A question in Hilbert spaces

Question

Let be $ X=C[-1,1]$ and define $\langle f,g \rangle = \int^1_{-1} f(t)g(t) dt$.

If $M=\{f \in\ C=[-1,1]\mid f\text{ is odd function}\}$, what is $ M^\perp$?

Answer 1

You can easily show that $$C([-1,1])=O([-1,1])\oplus E([-1,1])$$ where $O([-1,1])$ denotes the continuous odd functions on $[-1,1]$ and $E([-1,1])$ denotes the continuous even functions. (Hint: $f(t)=\frac{f(t)-f(-t)}{2}+\frac{f(t)+f(-t)}{2}$).

Next, you can show that odd and even functions are orthogonal to each other. Conclude that $O([-1,1])^{\perp}=E([-1,1])$.