Let $f :\mathbb{R} \to \mathbb{R} $.
Let $ \{f_n \}_{n=1}^{\infty}$ be periodic and continuous functions with period $P_n \gt0 $ for each $n \in N $.
Suppose $P_n \to 0 $ and $f_n \to f $ uniformly on $\mathbb{R}$. What can we say about $ f$ ?
I think that $f$ must be constant, but I'm not sure how to prove it.
Thanks for helping.
Consider two points $a$, $b\in{\mathbb R}$, and let an $\epsilon>0$ be given. Since the $f_n$ converge uniformly the function $f$ is continuous. There is a $\delta>0$ such that $|f(x)-f(a)|<\epsilon$ when $|x-a|<\delta$ and $|f(x)-f(b)|<\epsilon$ when $|x-b|<\delta$.
There is an $N\gg1$ such that the following are simultaneously true: $|f_N(x)-f(x)|<\epsilon$ for all $x\in{\mathbb R}$, and $P_N<\delta$. It follows that there are two points $\alpha\in\>]a-\delta,a+\delta[\>$ and $\beta\in\>]b-\delta,b+\delta[\>$ such that $f_N(\alpha)=f_N(\beta)$.
This implies $$\eqalign{|f(a)-f(b)|&\leq|f(a)-f(\alpha)|+|f(\alpha)-f_N(\alpha)|+|f_N(\alpha)-f_N(\beta)|\cr&\quad+|f_N(\beta)-f(\beta)|+|f(\beta)-f(b)|\cr&<4\epsilon\ .\cr}$$ Since $\epsilon>0$ was arbitrary we can conclude that $f(a)=f(b)$.