# Some functional analysis problem

In my undergraduate functional analysis course, there is this $1$ problem that I am stuck on, and it goes:

For $n = 1, 2, 3,...,$ let the functions $f_n : [0, 1] → [0, 1]$ satisfy $|f_n(x)−f_n(y)|≤|x−y|$ whenever $|x − y| ≥ \frac{1}{n}.$ Prove that the sequence ${f_n}$ has a uniformly convergent subsequence.

How do I start this proof? Can anyone give me some leads?

My Solution: If we let some function $\phi_n$ be continuous, which equates to $f_n$ at some $i/n$ for $i = 0,1,2,3,...,n.$ Denote the graph of $\phi_n$ be a line segment [(i-1)/n,i/n].Then it is not hard to see that $\phi_n \in [0,1]$ implies the $\phi_n$'s are bounded (uniformly).Using the assumption given in the problem regarding $f_n,$ we can calculate the absolute values of the slopes of $\phi_n$'s on the intervals: $$\bigg|\frac{\phi_n\big(\frac{i}{n}\big)-\phi_n\big(\frac{i-1}{n}\big)}{1/n}\bigg|=\bigg|\frac{f_n\big(\frac{i}{n}\big)-f_n\big(\frac{i-1}{n}\big)}{1/n}\bigg|\le 1$$

If $0 \le k/n \le x\le y \le (k+1)/n \le 1$, then it immediately follows that $|\phi_n(x)-\phi_n(y)|\le |x-y|.$ If on the other hand we consider the interval, $0\le x\le k/n \le m/n \le y \le 1,$ then we have the following computations: $$|\phi_n(x)-\phi(y)| \le \bigg|\phi_n (x) -\phi_n\bigg(\frac{k}{n}\bigg)\bigg|+\sum^{m-1}_{p=k}\bigg|\phi_n \bigg(\frac{p}{n}\bigg) -\phi_n\bigg(\frac{p+1}{n}\bigg)\bigg|+\bigg|\phi_n \bigg(\frac{m}{n}\bigg)-\phi_n(y)\bigg|$$$$\le |x-y|$$

From here take the $\epsilon$ to be $\epsilon = \delta$, then $\{\phi_n\}$ is equicontinuous. From here we can apply the Arzela - Ascoli's theorem, some subsequence $\{\phi_{n_{j}}\}$ converges uniformly on $[0,1]$ to a limit function $\phi$. Now simply fix some $x\in [0,1]$ , then for each $j$, we let $x_j$ be a point in $[0,1]$ of the form $i/n_j$ for $i = 0,1,2,...,n_j$ such that $1/n_j\le |x-x_j|\le 2/n_j.$ Now we can conduct the final step: $$|\phi(x)-f_{n_j}(x)|\le |\phi(x)-\phi_{n_j}(x)|+|\phi_{n_j}(x)-\phi_{n_j}(x_j)|+|f_{n_j}(x_j)-f_{n_j}(x)|\le ||\phi-\phi_{n_j}||_{\infty}+\frac{2}{n_j}+\frac{2}{n_j}\rightarrow 0$$

Hence clearly $\{f_{n_j}\}$ converges uniformly on $[0,1]$ to $\phi.$

• Have you covered the Arzela-Ascoli theorem? – David C. Ullrich Jun 5 '18 at 13:46
• Yes we learned that last month, indeed I think we should use that theorem, but there seems to be some technical details that we need to cover up i.e. some limit computations using epsilon delta, which confuses me alittle bit. – Aurora Borealis Jun 5 '18 at 13:54
• I think you mean $|x-y|<1/n$ not $|x-y|>1/n$. – Omran Kouba Jun 5 '18 at 14:03
• Hmmm I am not sure cause the problem is how it is stated here in my notes. So how does one start a proof for these kind of problems? – Aurora Borealis Jun 5 '18 at 16:09
• I posted my workings using some examples and notions from my lecture notes, but I have no idea if it is correct. – Aurora Borealis Jun 6 '18 at 2:28

First of all, notice that we have the following substitute property for continuity for $|x-y|<1/n$: in this case, $|f(x)-f(y)|\le 4/n$. This is immediate from comparing $f(x),f(y)$ with $f(z)$ for a $z$ with $|x-z|,|y-z|\ge 1/n$.
Now proceed as in the proof of A-A and make $f_n$ convergent at (let's say) all rational $q\in [0,1]$, by a diagonal procedure. If we now partition $[0,1]$ into $N\gg 1$ intervals $I_j$ and fix a rational $q_j$ in each of these, then it follows that $|f_n(x)-f_n(q_j)|\le \max \{ 4/n, 1/N\}$ for $x\in I_j$. So there is uniform (in $x$) control on how much $f_n(x)$ can differ from the $\liminf$ and $\limsup$ of this sequence, and these are at most $2/N$ apart. Since $N$ was arbitrary, uniform convergence follows now.