Limit of sum of periodic function

Question

Let $f_1,f_2,...,f_n$ are periodic functions,if $\lim\limits_{x\rightarrow\infty}\sum_{i=1}^n f_i(x)$ is existent and bounded.

How to show $\sum_{i=1}^n f_i(x)\equiv C$ ?

$C$ is a constant.

Answer 1

(this assumes continuity of the $f_i$, still thinking of how to remove it)

Suppose $F(x):=\sum_i f_i(x)\to C$, where $f_i(x+p_i)=f_i(x)$ for each $i$.

Now, find an sequence of positive reals $h_N$ increasing to $\infty$ which is simultaneously close to $p_i\mathbb{Z}$ for all $i$, i.e.

$$h_N=a_i^{(N)}p_i+\varepsilon_N^{(i)}, \text{where }z_N^{(i)}\in\mathbb{Z}, \varepsilon_N^{(i)}\to 0$$

as by noting Dirichlet's simultaneous approximation theorem applied to $\{\frac{1}{p_1},\frac{1}{p_2},...,\frac{1}{p_n}\}$, we can find integers $a_i^{(N)}$ and an integer $h_N\le N$ such that for each $i$:

$$\left\vert \frac{1}{p_i}-\frac{a_i^{(N)}}{h_N}\right\vert \le \frac{1}{h_N N^{1/n}}$$

Upon rearrangement, this becomes $\vert h_N - a_i^{(N)}p_i \vert \le \frac{p_i}{N^{1/n}}\to0$

Then, for any $x$:

$$\lim_{n\to\infty}F(x+h_n)=\lim_{x\to\infty}F(x)=C$$

$$\lim_{n\to\infty}F(x+h_n)=\lim_{n\to\infty}\sum_i f_i(x+h_n)=\lim_{n\to\infty}\sum_i f_i(x+\varepsilon_n^{(i)})=\sum_i f_i(x)=F(x)$$

So, $F(x)=C$ for all $x$.

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Alternative Proof

Let $P(n)$ be the statement "If a sum of $n$ periodic functions has a limit $C$, then this sum is equal to $C$ for all $x$".

1. If $f$ is $p$-periodic and tends to $C$, then for any $\varepsilon > 0$, there exists $N$ such that $x>N\implies \vert f(x) - C \vert < \varepsilon$. But periodicity gives that this is actually true for all $x$. As this is true for any $\varepsilon > 0$, we recover that $f=C$ for all $x$. So, $P(1)$ is true.

2. Suppose $P(1), P(n-1)$ are true, and consider a sum of $n$ periodic functions, $F(x)=\sum_1^n f_i(x)$ with limit $C$, where in particular, $f_n$ has period $p_n$. Then $F(x+p_n)-F(x)=\sum_1^{n-1} [f_i(x+p_n)-f_i(x)]$ is a sum of $(n-1)$ periodic functions, and converges to $0$, hence is equal to $0$ by $P(n-1)$.

So, $F$ is $p_n$-periodic, and converges to $C$, and $P(1)$ tells us that it is identical to $C$ as a result, i.e. $P(n)$ is true.

Thus, by induction, $P(n)$ is true for all $n$, and any finite sum of periodic functions with a limit at $\infty$ is constant.

Answer 2

I don't think it's true in the full generality of the question. The questioner needs to give more information about the functions domain an range. Here's a counterexample.

Define $\chi_n(m) = \begin{cases} 1 , \text{ if } m \in (n), \\ 0, \text{ if } m \notin (n) \end{cases},$

where $(n) \subset \Bbb{Z}$ is the ideal generated by $n \in \Bbb{N}$. Then each $\chi_n$ is periodic of period $n$. $\chi_m \chi_n = \chi_{\text{lcm}(m,n)}$ and the characteristic function for $(n) \cup (k)$ is $\phi = 1 - (1-\chi_n)(1- \chi_k) = \chi_n + \chi_k - \chi_k \chi_n$. In general for the finite set $n= \{(n_i)\}_{i=1..N}$ of ideals we can use the inclusion-exclusion principle to compute $\phi_{\cup n}$. For instance, $n = \{(a), (b), (c)\}$ and $\phi_{\cup n} = \chi_a + \chi_b + \chi_c - \chi_a\chi_b - \dots + \chi_a \chi_b \chi_c$.

Since $\phi_{\cup n}$ is a finite sum of periodic functions with integer period, its period is simply the lcm of everything which is simply $\text{lcm}(n_1, \dots, n_N)$.

Now let $\psi_n = \phi_{\cup \{(2), \dots, (n)\}}$. Then $\psi_n$ is a finite partial sum of periodic functions, and

$$ \lim\limits_{n\to \infty} \psi_n (m) = 1 $$

always for $x \in \Bbb{N}\setminus \{1\}$ (why?), yet its partial sums are not so constant!