# Periodic functions whose sum is null

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 $$f_1+\ldots+f_n=0$$ ?

• @Winther Wouldn't this require the $f_i$ to be continuous? Commented Sep 14, 2019 at 21:59
• It works if we assume the ratios of the periods are rational, so the sum will itself be periodic. Commented Sep 14, 2019 at 23:21
• There's a proof below for two functions (with no extra assumptions); I'll let you think about extending it to $n$ functions. Commented Sep 18, 2019 at 15:40

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.