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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.

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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?

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

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  • $\begingroup$ I was thinking onthe space of even functions, but now that you mention it, I'm kind of confused about the constant functions, because as they are both even and odd that means that they have to be in both subspaces, and that cannot be possible, as there is a theorem that assures that the intersection of a space and its orthogonal complement is empty... $\endgroup$ – Lessa121 Dec 9 '13 at 5:20
  • $\begingroup$ If you do not understand something please ask me, I´m Spanish and I don't really know if my English is understandable or not :) $\endgroup$ – Lessa121 Dec 9 '13 at 5:21
  • $\begingroup$ @LunaSage: All constant functions are even, but they're not all odd... for instance, if $f$ is the constant function with value $1$ then $f(-x)=1 \ne -1 = -f(x)$. Tu inglés es perfecto :) $\endgroup$ – Clive Newstead Dec 9 '13 at 5:30
  • $\begingroup$ P.S. The intersection of a space and its orthogonal space isn't empty, since the zero vector necessarily lies in both. $\endgroup$ – Clive Newstead Dec 9 '13 at 5:32
  • $\begingroup$ Thank you! Tu español también! So the space of even functions is the orthogonal complement? I think I almost have the proof but I want to make sure I didn´t messed it up at some point $\endgroup$ – Lessa121 Dec 9 '13 at 5:32

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