Inner product problem 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.
 A: 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:


*

*What do you think the answer should be?

*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

*Which functions are both even and odd?



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 $U_{\text{even}} = U_{\text{odd}}^{\perp}$.
