Showing the following sequence of functions are uniformly convergent

Question

Let $f_n:[0,1]\to \mathbb{R}$ be a sequence of continuously differentiable functions such that

$$f_n(0)=0,\:\: |f_n'(x)|\leq 1, \text{for all }n\geq 1, x\in (0,1).$$ Suppose further that $f_n(.)$ is convergent to some function $f(.)$. Show that $f_n(.)$ converges to $f(.)$ uniformly.

I tired to prove this problem, but I'm lost and I cannot find a correct approach.

Firstly, I don't know if $f(.)$ is continuous. There must be some way to show $f$ is continuous, but I can't and I'm not sure if I need it. Secondly, I've obtained that $(f_n(.))$ are uniformly bounded and this because of $$\text{for some }z\in(0,x)\subset(0,1),\:\:|f'_n(z)|=|\frac{f_n(x)-f_n(0)}{x-0}|\leq 1\to |f_n(x)|< 1.$$

I tried to show $(f_n(.)) $ is equicontinuous. Here is my work.

Let $x\in (0,1),$ and let $\epsilon>0$. As $|f_n'(x)|\leq 1$, we can write $\lim_{y\to x}|\frac{f_n(x)-f_n(y)}{y-x}|\leq 1$, hence $|f_n(x)-f_n(y)|\leq |x-y|$. So if we take $\delta=\epsilon$,then for any $y$ that $|x-y|<\delta$, we have $|f_n(x)-f_n(y)|<\epsilon.$ So $(f_n)$ is equicontinuous on $(0,1)$.

Also we know $f_n(0)=0$, so this sequence is equicontinuous on $x=0$. If I'm correct in each of these steps, probably I'm done with the proof by showing that (f_n(.)) is equicontinuous on $x=1$.

Also in my proof, I didn't use this fact that $(f_n())$ are continuously differnetiable. I also saw the following proof, for this quetsion, but honestly, I think, it's not totally correct, since they use the continuity of $f(.)$ without proving it. here is the link "Show that $f_n(\cdot)$ is uniformly convergent."

Answer 1

Let $\varepsilon>0$ be given, and set $\|\, f_k'\| = \sup \left[|\,f_k'(x)|: x \in (0,1) \right]$ $(k=1,2,\ldots)$. Since the collection of open balls $\mathcal{B}: = \{B(\, x, \frac{\varepsilon}{3}) : x \in [0,1] \}$ is a cover for $[0,1]$, we may find a finite subcover, say $B(\,x_1, \frac{\varepsilon}{3}), \, \ldots, \, B(\,x_M, \frac{\varepsilon}{3})$ (Heine-Borel Theorem). Since $f_n$ converges pointwise on $[0,1]$, for each point $x_j \: \left(\,j=1,\ldots, M \right)$ we may find a positive integer $N_j$ so that \begin{equation} \left|\, f_n(x_j) - f_m(x_j) \right| < \frac{\varepsilon}{3} \text{ whenever } n, m \geq N_j \,. \end{equation} Setting $N = \max [N_1, \ldots, N_M]$ shows that

\begin{aligned} \left|\,f_n(x)- f_m(x) \right| & \leq \left| \,f_n (x)- f_n(x_j) \right| + \left|\, f_n (x_j)- f_m(x_j) \right| + \left|\, f_m(x_j)- f_m(x) \right| \\ & < \| \,f_n'\||x-x_j| + \frac{\varepsilon}{3} + \|\,f_m'\| |x_j-x| \\ & \leq \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3}= \varepsilon \; \: \text{ whenever } \, n,m \geq N \text{ and } x \in [0,1] . \end{aligned}

Since $\mathbb{R}$ is complete (or rather $f(x):=\lim_{n \to \infty} f_n(x)$ exists for $x \in [0,1]$ ), it follows that the sequence of functions $\{\,f_n\}_{n=1}^\infty$ converges uniformly on $[0,1]$ (Cauchy Criterion).

Answer 2

The reasoning you give is correct. The bound on the derivative gives equicontinuity immediately and, as you show, the sequence is pointwise bounded.

Since $(f_{n})_{n \in \mathbb{N}}$ is pointwise bounded and equicontinuous, the Arzela-Ascoli Theorem implies $(f_{n})_{n \in \mathbb{N}}$ is pre-compact in $C([0,1])$. Since $f_{n}(x) \to f(x)$ pointwise, any uniformly convergent subsequence converges to $f$, and, thus, $(f_{n})_{n \in \mathbb{N}}$ must converge uniformly to $f$.