I would appreciate help with the first part of a problem I am self-studying:
Let $V=\mathcal P_2(\mathbb{R})$ prove $\varphi_j(p)=p(j)$ is a [dual] basis of $\mathcal P_2(\mathbb{R})'$
I think I should first show the $\varphi_j(p)$ are linear independent:
If $a_0\varphi_0+a_1\varphi_1+a_2\varphi_2=0$ then all the $a_i$ are $0$.
Taking $(a_0\varphi_0+a_1\varphi_1+a_2\varphi_2)p(j)=0$ I would like to show each $a_j=0$ but I don't see how to do that.
Maybe my whole approach is incorrect.
Thanks
To show linear independence, we need to show that if $a_0\phi_0 + a_1\phi_1 + a_2\phi_2 = 0$, then $a_i = 0$ for $i=0,1, 2$. We know that $\Phi =a_0\phi_0 + a_1\phi_1 + a_2\phi_2 $ is an element of $V^{'}$, the dual space of $V$, and so $\Phi$ is a map from $V$ to $\mathbb{R}$. We are assuming that $\Phi = 0$, or in other words, $\Phi(p) = 0$ for every $p\in V$.
Consider $p = (x-1)(x-2) = x^2 -3x + 2 \in V$. By our previous assumption we have \begin{eqnarray} 0 &=& \Phi(p) \\ &=& (a_0\phi_0 + a_1\phi_1 + a_2\phi_2)(p) \\ &=& a_0\phi_0(p) + a_1\phi_1(p) + a_2\phi_2(p) \\ &=& a_0p(0) + a_1p(1) + a_2p(2)\\ &=& a_0(2) + a_1(0) + a_2(0)\\ &=& 2a_0 \end{eqnarray}
This implies that $a_0 = 0$. So now we can play this same game (i.e., find other polynomials $q, r$) to show that $a_1 = 0$ and $a_2 = 0$. This would prove the linear independence of $\phi_0, \phi_1$, and $\phi_2$.