Orthogonal Complement of quadratic functions in L²([-1,1])

Question

How can we characterize $V^{\perp}$ where $ V= \{v \in L^2([-1,1]): v(x)=ax+bx^2,b\neq 0\}$ ? I've tried looking for $ \{f \in L^2([-1,1]): \int_{[-1,1]}{fv\ d\lambda}=0, \ v(x)=ax+bx^2 \}$ but I cannot figure out an explicit characterization of this space.

Answer 1

Keep in mind that if $f \in V^\perp$ then for all $a,b\in \mathbb{R}$ and $b\not= 0 $ we must have that $$\int_{-1}^{1}f(x)(ax+bx^2)dx = 0$$ which is equivalent to $$ a\int_{-1}^1 xf(x) dx = -b\int_{-1}^1 x^2f(x) dx$$ We choose $a = 0$ to see that since $b\not=0$ we must have that $\int_{-1}^1 x^2f(x) dx = 0$. Hence for all $a\in\mathbb{R}$ we also have that $a\int_{-1}^1xf(x) dx = 0$ which can only happen if $\int_{-1}^1 xf(x)dx = 0$ as well.

Let us dig a little further and decompose $f\in V^\perp$ into its even $(f_e)$ and odd $(f_o)$ components and use the symmetry of $[-1,1]$ to see if we learn anything else. Well the product of even functions is even and similarly for the product of odd functions. Further an even function times an odd function is odd. Since the integral of an odd function over a symmetric domain is $0$, then $$\int_{-1}^1 xf(x)dx = \int_{-1}^1 x f_e(x)dx +\int_{-1}^1 x f_o(x)dx = 0+\int_{-1}^1 x f_o(x)dx = 0$$ $$\int_{-1}^1 x^2f(x)dx = \int_{-1}^1 x^2 f_e(x)dx +\int_{-1}^1 x^2 f_o(x)dx = \int_{-1}^1 x^2 f_e(x)dx +0 = 0$$ Hence if we take any even function on $[-1,1]$ and orthogonal to $x^2$ on the same domain and any odd function on $[-1,1]$ and is orthogonal to $x$ on the same domain, then their sum is in $V^\perp$. This completely defines the space.

As Medo points out in comments, the only odd function orthogonal to $x$ on $[-1,1]$ is the $0$ function and the only even function orthogonal to $x^2$ on $[-1,1]$ is the again the $0$ function. Hence $f = 0$ is the only function in $V^\perp$.