TITLE: Sum of two periodic functions is periodic?
QUESTION [24 upvotes]: I have following paragraph taken from the Stanford's study material. 

Question: Is the sum of two periodic functions periodic?
Answer: I guess the answer is no if you are Mathematician, yes if you are an Engineer i.e. no if you believe in irrational numbers and
  leave it at that, and yes if you compute things and work with
  approximation.

This sounds something interesting to me. As an engineer, can I always assume that sum of periodic functions is always periodic?

REPLY [15 votes]: Let $f$ and $g$ two periodic function two non-constant periodic continuous functions of $\mathbb{R}\rightarrow\mathbb{R}$. Note $a>0$ the smallest period of f and $b$ the smallest period of g. 
Find a necessary and sufficient condition for f + g is periodic.

Notice that If $b$ is a multiple of $a$, then $f+g$ is clearly periodic.
Plus, she $\frac{a}b \in \mathbb{Q}$, where $\frac{a}b=\frac{p}q$ then $aq=bp$ is clearly a period of $\quad f+g$

This condition is actually necessary.

Proof.
By contradiction, assume $\frac{a}b \notin \mathbb{Q}$ and $f+g$ periodic and $c$ the smallest period of g, $\forall x\in \mathbb{R}$ we have,
$$
f(x+c)+g(x+c)=f(x)+g(x),
$$
What is better rewritten in the invariant form :
$$
f(x+c)-f(x)=g(x)-g(x+c)
$$
Denote $\gamma(x)$ the common value, then for $k,l$ integer we find
$$
 \gamma(x + ka + lb) = f (x + ka + lb + c) - f (x + ka + lb) 
$$
$$
= f (x + lb + c) - f (x + lb)
$$
$$
=  \gamma(x + lb) = g (x + lb + c) -g (x + lb) = g (x + c) -g (x) =  \gamma(x)
$$
Therefore, all real of $a\mathbb{Z}+b\mathbb{Z}$ is a  period of $\gamma$
But $a\mathbb{Z}+b\mathbb{Z}$ is dense in $\mathbb{R}$ because $\frac{a}b \notin \mathbb{Q}$.
Therefore $\gamma$ is $\epsilon$-periodic fo all $\epsilon$.
As $\gamma$ is continuous, he is necessarily constant :
$$
\gamma=\gamma_0
$$
Furthermore,
$$
f(Id+c) = f + \gamma_0,
$$
Then for all real x we have,
$$
f(x+nc)=f(x)+n\gamma_0
$$
$$
\Longrightarrow \gamma_0=0
$$
because $f$ is continuous and periodic thus $f$ is bounded.
Therefore, $c$ is a common period of $f$ and $g$.
Then $c$ is in $a\mathbb{N^*}\bigcap b\mathbb{N^*}$, but he is empty because $\frac{a}b \notin \mathbb{Q}$. QED