Find an explicit second-degree polynomial given a definition of an inner product on a polynomial vector space

Question

Let $\mathcal{P}_2$ denote the vector space of of all second-degree polynomials from $[a,b]$ to $\mathbb{R}$. Define the inner product on $\mathcal{P}_2$ to be:

$$\left< p,q \right> = \int_{-1}^{1} p(x)q(x) dx $$

How can I find a unique, explicit polynomial $q(x) \in \mathcal{P}_2$ such that for every $p \in \mathcal{P}_2$, we have:

$$p(1) = \int_{-1}^{1} p(x)q(x) dx $$

I started by defining $p(x) = a + bx + cx^2$ and $q(x) = \alpha + \beta x + \delta x^2$. Then I calculated $p(1)$, which equals to $a+b+c$. From this we get:

$$ a+b+c = \int_{-1}^{1} (a + bx + cx^2)(\alpha + \beta x + \delta x^2) dx $$

However, after this, no matter what I try, I cannot get explicit values for $\alpha, \beta$, and $\delta$. I tried plugging it into WolframAlpha, but each value depends on another value and the values keep changing from one case to another. Moreover, we cannot simply fix values for the constants because the above statement must hold for every polynomial $p \in \mathcal{P}_2$.

Answer 1

For $p(x)=1$ we get $1=\int_{-1}^{1} q(x) dx$.

For $p(x)=x$ we get $1=\int_{-1}^{1} xq(x) dx$.

For $p(x)=x^2$ we get $1=\int_{-1}^{1} x^2q(x) dx$.

Can you proceed ?

Answer 2

It is sufficient to define $q$ on a basis. A suitable basis is $e_k(x)= x^k$ for $k=0,1,2$.

Let $q(x) = q_0+ q_1x+q_2 x^2$.

Since $e_k(1) = 1$ for all $k$ we require $\langle e_k, q \rangle = 1$ for all $k$.

This gives the equations $2q_0 + {2 \over 3} q_2 = 1, {2 \over 3} q_1 = 1, {2 \over 3} q_0 + {2 \over 5} q_2 = 1$.

Hence $q_0 = -{3 \over 4}, q_1 = {3 \over 2}$, $q_2 = {15 \over 4 }$.