$2\pi$ periodic functions with null barycenter

Question

Let $f$ be a continuous $2\pi$-periodic (real-valued) function satisfaying:

$$\forall x \in [0, 2\pi),\ \forall n \ge 2,\ \sum_{k =1}^n f\left(x + \frac{2\pi}{n}k\right) = 0.$$

Can we conclude that there exist $A$ and $\phi$ such that $f(x) = A \cos(x - \phi)$?

Or are they other possible functions?

Answer 1

Let $$g_n(x)=\sum_{k =1}^n f\left(x + \frac{2\pi}{n}k\right)$$ Just like $f$, function $g_n$ is both continuous and $2\pi$ periodic, so it is square integrable on any interval of length $2\pi$. Thus it has Fourier series coefficients given by $$\begin{split} c_p(g_n) &= \frac 1 {2\pi} \int_{-\pi}^\pi e^{ipx}\sum_{k =1}^n f\left(x + \frac{2\pi}{n}k\right) dx\\ &= \sum_{k =1}^n \frac 1 {2\pi} \int_{-\pi}^\pi e^{ipx} f\left(x + \frac{2\pi}{n}k\right) dx\\ &= \sum_{k =1}^n \frac 1 {2\pi} e^{-\frac{2\pi} {n}kp}\int_{-\pi+\frac{2\pi}{n}k}^{\pi+\frac{2\pi} {n}k} e^{ipx} f(x) dx\\ &= \left (\sum_{k =1}^n e^{-\frac{2\pi} {n}kp}\right) c_p(f) \end{split}$$ where $c_p(f)$ is the $p$-th Fourier coefficient of $f$.

Now we know that for all $n\geq 2$, $g_n$ is identically $0$. Thus so are its Fourier coefficients.

So both sides are always $0$ if $n\geq 2$.

Now, notice that the sum of complex exponential in the right-hand side is zero, unless $n$ divides $p$.

• For $p=0$, that sum is non-zero, so $c_0(f)=0$
• For $|p|\geq 2$, it suffices to take $n=2p$ to make that sum non-zero, implying that $c_p(f)$ is zero.

Consequently, the only coefficients that are potentially non-zero are $c_p(f)$ for $p=\pm 1$. Because $f$ is real-valued, we have that $c_1 = c_{-1}^*$, and this means $f$ has the form that you suggested.