# Equicontinuous family of functions with $f_n(t)=f(nt)$

Suppose $f$ is a real continuous function on $\mathbb{R},$ $f_n(t)=f(nt)$ for $n\in \mathbb{N}$ and $\{f_n\}$ is equicontinuous on $[0,1]$. What conclusion can you draw about $f$?

Proof: Let $\hat{x}=0$ and $q\ge 2$ is an integer.

Let $\epsilon>0$ be given then $\exists \delta>0$ such that $\forall x,y \in [0,1]$ with $|x-y|<\delta$ and $\forall n\in \mathbb{N}$ we have $$|f_n(x)-f_n(y)|=|f(nx)-f(ny)|<\epsilon.$$ Putting $x=\hat{x}=0$ and $y=\frac{1}{qk}$ for $k\ge \dfrac{[1/{\delta}]+1}{q}$ (since $y<\delta$) then $\forall n\in \mathbb{N}$ we have $|f(ny)-f(0)|<\epsilon.$

1) Taking $n=k$ we have $|f(\frac{1}{q})-f(0)|<\epsilon$ since $\epsilon$ is arbitrarily then $f(\frac{1}{q})=f(0).$ Continuing for $n=2k$ we get that $f(\frac{2}{q})=f(0)$. Proceeding in this way we get that $f(\frac{p}{q})=f(0)$ for any $p>0$ where $q$ is fixed. Since $q$ is arbitrarily then $f(r)=f(0)$ for $r\in \mathbb{Q}_{>0}$.

Since $f$ is a continuous function and for every irrational number $\alpha$ exists $q_n\in \mathbb{Q}$ such that $q_n\to \alpha$ then by continuity $f(\alpha)=\lim \limits_{n\to \infty}f(q_n)=f(0)$. Thus $f\equiv f(0)$ on $[0,+\infty)$.

So conculison is the following: $f\equiv \text{const}$ on $[0,+\infty)$.

What did you think about above proof? Is my conclusion true?

• Your proof seems to be correct, but your proof has detoured slightly. You can modify it more directly. – Hanul Jeon Dec 11 '15 at 9:09
• @HanulJeon, Edited! – ZFR Dec 11 '15 at 9:17

First of all, your $f$ might be defined only for nonnegative real number instead of the whole $\mathbb R$. The equicontinuity of $f_n$ on $[0,1]$ says nothing about the value of $f$ on $(-\infty, 0)$.
Your proof is correct, but is over-complicated. You don't need to deal with the case of rational first: Let $\epsilon, \delta >0$ be chosen as in your proof and let $y >0$ be arbitrary. Then there is $n \in \mathbb N$ so that $y/\delta <n$, by the Archimedean property. Then $\frac yn <\delta$ and we have
$$\left|f_n(0) - f_n\left(\frac yn\right)\right|<\epsilon,$$ which is the same as $|f(0) - f(y)|<\epsilon$. As $\epsilon >0$ is arbitrary, $f(0) = f(y)$. As $y>0$ is arbitrary, $f$ is a constant function.
• Dear John! I have several questions: 1) Where did you use the continuity of $f$? 2) Why $f$ is not defined for negative reals? – ZFR Dec 11 '15 at 9:20
• 1) Somehow the continuity assumption is not used. But if you say $\{f_n\}$ is equicontinuous, then each $f_n$ is continuous and in particular $f$ is continuous. 2) Consider $$f(x) = \begin{cases} 0 & x\ge 0 \\ g(x) & x<0 .\end{cases}$$ Then this $f$ satisfies that $f_n = 0$ on $[0,1]$ for all $n$. But you cannot say anything about $g$. @RFZ – user99914 Dec 11 '15 at 9:26