While thinking about Fourier coefficients of periodic, integrable, complex-valued functions using various periods I thought to check that the definition of Fourier coefficients
$$\hat{f}(k)=\frac{1}{L}\int_{-L/2}^L f(x)e^{-2 \pi i k x/L}\,dx$$
actually agree for a function which is both 1-periodic and $\pi$-periodic. But this made me wonder if such a function must necessarrily in fact be constant. To me, it feels like it should be. It would suffice to show that for any $x,y\in\mathbb{R}$ we have $f(y)=f(x+m\pi+n)$ for some $m,n\in\mathbb{Z}$. I sketched a proof that the sequence $(x_m)_{m\in\mathbb{N}}$ defined $x_m \langle x + m\pi + n\rangle$ (where this denotes the fractional part of $x+m\pi+n$) is equidsitributed on $[0,1)$ by using Weyl's criterion and thus must be dense in $[0,1)$. I think this proves the claim if $f$ is continuous but I am unsure.
I am wondering if there is a better approach to proving this claim or if in fact the claim is not true at all.
Best Answer
The set $G$ of periods of $f$ is a subgroup of $\mathbb{R}$ and all the subgroups of $\mathbb{R}$ are either dense or of the form $\alpha\mathbb{Z}$ with $\alpha \geqslant 0$. As $\pi$ is irrational, it can't be of the form $\alpha\mathbb{Z}$ or else we would have $1 = \alpha n$ and $\pi = \alpha m$ for some integers $(n,m)$, so $\pi = m/n$. Therefore, $G$ is dense.
For all $L \in G$, $f(L) = f(0)$ because $t$ is a period of $f$. Therefore, if $f$ is continuous, we have $f = f(0)$ on $\mathbb{R}$, $f$ is constant. For a general locally $L^1$ periodic function, it is a bit harder. You are right about the fact that for all $n$, $\hat{f}_1(n) = \hat{f}_\pi(n)$ where for all $L,n$, $$ \hat{f}_L(n) = \frac{1}{L}\int_0^L f(t)e^{-2in\pi t/L} \, dt $$ Indeed, consider a rational sequence $(p_m/q_m)$ that converges toward $\pi$ from below and let $\varepsilon_m = \pi - p_m/q_m \rightarrow 0^+$. By $p_m$- and $q_m\pi$-periodicity, we have for all $n,m$, $$ \hat{f}_1(n) = \frac{1}{p_m}\int_0^{p_m} f(t)e^{-2in\pi t/p_m} \, dt, \qquad \hat{f}_\pi(n) = \frac{1}{q_m\pi}\int_0^{q_m\pi} f(t)e^{-2int/q_m} \, dt $$ Therefore, \begin{align*} \hat{f}_\pi(n) - \hat{f}_1(n) & = \frac{q_m\pi}{p_m}\hat{f}_\pi(n) - \hat{f}_1(n) + \mathrm{o}(1)\\ & = \frac{1}{p_m}\int_0^{q_m\pi} f(t)e^{-2int/q_m} \, dt - \frac{1}{p_m}\int_0^{p_m} f(t)e^{-2in\pi t/p_m} \, dt + \mathrm{o}(1)\\ & = \frac{1}{p_m}\int_0^{p_m} f(t)e^{-2int/q_m} \, dt + \frac{1}{p_m}\int_{p_m}^{q_m\pi} f(t)e^{-2int/q_m} \, dt - \frac{1}{p_m}\int_0^{p_m} f(t)e^{-2in\pi t/p_m} \, dt + \mathrm{o}(1)\\ & = \frac{1}{p_m}\int_0^{p_m} f(t)(e^{-2int/q_m} - e^{-2in\pi t/p_m}) \, dt + \frac{1}{p_m}\int_0^{p_m\varepsilon_m} f(t)e^{-2int/q_m} \, dt + \mathrm{o}(1), \end{align*} and \begin{align*} \left|\frac{1}{p_m}\int_0^{p_m\varepsilon_m} f(t)e^{-2int/q_m} \, dt\right| & \leqslant \frac{1}{p_m}\int_0^{p_m\varepsilon_m} |f(t)| \, dt\\ & \leqslant \frac{1}{p_m}\int_0^{\lceil p_m\varepsilon_m\rceil} |f(t)| \, dt\\ & = \frac{\lceil p_m\varepsilon_m\rceil}{p_m}\|f\|_{L^1([0,1])}\\ & \rightarrow 0, \end{align*} and \begin{align*} \left|\frac{1}{p_m}\int_0^{p_m} f(t)(e^{-2int/q_m} - e^{-2in\pi t/p_m}) \, dt\right| & \leqslant \frac{1}{p_m}\int_0^{p_m} |f(t)(e^{-2int/q_m} - e^{-2in\pi t/p_m})| \, dt\\ & \leqslant \frac{1}{p_m}\int_0^{p_m} |f(t)|\left|-\frac{2nt}{q_m} + \frac{2n\pi t}{p_m}\right| \, dt\\ & \leqslant 2n\left(\frac{\pi}{p_m} - \frac{1}{q_m}\right)\frac{1}{p_m}\int_0^{p_m} |f(t)| \, dt\\ & = \frac{2n}{p_m}\varepsilon_m\|f\|_{L^1([0,1])}\\ & \rightarrow 0, \end{align*} thus $\hat{f}_1(n) = \hat{f}_\pi(n)$. Since it is true for $1$ and $\pi$, we could use the same reasoning to prove that for all $L \in G\backslash\{0\}$ and all $n$, $\hat{f}_L(n) = \hat{f}_1(n)$ doesn't depend on the choice of $L$. We shall simply call this quantity $\hat{f}(n)$.
Now, recall Lebesgue theorem in dimension $1$, for all measurable locally integrable function, we have, $$ \forall x \in \mathbb{R} \textrm{ a.e.}, \frac{1}{2R}\int_{x - R}^{x + R} g(t) \, dt \underset{R \rightarrow 0}{\longrightarrow} g(x). $$ Thus, for almost all $x \in \mathbb{R}$ and for all $n$, \begin{align*} \left|\hat{f}(n)\right| & = \left|\frac{1}{L}\int_{x - L/2}^{x + L/2} f(t)e^{2in\pi/L} \, dt\right|\\ & \leqslant \frac{1}{L}\int_{x - L/2}^{x + L/2} |f(t)| \, dt\\ & \rightarrow |f(x)|, \end{align*} when $L \rightarrow 0$ (we can do it by density of $G$). For $n = 0$ (we actually only need this case), we obtain that for almost all $x$, $|f(x)| \leqslant \|f\|_{L^1([0,1])}$ and by integrating over $[0,1]$, $$ \|f\|_{L^1([0,1])} = \int_0^1 |f(t)| \, dt \leqslant \int_0^1 \|f\|_{L^1([0,1])} \, dt = \|f\|_{L^1([0,1])}. $$ It implies that the inequality is an equality, so we have almost everywhere $f(t) = \|f\|_{L^1([0,1])}$. $f$ is indeed constant.