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I have a curious question I stumbled upon 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:

  1. $\int_0^{2\pi} \frac{dx}{2\pi} f_i(x) = 0$

  2. $ \int_0^{2\pi} \frac{dx}{2\pi} f_i(x) f_j(x) = \frac{1}{3} \delta_{ij}$

  3. $f_1(x)^2 +f_2(x)^2 + f_3(x)^2 = 1$ 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.

**Edit: if continuity is replaced by smoothness (at least 1-differentiable), does anything change?

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  • $\begingroup$ You can of course do this easily with functions that just take the values $\pm 1/\sqrt{3}$ ("Rademacher functions"). Obviously, these will not be continuous, but it seems clear (?) that continuous modifications will also work. $\endgroup$ Jul 24, 2022 at 17:11
  • $\begingroup$ @ChristianRemling Thanks. I thought about those too functions. But it is not clear to me how continuous modifications will work? Also, I realized that I actually want smooth (at least once-differentiable) functions. $\endgroup$
    – nervxxx
    Jul 24, 2022 at 17:53

1 Answer 1

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I think we can do it smoothly. Think of $f=(f_1,f_2,f_3)$ as a map from the circle into $\mathbb R^3$. I need it to take values in the unit sphere. Mark the three obvious great circles on the sphere, and note the twelve quarter-circles in this picture.

Here is a continuous solution. As $x$ goes from $0$ to $\pi/6$, $f(x)$ goes from $(1,0,0)$ to $(0,1,0)$ following a great circle path at constant speed. As $x$ goes from $\pi/6$ to $\pi/3$, $f(x)$ follows another such path, from $(0,1,0)$ to, say, $(0,0,1)$. You can keep going like this, in a twelve-part path that eventually traverses each of those quarter-circles once. The average of $f_i$ is $0$ because four of the twelve terms are $0$ and the remaining eight cancel in pairs. For $i\neq j$ the average of $f_if_j$ is $0$ because eight of the twelve terms are $0$ and the remaining four cancel in pairs. The average of $f_i^2$ is independent of $i$.

This can be modified to a smooth solution by arranging for each of the twelve paths to be not a constant-speed path but rather one that travels along a quarter-circle beginning and ending with a brief interval of staying still, in a symmetrical way.

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  • $\begingroup$ This same idea works for more than three functions. $\endgroup$ Jul 25, 2022 at 20:32

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