Are continuous orthogonal functions always paired in even and odd symmetric combinations, with respect to their domain? So as an example, consider the set of orthogonal polynomials $\{1, \, x, \, x^2-\tfrac{1}{3}\}$ on $x \in [1, \, -1]$. So the first and third members are even, second one odd. If you look at their graphs it simply seems like the definition of orthogonal only allows functions that are either


*

*Constant on the domain.

*Even about the midpoint of domain and having and integral = 0, or

*Odd about the midpoint of the integration domain. 


Is this the only way two continuous elementary functions can be orthogonal? Or can we have two functions that seem to have no individual symmetry with respect to the domain but still be orthogonal?
Edit: I'm defining (tell me if this is acceptable or not), the inner product of two continuous polynomials f and g to be 
$$(f,g) = \int_{a}^{b}(fg)(x) \, dx $$  
and orthogonal if $ \, (f,g) = 0$.
 A: It depends on the inner product you define for polynomials $p_1,p_2$, which is usually induced by a norm, which you would also have to define. I suppose that you chose equal weighting for all x, that is, a constant kernel $\mu$ in a Riemann-Stieltjes integral:
$\left<p_1,p_2\right>:=\int_{-1}^1 p_1(x)\cdot p_2(x)\mathrm{d}\mu(x)$, with $\mu(x):=1$
or any other constant.
Obviously, you can come up with arbitrary kernels and contradict your idea.
But even for a constant kernel, it is easily disproved:
[Edit: The following might not be applicable to continuous functions as demanded in the question title! The examples below introduce discontinuities, but they produce at least piecewise-continuous functions.]


*

*Find a sequence of orthogonal functions for any kernel.

*Exchange any subset of function values on the domain for all functions alike (while preserving integrability).

*The obtained sequence is necessarily orthogonal as well.


The reason is the inner product's invariance towards the order of integration.
For example, on the interval $\left[0,1\right]$, replace $x^2-1/3$ with $(1-x)^2-1/3$ and $x$ with $1-x$, basically mirroring all functions on half of your domain. This will not change the inner product and hence, they are orthogonal, but not symmetric.
