Exercise 7.15

Question

Exercise 15: 
    Suppose $f$ is a real continuous function on $\mathbf R$, $f_n(t)=f(nt)$ for $n=1,2,3,\ldots,$ and $\{f_n\}$ is equicontinuous on $[0,1]$. What conclusion can you draw about $f$?

analambanomenos · Accepted Answer

\def\eps{\varepsilon} The function $f$ must be constant on $[0,\infty]$. Let $0\le x0$. Then there is a $\delta>0$ such that for all $n=1,2,3,\ldots,$ we have $\big|f_n(z)-f_n(w)\big|< \varepsilon$ if $0\le w,z\le 1$ and $|w-z|<\delta$. For large enough $N$, we have $$\bigg|\frac{y}{N}-\frac{x}{N}\bigg|=\frac{|y-x|}{N}<\delta,\quad\quad 0\le\frac{x}{N}<\frac{y}{N}\le1.$$ Hence $$\big|f(y)-f(x)\big|=\Bigg|f_N\biggl(\frac{y}{N}\biggr)-f_N\biggl(\frac{x}{N}\biggr)\Bigg|<\varepsilon.$$ Since $\varepsilon>0$ was arbitrary, this shows that $f(x)=f(y)$. By the way, it is easy to extend this argument to show that if $\{f_n\}$ were equicontinuous on $[-1,1]$, then $f$ would be constant on all of $\mathbf R$.

Exercise 7.15

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Exercise 7.15

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