Homework Help: Orthogonal complement of even functions

1. May 29, 2014

mahler1

The problem statement, all variables and given/known data.

Let the $\mathbb R$- vector space $C([-1,1])=\{f:[-1,1] \to \mathbb R\ : f \space \text{is continuous}\ \}$ with the inner product $<f,g>=\int_{-1}^1 f(t)g(t)dt$. Determine the orthogonal complement of the subspace of even functions (call that subspace $S$).

The attempt at a solution.

Straight from the definition of orthogonal complement, $g \in (C([-1,1]))^\bot$ if and only if $\int_{-1}^1 g(t)f(t)dt=0$ for all $f$ even function. By hypothesis, $f(-t)=f(t)$, so $\int_{-1}^1 g(t)f(-t)dt=0$ From this condition I don't know what else to conclude. I would appreciate any suggestions.

Last edited: May 29, 2014
2. May 29, 2014

micromass

Some things to try:

1) The function $f(x) = 1$ for all $x$ is an even function. We want an element of $S$ to be orthogonal to that, what does that imply for functions in $S$?

2) Can you give some (nontrivial) examples of functions in $S$?

3. May 29, 2014

mahler1

Hmm, for 1), the only thing I can think of is something that follows trivially, which is: $\int_{-1}^1 g(t)dt=0$.

2) Yes, I can think of $\sin(x)$, which is an odd function. Suppose $g$ is odd, then $\int_{-1}^1 g(-t)f(-t)dt=-\int_{-1}^1 g(t)f(t)dt=0$ since $g(-t)=-g(t)$ and $f(-t)=f(t)$, then the set of odd functions (subspace) is included in $S$.

Is there any other function which belongs to $S$? I don't know what you are trying to insinuate me in 1).

4. May 29, 2014

micromass

Forget about (1), it was just to give you some more intuition.

But it seems you now have a good conjecture. Is $S$ exactly the set of odd functions or not? I think you should try to prove that it is.

5. May 29, 2014

pasmith

Every function can be written as the sum of an even function and an odd function. Thus, given an arbitrary $g \in C([-1,1])$, we have $g = f + h$ where $f \in S$ and $h$ is odd. It would be worth looking at $$\int_{-1}^1 f(x)g(x)\,dx = \int_{-1}^1 f(x)^2\,dx + \int_{-1}^1 f(x)h(x)\,dx.$$

6. May 29, 2014

mahler1

I want to check if this is correct:

Taking your suggestion, $\int_{-1}^1 f(x)^2\,dx + \int_{-1}^1 f(x)h(x)\,dx=\int_{-1}^1 f(x)^2\,dx =\int_{-1}^0 f(x)^2\,dx + \int_0^1 f(x)^2\,dx$. Making the substitution $x=-t$, I get $\int_{-1}^0 f(x)^2\,dx + \int_0^1 f(x)^2\,dx=-\int_1^0 f(-u)du+ \int_0^1 f(x)^2\,dx=\int_0^1 f(u)du+ \int_0^1 f(x)^2\,dx=2\int_0^1 f(x)^2\,dx$

But this means $\int_0^1 f(x)^2\,dx=0 \implies f^2(x)=0 \implies f(x)=0$. From here it follows $g=h$, but $h$ is an odd function so $g$ is odd.