# 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.