Periodic functions whose sum is null

Question

If $f_1,\ldots,f_n:\mathbb{R}\rightarrow\mathbb{R}$ are periodic functions such that $$ \lim\limits_{x\rightarrow +\infty}{(f_1(x)+\ldots+f_n(x))}=0 $$ how can I prove that $f_1+\ldots+f_n=0$ ?

Answer 1

In a previous version (below) we showed it's true if the $f_j$ are continuous. Today it turns out it's true for two arbitrary periodic functions (probably the proof generalizes to finitely many periodic functions; I have to go to class).

Suppose that $f$ has period $1$, $g$ has period $p>0$, and $f+g$ tends to $0$ at $+\infty$.

$f$ has period $p$ (hence $f+g$ is periodic, hence $f+g=0$).

Indeed, since $f$ has period $1$ and $g$ has period $p$ it follows that $$f(a+p)-f(a)=(f+g)(a+p+n)-(f+g)(a+n)\to0\quad(n\to\infty).$$

Previous Result

I suspect it's true if the $f_j$ are arbitrary periodic functions; I can prove it if they're locally integrable:

Say a trigonometric polynomial is a (possibly non-periodic) linear combination of the functions $e_\omega$ ($\omega\in\Bbb R$), where $e_\omega(t)=e^{i\omega t}$.

Lemma 0. If $p$ is a trigonometric polynomial and $A\in\Bbb R$ then $\sup_{t\in\Bbb R}|p(t)|=\sup_{t>A}|p(t)|$.

This is clear from standard results about "almost periodic functions". Or, it follows easily from

Lemma 1. If $\omega_1,\dots,\omega_n\in\Bbb R$ there exists a sequence $t_k\to\infty$ such that $\lim_ke^{i\omega_jt_k}=1$ for $j=1,\dots,n$.

Proof. $\Bbb R$ is a vector space over $\Bbb Q$. Since any spanning set for any subspace contains a basis, we may assume, perhaps after reordering the $\omega_j$, that $\omega_1,\dots,\omega_k$ are $\Bbb Q$--independent and every $\omega_j$ for $j>k$ is a $\Bbb Q$-linear combination of $\omega_1,\dots,\omega_k$.

That famous thing about equidistribution on the torus shows that there exist $t_n\to\infty$ with $$\lim_ne^{i\omega_jt_n}=1\quad(1\le j\le k).$$

Choose an integer $K$ so that if $j>k$ then $K\omega_j$ is a linear combination of $\omega_1,\dots,\omega_k$ with integer coefficients. It follows that $$\lim_ne^{i\omega_jKt_n}=1\quad(1\le j\le n).$$(Because if $K\omega_j=\sum_{l=1}^k\alpha_l\omega_l$ with $\alpha_l\in\Bbb Z$ then $$e^{i\omega_jKt_n}=\prod_{l=1}^k(e^{i\omega_lt_n})^{\alpha_j}.)$$

Proof of Lemma 0. Say $$p(t)=\sum_{j=1}^dc_je^{i\omega_jt}.$$Choose $t_n\to\infty$ as in Lemma 1. Then $$p(t)=\lim_np(t+t_n).$$

Now assume $f_1,\dots,f_n$ are periodic, let $f=f_1+\dots+f_n$, and assume $f(x)\to0$ at $+\infty$.

Assume first that each $f_j$ is continuous. Each $f_j$ is the uniform limit of a sequence of (periodic) trigonometric polynomials, so $f$ is the limit of a sequence of (aperiodic) trigonometric polynomials. So Lemma 0 implies that $$|f(x)|\le\sup_{t>A}|f(t)|,$$so $f(x)=0$.

Finally suppose just that each $f_j$ is a locally integrable periodic function. Say $(\phi_k)\subset C_c(\Bbb R)$ is an approximate identity: $\phi_k\ge0$, $\int\phi_k=1$ and $supp(\phi_k)\to\{0\}$. Now $\phi_k*f_j$ is continuous and periodic, and $$\sum_j\phi_k*f_j(x)=\phi_k*f(x)\to0\quad(x\to\infty),$$so the previous paragraph implies that $\phi_k*f(x)=0$. But $\lim_k\phi_k*f=f$ almost everywhere, hence $f=0$ almost everywhere.