I am required to prove the following proposition.
Proposition. Let $V = C[-1,1]$ denote the set of continious real valued functions over the interval $[-1,1]$ and let $W_e$ and $W_o$ be the subspaces of even and odd functions over the interval $[-1,1].$ Show that $W_e^{\perp} = W_o$ with the inner product defined as follows $$\langle f,g\rangle = \int_{-1}^{1}f(t)g(t)dt$$
Now to prove $W_e^{\perp} = W_o$ we need to show that $W_e^{\perp}\subseteq W_o$ and $W_o\subseteq W_e^{\perp}$. I am confident that i can prove the later claim but i am not sure how i can prove the former.
What follows is my attempt to show that $W_e^{\perp}\subseteq W_o$. Please critique it and let me know if it is correct and if not please do suggest any corrections.
Proof. Assume that $f\in W_e^{\perp}$ to show that $f\in W_o$ we must show $f(-t) = -f(t),\forall t\in [-1,1]$. So let $t_o\in [-1,1]$ and consider the function $\delta(t-t_0)+\delta(t+t_o) = g\in W_e$ then since $f\in W_e^{\perp}$ it follows that $\langle f,g\rangle = 0$ consequently $$\langle f,g\rangle = \int_{-1}^{1}f(t)g(t)dt = \int_{-1}^{1}f(t)\cdot\left(\delta(t-t_0)+\delta(t+t_o)\right)dt$$ $$ = \int_{-1}^{1}f(t)\delta(t-t_0)dt+\int_{-1}^{1}f(t)\delta(t+t_o)dt = f(t_0)+f(-t_0) = 0$$ implying that $f(-t_0) = -f(t_0)$ since our choice of $t_0\in[-1,1]$ was arbitrary it follows that $f(-t) = -f(t),\forall t\in [-1,1]$ implying that $f\in W_o$.
$\blacksquare$
Here is perhaps a better way: First note that any continuous function can be written as a sum of an even and odd function: see here. Therefore, $W_e \bigoplus W_o = C[-1,1]$.
Now we just need to show the subspaces are orthogonal, which can be done as follows: let $f$ be even, $g$ odd. Then $fg$ is an odd function, so $\langle f,g \rangle = \int_{-1}^1fg= 0$.