Assume $f$ is some 'nice' real-valued function defined on $[0,\pi]$ and satisfies $$f(t)-f\left(t+\frac{\pi}{3}\right)+f\left(t+\frac{2\pi}{3}\right).$$
In this paper, it says f(t) must have the form $a(t)\cos(t)+b(t)\sin(t)$ with $a,b$ $\frac{\pi}{3}$-periodic. I'm assuming the 'niceness' (example: smooth, analytic) of $f$ will be reflected in the functions $a$, $b$.
However, I don't know how to prove even existence of such a representation. How would one begin?
On
Hint.
$$ f(t)-f(t+\delta)+f(t+2\delta) = 0 $$
or
$$ f\left(\delta\frac{t}{\delta}\right)-f\left(\delta\left(\frac{t}{\delta}+1\right)\right)+f\left(\delta\left(\frac{t}{\delta}+2\right)\right) = 0 $$
or making $F(t) = f\left(\delta t\right)$ and $\frac{t}{\delta} = u$
$$ F(u)-F(u+1)+F(u+2) = 0 $$
now this functional equation is easy to solve.
All solutions to the equation $(\dagger)$ $$f(t)-f(t+\frac{\pi}{3})+f(t+\frac{2\pi}{3}) = 0$$ can be characterized by: $$f(t) = a(t) \cos(t) + b(t) \sin(t)$$ where $a(t)$ and $b(t)$ are functions with period $\frac{\pi}{3}$.
WLOG I'm taking $f(t)$ to go from $\mathbb{R} \rightarrow \mathbb{R}$, but it could go to $\mathbb{R}^k$ or other stuff.
Proof:
First we prove that $f(t)$, $t \geq 0$ is fixed by its values in the interval $0 \leq t \leq \frac{2\pi}{3}$.
It suffices to show that $f(t+\frac{2\pi}{3}), f(t+\pi), f(t+\frac{4\pi}{3}), f(t+\frac{5\pi}{3}), f(t+2\pi)$ are fixed by $f(t)$ and $f(t+\frac{\pi}{3})$, and $f(t) = f(t+2\pi)$. Let $f(t) = x$ and $f(t+\frac{\pi}{3}) = y$. Iterating $\dagger$ gives: $f(t+\frac{2\pi}{3}) = y-x, f(t+\pi) = -x, f(t+\frac{4\pi}{3}) = -y, f(t+\frac{5\pi}{3}) = -y+x, f(t+2\pi) = x$.
We now prove that such a function is characterized by $f(t) = a(t) \cos(t) + b(t) \sin(t)$.
To do so, we show that for any set of values $f(t)$ defined over $0 \leq t \leq \frac{2\pi}{3}$, there exists viable $a(t)$ and $b(t)$ with period $\frac{\pi}{3}$.
To prove that, it suffices to show that for any $f(t) = c_1$ and $f(t+\frac{\pi}{3}) = c_2$, you can choose viable values for $a(t) = a(t+\frac{\pi}{3})$ and $b(t)= b(t+\frac{\pi}{3})$.
This comes down to solving the two systems of equations: $$a(t)\cos(t)+b(t)\sin(t) = c_1$$ $$a(t+\frac{\pi}{3})\cos(t+\frac{\pi}{3})+b(t+\frac{\pi}{3})\sin(t+\frac{\pi}{3}) = a(t)\cos(t+\frac{\pi}{3})+b(t)\sin(t+\frac{\pi}{3}) = c_2$$ for $a(t)$ and $b(t)$.
This is obviously doable.
So it follows that every solution for $\dagger$ can be written in the form $a(t)\cos(t)+b(t)\sin(t)$ where $a(t), b(t)$ have period $\frac{\pi}{3}$.
We now need to show that all functions of the form $a(t)\cos(t)+b(t)\sin(t)$ satisfy the problem's constraint.
Plugging and simplifying gets: $$f(t)-f(t+\frac{\pi}{3})+f(t+\frac{2\pi}{3})$$
$$a(t)\cos(t)+b(t)\sin(t)-a(t+\frac{\pi}{3})\cos(t+\frac{\pi}{3})-b(t+\frac{\pi}{3})\sin(t+\frac{\pi}{3})+a(t+\frac{2\pi}{3})\cos(t+\frac{2\pi}{3})+b(t+\frac{2\pi}{3})\sin(t+\frac{2\pi}{3})= $$
$$a(t)\cdot (\cos(t)-\cos(t+\frac{\pi}{3})+\cos(t+\frac{2\pi}{3})) + b(t)\cdot (\sin(t)-\sin(t+\frac{\pi}{3})+\sin(t+\frac{2\pi}{3}))= $$ $$a(t)\cdot 0 + b(t)\cdot 0 = 0$$ where the last statement follows from the identities $(\cos(t)-\cos(t+\frac{\pi}{3})+\cos(t+\frac{2\pi}{3})) = 0$ and $(\sin(t)-\sin(t+\frac{\pi}{3})+\sin(t+\frac{2\pi}{3}) = 0$.
And we're done.