Compute the Orthogonal Complement of $M$ in $\mathcal{H}$

Question

Let $\mathcal{H}:=C([-1,1])$ and now define the inner product on $\mathcal{H}$ as $\langle f,g\rangle:=\int_{[-1,1]}\overline{f(x)}g(x)dx$. Furthermore, let $M:=\{ f\in \mathcal{H}: f(x)=f(-x) \forall x[0,1]\}$

Compute $M^{\perp}$.

First I will consider the real case:

let $f \in M$ and $g \in M^{\perp}$: by definiton $0=\langle f,g\rangle=\int_{[-1,1]}f(x)g(x)dx=\int_{0}^{1}f(x)g(x)dx+\int_{-1}^{0}f(x)g(x)dx=\int_{0}^{1}f(x)g(x)dx+\int_{0}^{1}f(x)g(-x)dx=\int_{0}^{1}f(x)(g(x)+g(-x))dx$

so any $g \in C([-1,1])$ so that $g(x)=-g(-x)$ for any $x \in [0,1]$.

Is this truly my $M^{\perp}$? I am uncertain.

Answer 1

From the comments it seems that what remains to show that if $g \in M^\perp$ then $g$ is an odd function (since you know that odd functions are in $M^\perp$).

You have already rightly observed that if $f \in M$ and $g \in C([-1,1])$ then $$\int_{-1}^1 f(x) g(x) dx = \int_0^1 f(x) (g(x) + g(-x)) dx = 0$$ for all $f \in M$. We want to see that this implies that $g$ is an odd function.

Suppose not. Then there is an $x \in (0,1)$ such that $g(x) + g(-x) \neq 0$. Without loss of generality, $\delta = g(x) + g(-x) > 0$. Then, by continuity, there is an $\varepsilon >0$ such that $g(y) + g(-y) > \delta/2$ for all $y \in (x - \varepsilon, x + \varepsilon) \subseteq (0,1)$. Now just pick a continuous function $\tilde{f}: [0,1] \to \mathbb{R}$ that is $0$ outside of $(x - \varepsilon, x + \varepsilon)$, is non-negative and satisfies $\int_0^1 \tilde{f}(x) dx = 1$.

If $f$ is the even extension of $\tilde{f}$ to $[-1,1]$ then $f \in M$ and

$$\int_{-1}^1 f(x) g(x) = \int_0^1 f(x) (g(x) + g(-x)) \geq \delta / 2 \int_0^1 f(x) dx > 0$$ which is a contradiction. Hence $g \in M^\perp$ implies that $g$ is odd.

Answer 2

If $g \in M^\perp$ then $f(x) = g(x)+g(-x) \in M$ so

$$0 = \int_{-1}^1 f(x)g(x)\,dx = \int_0^1f(x)(g(x)+g(-x))\,dx = \int_0^1(g(x)+g(-x))^2\,dx $$

which implies $g(x)+g(-x) = 0$ for all $x \in [0,1]$.