Inner product and orthogonal

Question

Let = [−1,1] with the inner product 〈, 〉 = ∫()() (-1,1)

= { ∈ | (−) = ()}.

Show that ⊥ = { ∈ | (−) = −()}.

Anyone can help me?

I tried this

〈, 〉 = ∫()() (-1,1)

= ∫h(x) (-1,1) =∫h()dx(-1,0) + ∫h()(0,1)

=-∫h()dx(0,-1) + ∫h()(0,1)

Answer 1

Let $g \in W^\top$. $g$ is the sum of its even and odd part: $$ g(x)=\frac{g(x)+g(-x)}{2} + \frac{g(x)-g(-x)}{2}. $$ We get $$ 0=\int_{-1}^1 g(x) \frac{g(x)+g(-x)}{2} dx $$ $$ =\int_{-1}^1\left(\frac{g(x)+g(-x)}{2}\right)^2 + \frac{g(x)^2-g(-x)^2}{4} dx $$ $$ =\int_{-1}^1\left(\frac{g(x)+g(-x)}{2}\right)^2 dx. $$ Hence $g(x)+g(-x)=0$. Note that $$ \int_{-1}^1 \frac{g(x)^2-g(-x)^2}{4} dx = 0 $$ as the integrand is odd. Reversely if $g$ ist odd and $f$ is even, then $gf$ is odd and $$ \int_{-1}^1 g(x)f(x) dx = 0 $$