Complete orthonormal system in P2(R)

Question

Does someone knows a complete orthonormal system on $$P_2(R) := \{ax^2 + bx + c : a,b,c ∈ R\}$$ equiped with scalar product $$⟨p,q⟩ = p(−2)q(−2) + p(0)q(0) + p(2)q(2)$$ ?

Answer 1

Start with $f(x)=x^2+dx+e$. Then $$\begin{align}\langle1,f\rangle&=4-d+e+e+4+d+e=3e+8=0\\ \langle x,f\rangle&=-8+2d-2e+0+8+2d+2e=4d=0\end{align}$$ So we arrive at $e=-\frac83$ and $d=0$, so $$p_2(x)=a\left(x^2-\frac83\right)$$ Then $$\langle p,p\rangle=a^2\left(\left(\frac43\right)^2+\left(-\frac83\right)^2+\left(\frac43\right)^2\right)=\frac{32}3a^2=1$$ So we may let $a=\frac{\sqrt6}{8}$ and then $$p_2(x)=\frac{\sqrt6}8\left(x^2-\frac83\right)$$ Oh, but you wanted a complete orthonormal system. We already can see that $\langle1,x\rangle=0$, so we only need worry about normalization. Simlilar to what we did above for $p_2(x)$ we can show that $$p_0(x)=\frac1{\sqrt3}$$ $$p_1(x)=\frac1{\sqrt8}x$$ If course we can apply any orthogonal transformation to get other solutions. Another way is with the Lagrange basis: $$\begin{align}q_{-2}(x)&=\frac{(x-0)(x-2)}{(-2-0)(-2-2)}=\frac18x(x-2)\\ q_{0}(x)&=\frac{(x+2)(x-2)}{(0+2)(0-2)}=-\frac14\left(x^2-4\right)\\ q_{2}(x)&=\frac{(x+2)(x-0)}{(2+2)(2-0)}=\frac18x(x+2)\end{align}$$