Question:
Let two $\varphi(x) $ and $\psi(x)$ periodic and continous functions such that $$ \lim_{ x\to\infty}(\varphi(x)-\psi(x))=0, \quad x\in \mathbb{R}. $$ Prove that
$$ \varphi(x)\equiv \psi(x), \quad x\in \mathbb{R}. $$
It seems easy at first glance to prove,however i have seen one a long proof and didn't like it and i need to your help,who can prove it by short way.
Thanks in advance!
In fact, continuity of only one of the two functions is sufficient. Suppose $\varphi$ is continuous.
Pick $x_0\in\Bbb R$ and $\epsilon>0$. Let $S$ and $T$ be the periods of $\varphi$ and $\psi$ respectively. Take $\delta>0$ such that if $|x-x_0|<\delta$ then $|\varphi(x)-\varphi(x_0)|<\epsilon/2$ and let $N\in\Bbb N$ be such that if $x>N$, then $$|\varphi(x)-\psi(x)|<\epsilon/2.$$ By Dirichlet Approximation Theorem, there exists $n,k\in\Bbb N$ such that $|nS-kT|<\delta$. Moreover, by dividing $\delta$ by some constant depending on $x_0,N$ and $T$, we can make sure that $k$ is large enough so that $$x_0+kT>N.$$ Now, we have $$|(x_0+nS)-(x_0+kT)|=|nS-kT|<\delta,$$ and hence $$ \begin{align} |\varphi(x_0)-\psi(x_0)| &= |\varphi(x_0+nS)-\psi(x_0+kT)| \\ &\leq|\varphi(x_0+nS)-\varphi(x_0+kT)|+|\varphi(x_0+kT)-\psi(x_0+kT)| \\ &<\epsilon/2+\epsilon/2\\ &=\epsilon \end{align} $$ Since $x_0$ and $\epsilon>0$ are arbitrary, we conclude that $\varphi\equiv\psi$.