How to prove $f_n:[0,1] \to \mathbb{R}$ converge uniformly? (with conditions)

Question

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

• $\forall n\in \mathbb{N}: f_n$ is a derivation function
• $\forall n\in \mathbb{N},x\in [0,1]: \left| f_n'(x)\right|\le 3 $
• $\exists f:[0,1]\to \mathbb{R} \ \forall x\in [0,1]:\lim_{n \to \infty} f_n(x) =f(x)$

Is $f_n \xrightarrow[]{u}f$?

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EDIT:

We know that:

1. $\forall n:f_n$ is continuous.
2. $\forall \varepsilon: \exists \delta :\forall x,y\in [0,1]:|x-y|<\delta \implies |f_n(x)-f_n(y)|<\varepsilon $
3. $\forall x,y \in [0,1]: \exists c=c_{x,y}\in(x,y): \frac{f_n(x)-f_n(y)}{x-y}=f'(c) \implies |f_n(x)-f_n(y)|<3|x-y|$
4. $\forall x\in [0,1]\forall \varepsilon: \exists N :\forall n>N : |f_n(x)-f(x)|<\varepsilon $

First of all, I feel overwhelmed with all these given information. I'm not sure what to use and why. I'd appreciate if someone could solve it and most importantly explain their draft.

When I look on this, I know I need to find $N$ s.t $\forall n>N \ \forall x\in [0,1]: |f_n(x)-f(x)|<\varepsilon $. From $(4)$ I know I can get such $N$ for all $x$. Who says it's the same $N$? So I better use the delta somehow, maybe divide $[0,1]$ to subintervals of length $\delta$. Then I'm not sure where I need to use $(3)$.

Thanks in advance for any help!!!

Answer 1

You can prove it by contradiction. The idea is that if $(f_n)$ doesn't converge uniformly toward $f$, then there is a sequence $(x_n)$ of $[0,1]$ such that $|f_n(x_n) - f(x_n)| \geqslant \varepsilon$ for all $n$ (up to extraction), for some $\varepsilon > 0$. Then, still up to extraction, we can assume that $(x_n)$ converges toward some $x_\infty \in [0,1]$ and finally, we can use the fact that each $f_n$ is $3$-Lipschitzian to control $|f_n(x_n) - f(x_n)|$ in function of $|f_n(x_\infty) - f(x_\infty)|$ (which converge toward $0$ by simple convergence of $(f_n)$ toward $f$) and reach a contradiction.

Let us begin by showing that $f$ is continuous. Indeed, for all $(x,y) \in [0,1]^2$, $|f_n(x) - f_n(y)| \leqslant 3|x - y|$ so when $n \rightarrow +\infty$, the simple convergence of $f_n$ toward $f$ implies $|f(x) - f(y)| \leqslant 3|x - y|$. $f$ is $3$-Lipschitzian hence continuous.

Then, $f_n \rightarrow f$ uniformly means that $\sup_{[0,1]}|f_n - f| \rightarrow 0$. If it is not the case, then, up to extraction, you can find an $\varepsilon > 0$ such that for all $n$, $\sup_{[0,1]}|f_n - f| \geqslant \varepsilon$. Now, consider $x_n$ such that $|f_n(x_n) - f(x_n)| = \sup_{[0,1]}|f_n - f| \geqslant \varepsilon$ (such an $x_n$ exists because $[0,1]$ is compact and both $f_n$ and $f$ are continuous).

Up to an other extraction, we may assume that $(x_n)$ converges (because $[0,1]$ is compact) toward some $x_\infty \in [0,1]$. And we have, \begin{align*} |f_n(x_n) - f(x_n)| & \leqslant |f_n(x_n) - f_n(x_\infty)| + |f_n(x_\infty) - f(x_\infty)| + |f(x_\infty) - f(x_n)|\\ & \leqslant 3|x_n - x_\infty| + |f_n(x_\infty) - f(x_\infty)| + 3|x_\infty - x_n|\\ & = 6|x_n - x_\infty| + |f_n(x_\infty) - f(x_\infty)|\\ & \rightarrow 0. \end{align*} It contradicts $|f_n(x_n) - f(x_n)| \geqslant \varepsilon$.

Answer 2

An application of Mean value Theorem shows that $(f_n)$ is equi-continuous. It is also bounded at $0$, for example. By Arzela Acoli Thoerm this implies that every subsequence of $(f_n)$ has a further subsequence converging uniformly. But all subsequential limits are the same because $f_n$ converges pointwise to $f$. This implies that the whole sequence $(f_n)$ converges uniformly to $f$.

The answer by Sangchul Lee here shows that pointwise boundeness and uniform boundeness are equivalent for equi-continuous families: Arzela-Ascoli Theorem: Is only pointwise boundedness required?