# Show that $(f_n)$ is equicontinuous, given uniform convergence

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 $n∈N$
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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:

$$|f_n(y)-f_n(x)| \leq |f_n'(c_n)| \cdot |x - y| < M \cdot \epsilon/M = \epsilon$$

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Hint

$$|f_n(x)-f_n(y)|=|(f_n(x)-f(x))+(f(x)-f(y))+(f(y)-f_n(y))|$$

$$\leq |f_n(x)-f(x)|+|f(x)-f(y)|+|f(y)-f_n(y)| \,.$$

Now, use the assumptions you have been given.

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