Example / Counterexample of Equicontinuity on the rational set

Question

Let $h:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function. Suppose $(f_{n})$ is an equicontinuous sequence of real functions defined in $[1,2]$, such that $|f_{n}(x)|\leq |h(x)|$ for all $x\in[1,2]$. $A$ is the set of rationals in $[1,2]$, $A=\Bbb Q\cap [1,2]$.

For the three statements below provide a worked example or counter example proving if is true or false:

1. If $(f_{n})$ converges pointwise in $A$, then it converges pointwise in $[1,2]$.
2. If $(f_{n})$ converges pointwise in $A$, then it converges uniformly in $[1,2]$.
3. It is always possible to extract from $(f_{n})$ a subsequence $(g_{n})$ that converges uniformly in $[1,2]$.

The defintion of EQ and UEQ I am working with are as follow:

*Point equicontinuity: A family of functions {${f_n}$} is equicontinuous at $x_{0}\in A$ if for any $\epsilon>0$, $\exists\ \delta>0$ so that for all $n\in\Bbb N$ one has $|f_{n}(x)-f_{n}(x_{0})|<\epsilon$ whenever $x\in A$ with $|x-x_{0}|<\delta$.

*Uniform Equicontinuity: A family of functions {${f_n}$} is equicontinuous over $A$ if for any $\epsilon>0$, $\exists\ \delta>0$ so that for all $n\in\Bbb N$ one has $|f_{n}(x)-f(z)|<\epsilon$ for all $x,z\in A$ so that $|x-z|<\delta$.

Answer 1

To show (1) it is sufficient to show that $f_n$ converging pointwise on a dense subset implies that it $f_n(x)$ is Cauchy on the entire space, this will be an $\epsilon/3$ proof based on equicontinuity:

1. Let $x\in I-A$ and $\epsilon>0$. From equicontinuity you have a $\delta$ so that $|f_n(x)-f_n(a)|<\epsilon/3$ whenever $|x-a|<\delta$. Let $a\in A$ be such an element. Consider: $$|f_n(x)-f_m(x)|≤|f_n(x)-f_n(a)|+|f_n(a)-f_m(a)|+|f_m(a)-f_m(x)|$$ Since $f_n(a)$ converges it must be a Cauchy sequence and there exists an $N$ so that whenever $n,m>N$ $|f_n(a)-f_m(a)|<\epsilon/3$. Combine this with the equicontinuity to get: $$|f_n(x)-f_m(x)|<\epsilon$$ for all $n,m>N$. So $f_n(x)$ is Cauchy and thus converges. It follows $f$ converges pointwise.

So from (1) we have learned that pointwise convergence on a dense set implies pointwise convergence everywhere if we have equicontinuous functions. Number two can now be recast into a "standard result", namely that on a compact space an equicontinuous limit of continuous functions is a uniform limit. The proof of it is this:

2. Let $\epsilon>0$, $x\in I$. From pointwise convergence it follows that there exists an $N_x\in \Bbb N$ so that $n>N_x$ implies $|f(x)-f_n(x)|<\epsilon/3$. Since the equicontinuous limit of continuous functions is continuous we have a $\delta>0$ so that $y\in B_\delta(x)$ implies both $|f(x)-f(y)|<\epsilon/3$ and $|f_n(x)-f_n(y)|<\epsilon/3$ (the last is possible due to equicontinuity). Putting it together gives: $$|f(y)-f_n(y)|≤|f(y)-f(x)|+|f(x)-f_n(x)|+|f_n(x)-f_n(y)|<\epsilon$$ The balls $B_{\delta_x}(x)$ cover $I$, but $I$ is compact so you can take finitely many of them, let their centers be $\{x_1,..,x_n\}$. Define $N:=\max\{N_{x_1},...,N_{x_n}\}$. It follows that every $y$ of $I$ lies in a $B_{\delta_i}(x_i)$ and so whenever $n>N$ the above inequality holds for any $y\in I$. This means the limit is uniformly continuous.

The final statement is the theorem of Azerlá Ascoli:

Theorem If $f_n$ is a uniformly bounded (meaning $\exists M, |f_n(x)|

Strictly speaking Azerlá Ascoli has another part, namely

If $f_n$ is a sequence of continuous functions on a compact subset of $\Bbb R^n$, so that every subsequence has a convergent subsequence, then $f_n$ is equicontinuous and uniformly bounded.

But the second part is not relevant here.