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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!

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

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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 $(f_n)_n$.

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  • $\begingroup$ $f_n$ 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

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