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.

  • $\begingroup$ Do you assume that the $f_i$ have a common period (apart from $0$), or are the periods allowed to be unrelated? $\endgroup$ Jan 22, 2016 at 10:38
  • $\begingroup$ @DanielFischer I think it is not necessary. I feel it is enough that $f_i$ is periodic function. But I don't know how to show it . $\endgroup$
    – lanse7pty
    Jan 30, 2016 at 4:40

2 Answers 2


(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 $x$:


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

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

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.

  • $\begingroup$ Thanks, but I still think there is real difficulty what is not solved in your answer. $\endgroup$
    – lanse7pty
    Feb 19, 2016 at 5:33
  • $\begingroup$ In removing the continuity assumption, finding the sequence, or both? $\endgroup$
    – πr8
    Feb 19, 2016 at 5:44
  • $\begingroup$ Finding the sequence . $\endgroup$
    – lanse7pty
    Feb 19, 2016 at 6:00
  • $\begingroup$ For $n = 2$, at least, the sequence is guaranteed by Weyl's equidistribution theorem or general ergodic theory on $\mathbb{R}/\mathbb{Z}$. (Assuming the two periods have irrational ratio, but the result is trivial otherwise.) $\endgroup$
    – anomaly
    Feb 19, 2016 at 6:08
  • $\begingroup$ Added in an alternative proof that avoids the continuity issue and is a lot briefer. $\endgroup$
    – πr8
    Feb 19, 2016 at 7:22

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!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.