Inner product problem

Question

I'm having trouble with this excercise:

Determine the orthogonal space of odd continuous functions in the space of continuous functions. (In the interval $ [-1,1] $)

I assume that the inner product is $$ \int_{-1}^1f(x)g(x)dx\ $$

but I still don´t how how to solve this. If somebody can give me a hint or something, it would be great. Thank you! Please ask me if you don't understand something of the problem.

Answer 1

The orthogonal complement is the collection of all functions $f : [-1,1] \to \mathbb{R}$ such that, given any odd function $g : [-1,1] \to \mathbb{R}$, you have $$\int_{-1}^1 f(x)g(x)\, dx=0$$ I'm not going to do this for you, but I'll give a few hints:

1. What do you think the answer should be?
2. Every function $f$ can be written as $f_{\text{even}} + f_{\text{odd}}$, where $f_{\text{even}}$ is even and $f_{\text{odd}}$ is odd
3. Which functions are both even and odd?

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Added: You've correctly identified that the orthogonal complement is the space of even functions. All you have to do now is prove it!

To fix notation, let $V = \{ \text{continuous functions}\ [-1,1] \to \mathbb{R} \}$, and let $U_{\text{odd}}$ and $U_{\text{even}}$ denote the subspaces of odd and even functions, respectively. You need to show:

• If $f \in V$ then there exist $f_{\text{odd}} \in U_{\text{odd}}$ and $f_{\text{even}} \in U_{\text{even}}$ such that $f = f_{\text{odd}} + f_{\text{even}}$. This proves $V = U_{\text{odd}} + U_{\text{even}}$.

• If $f \in U_{\text{odd}} \cap U_{\text{even}}$ then $f=0$. This proves $V = U_{\text{odd}} \oplus U_{\text{even}}$.

• If $f \in U_{\text{even}}$ and $g \in U_{\text{odd}}$ then $\int_{-1}^1 f(x)g(x)\, dx = 0$. This proves that $U_{\text{even}} = U_{\text{odd}}^{\perp}$.