2
$\begingroup$

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

$$f_n(0)=0,\:\: |f_n'(x)|\leq 1, \text{for all }n\geq 1, x\in (0,1).$$ 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."

$\endgroup$
  • $\begingroup$ The work you give shows $(f_{n})_{n \in \mathbb{N}}$ satisfies the criteria for pre-compactness given in the Arzela-Ascoli Theorem. Do you understand my answer to your question? I didn't address how we know $f$ is continuous and I wasn't super specific about how we know $(f_{n})_{n \in \mathbb{N}}$ converges uniformly to $f$, but I can flesh out my answer if necessary. $\endgroup$ – fourierwho Jul 22 '17 at 4:36
1
$\begingroup$

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).

$\endgroup$
  • 1
    $\begingroup$ Thanks, you have shown the sequence is uniformly Cauchy. $\endgroup$ – Parisina Jul 22 '17 at 1:28
1
$\begingroup$

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 $f$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.