Operator contained in a space of polynomials?

Question

Consider $L^{2}([-1,1], \mathbb{C}),$ endowed with the inner product

$\langle f \mid g\rangle=\int_{-1}^{1} f(x) \overline{g(x)} d x$

and with the associated norm.

$\mathcal{P}$ is the space of polynomial of degree inferior to $2,$ taken as a subspace of $L^{2}([-1,1], \mathbb{C})$

For all $f$ de $L^{2}([-1,1], \mathbb{C})$ and all $x \in[-1,1] $ let

$A(f)(x)=\int_{-1}^{1}\left(x^{2}+x y+y^{2}\right) f(y) d y $

1. Verify we define $A$ from $L^{2}([-1,1], \mathbb{C})$ in itself.

I know that $f$ is $L²$ so is the polynomial (as a continuous on a compact), how do I prove their product is $L² $ too?

2. Show the image of $A$ is contained in $\mathcal{P}$ and that $A f=0$ for all $f$ orthogonal to $\mathcal{P}$

How do I show such an integral is a polynomial when $f$ is any $L²$ function?

Answer 1

a) What did you try? (this is not a site to do your homeworks)

b) Hints:

• Question (1) means that assuming that $f∈ L^2$, you need to prove $A(f)∈ L^2$ (Cauchy-Schwarz ...).

• Question (2), remark that you just need to show that it is a polynomial in the $x$ variable! (the $y$ variable is just a dummy variable of integration ... for example $∫ y^2 f(y)\,\mathrm{d}y$ is a constant, so a polynomial of order $0$ ...)