3
$\begingroup$

Let the function $f_n$ : $[0,1] \rightarrow [0,1]$ satisfy \begin{align*} \vert f_n(x)-f_n(y)\vert \leq \vert x-y \vert \textrm{ whenever } \vert x-y \vert \geq \frac{1}{n} \end{align*} Show that the sequence $\{f_n\}$ has a uniformly convergent subsequence.

I try to use the Arzela-Ascoli Theorem. Thus, I check to $\{f_n\}$ is uniformly bounded and equicontinuous. But, I don't know how to prove that when $\vert x - y \vert <\frac{1}{n}$

Any help is appreicated...

Thank you!

$\endgroup$
  • $\begingroup$ $$f_n(x) - f_n(y) = \bigl(f_n(x) - f_n(x-1/n)\bigr) + \bigl(f_n(x-1/n) - f_n(y)\bigr)$$ if $1/n \leqslant x < y$. $\endgroup$ – Daniel Fischer Jul 18 '18 at 14:44
1
$\begingroup$

Although functions $f_n$ are not equi-continuous (and not even continuous), one can still apply the Arzela-Ascoli diagonalization argument, with a slight preparation.

Let $n\in \mathbb{N}$ be fixed. Take any $x_1\in (0,1)$ and consider the set $$U_{x_1} = (x_1 - \frac 2n, x_1 + \frac 2n) \setminus [x_1-\frac 1n,x_1 + \frac 1n].$$ Then $U_{x_1}$ is an open set, and for any $x\in U_{x_1}$ we have, by the condition of the problem, that $$ | f_n(x) - f_n(x_1) |\leq \frac 2n. $$ Hence, for any $x,y \in [0,1]$ with $|x-x_1|, |y-x_1| \leq 1/n$ we get $$ \tag{1} |f_n(x) - f_n(y)| \leq |f_n(x) - f_n(x_1) | + |f_n(y) - f_n(x_1)| \leq \frac 4n. $$

Let $x_1$ run over rationals $Q$ (say), then $\{(x_1 - 1/n, x_1 + 1/n)\}_{x_1 \in Q}$ will serve as an open cover of $[0,1]$. Extracting a finite sub-cover and applying $(1)$ to pass from one interval of length $2/n$ to its neighbor, we get $$ \tag{2} |f_n(x) - f_n(y) |\leq \max\left\{ \frac 8n, |x-y| \right\}. $$

Now, $(2)$ shows "asymptotic equi-continuity", the oscillations of $f_n$ decay uniformly as $n\to \infty$, and we are now ready to apply the argument in the proof of Arzela-Ascoli using $(2)$ instead of the usual equi-continuity.

$\endgroup$
2
$\begingroup$

Warning: this solution assumes that the functions $f_n$ are continuous on $[0,1]$. In general Arzelà-Ascoli is not applicable here.

Let $\varepsilon > 0$. Pick $n \in \mathbb{N}$ such that $\frac5n < \varepsilon$. For $|y-x| < \frac1{n}$ and $k \ge n$

$$\left|\left(x-\frac2n\right) - x\right| = \frac2n \ge \frac1n \ge \frac1k$$

$$\left|y-\left(x-\frac2n\right)\right| \ge \frac2n- |y-x| \ge \frac1{n} \ge \frac1k$$

so

\begin{align} |f_k(x) - f_k(y)| &\le \left|f_k(x) - f_k\left(x-\frac2n\right)\right| + \left|f_k\left(x-\frac2n\right) - f_k(y)\right|\\ &\le \left|\left(x-\frac2n\right) - x\right| + \left|y-\left(x-\frac2n\right)\right| \\ &\le \frac2n + |y-x| + \frac2n\\ &< \frac5{n}\\ &< \varepsilon \end{align}

Each $f_j$ is uniformly continuous on $[0,1]$ so there exists $\delta_j > 0$ such that $|y-x| < \delta_j$ implies $|f_j(x) - f_j(y)| < \varepsilon$.

Therefore for $|y-x| < \delta =\min\left\{\delta_1, \ldots, \delta_{n-1}, \frac1n\right\}$ we have $|f_j(x) - f_j(y)| < \varepsilon, \forall j \in \mathbb{N}$.

Hence $(f_n)_n$ is equicontinuous. $(f_n)_n$ is also uniformly bounded by $1$.

By Arzelà-Ascoli, there exists a uniformly convergent subsequence of $(f_n)_n$.

$\endgroup$
  • $\begingroup$ $f_n$ is not necessarily continuous on $[0,1]$, the index $k$ in your argument depends on $\varepsilon$ via $n$. $\endgroup$ – Hayk Jul 18 '18 at 16:39
  • $\begingroup$ @Hayk You're right, I assumed that they are continuous. Arzela-Ascoli is not applicable here. $\endgroup$ – mechanodroid Jul 18 '18 at 16:46

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.