1
$\begingroup$

Let $f_n : [0,1] \to \mathbb{R}$ be a sequence of continuously differentiable functions such that $$f_n(0) = 0, \ \ |f'_n(x)| \leq 1, \ \ \forall n \in 1, x \in (0,1).$$ Suppose further that $f_n(\cdot)$ is convergent pointwise to some function $f(\cdot)$. Show that $f_n(\cdot)$ converges to $f(\cdot)$ uniformly.

We must show that $\forall \epsilon > 0$ there exists an index $N(\epsilon)$ independent of $x$ such that $$|f_n(x) - f(x) | < \epsilon \ \forall n \geq N(\epsilon).$$

We know that $\forall \epsilon > 0$ there exists an index $N(\epsilon, x)$ dependent on $x$ such that $$|f_n(x) - f(x) | < \epsilon \ \forall n \geq N(\epsilon,x).$$ We also know that since the functions are continuously differentiable, they are continuous. Since they are continuous, each function also achieves a maximum and minimum, and are bounded.

I'm not sure where to go with all of this information. How can the bounded derivative help me?

$\endgroup$
3
  • $\begingroup$ Do you know the Arzelà–Ascoli theorem? $\endgroup$
    – user251257
    Mar 26, 2017 at 18:05
  • $\begingroup$ We just learned the one about equicontinuity. Should I try to show that the family of functions is equicontinuous? $\endgroup$
    – user389056
    Mar 26, 2017 at 18:10
  • $\begingroup$ if you haven't had arzela ascoli, you should prove it directly. In your particular case it would be simpler. $\endgroup$
    – user251257
    Mar 26, 2017 at 18:16

3 Answers 3

1
$\begingroup$

A direct proof:

Let $\epsilon>0$. Take $(x_{l})_{l=1}^{m}\subseteq [0,1]$ such that $\bigcup_{l=1}^{m}\{x\in[0,1]:|x-x_{l}|<\epsilon\}=[0,1]$. If $n$ is large enough, then $|f(x_{l})-f(x_{l})|<\epsilon$, for all $l\in [m]$. Take $x\in [0,1]$ and $j\geq n$ such that $|f_{j}(x)-f(x)|<\epsilon$. We have: $$ |f(x)-f_{n}(x)| \leq |f(x)-f_{j}(x)|+|f_{j}(x)-f_{j}(x_{l})|+|f_{j}(x_{l})-f_{n}(x_{l})|+|f_{n}(x_{l})-f_{n}(x)|$$ Then, $$|f(x)-f_{n}(x)|<\epsilon +\|f'_{j}\||x-x_{l}|+2\epsilon+\|f'_{n}\||x-x_{l}|$$ Since $\|f'_{k}\|<1$, for all $k$, if we choose $x_{l}$ close to $x$, we finish the proof.

Edit: $\|f'_{k}\|=\sup_{x\in[0,1]}|f'_{k}(x)|$

$\endgroup$
1
$\begingroup$

Hint: By the mean value theorem, $|f_n(y)-f_n(x)| \le |y-x|$ for each $n.$ This implies, just from pointwise convergence, that $|f(y)-f(x)| \le |y-x|$ as well. Given $\epsilon >0$ think about a partition of $[0,1]$ into subintervals of length less than $\epsilon$ and use the pointwise convergence at the partition points.

$\endgroup$
0
0
$\begingroup$

Hints:

  1. Show that $f$ need to be Lipschitz continuous with constant $1$. Thus, you may assume without loss of generality that $f=0$ (if you assume $f_n$ has Lipschitz constant 2).

  2. Fix $m$. Let $x_0=0$ and let $x_k = k/m$ for $k=1,\dotsc,m$. Then, for $x\in [x_{k-1}, x_k]$ it follows $$ |f_n(x)| \le |f_n(x_k)| + |f_n(x_k) - f_n(x)| \le |f_n(x_k)| + \frac2m. $$ Show that the right hand side converges to $0$.

It is basically the proof of Arzela Ascoli.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .