Question about inner product.

Question

Let $V=C([-1, 1])$ and $$\langle f, g\rangle=\int_{-1}^1 f(x)g(x)dx$$ Let $W=\{f \in V \mid f\text{ is even}\}$. Find $W^\perp$.

Progress: I know that every odd function belongs to $W^\perp$ and I suspect $W^\perp=\{f \in V \mid f\text{ is odd}\}$.

Answer 1

Let $g$ be an odd function. Then $fg$ is odd for all even functions $f$ in $C([-1, 1])$, therefore $\langle f, g\rangle=0$ for all $f \in C([-1, 1])$ Now suppose $g$ isn't odd. Let:
$$u(x)=\frac{g(x)+g(-x)}{2}$$ $$v(x)=\frac{g(x)-g(-x)}{2}$$ It's clear that $g=u+v$, $u$ is even and $v$ is odd. Since $g$ isn't even, $u\neq0$. Finally:
$$\langle u, g\rangle=\int_{-1}^1 u(x)g(x)dx=\int_{-1}^1 u(x)v(x)dx+\int_{-1}^1 u(x)u(x)dx=0+\int_{-1}^1 u(x)^2dx>0$$
We have shown that $g \in W^\perp \Leftrightarrow g \text{ is odd}$.

Thanks to: vadim123