I have a curious question that may be interesting to some.
Consider real-valued continuous functions on the circle $f_1(x),f_2(x),f_3(x)$ (so they are periodic in $x \mapsto x+2\pi$).
They have the following properties:
$\int_0^{2\pi} \frac{dx}{2\pi} f_i(x) = 0$
$ \int_0^{2\pi} \frac{dx}{2\pi} f_i(x) f_j(x) = \frac{1}{3} \delta_{ij}$
$f_1(x)^2 +f_2(x)^2 + f_3(x)^2 = 1$ for all $x$.
Considered as square-summable vectors on the circle, the first two conditions mean that $f_i(x)$ are orthogonal to the constant function $1$, and they are mutually orthogonal, each with norm $\frac{1}{\sqrt{3}}$. There are infinitely many functions that satisfy properties 1 and 2.
But is there a solution now also requiring property 3 to hold?
If yes, what is an example of such a set of functions? (Or more generically, how to construct them in general?) If no, how do we rule them out?
Thanks.
Partial Answer:
Observe that: $$(f_1+f_2+f_3)^2 = f_1^2+f_2^2+f_3^2+2f_1f_2+2f_1f_3+2f_2f_3$$ Therefore: $$\frac{1}{2\pi} \int_{0}^{2\pi} (f_1+f_2+f_3)^2 \ dx = \frac{1}{2\pi} \int_{0}^{2\pi} f_1^2+f_2^2+f_3^2 \ dx = 1$$
Since each $f_i$ is continuous on a compact interval $[0,2\pi]$, they are all uniformly bounded by some constant $M > 0$. In particular: $$-M \leq f_i \leq M$$ It follows that: $$-3M \leq f_1+f_2+f_3 \leq 3M$$
If we include the additional assumption that $f_1+f_2+f_3 \geq 0$, then the following holds: $$-3M(f_1+f_2+f_3) \leq (f_1+f_2+f_3)^2 \leq 3M(f_1+f_2+f_3)$$ Integrating throughout and using property $(1)$, we find that: $$\int_{0}^{2\pi} (f_1+f_2+f_3)^2 \ dx = 0$$
which is a contradiction. Hence, a triple of such functions satisfying the given properties along with the assumption I've included above won't exist.