basis of $\phi$ functions and from that point I would compute the coordinates of $p$ in that basis which would give the answer. But this solution seems too complex and too hard to compute. Is there a simpler one? Also, are my earlier arguements correct?

• do you already experimented on ${\Bbb{R}}[x]_2$? – janmarqz Jan 6 '15 at 2:39
• what do you mean? – qiubit Jan 6 '15 at 2:57
• that you are trying to solve for ${\Bbb{R}}[x]_4$, try for ${\Bbb{R}}[x]_2$ or/and ${\Bbb{R}}[x]_3$ – janmarqz Jan 6 '15 at 3:03

Firstly, you want to show that the functionals are linearly independent. Suppose $\phi_c = \sum_i c_i \phi_i = 0$, then $\phi_c( x (x-1) (x-2) (x+2) ) = 0$ but also $\phi_c( x (x-1)(x_2)(x+2) ) = c_0$, by directly checking explicitly what it evaluates for the other basis functionals. So $c_0 = 0$. Similarly conclude that the other $c_i = 0$.
Finally, if $t = \sum_i a_i p_i(t)$, then $\phi_j(t) = a_j$ by the fact that $\phi_j (p_i) = \delta_{ij}$. So just evaluate the functionals on $t$.
You might also try to calculate explicitly the dual basis for practice. For example $p_0 = \frac{-1}{6} x(x-1)(x-2)(x+2)$ evaluates to zero at the right places, and is scaled so that $\phi_0(p_0) = 1$. Then you can check explicitly if the $a_i$ are correct.