# Orthogonality of functions over complex field

I found this problem on a teacher's handout:

Prove that the decomposition into even and odd functions over $C^0([-a,a],\mathbb{C})$ is orthogonal if we use the inner product: $$\langle f,g \rangle = \int^a_{-a} f(t) \overline{g(t)} dt$$

I'm having trouble understanding what this means:

$C^0([-a,a],\mathbb{C})$

.

Edit: Also, what does the zero superscript mean?

In general, how do I use that notation for any other vector spaces?

Is the expression before the comma a domain? Is the $\mathbb{C}$ after the comma a field?

• Do you know what even and odd functions are? You can uniquely write any function as the sum of an even function and an odd function. That's a good exercise as a starting point. – Ted Shifrin Apr 9 '18 at 1:06
• I think instead of decomposing any given function, the question asks for a proof that the whole vector space $C^0([-a,a],\Bbb C)$ can be decomposed into an internal direct sum $E\oplus O$, with $E$ the space of all even continuous functions, $O$ the space of all odd continuous functions, and $E$, $O$ are orthogonal to each other. – edm Apr 9 '18 at 1:18
• Thank you @TedShifrin and edm for the comments. I edited my post to clarify my question – evaristegd Apr 9 '18 at 2:12
• $C^0([-a,a],\mathbb C)$ is the set of continuous, complex-valued functions defined on the interval $[-a, a]$ – saulspatz Apr 9 '18 at 2:12
• Thank you @saulspatz . Please feel free to check the more specific questions I just added on the edit – evaristegd Apr 9 '18 at 2:22

## 1 Answer

Stripped of the verbiage, the problem is simply to show that if $f$ is odd and $g$ is even, or vice-versa, then $\int_{a}^{a}{f(t)\overline{g(t)} \text{dt}}=0$

The complex conjugate of an even function is even, and likewise, the conjugate of an odd function is odd, so in any event, the integrand is the product of an odd function an an even function. Now the product of an odd function and an even function is odd, for suppose $h$ is even and $k$ is odd. Then $$hk(-z)=h(-z)k(-z) = h(z)(-k(z))=-h(z)k(z)=-hk(z)$$

Since the integrand is odd, the integral is $0$.

• On real-valued functions, that integral would be zero. In this case, It would be the product of and odd function and the conjugate of an even function, right? – evaristegd Apr 9 '18 at 2:35
• Okay, what can you say about the conjugate of an even function? – saulspatz Apr 9 '18 at 2:39
• That it's equal to the conjugate of the same function evaluated on -t , i.e. evaluated on the negative of the argument (?) – evaristegd Apr 9 '18 at 2:44
• If $\overline{g(z)} = \overline{g(-z)}$ isn't $\overline{g}$ an even function? – saulspatz Apr 9 '18 at 2:58
• Yes, I just realized that. Thank you! – evaristegd Apr 9 '18 at 3:01