Consider $L^2([-1,1])$ and $$ V= \{f(x) = ax + bx^2 : x \in [-1,1], a,b \in \mathbb{R}\}$$ Find $V^\perp \subset L^2([-1,1])$.
I'm having issues answering this question. Is there a way to find $f(x)$ in a "explicit way" such that it belongs to $V^\perp$?
Hint:
The Legendre polynomials $$P_n(x)=\frac{\sqrt{2n+1}}{n!2^n\sqrt{2}}D^n((1-x^2)^n),$$ where $D$ is the derivative operator, form a complete orthonormal system for $L_2([-1,1])$. The first three Legendre polynomials are $P_0(x)=1$, $P_1(x)=x$ and $P_2(x)=\frac12(3x^2-1)$. Then $\operatorname{span}\{P_1,P_2+\frac12P_0\}=V$, where $V$ is as in your problem. Then, the span of $\{P_n:n>2\}\cup\{P_2-P_0\}$ is $V^\perp$, since $$(P_2+\frac12P_0,P_2-P_0)=(P_2,P_2)+\frac12(P_0,P_2)-(P_2,P_0)-(P_0,P_0)=1-1=0$$
The Legendre polynomials may be obtained by considering the collection of polynomials $\{x^n:n\in\mathbb{Z}_+\}$ ($\mathbb{Z}=\mathbb{N}\cup\{0\}$), whose span is dense in $L_2([-1,1])$ (by the Stone-Weirstrass theorem), and apply the Gram-Schmidt orthogonalization procedure.
Another way to analyze your problem, is to order the polynomials $\{x^n:n\in\mathbb{N}\}$ as $\{x, x^2,1,x^3,\ldots\}$ and apply the Gram-Schmidt process. $V=\operatorname{span}\{x,x^2\}$.