Let $f_n: [a,b] \rightarrow, n \in \mathbb{N}$, be a sequence of functions converging uniformly to $f: [a,b] \rightarrow \mathbb{R}$ on $[a,b]$. Suppose that each $f_n$ is continuous on [a,b] and differentiable on (a,b), and that the sequence of derivatives $(f'_n)$ is uniformly bounded on (a,b). This means that there exists an $M>0$ such that $|f'_n(x)| \le M$ for all $x \in (a,b)$ and all $n \in \mathbb{N}$
Question: Show that $(f_n)$ is equicontinuous.
Known definitions:
- A sequence of functions $(f_n)$ converges uniformly to a limit function $f$ on a set $A$, if, for every $\epsilon >0$ , there exists an $N\in \mathbb{N}$. such that$|f_n(x) – f(x)| < \epsilon$ whenever $n \ge N$ and $x \in A$
- Cauchy Criterion for Uniform Convergence: A sequence of functions $(f_n)$ converges uniformly on a set $A$, if and only if, for every $\epsilon >0$ , there exists an $N\in \mathbb{N}$. such that$|f_n(x) – f_m(x)| < \epsilon$ for all $n,m \ge N$ and all $x \in A$
- A sequence of functions $(f_n)$ defined on a set $E$, is called equicontinuous if for every $\epsilon >0$ , there exists a $\delta>0$ such that $N\in \mathbb{N}$. such that$|f_n(x) – f_n(y)| < \epsilon$ for all $n \in N$ and $|x-y| \lt \delta$ in $E$
- A sequence of derivatives $(f_n')$ is uniformly bounded on (a,b) if there exists an $M>0$ such that $|f_n'(x)|≤M$ for all $x∈(a,b)$ and all $n∈N$
Best Answer
You only need the uniform boundedness of the derivatives. Let $\epsilon > 0$, and choose $\delta = \epsilon/M$. Then for all $x<y \in [a,b]$ such that $|x-y| < \delta$, for every $n$, by the mean value theorem, there exists $c_n \in (x,y)$ such that $f_n(y) - f_n(x) = f_n'(c_n) \cdot (y - x)$. Therefore:
$$|f_n(y)-f_n(x)| \leq |f_n'(c_n)| \cdot |x - y| < M \cdot \epsilon/M = \epsilon$$